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Bruce H. Edwards

Complete Solutions

Guide

Volume

. X^iKT''-^'^',*

Calculus

Seventh Edition

^jl i- •' ■■^"WJ :.--■?':'

Larson • Hostetler • Edwards

For use with

Calculus with Analytic Geometry, Seventh Edition

Calculus of a Single Variable, Seventh Edition

Digitized by the Internet Archive

in 2011 with funding from
Lakewood Park Christian School

http://www.archive.org/details/calculuscomplete01edwa

PART I

CHAPTER P
Preparation for Calculus

Section P.l Graphs and Models 2

Section P.2 Linear Models and Rates of Change 7

Section R3 Functions and Their Graphs 14

Section P.4 Fitting Models to Data 18

Review Exercises 19

Problem Solving 23

CHAPTER P
Preparation for Calculus

Section P.l Graphs and Models

Solutions to Odd- Numbered Exercises

1. >> = —jx + 2
x-intercept: (4, 0)
y-intercept: (0, 2)
Matches graph (b)

3. y = 4 - x^

x-intercepts: (2,0), (-2,0)
y-intercept: (0, 4)
Matches graph (a)

5. >> = 5X + 1

7. >- = 4 - x2

X

-4

-2

0

2

4

y

-5

-2

1

4

7

X

-3

-2

0

2

3

y

-5

0

4

0

-5

(3,-5)

9. .V = U + 21

X

-5

-4

-3

-2

-1

0

1

y

3

2

1

0

1

2

3

11. y = 7x - 4

X

0

1

4

9 16

>"

-4

-3

-2

-1 0

Section P. I Graphs and Models 3

13.

Xmin = -3
Xmax = 5
Xscl = 1
Ymin = -3
Ymax = 5
Yscl = 1

Note that y = A when x = 0.

15.

(-4.f«. 3,

i:. 1.73)

(a) (2,.v) = (2, 1.73) (y = V5 - 2 = 73 = 1.73)

(b) (x, 3) = (-4, 3) (3 = 75 - (-4))

\1. y = x^ + X- 2

y-intercept: y = 0"^ + 0 — 2
y=-2;(0, -2)

0 = x~ + X - 2

jc-intercepts:

0 = (x + 2){x - 1)

X = -2, 1; (-2,0), (1,0)

21. V =

3(2 - V^)

y-intercept: None, x cannot equal 0.

JT-intercepts:

0 = 2 - Vx
.1 = 4: (4, 0)

19. y = xV25 - .t'

y-intercept: y = 0^725 - 0^
y = 0; (0, 0)

jc-intercepts:

; 0 = .r=v/25^^

0 = .r=7(5 - .x)(5 + x)
x = 0, ±5: (0,0): (±5,0)

23. .r^' - x^ + Ay = 0
y-intercept:

O^Cv) - 02 -(- 4y = 0

y = 0: (0. 0)
-T-intercept:

x-{Q) - X- + 4(0) = 0

x = Q; (0, 0)

25. Symmetric with respect to the y-axis since

y = (— x)- — 2 = X- — 2.

27. Symmetric with respect to the x-axis since
(-y)- = y^ = x^ - 4x.

29. Symmetric with respect to the origin since
(-x)(-y) = xy = 4.

31. y = 4 - 7x-l- 3

No symmetry with respect to either axis or the origin.

33. Symmetric with respect to the origin since
—X

' i-xf + 1

X

^ x^ + r

37.

y=-3x + 2

Intercepts:

(i0),(0,2)

Symmetry: none

35. y = |.T^ + x| is symmetric with respect to the y-axis

since y = \(-xy + (-.r)| = \-(x^ + x)\ = \x^ + x\.

4 Chapter P Preparation for Calculus

39. y = |.v - 4

Intercepts:

(8,0), (0,-4)
Symmetry: none

45. )> = -x^ + 2
Intercepts:

(-3^,0), (0,2)
Symmetry: none

\ 1 \-^x

41. y = 1 - x^
Intercepts:

(1,0), (-1,0), (0,1)
Symmetry: y-axis

47. y = xjx + 2
Intercepts:

(0,0), (-2,0)
Symmetry: none
Domain: x > ~2

43. y = (x + 3)2
Intercepts:

(-3,0), (0,9)
Symmetry: none

-10 -8 -6 (-3,0)

49. ;c = /

Intercepts: (0,0)
Symmetry: origin

Intercepts: none
Symmetry: origin

53. y = 6 - |.y|
Intercepts:

(0,6), (-6,0), (6,0)
Symmetry: >'-axis

55. y^

- x=9

y^ = x + 9
y = ±Jx + 9

Intercepts:

(0,3),(0, -3), (-9,0)
Symmetry: jr-axis

57. X + 3y2 = 6

3y2 = 6 - ;t

Intercepts:

(6, 0), (O, V2), (O, - V2)
Symmetry: A-axis

Section p.] Graphs and Models 5

59. y = (x + 2){x - 4){x - 6) (other answers possible)

61. Some possible equations:

y = x

y = x^

y = 3x^ — X

„ = 3/;

63. x+y = 2=>y = 2-x
2x — y = I =^ y = 2x — 1

2 -;c = 2x- 1
3 = 3;c
1 =x
The corresponding >>- value \sy = 1.
Point of intersection: (1,1)

65. X + y = 1 =^ y = 1 — X

Zx - 11

\A-lx='ix- 11
-5x= -25
;c = 5
The corresponding >>- value is >> = 2.
Point of intersection: (5, 2)

61. x'^+y = 6^y = 6- x'^

x + y = A=!>y = A — X

6- x^ = A- X

0 = x^ - x~2

Q = {x- 2)(x + 1)

;c = 2, - 1

The corresponding y- values are y = 2 (for x = 2)
and y = 5 (for x = — 1 ).

Points of intersection: (2, 2), (—1,5)

69. ;c2 + y2 = 5 => y2 = 5 - ;c2
jr — y= \ => y = X — 1
5- x- = {x- 1)2

5-x^ = x--2x+\

Q = 2x~ -2x- A = 2(x+ \){x - 2)

X = - 1 or .r = 2
The corresponding y- values are y - - 2 and y = 1 .
Points of intersection: (- 1, -2), (2. 1)

71. y = x3

y = X
x? = X
:^ - x = Q

x{x + \)(x - 1) = 0
X = Q,x = — 1, orx = 1

The corresponding y- values are v = 0, y = - 1, and
y=l.

Points of intersection: (0, 0), (- 1, - 1), (1, 1)

73.

>• =

x> -Ztt + x- \

y = -jr + 3.t - 1

^ -2x- + x- \ = -X- + 3.r - 1

.r3 - .^2 - It = 0

x(x - 2)(x + 1) = 0

.r= -1,0.2

(-1, -5),(0. -1).(2. 1)

l,^^:dt

7^

^,2.U

(O.-in

\

-

' 1 '"^^

3

6 Chapter P Preparation for Calculus

75. 5.5v^ + 10,000 = 3.29x

(5.5^)^ = (3.29x - 10,000)2

30.25;c = 10.8241JC2 - 65,800x + 100,000,000

0 = 10.8241^2 - 65,830.25;c + 100,000,000 Use the Quadratic Formula.
X = 3133 units

The other root, x = 2949, does not satisfy the equation R = C.

This problem can also be solved by using a graphing utility and finding the intersection of the graphs of C and R.

77. (a) Using a graphing utility, you obtain

y = -0.0153f- + 4.9971f + 34.9405

(c) For the year 2004. J = 34 and
y = 187.2 CPl.

(b)

79. 400

If the diameter is doubled, the resistance is changed by approximately a factor of (1/4). For instance, ^(20) ~ 26.555 and
. ^(40) = 6.36125.

81. False; j:-axis symmetry means that if (1, -2) is on the graph, then (1, 2) is also on the graph.

83. True; the x-intercepts are

-b± JlP- - 4ac

2a

, 0

85. Distance to the origin = K x Distance to (2, 0)

Jx- + y^ = A:V(x - 2)2 + f, K + 1
x^ + y'^ = K-{x^ - Ax + A + y-)
(1 - K'^)x^ + (1 - K^)y^ + 4Kh - 4K^ = 0
Note: This is the equation of a circle!

Section P.2 Linear Models and Rates of Change 7

Section P.2 Linear Models and Rates of Cliange

\. m=\

3. m = 0

5. m= -\2

7,

9. m

2 - (-4)
5-3

(3, -4)

11. m

5 - 1

2-2

^4
0

undefined

(2,5)

(2,1)

13. m =

2/3 - 1/6
1/2 -(-3/4)

111
1/4

15. Since the slope is 0, the line is horizontal and its equation is >■ = 1. Therefore, three additional points are (0, 1), (1, 1),
and (3, 1).

17. The equation of this line is
y-l = -3{x-\)
y= -3x + 10.
Therefore, three additional points are (0, 10), (2, 4), and (3, 1).

19. Given a line L, you can use any two distinct points to calculate its slope. Since a line is straight, the ratio of the change in
y-values to the change in jr-values will always be the same. See Section P.2 Exercise 93 for a proof.

8 Chapter P Preparation for Calculus

21. (a)

E

= 260- ■

I I I I I I I I l>
123456789

Year (0<-» 1990)

(b) The slopes of the line segments are
255.0 - 252.1

2 - 1

257.7 - 255.0
3-2

260.3 - 257.7
4-3

262.8 - 260.3
5-4

265.2 - 262.8
6-5

267.7 - 265.2
7-6

270.3 - 267.7
8-7

2.9

2.7

= 2.6

= 2.5

2.4

2.5

2.6

The population increased most rapidly from 1991 to 1992.

(m = 2.9)

23. x + 5y--20

Therefore, the slope is /n =
(0,4).

and the y-intercept is

25. ;c = 4

The line is vertical. Therefore, the slope is undefined and
there is no y-intercept.

27. y = |.r + 3
Ay = -ix+ 12
0 = 3;c - 4y + 12

31. y + 2 = 3(;c - 3)

y + 2 = 3x - 9
y = 3.r - 11
y - 3a:+ 11 = 0

33. m

6-0

= 3

2-0

y - 0 = 3U - 0)

y = 3x

35. m

1 - (-3)

37. m

8

-0

8

2

- 5

3

y -

0

= -1-

-5)

y

= -|-

40
3

3y + 8jc - 40 = 0

Section P.2 Linear Models and Rates of Change 9

39. m

1

Undefined.

5-5
Vertical line jr = 5

(5.8)

(5.1)

41. m

1/1 - 3/4 11/4 _ 11

1/2

- 0

1/2

y-

3 _
4

^U - 0)

y =

11 ^3

T^ + i

llx' Ay + ?, = Q

43. .r = 3

jc- 3 = 0

(3,0)

45. ^ + f=l

2 3

3;c + 2j' - 6 = 0

47.^ + ^=1
a a

1 + ^=1
a a

- = 1
a

a = 3=»jr + y = 3
.X + V - 3 = 0

49. y = -3

y + 3 = 0

12 3 4 5

51. y=-lx+\

53. y - 2 = fU - 1)

y = 2.t + 2
2y - 3x - 1 = 0

55. 2t - y - 3 = 0

y = 2x - 3

10 Chapter P Preparation for Calculus

57.

The lines do not appear perpendicular.

The lines appear perpendicular.

The lines are perpendicular because their slopes 1 and — 1 are negative reciprocals of each other.
You must use a square setting in order for perpendicular lines to appear perpendicular.

59. 4a: - ly = 3

y = 2x - 2
/n = 2

(a) y - 1 = 2{x - 2)
y - \ =2x- 4

2x - y - 3 = 0

(b) J - 1 = -\{x - 2)
2>' - 2 = -;c + 2

X + 2y - A = 0

61, 5;c - 3y = 0

y = h

m = l

(a) y-l

= 1(^-1)

24y - 21

= 40;c - 30

24y - 40jc + 9

= 0

(b) y-l

= -f(- - !)

40y - 35

= -24a; + 1

40y + 24;c - 53

= 0

63. (a) x = 2 => X - 2 = 0
0})y = 5=>y-5 = 0

65. The slope is 125. Hence, V = 125(f - 1) + 2540

= 125? + 2415

67. The slope is -2000. Hence, V

-2000(f - 1) + 20,400
-2000/ + 22,400

69.

You can use the graphing utility to determine that the points of intersection are (0, 0) and (2, 4). Analytically,
x^ = 4x - x^
2x2 - 4x = 0
2x{x - 2) = 0

x = 0=^y = 0^>{0,0)
X = 2 => >- = 4 ^ (2, 4).
The slope of the line joining (0, 0) and (2, 4) is m = (4 - 0)/(2 — 0) = 2. Hence, an equation of the line is
>- - 0 = 2(x - 0)
y = 2x.

Section P.2 Linear Models and Rates of Change 11

71. m, =

1 -0

-2-(-l)

-2- 0 _ _2
"^"l-C-D" 3

The points are not collinear.
73. Equations of perpendicular bisectors:

c a — b

c a + b

a + b

b- a

Letting ;<: = 0 in either equation gives the point of inter-
section:

0,

2c j

This point lies on the third perpendicular bisector, ^ = 0.

(-a,0)

75. Equations of altitudes:
a- b,

-ix + a)

x = b

a + b. ,

y = U - a)

Solving simultaneously, the point of intersection is
.2_ ^\

[-'-

77. Find the equation of the line through the points (0, 32) and (100, 212).

180 9
"^ ~ 100 ~ 5

f - 32 = f (C - 0)

f = f C + 32

5F - 9C - 160 = 0

For F = 72°, C = 22.2°.

79. (a) W, = 0.75j: + 12.50

W. = 1.30.t + 9.20

(c) Both jobs pay $17 per hour if 6 units are produced.
For someone who can produce more than 6 units per
hour, the second offer would pay more. For a worker
who produces less than 6 units per hour, the first oifer
pays more.

(b) 50

Using a graphing utility, the point of intersection is
approximately (6. 17). Analytically,

0.75.x- + 12.50 = 1.30.V + 9.20

3.3 = 0.55.V => .V = 6

y = 0.75(6) + 12.50 = 17.

12 Chapter P Preparation for Calculus

81. (a) Two points are (50, 580) and (47, 625). The slope is

625-580
'"= 47-50 =-''■

p - 580 = - \5{x - 50)

p= - \5x + 750 + 580 = - \5x + 1330

or .r = 15(1330 - p)

(b) =0

Ifp = 655, ;t = -^(1330 - 655) = 45 units,
(c) \£p = 595, X = n(1330 - 595) = 49 units.

83. 4r + 3v - \Q = Q^>d =

_ |4(0) + 3(0) - 10| _ 10 _

J A- + r-

i5. x-y-2 = Q:

|l(-2) + (-l)(l)-2| ^^

VI- + r- V2

5./2

87. A point on the line x + y = 1 is (0, 1). The distance from the point (0, 1) to x + y — 5 = 0 is

11(0) + 1(1) -51 11-51 4 ,-

d =

JV- + 1-

v/2 ./2

89. If A = 0, then fiy + C = 0 is the horizontal line y = — C/B. The distance to {x^, yj is

|5y, + C\ JAri + By, + C\

d =

yi

-c

B

\B\ J A- + B-

If B = 0, then Ar + C = 0 is the vertical line x = — C/A. The distance to (xj, yj is

|Aci + C| |Aci + Byi + C|

d =

-(^

|A| ^M2 + B2

(Note that A and B cannot both be zero.)

The slope of the line Ar + By + C = 0 is —A/B. The equation of the line through ix^^, y^ perpendiailar
to Ac + By + C = 0 is:

y~yi = -^yx - x{\

Ay — Ayi = Bx — Bxj

BX[ — .4y| = Bx — Ay
The point of intersection of these two lines is:

Ac + By = -C =^ A-X + ABy = -AC (1)

Bx - Ay = Bx^ - Ay^ => B-x - .4By = B-x, - .4Byi (2)

(A' + B-)x = -AC + B-Xi - AByj (By adding equations (1) and (2))
-AC + B-x^ - AByi

A- + B^
Ac + By =-C => ABx + B2y=-BC (3)

Bx - Ay = Bx, - Avi^ -AB^ + A-y = -ABX) + A-y, (4)

(A- + B-)y = -BC - ABx^ + A^y, (By adding equations (3) and (4))
-BC - ABxi + Ahi^

y =

A- + s-

— CONTINUED—

Section P.2 Linear Models and Rates of Change 13

89. — CO>mNUED—

(-AC + B-x.- ABv, -BC - ABx.+ A'^.\ . ^.

[ ^2 + gi ' ^2 + g2 j point of intersection

The distance between (.r,, y,) and this point gives us the distance between (x,, y\) and the line Ax + By + C = 0.

d= ,

-AC + B-x, - ABy^ _

A^ + S^

-BC - ABxj + Ay

1 _

A- + B-

-AC - ABy, - A-.t,
A= + B-

-BC - ABXj - Bh\

A~ + B-

-A(C + flvi + AC|)
A- + B'

(A^ + B-)[C + At, + By,Y

{A- + B-y

-B(C + /U) + By,)
/12 + B-

^ lAx, + gy, + C\

91. For simplicity, let the vertices of the rhombus be (0, 0),
(a, 0), (b, c), and (o + b, c), as shown in the figure. The
slopes of the diagonals are then

and in.

a + b

b - a

Since the sides of the Rhombus are equal, a^ = ir + c^,
and we have

ib.c) (a+b.c)

(0.0) (a.O)

c c c c

iinu - — — - • 7 - -7 ^ - ^ = - 1.

a + b b - a b — a- - c-

Therefore, the diagonals are perpendicular.

93. Consider the figure below in which the four points are
collinear. Since the triangles are similar, the result imme-
diately follows.

yi - yC yi - >'i

"^ -^1 -^2

95. True.

ax + by = c, => V = -"T-t "*" T"
b b

bx — a\ = C-. => V = — .r '

a a

n,, = --

nu= —

1

14 Chapter P Preparation for Calculus

Section P.3 Functions and Their Graphs

1. (a)/(0) = 2(0)-3 = -3
(b)/(-3) = 2(-3)-3 = -9

(c) fib) = 2i - 3

(d) f(x - 1) = 2(.r - 1) - 3 = It - 5

3. (a) g(0) = 3 - 02 = 3

(b) g(V3) = 3 - (^3)' = 3-3 = 0

(c) g(-2) = 3-(-2)2 = 3-4= -1

(d) g(r - 1) = 3 - (f - 1)2 = -f2 + 2t + 2

5. (a) /(O) = cos(2(0)) = cos 0 = 1

(b,/(-f =c.s2-f =

cosi -yl = 0

'"A!) -4(f)) --3

2£= _i

2

Ai Ax Ax -•(,;.

9.

/(x)-/(2) (i/v^rni-i)

X - 2

X - 2

1 - Vx - 1 1 + Vx - 1

2 -X

(x - 2)Vx - 1 1 + Vx- 1 (x - 2)Vx - 1(1 + Vx - 1) Vx - 1(1 + Vx - 1)

X5t2

11. h{x) = - Vx + 3

Domain: x + 3 > 0 => [-3, 00)
Range: (—00, 0]

13. fit) = sec

77t

Domain: all r t^ 4A' + 2, ^' an integer
Range: (— 00, — 1], [1, 00)

15. fix) = -

X

Domain: (— 00, 0). (0, 00)
Range: (-00, 0), (0, 00)

17. fix) =

2r + l,x < 0
2x + 2, X > 0

(a) /(-1) = 2(-1) + 1 = -1

(b) /(O) = 2(0) + 2 = 2

(c) /(2) = 2(2) + 2 = 6

(d) fif- + 1) = 2(r2 + 1) = 2?2 + 4
(Note: t2 + 1 > 0 for all f)
Domain: (—00, 00)

Range: ( — do, 1), [2, 00)

19. fix) =

\x\ + l,x < 1
-X + l,x > 1

(a) /(-3)= |-3| + 1=4

(b) /(1)= -1 + 1=0

(c) /(3) = -3 + 1 = -2

(d) /(fe'+l)= -(^+ 1)+ 1
Domain: (—00, 00)

Range: (- 00, 0] U [1, 00)

-fc2

Section P.3 Functions and Their Graphs 15

21. fix) = 4- X

Domain: (-00,00)
Range: (-00, 00)

23. h(x) = TT^n"
Domain: [1, oc)
Range: [0, ocj

25. fix) = V9 - x'^
Domain: [—3, 3]
Range: [0,3]

27. ^(f) = 2 sin vt
Domain: ( — oo. oc)
Range: [-2,2]

29. jc - y 2 = 0 => y = ± VI

y is not a function of .t. Some vertical lines intersect
the graph twice.

33. x^ + /

•y

±Va^.

y is not a function of .r since there are two values of v for
some X.

31. V is a function of x. Vertical lines intersect the graph
at most once.

35. ^^ = X-- \

■y = ±v.T-

y is not a function of x since there are two values of >■ for
some X.

37. fix) = |x| + \x-2\

If -T < 0, then/(.r) = -x - U - 2) = -2x + 2 = 2(1 - x).

If 0 < .t < 2, then/(A:) = .t - (.t - 2) = 2.

If X > 2, then/(x) = x + (x - 2) = Ir - 2 = 2fx - 1).

Thus,

[2(1 - x), X < 0

fix) =2, 0 < X < 2.

l2(x - 1), X > 2.

39. The function is ^x) = car. Since (1, —2) satisfies the
equation, c = - 2. Thus, gCr) = - 2xr.

41. The fimction is r{x) = c/x. since it must be undefined at
X = 0. Since (1, 32) satisfies the equation, c = 32. Hius.
rix) = 32/x.

43. (a) For each time t. there corresponds a depth d.
(b) Domain: 0 < r < 5
Range: Q < d < 30

12 3 4 5 6

16 Chapter P Preparation for Calculus

47. (a) The graph is shifted
3 units to the left.

(c) The graph is shifted
2 units upward.

(e) The graph is stretched
vertically by a factor of 3.

r-H 1—
2 4

(b) The graph is shifted
1 unit to the right.

(d) The graph is shifted
4 units downward.

(f) The graph is stretched
vertically by a factor
ofi

49. (a) y = v^x + 2

12 3 4

Vertical shift 2 units upward

(b) y = - v^

Reflection about the ;c-axis

(c) y = V7^^

Horizontal shift 2 units to the
right

51. (a) 71(4) = 16°, 7tl5) - 23°

(b) li H(t) = T{t — 1), then the program would turn on (and off) one hour later.

(c) If //(f) = T{t) - 1, then the overall temperature would be reduced 1 degree.

53. fix) = x\ g{x) = v^

(/ ° g)U) = figix)) = f{V^) = i^f = X, x>0
Domain: [0, oo)

(g 'f)(x) = g{f(x)) = g(;c2) = v<? = |;t|
Domain: (— oo, oo)
No. Their domains are different. {f°g) = (g -/) for x > 0.

55. fix) = -, gix) = ^2 - 1

if'g)i^=figix))=fix^-\)

Domain: all j: # ± 1

(.^/)(.)^.(/(.)) = .g) = gf- 1=1-1=^

Domain: all jc ^^ 0

No,/"g#go/.

Section P.3 Functions and Their Graphs 17

57. {A ■> r)it) = A{r{t)) = A(0.6f) = TT(0.6f)2 = 0.367rt2
{A ' r}{t) represents the area of the circle at time t.

59. f{-x) = (-x)H4 - i-xY) = .t2(4 - x^) =/(jc)
Even

61. f{-x) = (-x)cos(-x) = -jfcosjr = -/W
Odd

63. (a) If/ is even, then (f , 4) is on the graph.

65. f(-x) = a^^ti-x)^"^' + • • • + a3{-jc)3 + a,(-x)
= -K + i^""^' + • ■ ■ +a,x^ + a,x]
= -fix)
Odd

67. Let F(;c) = f(x)g{x) where/ and g are even. Then

F{-x) = /(-.r)5(-.r) = /WgW = F (x).
Thus, F(x) is even. Let FU) = f{x)g{x) where/and g are odd. Then

F(-x) =f{-x)g(-x) = [-f(x)l-g(x)] =f(x)g(x) = Fix).
Thus, F(x) is even.

(b) If/ is odd, then (j, —4) is on the graph.

69. fix) = jc' + 1 and gix) = jt^ are even.
fix)gix) = ix^ + l)(j^) = x« + x^ is even.

5

/(x) = a:^ — x is odd and g{x) = x- is even.
/WgW = ix^ - x)(x^) =x^ - x^h odd.

71. (a)

X

length and width

volume V

1

24 - 2(1)

484

2

24 - 2(2)

800

3

24 - 2(3)

972

4

24 - 2(4)

1024

5

24 - 2(5)

980

6

24 - 2(6)

864

(b)

The maximum volume appears to be 1024 cm',
(c) V = x(24 - 2x)- = 4.t(12 - x)-
Domain: 0 < a < 12

Yes, V is a function of .v.

(d) "00

Maximum volume is V = 1024 cm' for box having
dimensions 4 x 16 x 16 cm.

73. False; let/(x) = x^.

Then/(-3) =/(3) = 9, but -3 9^ 3.

75. True, the function is even.

18 Chapter P Preparation for Calculus

Section P.4 Fitting Models to Data

1. Quadratic function

3. Linear function

5. (a), (b)

250--
200- -

H i 1 1 h*-'

3 6 9 12 15

Yes. The cancer mortality increases linearly with
increased exposure to the carcinogenic substance.

(c) If.:i: = 3, then>'== 136.

7. (a) d = 0.066For F = 15. W + 0.1

(b) 125

The model fits well,
(c) If F = 55, then d » 0.066(55) = 3.63 cm.

9. (a) Let x = per capita energy usage (in millions of Btu)

y = per capita gross national product (in thousands)

y = 0.0764a; + 4.9985 = 0.08x + 5.0

r = 0.7052

(b) «

(c) Denmark, Japan, and Canada

(d) Deleting the data for the three countries above,
y = 0.0959X + 1.0539

(r = 0.9202 is much closer to L)

11. (a) y, = 0.0343f3 - 0.3451^2 + 0.8837t + 5.6061
y2 = 0.1095r + 2.0667
y^ = 0.0917? + 0.7917

(b) 15

y'i^y.

*h

N\ _^.r^^^

nn ^

^

■ ■■'■■

For 2002, t = 12 and ^i + jj + jj = 3 1.06 cents/mile

13. (a) y^ = 4.0367r + 28.9644

^2 = -0.0099;' + 0.5488?^ + 0.2399f + 33.1414
(b) ™

>! =4.04r+29.0

^

^

W^

25

>, = -O.OlOl^ + 0.549l^ + 0.24( + 33. 1

(c) The cubic model is better.

(d) ^3 = 0.4297f2 + 0.5994t + 32.9745

(e) The slope represents the average increase per year
in the number of people (in millions) in HMOs.

(f) For 2000, r = 10, and yj = 69.3 million. Oinear)
yj ~ 80.5 million (cubic)

Review Exercises for Chapter P 19

15. (a) y = - 1.81;c3 + 14.58;t2 + 16.39x + 10

(b) 300

(c) Ifx = 4.5, y = 214 horsepower.

17. (a) Yes, > is a function of t. At each time t, there is one
and only one displacement y.

(b) The amplitude is approximately

(2.35 - 1.65)/2 = 0.35.

The period is approximately

2(0.375 - 0.125) = 0.5.

(c) One model is y = 0.35 sin(4Trf) + 2.
(d)

19. Answers will vary.

Review Exercises for Chapter P

1. >- = 2x - 3

;c = 0 ^> >■ = 2(0) - 3 = -3 =* (0, -3) y-intercept

y = 0:^0 = 2x-3=>jt

(5, 0) .r-intercept

3.y =

X - 1

x-2

X = 0:

y = 0=4>0 =

0 - 1 _ ]_
0-2 2

X - 1

0, -jr I y-intercept

V = 1 =» ( 1 , 0) .T-intercept

5. Symmetric with respect to y-axis since
{-xYy - (-x)' + 4y = 0
xry — X- + 4y = 0.

1 1 _i_ 3

7. y = — 2.V + 2

- vt + fi.V = 1

JA- + y = T

11. V = 7 - dt - X-

y = f .r +

Slope:

y-intercept: j

20 Chapter P Preparation for Calculus

13. y = 75 -;c

Domain: (— oo, 5]

15. y = Ax^- 25

Xmin = -5
Xmax = 5
Xscl = 1
Ymin = -30
Ymax = 10
YscI = 5

17.

3a:-

Ay =

8

Ax +

Ay =

20

Ix

=

28

X =

4

y =

1

Point

: (4, 1)

19. You need factors (x + 2) and {x — 2). Multiply by x to obtain origin symmetry

y = x(x + 2)(x - 2)_
= x^ — 4x.

21.

12 3 4 5

Slope

(5/2) - 1 ^ 3/2 _ 3
5 - (3/2) 7/2 7

23.

1 - t

1 - 5

1-0 1 - (-2)
l-r=-^

25. 3,_(_5) = i(;,_o)

y = F ~ 5

2>' - 3x + 10 = 0

27. y - 0 = -f(x - (-3))

y = -3X - 2
3y + 2x + 6 = 0

Review Exercises for Chapter P 21

29. (a) y - 4 = ~(x + 2)
lo

16y - 64 = 7x + 14

0 = 7.ir - \6y + 78
.X 4-0

y = -2x
2x + y = Q

(b) Slope of line is — .

>- - 4 = jU + 2)

3v - 12 = 5j: + 10

0 = 5;t - 3 V + 22

(d) x=-l

x + 2 = 0

31. The slope is -850. V = -850f + 12,500.
V(3) = -850(3) + 12,500 = $9950

33. a: - r = 0

y = ±Vx

Not a function of j: since there are two values of >> for
some X.

35. y = x^-2x

Function of j: since there is one value oiy for each x.

37. fix) = -

(a) /(O) does not exist.

1

(b)

/(I + Ax) -/(I) ^ 1 + Ax 1 ^ 1 - 1 - A.X:
Ai- Ar (1 + At) Ax

-1

1 + Ar

, Ax9t -1,0

39. (a) Domain: 36 - .v- > 0 ^. -6 < x < 6 or [-6, 6]
Range: [0, 6]

(b) Domain: all x =^ 5 or (-oc. 5). (5, ex:)
Range: all >• ^ 0 or (- oo. 0), (0, oc)

(c) Domain: all x or (-oc, oc)
Range: ally or ( — oc, oo)

41. (a) fix) = x^ + c,c = -2, 0, 2

c = 0

(b) /(x) = (x - c)-\ c = -2,0.

—CONTINUED—

22 Chapter P Preparation for Calculus

41. —CONTINUED—

(c) fix) = ix- 2)3 + c,c

-2,0,2

(d)/(;c) = cx\c= -2,0,2

43. (a) Odd powers: fix) = x, gix) = x^,hix) = x^

r

/

Even powers: fix) = x^, gix) = x*, hix) = j:*

-rA-

The graphs of/, g, and h all rise to the right and fall to The graphs of/, g, and h all rise to the left and to the

the left. As the degree increases, the graph rises and right. As the degree increases, the graph rises more

falls more steeply. All three graphs pass through the steeply. All three graphs pass through the points (0, 0),

points(0,0), (1,1), and (-1,-1). (1, 1), and (- 1, 1).

(b) y = x'' will look like hix) = j^, but rise and fall even more steeply.

y = x^ will look like hix) = x^, but rise even more steeply.

45. (a)

(b) Domain: 0 < jc < 12

2jc + 2>' = 24 . , ,

- )- = 12 - X ,. . ;

A = xy = x(12 - x) = I2x - x^

47. (a) 3 (cubic), negative leading coefficient

(b) 4 (quartic), positive leading coefficient

(c) 2 (quadratic), negative leading coefficient

(d) 5, positive leading coefficient

(c) Maximum area is /I = 36. In general, the maximum
area is attained when the rectangle is a square. In this
case, X = 6.

49. (a) Yes, y is a function of t. At each time t, there is one
and only one displacement y.

(b) The amplitude is approximately

(0.25 - (-0.25))/2 = 0.25.

The period is approximately 1.1.

(c) One model is y

(d) 05

^cos(Yjrj»^cos(5.7f)

Problem Solving for Chapter P 23

Problem Solving for Chapter P

1- (a) ;,2 _ 6;c + y2 _ 8>' = 0

U2 ~ 6x + 9) + {y^-8y+ 16) =9 + \6

U - 3)2 + (y - 4)2 = 25

Center: (3,4) Radius: 5

4-0 4

(c) Slope of line from (6, 0) to (3, 4) is 7 = --.

3 — 6 3

Slope of tangent line is -. Hence,

3 3 9

>" — 0 = —(x — 6) => y = T-T - - Tangent line

(b) Slope of line from (0, 0) to (3, 4) is - Slof>e of tangent line

is -^ Hence,

4

(d)

y-0
3

Hx - 0)

y = --X Tangent line

':x = ':x — —

4 4 2

3

7

3 9
::x = —

2 2

X = 3
Intersection: I 3,

3. H(x) =

1 ;t > 0
0 ^ < 0

I I I 1— ^— i 1 1 h-

" " 12 3 4

(a) H(x) - 2

-4 -3 -2 -1

(b) H(x - 2)

4--
3--

2
I-

-2--
-3--

(c) -H{x)

-2--
-3

-4— I 1 I II— I 1 1 — 1-»-

-4-3-2-1 - - -

(d) H(-x)

4- ■
3--

2- ■

^ — I — I — 6 I — f

J _T _T _l II

-4 -3 -2 -1

-2
.3..

12 5 4

(e) WU)

(f) -H(x - 2) + 2

4-

-4 -3 -2 -1

I I II 1 1 1 >-*-:

-2-
-3-

H 1 — 1 h^

-2-
-3--

24 Chapter P Preparation for Calculus

5. (a) X + ly = 100 =* y =

100 -a:

/100-x\ x-

A(x) = xy = x\ ^ 1 = -— + 5Qx

Domain: 0 < x < 100

(b) 1600

7. The length of the trip in the water is V2^ + x^, and the
length of the trip over land is Vl + (3 - x)^. Hence,
the total time is

_ VaTI? , VI + (3 - xY ,

T = r 1 hours.

Maximum of 1250 m- at j: = 50 m, >> = 25 m.
(c) A{x) = -|(x2 - mx)

= -|(x2 - lOOx + 2500) + 1250

= -|U - 50)2 + 1250
A(50) = 1250 m- is the maximum, x = 50 m, y = 25 m.

9-4
9. (a) Slope = = 5. Slope of tangent line is less than 5.

4 - 1

(b) Slope = _ = 3. Slope of tangent line is greater than 3.

4.41 - 4

(c) Slope = -r-j — = 4. 1 . Slope of tangent line is less than 4. 1 .

(d) Slope

2.1 - 2

/(2 + ;/)-/(2)

{2 + h) -2

(2 + hf- A
h

^ Ah + h^

" h .

= A + h,hi-0

(e) Letting h get closer and closer to 0, the slope approaches 4. Hence, the slope at (2, 4) is 4.

11. (a) At X = 1 and x = — 3 the sounds are equal.
/ 2/

(h)

vVT7 Jix - 3)2 + y-
U - 3)2 + y2 = 4(^2 + y2)

3^2 + 3y2 + 6x = 9

^2 + 2x + / = 3

(X + 1)2 + y2 = 4

Circle of radius 2 centered at (— 1, 0)

Problem Solving for Chapter P 25

13. d^d^ =

[U + 1)^- + ylS.x - \Y + f] =

{x + DHx - 1)2 + y\{x + 1)= + (jc - 1)2] + / =

{x^ - 1)2 + y^[2x^ + 2] + / =

x'^ -2x- + \ + 2xY + 2v2 + / =

(x2 + /)2 = 2(;c2 - /)
Let V = 0. Then x* = 2x- =^ x = 0 or x- = 2.
Thus, (0, 0), {V2, O) and (- V2, o) are on the curve.

(->/2.0)
\

ee

/

(0.0)

CHAPTER 1
Limits and Their Properties

Section 1.1 A Preview of Calculus 27

Section 1.2 Finding Limits Graphically and Numerically 27

Section 1.3 Evaluating Limits Analytically 31

Section 1.4 Continuity and One-Sided Limits 37

Section 1.5 Infinite Limits 42

Review Exercises 47

Problem Solving 49

CHAPTER 1
Limits and Their Properties

Section 1.1 A Preview of Calculus

Solutions to Odd-Numbered Exercises

1. Precalculus; (20 ft/sec)( 15 seconds) = 300 feet

3. Calculus required: slope of tangent line at.ic = 2 is rate of
cliange, and equals about 0.16.

5. Precalculus: Area = \bh = \{5){3) = ysq. units

7. Precalculus: Volume = (2)(4)(3) = 24 cubic units

9. (a)

(b) The graphs of yj are approximations to the tangent line to Vi at j: = 1.

(c) The slope is approximately 2. For a better approximation make the list numbers smaller:

{0.2,0.1,0.01,0.001}

11. (a) D^ = V(5 - 1)' + (1 - 5)- = 716 + 16 « 5.66

(b)D, = yrr(f + 7i + (f-fr + 7i + (f-!r + Vi + (i-ir

« 2.693 + 1.302 + 1.083 + 1.031 =6.11
(c) Increase the number of line segments.

Section 1.2 Finding Limits Graphically and Numerically

1.

X

1.9

1.99

1.999

2.001

2.01

2.1

/w

0.3448

0.3344

0.3334

0.3332

0.3322

0.3226

Iim3

x->2 XT — X — l

~ 0.3333 (Actual limit is |.)

X

-0.1

-0.01

-0.001

0.001

0.01

0.1

fix)

0.2911

0.2889

0.2887

0.2887

0.2884

0.2863

lim ^•^"^^ ^ « 0.2887 (Actual limit is 1/(2^1).)

x-fO X

27

28 Chapter 1 Limits and Their Properties

X

2.9

2.99

2.999

3.001

3.01

3.1

fix)

-0.0641

-0.0627

-0.0625

-0.0625

-0.0623

-0.0610

lim ^^/('+jy^-(y^) ^ _o.0625 (Actual limit is -j-,)

7.

X

-0.1

-0.01

-0.001

0.001

0.01

0.1

fix)

0.9983

0.99998

1.0000

1.0000

0.99998

0.9983

lim = 1.0000 (Actual limit is 1.) (Make sure you use radian mode.)

x-»0 X

9. lim (4 - ;c) = 1

x~*3

11. lim/W = lim (4 - jc) = 2

jr-»2 x->2

13. lim

\x- 5\

does not exist. For values of j: to the left of 5, \x — 5|/(j: — 5) equals - 1,

5 X — 5

whereas for values of a- to the right of 5, \x — 5\/ix - 5) equals 1 .

15. lim tan x does not exist since the fimction increases and 17. lim cos(1/j:) does not exist since the function oscillates

decreases without bound as x approaches tt/2.

between — 1 and 1 as x approaches 0.

19. C(r) = 0.75 - 0.50|-(/- 1)1
(a)

(b)

t

3

3.3

3.4

3.5

3.6

3.7

4

C

1.75

2.25

2.25

2.25

2.25

2.25

2.25

lim Cit) =

r->3.5

2.25

(c)

t

2

2.5

2.9

3

3.1

3.5

4

c

1.25

1.75

1.75

1.75

2.25

2.25

2.25

lim Cit) does not exist. The values of C jump from 1.75 to 2.25 at r = 3.
r->3

21. You need to find 5 such that 0 < |x - 1 1 < 5 implies

l/W-i| =

-- 1

X

< 0.1. That is,

-0.1 < -- 1 < 0.1

X

1 - 0.1 < - < 1 + 0.1

X

A 1 ii

10 ^ ;c "^ 10
10 10

So take 5 = —-. Then 0 < |x - 1 1 < 5 implies

1 , 1

"Tf < -'^ - 1 < 9-

Using the first series of equivalent inequalities, you obtain
1

l/W

< e < 0.

'0 , , 10 ,

1 , 1

9>--l>-H-

Section 1 .2 Finding Limits Graphically and Numerically

29

23. lim (3;c + 2) = 8 = Z,

|(3a + 2) - 8| < 0.01

\'ix - 6| < 0.01

3|jt - 2| < 0.01

0 < |.T - 2| < ^ » 0.0033 = 5

Hence, if 0 < |a; - 2| < S

3|x- 2| < 0.01

\3x - 6| < 0.01

|(3.v + 2) - 8| < 0.01

\f{x) - L\< 0.01

0.01

you have

25. lim (jc^ - 3) = 1 = L

x—*2

|U2 - 3) - 1| < 0.01

\x'^ - 4| < O.OI

|(* + 2){x - 2)1 < 0.01

\x + 2| |.r - 2| < 0.01
0.01

' ' k + 2|

If we assume 1 < x < ?>, then 5 = 0.01/5 = 0.002.
Hence, if0< \x - 2\ < S = 0.002, you have

1^ - 2| < 0.002 = y(0.01) < , I ^,(0.01)

|jc + 2||a:- 2| < 0.01

\x"- - 4| < 0.01

\{x^ - 3) - 1| < 0.01

\f{x) - L\ < 0.01

27. lim (x + 3) = 5

jc->2

Given e > 0:

\{x + 3) - 5| < e

|.r -2| < e= 6

Hence, let S = e.

Hence, if 0 < |.r - 2 1 < 5 = e. you have

\x - 2\ < €

\{x + 3) - 5| < 6
\f(x) -L\<e

29. lim i^x- l) =i(-4)-l = -3

J— »-4

Given e > 0:

\{\x- l)-(-3)| < e

k.+

2 < 6

k\x-{-4)\ < e
\x - (-4)1 < 2e
Hence, let 6 = 2e.

Hence, if 0 < |.r - (-4)| < S = 2e, you have
|;c- (-4)1 < 26
lit + 2| < e
\{kx - l) + 3| < 6
l/W -L\<e

31. lim 3 = 3

x-^6

Given e > 0:

|3-3| < 6
0 < e
Hence, any 8 > 0 will work.
Hence, for any 5 > 0, you have

|3-3|<e
l/W -L\<e

33. Iim4/x = 0

.t-.0

Given e > 0: |.i^ - o| < e
\Vx\ <e

\x\ < e^ = 5
Hence, let 6 = e\

Hence for 0 < |.v — 0| < 6 = e\ you have
|.r| < r'

\^-0\ <€
1/U) -L\<e

30 Chapter 1 Limits and Their Properties

35. lim \x - 2| = |(-2) - 2| = 4
Given e > 0:

\\x - 2| - 4| < 6
|-(;c-2) -4| < e (x - 2 < 0)
\-x-2\ = |;«: + 2| = |;c - (-2)| < e
Hence, 8 = e.

Hence forO < |jr - (-2)| < 5 = e, you have

\x + 2\ < e

\-{x + 2)\ <6

|-U-2)-4| < e

||x - 2| - 4| < e (because a: - 20)

l/W - L| < e

37. lim (x^ + 1) = 2
Given e > 0:

\(x^ + 1) - 2| < e

\x2- 1| < 6

\ix + l)(x - 1)1 < e

U - 1 <

|x+l|
If we assume 0 < a: < 2, then 8 = e/3.

Hence for 0 < |jr - 1 1 < 5 = -, you have

II 1 1

1 < 6

|U= + 1) - 2| < e
l/W - 2| < e

^, , Vx + 5 - 3
^(^^- .-4

lim/W = 1

0

s

The domain is [-5, 4) U (4, oo).

The graphing utility does not show the hole at (4, g)

41. f(x) =

x-9

lim /(a;) = 6

JT— ♦Q

The domain is all x > 0 except x = 9. The graphing
utility does not show the hole at (9, 6).

43. lim/(jc) = 25 means that the values of/ approach 25 as jc gets closer and closer to 8.

45. (i) The values of/ approach different
numbers as x approaches c from
different sides of c:

-4 -3 -2 -1

H — I — I — 1-»-
12 3 4

(ii) The values of/ increase with-
out bound as x approaches c:

(iii) The values of/oscillate

between two fixed numbers as
x approaches c:

4-.
3--

, 47. fix) = (1 + xY

/x

lim(l +xy'' = e« 2.71828

x

/w

X

fix)

-0.1

2.867972

0.1

2.593742

-0.01

2.731999

0.01

2.704814

-0.001

2.719642

0.001

2.716942

-0.0001

2.718418

0.0001

2.718146

-0.00001

2.718295

0.00001

2.718268

-0.000001

2.718283

0.000001

2.718280

12 3 4 5

Section 1.3 Evaluating Limits Analytically 31

49. False; /(x) = (sinx)/jcis
undefined when x = 0.
From Exercise 7, we have

hm = 1.

j:->0 X

51. False; let

/w =

/(4) = 10

S3. Answers will vary.

X- - 4x, X ^ 4
la X = 4'

lim/(x) = lim (x^ - 4x) = 0 i= 10

55. If lim/(x) = L, and lim/(x) = L^, then for every e > 0, there exists S, > 0 and 8-, > 0 such that Ix - cl < 5,

.i->c x->c '

|x — c| < So => |/(x) — LjI < 6. Let S equal the smaller of 5, and Sj. Then for |x — c| < 5, we have

|Li - L2I = |L, -/(x) +/(x) - L2I < |L, -/(x)| + |/(x) - L2I < e + e.

H/(xJ-L,| < eand

Therefore, |Li - L2I < 2e. Since e > 0 is arbitrary, it follows that Lj = Z^.

57. lim [fix) - Z,] = 0 means that for every e > 0 there exists S > 0 such that if

0 < Ix - cl < 5,

then

|(/(x) - L) - 0| < 6.

This means the same as |/(x) - L\ < e when
0 < |x - c| < 5.

Thus, lim/(x) = L.

Section 1.3 Evaluating Limits Analytically

\

1

J

(a) lim h(x) = 0

-v— *5

(b) lim li(x) = 6

13 .t->-l

4

(a) lim/Cx) = 0

x—*0

(b) lim f(x) - 0.524

.t— >Tr/3

(=f)

h(x) = X- - 5x
5. lim.T^ = I'' = 16

fix) = X cos X

7. lim (2x - 1) = 2(0) - 1 = - 1

x->0

9. lim (x^ + 3x) = (-3)2 + 3(-3) = 9-9 = 0

11. lim (2x2 + 4x + 1) = 2(-3)2 + 4(-3) + 1 = 18 - 12 + 1 = 7

13. lim - = I

jr-»2X 2

15. lim

X - 3 1-3

.v^ix= + 4 1- + 4 5

17. lim

_5x 5(7) ^ 35 ^ 35

"" ./FT2 ~ 77 + 2 79 3

19. lim v'x + 1 = V 3 + 1 = 2

-V-.3

32 Chapter 1 Limits and Their Properties

21. lim (x + 3)- = (-4 + 3)- = 1

x-»-4

23. (a) liin/(x) = 5-1=4

(b) lim g{x) = 43 = 64

(c) limg(/W) = g(/(l)) = g(4) = 64

x->l

25. (a) lim/(x) =4-1=3

(b) lim g{x) = 73 + 1 = 2

(c) lim g{f{x)) = g(3) = 2

X->1

27. lim sin jc = sin — = 1

x->7r/2 2

_„ ,- "fx 7r2 1

29. lim cos -r- = cos -^r- = — —

x^2 3 3 2

31. lim sec 2j: = sec 0 = 1

jr->0

„- ,. . . Stt 1

33. lim sm X = sm -7- = —

JC— *5ir/6 6 2

35. lim tan — - = tan — —

x-,3 \ 4 4

= -1

37. (a) lim[5gW] = 5 Urn gix) = 5(3) = 15

(b) lim L/U) + g{x)] = lim/U) + lim gW = 2 + 3 = 5

(c) lim lf{x)g{x)] = riim/(x)]riim g{x)] = (2)(3) = 6

x~^c Ix-^c ilx—*c J

(d) lim ^ = ^^^-7T = ^
x-.cgU) hmgU) 3

39. (a) lim [f{x)f = [lim/(x)? = (4)^ = 64

(b) lim v{70c) = /lim/W = v^ = 2

j:->c v x->c

(c) lim [3/W] = 3 lim/(x) = 3(4) = 12

X— >c x—*c

id) lim[/(x)?^^ = riim/(x)f ^ = (4)V2 = 8

x—*c Lj:— >c J

— 2x^ + X
41. /(x) = - 2jc + 1 and g(j:) = agree except at

jc = 0. ^

(a) lLmg(x) = lim/(x) =1

(b) lim g(x) = lim f(x) = 3

X— > — I X— >-l

43. fix) = xix + 1) and gix) = — — - agree except at x = 1.

(a) lim gix) = lim/(x) = 2
x— >i x— >i

(b) lim g(jc) = lim fix) = 0

X—¥—l X^*-l

X- - 1
45. fix) = — — — and gix) = x - I agree except at x

lim /(jc) = lim gix) = —2

x-*—l x-*-l

x^ - 8
47. fix) = — and gix) = x- + 2j: + 4 agree except at

x = 2.

Vim fix) = lim gix) = 12

X— >2 .r— >2

49. lim

X- 5

lim-

X- 5

x.^5 x^ - 25 X-.5 (x + 5)(x - 5)

= lim

1

1

5X + 5 10

CI r x^ + x-6 ,. ix + 3)(x - 2)
51. lim — , — = Iim

x^-i x^-9

-3 ix + 3)ix - 3)

,. x-2 -5 5
= lim = —7 = 7

Jr-»-3 JC — 3 —6 6

Section 1.3 Evaluating Limits Analytically 33

., ,. V.r + 5 - Vs ,. VxTl - J5 V^ + 5 + 75
53. lim = hm • , 1=

'^0 X x->0 X Jx + 5 + vO

= lim

U + 5) - 5

lim

1

1 75

x^Ox{jx + 5+75) -r->o 7;c + 5 + 75 275 10

,. ,. 7jc + 5 - 3 ,. Jx + 5 - 3 jm + 3
55. lim = hm • ,

x-^i X - 4 x^i X - 4 7^ + 5 + 3

U + 5) - 9 ,. 1

x^4 (x - 4)(7.r + 5 + 3) x^4 Jx + 5 + 3 79 + 3 6

1 1

2 - (2 + x)

._ ,. 2+ x 2 ,. 2(2 +;<:) ,. -1 1

57. hm = hm — ^^ — = lim --; r = --

x-*o x .T-*o X x~*o 2(2 + x) 4

._ ,. 2{x + Ax) - 2x ,. 2x + 2Ax - Ix ,. ^ ,

59. hm -^^ —^ = hm 1 = hm 2 = 2

^x->o A.V A.v->o A.r \x->o

^, ,. {x + Ax)- - 2{x + Ax) + 1 - (.t- - 2;c + 1) ,. x^ + 2xAx + (Ax)- - 2v - 2Ajc + 1 - x= + 2x - 1
61. hm : = lim :

Ai->0 Ax A.x->0 Ax

= lim (2x + Ar - 2) = 2x - 2

„ ,. 7x + 2 - 72 -,,.
63. lim « 0.354

Jr->0 X

X

-0.1

-0.01

-0.001

0

0.001

0.01

0.1

fix)

0.358

0.354

0.345

?

0.354

0.353

0.349

A , • „ >■ v^^"+^- 72 ,. 7^rr2- 72 v^m+ v/2

Analytically, hm = hm • , - — —

x^o X x^o X J7T2 + 72

,. X + 2 - 2 ,.

= lim —, — , ^r = hm

1

./2

0 x(7x + 2+72) --^^o Vx + 2 + v'l 2v'2 4

= 0.354

1 1

65. lim

.t->0

2 +x

X

-0.1

-0.01

-0.001

0

0.001

0.01

0.1

fix)

-0.263

-0.251

-0.250

7

-0.250

-0.249

-0.238

^

1 1

Analytically, lim

2 + X 2

x-*0 X

2-(2+.r) 1 ,. -X 1 ,. -1 1

hm — Z7Z ; — ■ ~ = hm —rz r • — = lim ttt r = — -7-

X .«-^o2(2+.v) X .t-K)2(2 + .x) 4

% 2(2 +.v)

34 Chapter 1 Limits and Their Properties

.„ ,. sin a; ,.
67. lim —1 — = lim
.t->o 5jc j->o

sinx

X

%-

».|i4

„ ,. sin 41 - cos a:)

69. hm — T — — - = lim

1_ sin a: 1 - cos x\

2 X X \

= |(1)(0) = 0

_, ,. sin-x

71. lim = lim

;t-»0 X j:-»0

sin a: = (1) sin 0 = 0

„ ,. (1 - cosh)^ ,.
73. hm ^ ; = hm

h->0 h h-tO

1 - cos h

(1 - cosh)\

(0)(0) = 0

_. ,. cos:r ,. . ,

75. hm = lim sin jc = 1

j:->7r/2 cot X x-^Tt/l

__ ,. Sin 3r ,. / sin 3f
77. hm —r — = hm -— —

,_»o It 1^0 \ 3f

i^-nm-i

2 2

79. m

sin3f

t

-0.1

-0.01

-0.001

0

0.001

0.01

0.1

m

2.96

2.9996

3

7

3

2.9996

2.96

sin 3r ,. ,/sin3f\ ,,,,

Analytically, hm = hm 3 — ;— = 3(1) = 3.

■' t^o t 1^0 \ 3t J

■v^jyj V/Y/~v^

The limit appear to equal 3.

81. fix) =

X

-0.1

-0,01

-0,001

0

0.001

0.01

0.1

fix)

-0.099998

-0.01

-0.001

?

0.001

0.01

0.099998

. , • 1, ,• smj:' ,. /sinr^\ „,,,
Analytically, hm = hm j: — j—i = 0(1) = 0.

im^i \/wvw

o, ,. fix + h)- fix) ,. 2ix + h) + 3 - i2x + 3) ,. 2.x + 2/2 + 3 - 2;c - 3 ,. 2h ,

83. hm-^-^^ r — = lim "^ ; ^ = 1™ ; = hm -- = 2

A-»o h h->o h h^o h '!->o h

„^ ,. fix + h)- fix) ,. x + h X ,. 4x-4ix + h) ,. -4 -4

85. hm-"-!^ f — ^-^ = hm r = hm , ',. , = hm , ^ , , = -^

h-M h h->o h /.->o {x + h)xh /i->o (x + h)x x^

87. lim iA- }?■) < Wm fix) < lim (4 + x^)

x->0 x^O-" x-^0

4 < lim/(;c) < 4

x~*0

Therefore, lim/(j:) = 4.

x-tO

89. fix) = X cos X

lim ix cos jc) = 0

x->0

Section 1.3 Evaluating Limits Analytically 35

91. fix) = \x\ sin X

lim \x\ sinx = 0

;r— »0

93. f(x} = JT sin -

lim I ;c sin - 1 = 0

Ar-»0 V xl

95. We say that two functions /and g agree at all but one
point (on an open interval) if f(x) = g(x) for all jc in the
interval except for x = c, where c is in the interval.

97. An indeterminant form is obtained when evaluating a limit
using direct substitution produces a meaningless fractional
expression such as 0/0. That is,

Jr-.c g(x)

for which lim/(jr) = lim g(x) = 0

99. fix) = X, gix) = sin x, hix) =

S h

^^

'Y

^

When you are "close to" 0 the magnitude of/ is
approximately equal to the magnitude of g.
Thus, |g|/|/| ~ 1 when x is "close to" 0.

101. sit) = -16f2 + 1000

,. si5) - sit) ,. 600 - (-16r= + 1000) ,. 16(r + 5)(/ - 5) ,. ,^, ,^,

hm ^ — = lim —1 ' = lim — ^ — , ' , — - = lim - 16(r + 5) =

t^5 5 - t '->5 5 - r ;->5 -(r - 5) f->5

Speed = 160 ft/sec

103. j(f) = -4.9r2 + 150

,. j(3) - sit) ,. -4.9(3-) + 150 - (-4.9r + 150) ,. -4.9(9 - r)

lim — = lim = lim

1^3 3 - t /-:3 3 - r i->3 3 - f

= lim ""^-^^^ ~ '^^^ '^ '' = lim -4.9(3 + r) = -29.4 m/sec

I-»3 3 - r Jr->3

160 ft/sec.

105. Let/(.r) = l/.v and gix) = - l/.t. lim /(a) and lim ^(.t) do not exist.

.r-*0 -v— +0

lim [fix) + gix)] = lim

.r— >0 .v^O

.V \ X

= lim [0] = 0

.V-.0

107. Given fix) = b. show that for every e > 0 there exists a 5 > 0 such that |/(.v) - b\ < 6 whenever |.v - c\ < 5. Since
|/(.v) - b\ = \b - b\ = 0 < efor any e > 0. then any value of 5 > 0 will work.

109. If fo = 0, then the property is true because both sides are equal to 0. If fc ^ 0. let e > 0 be given. Since lim/(.v) = L.
there exists 5 > 0 such that |/(.t) - L\ < e/\b\ whenever 0 < \x - c\ < S. Hence, wherever 0 < |.v - c-| < 5.
we have

\b\\fix) - L\ < e or \bfix) - bL\ < e
which implies that lim [Z:'/(.v)] = bL.

36 Chapter 1 Limits and Their Properties

111. -M|/(x)| < f{x)g{x) < M\f(x) I

lim(-M|/W|) < limf(x)gix) < lim(M|/W|)

x^c x—*c x—*c

-M(0) < lim f{x)g{x) < M(0)

x—>c

0 < lim f{x)g{x) < 0

x—*c

Therefore, lim f{x)g{x) = 0.

M=_,

113. False. As x approaches 0 from the left, -•— '^ = — 1.

X

I
I

115. True.

117. False. The limit does not exist.

119. Let

fix)

[ 4, if.? > 0
[-4. if j: < 0

lim \f{x)\ = lim4 = 4.
j-»o '■' ' .x->0

lim/(x) does not exist since for a: < 0,/(x) =

l->0

-4andforx > 0,/(x) = 4.

121. f(x) =
g{x) =

0, if x is rational

1, if jc is irrational

0, if x is rational

X, if X is irrational

lim/(x) does not exist.

x—*Q

No matter how "close to" 0 x is, there are still an infinite number of rational and irrational numbers so that lim/(x) does not
exist.

lim g{x) = 0.

x—*0

When X is "close to" 0, both parts of the function are "close to" 0.

123. (a) lim

1 — cos X

1 - cosx 1 + cosx

J-»0 XT

lim ,

jr-»0 jr

1 + cos X

(b) Thus,

1 - cosx 1

1 — cos X ~ —x^

. 1 - cos^x

j:->OX^(1 + COSx)

sin'x 1

■»o jt 1 + cosx

COS X ~ I - -x^ for X = 0.

(c) cos(O.l) = 1 - |(0.1)2 = 0.995

= <=i

(d) cos(O.l) = 0.9950, which agrees with part (c).

Section 1.4 Continuity and One-Sided Limits 37

Section 1.4 Continuity and One-Sided Limits

1. (a) lim fix) = 1

(b) lim fix) = 1

j:->3

(c) lim/W = 1

The function is continuous at
x = 3.

3. (a) lim fix) = 0

(b) lim /W = 0

(c) lim/U) = 0

The function is NOT continuous at
jc = 3.

5. (a) lim fix) = 2

(b) lim /(j:) = -2

X— #4

(c) lim /U) does not exist

J— »4

The function is NOT continuous
at j; = 4.

_ ,. X - 5 1

7. lim ^i -— = lim

x^s- r= - 25

5- ;c + 5 10

without bound as .r

does not exist because

7F^^

grows

11.

,. \x\ ,. -;
lim ■■— ^ = lun —

x->0- X x->0- X

1 1

13.

X + \x X

Ax^O" Ajc

X - ix + A.r) 1
lim — , . . , • -:— = lim

-A.t

1

Ai-»o- A'tr + A.r) zir Ax^o~ .t(.x + Ajc) Lx
= lim -7 — — TT

1

1

a:U + 0)

X + 2 5
15. lim fix) = lim ^ — = -

jc-»3 j:->3 Z Z

17. lim /(.r) = lim (x + 1) = 2

.r— ^1^ x-^l"*"

lim/(.r) = limj.x^ + 1) = 2

lim/W = 2

jr— >1

19. lim cot X does not exist since

X—*TT

lim cot x and lim cot x do not exist.

21. lim (3W - 5) = 3(3) -5 = 4

-t— »4

(H = 3 for 3 < -r < 4)

23.

lim (2 -

x->3

\-

xfj does not exist

because

lim (2 -

-r-*3

-i-

-.^1) =

- 2 -

- (-3) =

5

and

lim (2 -
x—*y

-1-

-.rl) ■

- 2 ■

- (-4) =

= 6.

29. gix) = V25 - .t- is continuous
on [-5, 5],

25. fix)

X- - 4

has discontinuities at .t = — 2 and
.r = 2 since/(-2) and/(2) are not
defined.

31. lim fix) = 3 = lim f(x).

x->0- .t->0*

/is continuous on [— 1, 4].

27./W = H + ..

has discontinuities at each integer
k since lim f[x) ~ lim /(jr).

x—^k' x—*k'

33. /(.t) = .t- - It + 1 is continuous
for all real x.

38 Chapter 1 Limits and Their Properties

35. f(x) = 3x — cos X is continuous for all real x.

37. fix) = -^ is not continuous at x = 0, 1 . Since

I

, ,■ for jc =^ 0, ;: = 0 is a removable

X'^ — X X — \

discontinuity, whereas x = 1 is a nonremovable
discontinuity.

39. fix)

x^+ 1

is continuous for all real x.

41. fix)

x + 2

ix + 2)ix - 5)

has a nonremovable discontinuity at j: = 5 since lim/(x)
does not exist, and has a removable discontinuity at
X = —2 since

lim fix) = lim

\x + 2|

43. fix) = -' — — r"- has a nonremovable discontinuity at j: = - 2 since lim fix) does not exist.

45. fix) =

X, X < \

X?-, X > \

has a possible discontinuity at x = 1.
1. fix) = 1

lim /W = lim X = r
lhn/(jc) = limx^ = 1.

3. /(I) = lim/(jc)

lim/W = 1

/is continuous alx = 1, therefore, /is continuous for all real x.

, ^ r^ + 1, X <2

47. /U) = \ ^ has a possible discontinuity at x = 2.

l3 - X, X > 2

1. /(2) =-+1=2

lim/(.r) = lim (^+ 1) = 2]
2_ Mim/(x) does not exist.

lim fix) = lim (3 - x) = 1 J"""*
Therefore, /has a nonremovable discontinuity atx = 2.

49./(x) = r"4-

tan

^. U < 1

tan^. -1 <x < 1

1. /(- 1) = - 1

2. lim fix) = - 1

X— ♦ — 1

3./(-l)= lim/(x)

X—*~l

/(I) = 1
lim/(x) = 1

/(I) = lim/(x)

x—*l

X < — 1 or X > 1

has possible discontinuities atx= — l,x= 1.

/is continuous at x = ±1, therefore, /is continuous for all real x.

Section 1.4 Continuity and One-Sided Limits 39

51. f(x) = CSC 2x has nonremovable discontinuities at integer
multiples of ir/2.

53. fix) = [jc - ll has nonremovable discontinuities at each
integer k.

55. lim f(x) = 0

x—*0*

lim fix) = 0
/is not continuous at x = -2. -s

\

/

c/

57. /(2) = 8

Find a so that lim or = 8

^ = ^ = 2.

59. Find a and b such that lim (ax + ^) =

a - fo= -2

(+)3a + fe = -2

4a = -4

a= -I

b= 2 + (-1) = I

-a + b = 2 and lim (ox + fc) = 3a +

2, .t < - 1

/U) = |-;c + 1, -1 < X < 3
1-2, jc > 3

-2.

61. /(gU)) = ix - \Y

Continuous for all real x.

63. figix))

1

U- + 5) - 6 ;c- - 1
Nonremovable discontinuities at ;c = ± 1

65. y = ix\-x

Nonremovable discontinuity at each integer

^;mx

67. fix) =

2x - 4, .r < 3

Nonremovable discontinuity at .v = 3

5

I

/

/

/

69. fix)

X- + 1
Continuous on (— oo, oo)

71. /U) = sec^

Continuous on:

. . .,(-6. -2), (-2. 2), (2. 6), (6. 10),

73. fix) =

3

75. fix] = -^xf' - .v-^ + 3 is continuous on [l. 2].

/(I) = ^ and /(2) = -4. By the Intermediate Value
Theorem, /(c) = 0 for at least one value of c between
1 and 2.

The graph appears to be continuous on the interval
[-4, 4]. Since /(O) is not defined, we know that/has
a discontinuity at .v = 0. This discontinuity is removable
so it does not show up on the graph.

40

Chapter 1 Limits and Their Properties

77. f{x) = jc^ — 2 — cos X is continuous on [0, it].

/(O) = -3 and /(it) = 77^ - 1 > 0. By the Intermediate
Value Theorem, /(c) = 0 for the least one value of c
between 0 and -tt.

79. f{x)=x^ + X- \

f(x) is continuous on [0, 1].

/(O) = - 1 and/(l) = 1

By the Intermediate Value Theorem, f{x) = 0 for at least
one value of c between 0 and 1 . Using a graphing utility,
we find that j: = 0.6823.

81. git) = 2 cos r - 3r

g is continuous on [0, 1].
g{0) = 2 > Oandg(l)«

1.9 < 0.

By the Intermediate Value Theorem, g(t) = 0 for at least
one value c between 0 and 1. Using a graphing utility, we
fmd that t ^ 0.5636.

83. fix) = x^ + X- I

f is continuous on [0, 5].
/(O) = - 1 and/(5) = 29

-1 < 11 < 29
The Intermediate Value Theorem applies.
a:2 + jc - 1 = 11
x^ + X - 12 = 0
ix + 4)(;c - 3) = 0
x=—4oTX = 3
c = 3 {x = —4 is not in the interval.)
Thus,/(3) =11.

85. fix) = x^-x^ + x-2
/is continuous on [0, 3].
/(0) = -2and/(3) = 19 ,, ■

-2 < 4 < 19
The Intermediate Value Theorem applies.
x^-x^ + x-2 = 4

x^ + x

0

ix - 2)ix^ + ;c + 3) = 0
X = 2

(x^ + X + 3 has no real solution.)
c = 2
Thus,/(2) = 4.

87. (a) The limit does not exist at x = c.

(b) The function is not defined at x = c.

(c) The limit exists at -"^ ~ i^' but it is not equal to the
value of the function di^ ^ ^■

(d) The limit does not exist at x = c.

89.

The function is not continuous at x = 3 because
lim /(x) = 1 # 0 = lim fix).

91. The functions agree for integer values of x:

gix) = 3 - I-xl = 3 - (-x) = 3 + X "
fix) = 3 + W = 3 + X

for X an integer

However, for non-integer values of x, the functions
differ by 1 .

fix) = 3 + M = g(x) - 1 = 2 - I-xl.

For example, /(j) = 3 + 0 = 3, g{^ = 3 - (- 1)

Section 1.4 Continuity and One-Sided Limits 41

93. Mf) = 25 2

f + 2

- t

t

0

1

1.8

2

3

3.8

N{t)

50

25

5

50

25

5

Discontinuous at every positive even integer.Tlie
company replenishes its inventory every two months.

; 4 6 8 10 12
Tune (in months)

95. Let V = 3 TTr^ be the volume of a sphere of radius r.

V(l) =|7r«4.19

V(5) =3tt(53)-523.6

Since 4.19 < 275 < 523.6, the Intermediate Value Theorem implies that there is at least one value r between 1 and 5 such
that V{r) = 275. (In fact, r « 4.0341.)

97. Let c be any real number. Then \imf(x} does not exist since there are both rational and

x—*c

irrational numbers arbitrarily close to c. Therefore, /is not continuous at c.

-1, if.v < 0

99. sgn(.T) = 1 0, if a: = 0

1, ifx > 0

(a) lim sgnU) = - 1

(b) lim^ sgn(jc) = 1

(c) lim sgn(jr) does not exist.

x—^Q

4--
3--
2

1*

-4 -3-2-1

H 1 1 — 1-»-

12 3 4

101. True; iff(x) = g(x), x i- c, then lim/U) = lim 2,{x) and

X~*C .1— .c

at least one of these limits (if they exist) does not equal
the corresponding function s^x = c.

103. False: /(I) is not defmed and lim/(.r) does not exist.

X— »I

105. (a) f{x) =

(0 0 < .X < fc
\b b < x <2b

NOT continuous a.tx = b.

(b) g{x) =

0 < X < b

b -- b <x <2b

7

Continuous on [0. 2b].

42 Chapter 1 Limits and Their Properties

107. f{x) = ^''^'^ £, c > 0

Domain: x + (P->Q=^x> -c^andxT^o, [-c^, 0) U (0, oo)

•Jx + c~ — c ^x + c~ + c

,. Jx + c - c

lim = iim ■

x->0 X x^O X

JW? + c
lim •

U + c^) -c^ .. 1

•'-»'' \jx + c^ + c] '^0 Va: + c^ + c Ic

Define /(O) = l/(2c) to make /continuous at jc = 0.

109. h[x) = x\x\

h has nonremovable discontinuities at
x = ±l,±2,±3, ... .

"^

Section 1.5 InOnite Limits

1.

lim 2

, ,. TTX

3. lim tan — -

j:->-2* 4

lim 2

-»-2-

x^ -A

TTX

liin tan — - = oo

;c-»-2- 4

5. /W =

X

-3.5

-3.1

-3.01

-3.001

-2.999

-2.99

-2.9

-2.5

fix)

0.308

1.639

16.64

166.6

- 166.7

- 16.69

-1.695

-0.364

lim f(x) = oo

j:->-3

lim /U) = -oo

j:-*— 3

7. /U)

;c2-9

x

-3.5

-3.1

-3.01

-3.001

-2.999

-2.99

-2.9

-2.5

fix)

3.769

15.75

150.8

1501

-1499

- 149.3

- 14.25

-2.273

Vim fix) = oo

lim fix) = -oo

x—*-3*

Section 1.5 Infinite Limits 43

9. lim ^ = oo = lim -^

jr-»0* X^ i-»0~ X^

Therefore, at = 0 is a vertical asymptote.

"•ii.'?^(x-2)U+l) = °°

Therefore, j: = 2 is a vertical asymptote.

.-l'?r(;c-2)U+l) = °°

;t2-2

lim

-i-U-2)U+ 1)
Therefore, x = — 1 is a vertical asymptote.

x^ X-

13. lim -:; 7 = oo and lim

Jr->-2- X' — 4

' Jr-U"-2+ X^ - 4

Therefore, x = — 2 is a vertical asymptote.

T-2 r2

lim

-oo and lim

x->2- x^ — 4 """ -t'-^r jr2 - 4

Therefore, jt = 2 is a vertical asymptote.

15. No vertical asymptote since the denominator is never zero.

17. fix) = tan 2x = — has vertical asymptotes at

cos It

{2n + \)tt tt n-TT
X = = — + — , n any mteger.

19. lim 1 - -

r->0* V t-

-OO = lim 1 :;

1^0- V r-

Therefore, r = 0 is a vertical asymptote.

21. lim

x^-2- (x + 2)(x - 1)
lim 7 TTT TT = -oo

x-»-2- (x + 2)(x - 1)

Therefore, x = - 2 is a vertical asymptote.

lim -; — 7 -r = oo

x-^r (x + 2)(x - 1)

lim '. TT-, rr = — oc

x-.r (x + 2)(x - 1)

Therefore, x = 1 is a vertical asymptote.

23. Ax) =

A^ + 1 _ (.T-t- DU^ --r + 1)

X + 1

X + 1

has no vertical asymptote since

lim fix) = lim ix- - .r + 1) = 3

X— *— 1 J—*- I

25. fix.

5)(.r + 3) _ x + 3

X* 5

ix - 5)(a:- +1) X- + V
No vertical asymptotes. The graph has a hole at x = 5.

27. sit) = — — has vertical asymptotes al t = mr. n
smt

a nonzero integer. There is no \ertical asymptote at
t = 0 since

lim^— = 1.
t->o sm t

44 Chapter 1 Limits and Their Properties

29. lim i-— f- = Urn [x - 1)

jr-»-l X + \ Jr-»-l

31. lim — —r = oo
;t-»-l+ X + 1

;t^+ 1

lim

\- x+ \

Vertical asymptote at
x= -1

Removable discontinuity at a: = — 1

33. lim r-

x-,2* X - 1

35. lim

x^-i- (x - i)(x + 3)

,_ ,. x^ + 2x- ^ ,. X- \ 4
37. lim -^5— — = lim = 7

x-^-r XT + X — it j:-»-3 X — I 5

A — X X 1

x^l (x + l)(x - 1) x-H X^ + I 2

41. lim 1 + - = -00

43. lim -: — = 00

jr-»o+ sin X

45. lim = lim (v^sin;c) = 0

.r-»irCSCJ: x^-ir

47. lim X seclTTx) = 00 and lim x sec(irx)

^-^(1/2)- xMU2r

Therefore, lim x secfTTj:) does not exist.

x-*(l/2)

49. f(x) = -^^-j-

51. fix) =

x^- 25

lim f(x) = lim = 00

lim fix) = -00

x-*5

J

L

r

"^i

53. A limit in which fix) increases or decreases without
bound as x approaches c is called an infinite limit. 00 is
not a number. Rather, the symbol

lim fix) = 00

x-*c

says how the limit fails to exist.

55. One answer is fix) =

x-3

X- 3

ix - 6)(;c + 2) x^--4x- 12'

57.

59. 5 = -; , 0 < Irl < 1. Assume k i= 0.

1 - r

lim S = lim

r->r r->i- 1 — r

= 00 (or — 00 if fc < 0)

Section 1.5 Infinite Limits 45

^, ^ 52&X „

61. C = -— , 0 < X < 100

100 - X

(a) C(25) = $176 million

(b) C(50) = $528 million

(c) C(75) = $1584 million

(d) lim

528

-.100- 100 - ;c

OD Thus, it is not possible.

2(7) 7

63. (a) r = , ' = — ft/sec

(b) r-

(c) lim

7625 - 49 12

2(15) 3

V625 - 225 2

ft/sec

--^^25- V625 - x'-

65. (a)

X

1

0.5

0.2

0.1

0.01

0.001

0.0001

fix)

0.1585

0.041 1

0.0067

0.0017

= 0

= 0

= 0

X - sm.t
lim —■ = 0

j:->0* X

(b)

X

1

0.5

0.2

0.1

0.01

0.001

0.0001

fix)

0.1585

0.0823

0.0333

0.0167

0.0017

«0

«0

;c - smx
lim :; = 0

(c)

X

1

0.5

0.2

0.1

0.01

0.001

0.0001

fix)

0.1585

0.1646

0.1663

0.1666

0.1667

0.1667

0.1667

0.25

1.5

lim - — ^ = 0.1167 (1/6)

x->0* XT

(d)

X

1

0.5

0.2

0.1

0.01

0.001

0.0001

fix)

0.1585

0.3292

0.8317

1.6658

16.67

166.7

1667.0

X - sin.T

Iim ; = oo

.1^0* .V-*

^ , ,. .V - sin.v

For n > i, lim ;; = oo.

x-^O* X"

46

Chapter 1 Limits and Their Properties

67. (a) Because the circumference of the motor is
half that of the saw arbor, the saw makes
1700/2 = 850 revolutions per minute.

(c) 2(20 cot </)) + 2(10 cot (^): straight sections.
The angle subtended in each circle is

277

2(f-^

77+ 2</).

Thus, the length of the belt around the pulleys is
20(77 + 24>) + 10(77 + 24>) = 30(77 + 24>).
Total length = 60 cot ^ + 30(77 + 2</))

Domain: ( 0, —

69. False; for instance, let

f(x) = r or

six)

X - 1

x^+ 1

(b) The direction of rotation is reversed,
(d)

(e) 450

<!>

0.3

0.6

0.9

1.2

1.5

L

306.2

217.9

195.9

189.6

188.5

(0 lim L = 6077 == 188.5

«^(t/2)-

(All the belts are around pulleys.)

(g)

lim L =

00

71.

False; let

f{x) =

§

;c # 0
;c = 0.
The graph of /has a vertical asymptote at x = 0, but

m = 3.

73. Given lim/(j:) = cxd and lim g{x) = L:

x—*c x~*c

(2) Product:

If L > 0, then for e = L/2 > 0 there exists 5; > 0 such that \g{x) - L\ < L/2 whenever 0 < |a: - c| < S,. Thus,
L/2 < g{x) < 3L/2. Since lim/(j:) = 00 then forM > 0, there exists 5, > 0 such that /(a:) > M(2/L) whenever
|jc - c| < S;. Let 5 be the smaller of 5i and Sj- Then for 0 < |a: - c| < S, we have/(;t)g(x) > M(2/L){L/2) = M.
Therefore Vim f(x)g{x) = 00. The proof is similar for L < 0.

(3) Quotient: Let e > 0 be given.

There exists S, > 0 such that/(jr) > 3L/2e whenever 0 < \x — c\ < S^ and there exists 63 > 0 such that \g{x) ~ L\ <
L/2 whenever 0 < |x - c| < S,. This inequality gives us L/2 < g{x) < 3L/2. Let S be the smaller of 5, and Sj. Then
for 0 < \x — c\ < d, v/e have

g(x)

Ax)

3L/2
3L/2e

e(x)
Therefore, lim 7H = 0.

x-*cf(x)

75. Given lim -7-, = 0.

x^cf(x)

Suppose lim/(j:) exists and equals L. Then,

x-*c

liml

1

1

x->cf{x) lim/(;c) L

This is not possible. Thus, Umf(x) does not exist.

Review Exercises for Chapter 1 47

Review Exercises for Chapter 1

1. Calculus required. Using a graphing utility, you can estimate the length to be 8.3.
Or, the length is slightly longer than the distance between the two points, 8.25.

3.

X

-0.1

-0.01

-0.001

0.001

0.01

0.1

fix)

-0.26

-0.25

-0.250

-0.2499

-0.249

-0.24

lim/W « -0.25

5. h{x)

)r - 2x

(a) \\mh{x) = -2

x-»0

(b) lim h{x) = -3

7. lim (3 - a:) = 3 - 1 = 2

Let e > 0 be given. Choose 6 = e. Then for
0< |x-l| <5=6, you have

\x-\\<e

\\-x\<e

1(3 -x)-2\<e

\f(x) -L\<e

9. lim (x- - 3) = 1

Let e > 0 be given. We need |r= - 3 - 1 1 < e => |.t- - 4| = \{x - 2)(a: + 2)| < e ^ |.r - 2| <
Assuming, 1 < .v < 3, you can choose 6 = e/5. Hence, for 0 < |.x — 2| < S = e/5 you have

x + 2

X- 2 <T <

1

5 \x + 2\
\x- 2\\x + 2\ < e
\x- - 4| < e
l(x^-3)- 1| < e

l/W - L\<e

11. lim Vm = V4 + 2 = V6 = 2.45

<-»4

13. lim "5 7 = lim = —-

r->-2 r — 4 r->-2 t - 2 4

Jx - 2
15. lim — '■ = lim

Vi-2

-.4 X - 4

■4(v3^-2)(^ + 2)

lim ^ = — F = -

x-*'i Vx + 2 V4 + 2 4

„ ,. [l/(.v + D] - 1 ,. 1 - Ct + 1)

17. lim ^= = lim — : — . .,

.t->0 X Jr->0 x(x + 1)

1

= lim

i^>ox + 1

= -I

,n r ^ + 125 ,. (.V + 5)(.r - 5.t + 25)
19. lim — r- = lim

-5 X + 5

X + 5

,, ,. 1 - COS.V ,. / X \(l -COS.tA ,,Mr,\ rv

21. hm : = lim ^ — = (1)(0) = 0

x-»o sm.v x-'0\,sina:/\ .t /

lim (r - 5x + 25)

= 75

48 Chapter 1 Limits and Their Properties

,, ,. sin[(7r/6) + Ajc] - (1/2) ,. sin(Tr/6) cos A;c 4- cos(7r/6) sin Ax - (1/2)
23. lim : = lim :

Ax-»0 Ar Ax->0 A.t

1 (cos Ajc- 1) ,. V^ sin Ax

= lim - • -. H lun —T- • — : —

Aj-»o 2 Ax Ax-*o 2 Ax

25. lim[/(x)-g(.x)] = (-|)(f)=-|

27./(x) = ^-j

(a)

j;

1.1

1.01

1.001

1.0001

fix)

0.5680

0.5764

0.5773

0.5773

lim + 1 V3 ^ Q^^^ (Actual limit is ^3/3.^

J— ♦l^ ^ — 1

,, ,. J2x +1-73 ,. V2x + 1 - 73 72x +1 + 73

(c) lim = lim ; • , p

j:-i* .X - 1 ^-»i* x-1 72x +1 + 73

(2x + 1) - 3
~ x-^v (x - l)(72x + 1 + v^)

2
= hm — , =

^^'" 72x+ 1 + 73

2 1 _^

2VS 73 3

(b)

-_ ,. s{a) - s{t) ,. (-4.9(4)- + 200) - (-4.9r + 200)

29. lim = lim

r-»o a — t »->4 4 — t

,. 4.9(r - 4)(f + 4)
= lim :

r-*4 A — t

= lim -4.9(r + 4) = -39.2 m/sec

I->4

31. lim

\x- 31

j:^3- X: — 3

lim^^=-l

Jr^3- X — 5

33. lim/(x) = 0

x-*2

35. lim /!(f) does not exist because lim h{i) = 1 + 1=2 and

r— >I r— »1"

lim h(t) = i(l + 1) = 1.

37. fix) = Ix + 31

lim [x + 31 = /c + 3 where ^ is an integer.

lim |x + 3]1 = /c + 2 where A: is an integer.

Nonremovable discontinuity at each integer k
Continuous on {k, k + I) for all integers k

39. fix) =

3x- - x-2 (3x + 2)(x: - 1)

1

1

lim/(x:) = lim (3.x + 2) = 5

J— » I Jr— > 1

Removable discontinuity at x = 1
Continuous on (— oo, 1) U (1, oo)

41. fix)

lim

1

ix - 2)2
1

x^2 (;c - 2)2

Nonremovable discontinuity at x = 2
Continuous on (-co, 2) U (2, oo)

43. fix) =

X + 1
lim /(x) = — oo

JC-+1

lim /(x:) = oo
jif-*i*

Nonremovable discontinuity at x: = — 1
Continuous on (-oo, — 1) U (- 1, oo)

Problem Solving for Chapter 1 49

45. f{x) = CSC —

Nonremovable discontinuities at each even integer.
Continuous on

(2k, 2k +2)

for all integers k.

49. /is continuous on [l, 2]. /(I) = - 1 < 0 and

/(2) = 13 > 0. Therefore by the Intermediate Value
Theorem, there is at least one "alue c in (1, 2) such
that 2c5 - 3 = 0.

47. /(2) = 5

Find c so that lim (ex + 6) = 5.

c(2) + 6 = 5

2c = -1

1
c=--

51. fix) = ^^ ={x + 2)

\ X -2

Lk-2|

(a) lim/U) = -4

jr-»2

(b) lim^/W = 4

(c) lim/(jc) does not exist.

53. gW = 1 + -

Vertical asymptote at x = 0

2x^ + X+ \

57. lim

X + 2

55. /(.r

(x - 10)=
Vertical asymptote at x = 10

59. lim 4-^-

1

J-*-!* X-

_ ,. .r- + 2,t + 1

61. lim ;

j:-»r j; - 1

63. lim [x J

sin 4x
65. lim — - — = lim

.r->0* 5x X-.0*

4/ sin 4a:

5V 4x

„ ,. csclv ,. 1

67. lim = lim — ^ — — = oo

.v->o* X X ->o* X Sin Zr

69.C = |^,0.0<100
100 — p

(a) C(15) == $14,117.65 (b) C(50) = $80,000

(c) C(90) = $720,000

(d) lim tt:: = oo

p->ioo' 100 - p

Problem Solving for Chapter 1

1. (a) Perimeter APAO = Jx- + (v - 1)= + Jx- + y- + 1 (b) r(x)

Jx- + (.V-2 - 1): + Jx- + X* + 1

Perimeter APBO = V(.t - 1)- +r + n/^+T + 1

V(a- - IP + A^ + 7^2 + A-^ + 1

(c) hm r(x) = . , „ , . = T = 1
.v->o+ 1+0+1 2

x- + ix' - \)- + ^x~ + .1-' + 1

Jix - ly- + x" + ^'x^ + .x-* + 1

.X

4

2

1

0.1

0.01

Perimeter AP.-^O

33.02

9.08

3.41

2.10

2.01

Perimeter APBO

33.77

9.60

3.41

2.00

2.00

r{x)

0.98

0.95 1

1.05

1.005

50 Chapter 1 Limits and Their Properties

3. (a) There are 6 triangles, each with a central angle of
60° = 77-/3. Hence,

Area hexagon = 6

■bh

2
3V3

2(1) Sin -

- 2.598.

h = sme

Error: tt r — = 0.5435.

(b) There are n triangles, each with central angle of
6 = iTr/n. Hence,

An = n

bh

1,,, . Itt] nsmilv/n
:r(l sin — =- — ^r-^
2 « J 2

n

6

12

24

48

96

An

2.598

3

3.106

3.133

3.139

(c)

(d) As n gets larger and larger, In/n approaches 0.

Letting x = 27r/n,

sin(2ir/«) sin(2Tr/n) sinj:

An = — —. = , , , 77 = 77

2/n (277/m) j:

which approaches (1)77 = 77.

5. (a) Slope = —

12

(b) Slope of tangent line is

y+l2 = —{x- 5)
169

12'

y = — JT - — Tangent line

(c) Q=(x,y) = {x,^l69-x^)

- V169 - jc2 + 12
X — 5

,. 12 - V169 - x^ 12 + V169 - x^

(d) lim m^ = lim ;: • ,

-t^s ^ ;t^5 X - 5 12 + Vl69 - ;c2

^ 144 - (169 - x^)

" ^™ (;c - 5)(l2 + ^169"^^)

= lim

x^ - 25

= lim

^5 (x - 5)(l2 + V169 - x^)
ix + 5)

5 12 + Vl69 - x^
10 5

12 + 12 12
This is the same slope as part (b).

7. (a) 3 + ;c'/3 > 0
a:'/3 > -3
;c > -27
Domain: x > —27,xi= 1

(b)

(c) lim J{x) =

V3 + (-27)'/3 - 2

-27 - 1
-2 1

-28 14

0.0714

. ,, ,. ., , ,. V3 + -t'/^ - 2 V3 + x'/3 + 2
(d) lim/(x) = lim ; • . ,,:

= lim

3 + ;c'/3 - 4

= lim

1 (x - l^v/sT^iTs + 2)
;c'/3 - 1

^-^1 (xi/3 - l)(;c2/3 + ^1/3 + i)(V3 + x^/i + 2)

1

= lim

^->l (x2/3 + ;t'/3 + 1)(V3 + ;c'/3 + 2)

1

(1 + 1 + 1)(2 + 2) ~ 12

9. (a) lim/W = 3: ^„ ^4
fb) /continuous at 2: g,

(c) lim fix) = 3: g„ g,, g^

X— ♦2

Problem Solving for Chapter J 51

11.

-2--
-3--

13. (a) y

a b

(a) /(1) = I11 + I-]1= 1 +(-l) = 0
/(O) = 0

/(i) = 0 + (-l)=-l
/(-2.7) = -3 + 2 = -1

(b) \\mf(x) =

-1

lim fix) =

Jr->1*

-1

Junjix) =

-1

(c) /is continuous for all real numbers except
X = 0,±l,+2,±3, . . .

(b) (i) lim^ P„ ,{x) = 1

x—*a

(ii) lim P„ ,{x) = 0
(iii) lim P„ ,{x) = 0

X— >o

(iv) lim P^,(x) = 1

x—*b

(c) Pn ^ is continuous for all positive real numbers
except x = a,b.

(d) The area under the graph of u,
and above the j:-axis, is 1.

CHAPTER 2
Differentiation

Section 2.1 The Derivative and the Tangent Line Problem ... 53

Section 2.2 Basic Differentiation Rules and Rates of Change . 60

Section 23 The Product and Quotient Rules and

Higher-Order Derivatives 67

Section 2.4 The Chain Rule 73

Section 2.5 Implicit Differentiation 79

Section 2.6 Related Rates 85

Review Exercises 92

Problem Solving 98

CHAPTER 2
Differentiation

Section 2.1 The Derivative and the Tangent Line Problem

Solutions to Odd-Numbered Exercises

1. (a) m = 0
(b) m = -3

3. (a), (b)

(c).= ^^f^U -!)-/(!)

:(x- l) + 2

l(x - 1) + 2
X + I

12 3 4 5 6

5. fix) = 3 — 2j: is a line. Slope

= lim

AI-.0

/(0 +

Af)-/(0)

Af

= lim

A/->0

3(Ar)

- (Af)' - 0

Af

= lim

M->0

(3-

Ar) = 3

13. f(x) = -5x

fix) = lim

= lim -

A;t-»0

/(.t+ Ax)-/(a:)
zSa

-5U + .A-v) - (-5.t)

A.r

lim — 5 = -5

Al->0

7.

Slope

at (1,-3) =

= lim

Aj-»0

g(i + ^x) -

\x

gi\)

=

= lim

(1 + ^xy- -

A.r

4 -(-3)

=

= lim

1 + 2(A.t) +
Ax

(Ax)2 - 1

=

= lim

Aj:->0

[2 + 2(^x)']

= 2

11.

/(.O

= 3

ru)

= lim f^^

Ai->0

+ A.r)
A.r

-fix)

= lim — —

Ai->0 Ax

3

= lim 0 =

0

15

/i(i) =

= 3.f.

h'is)

= lim -^

+ ^s)

-his)

3+ j(5 +

ls)-{i^

f>)

A5^0

Is

= lim -;—

As-KI Aj

2
"I

53

54 Chapter 2 Dijferentiation

17. f[x) = 2.t2 + X - 1

f(x+ ^x)-f{x)

J \X)

— um

A.r

= lim

A1-.0

[2(;c + ^xf + [x + hx) -

-1]-

[2r + X - 1]

^x

= lim

{Ix" + 4x^x + liH^f +

;c + ^

- l)-

(2^2 +

X- 1)

Aa

= lim

Ax->0

Ax^x + 2(Ax)^ + Ajc ,. ,.

-, — = lim (Ax

Ajc ax-»o

+ 2Ax + 1) =

4x+ 1

19. fix)

= X3-

12x

fix)

= lim

f{x + ^x)-f{x)

Ax

= lim

Ax->0

[(x + AxY - \2(x + Ajc)

1-U'

-12x]

Ajc

= lim

x^ + 3x^Ax + 3a;(A;c)2 +

(AxP-

12x-

12A;c-

- x3 + 12x

Ax

= lim

Ax^O

3;c2Ax + 3;c(A;c)2 + (A;c)3
Ax

- 12Ax

= lim (3x2 + 3^^ + (^)2 _ j2) = 3^2 - 12

Ax->0

21. fix) '

/'(x) = lim

X- 1

/(x + Ax)-/(x)

Ax->0 Ax

1 1

x + Ax-1 X — 1

= lim 1

Ax-»0 Ax

U - 1) - (x + Ax - 1)
~ iJc^o Ax(x + Ax - l)(x - 1)

= r -Ax

iJ™oAx(x + Ax- l)(x- 1)

" iji^o (x + Ax - l)(x - 1)
^ 1

u-ip

23. /(x) = Vx+ 1

/'(x) = lim ^<-^y-^(-)

•^ Ax^O Ax

Vx + Ax +T - Vx + 1 _ /Vx + A.r + 1 + Vx + 1\
^^0 Ax VVxTAxTT + Vx + 1/

Ax-

(x + Ax + 1) - (x + 1)

= lim

Ax-*o Ax[Vx + Ax + 1 + Vx + 1]

lim , ^ ,

■^-'0 Vx + Ax + 1 + Vx + 1

1 ^ 1

Vx + 1 + Vx + 1 2Vx + 1

Section 2.1 The Derivative and the Tangent Line Problem 55

25. (a) fix) =x^ + \
fix) = lim

fix + Ax) -fix)
Ax

^ ^.^^^ [jx + Ax)^ +\]-[x'+l]

Ax-»0 AjC

,. 2xA;c + (A.r)^
= lim ;

A.t->0 Ax

= lim (Ix + A.x) = 2x

At (2, 5), the slope of the tangent line is

m = 2(2) = 4. The equation of the tangent line is

y-5

= 4(.t

-2)

>'-5

= 4x

- 8

y

= 4x

- 3.

27.

(a) fix) =

;c3

fix) =

lim

Ax->0

fix

+ A.x)
Av

-fix)

lim

ix +

Axf-
A.X

-x"

lim

A.v->0

3x^Ax + 3.r(Ax)^ +

(Ax)^

Ax

= lim (3x^ + 3xAx + (Ax)^) = 3x^

At (2, 8), the slope of the tangent \s m = 3(2)- = 12.
The equation of the tangent line is

> - 8 = 12(;t - 2)

V = 12;>: - 16.

(b)

V

\/-

/

(b)

r

/

1

29. (a) fix) = Vx

■' AX-.0 Ax

= lim

Vx + Ax - V^ Vx + Ax + Vx
z^mo Ax Jx + Ax + J~x

(x + Ax) - X

= lim

Aat-^o A,x( V.x + Ax + Vx)

= lim

1

A:c->o Vx: + Ax + Jx iji
At (1, 1), the slope of the tangent line is

1 1

m = — 1= = -.

2j\ 2

The equation of the tangent line is
y-l=\ix-i)

1 1

V = tj: + T.
■22

(b)

56 Chapter 2 Differentiation

31. (a)/(;c) =x + -

f'ix) = lim

f{x+ ^x)-f{x)

Ax

= lim

Ax->0

{x + Ax) + ——r - {x + -
x + Ax \ X

= lim

lim

Aj:->0

Ax

x{x + Ax)(x + Ax) + 4x - x\x + Ax) - 4(.t + A^-)
x{Ax)(x + Ax)

x^ + 2x^(Ax) + x(Ax)^ - x^ - x\Ax) - A(Ax)
x{Ax)(x + Ax)

x^(Ax) + ^(Ajc)^ - 4(M
Ajuo x{Ax){x + Ax)

,. x^+x{Ax)-A

Imi — ; — T —

Ajv^O x(x + Ax)

= 1

4

X X

At (4, 5), the slope of the tangent line is

,43 : :.

16 4

The equation of the tangent line is

(b)

y

1 J/'

X^\

y - 5 = -(x - 4)

y = -^x + 2.

33. From Exercise 27 we know that/'U) = 3x^. Since the
slope of the given line is 3, we have

3^2 = 3

x = ±\.

Therefore, at the points (1,1) and (- 1, - 1) the tangent
lines are parallel to 3x - > + 1 =0. These lines have
equations

and y + 1 = 3(.t + 1)

y-l=3{x- 1)
y = 3x - 2

y = 3x + 2.

35. Using the limit definition of derivative,
■1

fix) =

2xVx'

Since the slope of the given line is —5, we have

1_^ _1

2x^/x 2

x= 1.

Therefore, at the point (1, 1) the tangent line is parallel to
x + 2y - 6 = 0. The equation of this line is

1,

y -

1 = --{x - 1

y -

1 1

l = -2- + 2

1 ^3

37. ^(5) = 2 because the tangent line passes through (5, 2).

gXi)

2-0 _ 2
5 - 9 ~ -4

39. f{x) = X =>f'ix) = 1 matches (b)

Section 2. 1 The Derivative and the Tangent Line Problem 57

41. f(x) = Jx^> f'{x) matches (a)
(decreasing slope as jr — > oo)

43.

Answers will vary.
Sample answer: y = —x

45. (a) If /'(c) = 3 and /is odd. then/'(-c) =/'(c) = 3
(b) If /'(c) = 3 and/ is even, then/'(-c) = -/(c) = -3

47. Let (.rg, y^) be a point of tangency on the graph of/. By the limit definition for the derivative, /'(j;) = 4 - Ir. The slope of the
line through (2, 5) and (xq, Vq) equals the derivative of/ at x^.

5 - v„

= 4 - 2Xn

1- Xq

5 - Vo = (2 - x^){A - 2xo)

5 — {4xq — Xff) = 8 — 8xg + Zkq-

0 = Xq- - 4xo + 3

0 = (.Vo - l){xo - 3) =

J^o = 1.3

Therefore, the points of tangency are (1, 3) and (3, 3), and the corresponding slopes are 2 and -2. The equations of the tangent
lines are

y-5 = 2(.r-2) y - 5 = -2(.i: - 2)

y = 2.V + 1 y = -2x + 9

49. (a) g'iO) = -3

(b) ^'(3) = 0 ■ • ■

(c) Because g'(l) = -3, g is decreasing (falling) at x = 1.

(d) Because g'(-4) = 3, g is increasing (rising) at x = —4.

(e) Because g'{4) and g'(6) are both positive, g(6) is greater than g{4), and g(6) - g(4) > 0.

(f) No, it is not possible. All you can say is that g is decreasing (falling) at x = 2.

51. fix) = ir^

By the limit definition of the derivative we have/'(.r)

X

— 2

-1.5

-1

-0.5

0

0.5

1

1.5

2

fix)

-2

27

32

1

4

1

32

0

1

32

1
4

27
32

2

fix)

3

27
16

3
4

3
16

0

3
16

3
4

27
16

3

2

58 Chapter 2 Differentiation

53. gix)

fix + Om)-f{x)
0.01
= {2(x + 0.01) - {x + 0.01)2 _ 2x + a;2) + 100

3

\

^^

' /

\\

The graph of g(x) is approximately the graph off'{x).

55. /(2) = 2(4 - 2) = 4, /(2.1) = 2.1(4 - 2.1) = 3.99

T.qq — A
f'{2) - Yl^ = ~°-^ [Exact: /'(2) = 0]

57.f{x)=^andf'(x)=:^.

As X —> oo, / is nearly horizontal and thus /' ~ 0.

59. fix) = 4 - (;c - 3)2

/(2 + Ax)-/(2)

■Sa^ (x) =

iu

{x-2)+f{2)

4 - (2 + Ax - 3)2 - 3, ^, ., 1 - (Ax - 1)2 ^^ , , , ^., ^. ,
^ (jc - 2) + 3 = 4 H-t - 2) + 3 = i-^x + 2){x - 2) + 3

Ajc

Ax

(a) Ax = 1: 5^ = (x - 2) + 3 = X + 1
Ac = 0.5: S

^- (-j(x-2) + 3 =-x

19

Ax = 0.1:5,,= ^-j(x-2) + 3=-x--
(b) As Ajc->0, the line approaches the tangent line to/at (2, 3).

>

/^

^/ \

in. -'

61. /(x) =x2 - 1,C = 2

f'Oi 1- /W-/(2) ,. (x2 - 1) - 3 ,. (x - 2)(x + 2)
/ (2) = lim = lim — ■ — — = lim —

x->2 X — 2 x->2 X — 2 x->2 X — 2

= lim (x + 2) = 4

x->2

63. /(x) =x3 + 2x2+ l,c= -2

/(x) -/(-2) ,. x2(x + 2) ,. (x^ + 2x2 + 1) - 1

/'(-2) = lim ^^-^^^^ — ^-r = lim "" ^"" ' ' = lim

x->-2 X + 2 .r->-2 X + 2 x-4-2

x + 2

= lim x2 = 4

x^-2

65. g(x) = Vjlf, c = 0

pfr) - p(0) VlxT

^'(0) = lim ^^ fr^ = lim —!-!■. Does not exist.

J-^O X — 0 x->0 X

As X ^ 0 , = — p — > - oo

^ Vx

yui 1

Asx^ 0-^,—'-^ = ^^ oo
X Vx

67. /(x) = (x - 6)2/3, c = 6

j:— *6 X ~ O

,. (X - 6)2/3 _ 0

= am ;:

x-^6 X — 6

= lim-

1

""e (x - 6)'/3
Does not exist.

Section 2. 1 The Derivative and the Tangent Line Problem 59

69. h{x) = U + 5),c = -5

V(-5)= lim^-W^^

-t->~5 X - ( — 5)

,. |;c + 5| - 0
= lim ■■ ■— —

x->-5 X + 5

= lim

k + 5|

-5 X + 5

Does not exist.

71. fix) is differentiable everywhere except at ;<:
(Sharp turn in the graph.)

-3.

73. fix) is differentiable everywhere except at j: = - 1 .
(Discontinuity)

75. fix) is differentiable everywhere except at j: = 3.
(Sharp turn in the graph)

77. fix) is differentiable on the interval (1, oo).
(At j: = 1 the tangent line is vertical)

79. fix) is differentiable everywhere except at .r = 0.
(Discontinuity)

81./(;c)= |;t- 1|

The derivative from the left is

lim^W^=lim^^
The derivative from the right is

lim ^«^ = lim ^^

Jr-»1* X — I Jr-»1* X — 1

= 1.

The one-sided limits are not equal. Therefore, /is not
differentiable atx = 1 .

r<^ + 1, r < 2
85. Note that/is continuous at x = 2. fix) = \ ^

[4x - 3, -v > 2

83. fix)

ix - 1)3, X < 1

[x - 1)-, X > 1

The derivative from the left is

lim

/U)-/(i)

X - I

lim

ix - 1)

X - 1

lim ix - 1)= = 0.

The derivative from the right is

j:-»1* .t - 1 j:-»1" X - 1

= lim ix - 1) = 0.

Jr->1*

These one-sided limits are equal. Therefore. /is
differentiable at .t = 1. (/'(I) = 0)

The derivative from the left is lim ■'^-^ — {^ = lim — = lim (:t 4- 2) = 4.

Jr->2- X — 2 jr-»;- A" — 2 jr->2-

The derivative from the right is lim — '■ — = lim — '■ = lim 4 = 4.

X-.2* X - 2 x-*2* X - 2 .r->2*

The one-sided limits are equal. Therefore, /is differentiable at v = 2. (/'(2) = 4)

87. (a) The distance from (3, 1) to the line /?ir — >> -I- 4 = 0 is

|Ati + By, + C\
J A- + B'

|/7i(3) - 1(1) -F 4| |3m4-3|

(b)

JirF+l.

Jm- + 1

The function d is not differentiable at m = — 1. This corresponds to the line
y = —X + 4, which passes through the point (3, 1).

60 Chapter 2 Differentiation

89. False. The slope is lim

A;e->0

/(2 + Ax)-/(2)
^x

91. False. If the derivative from the left of a point does not equal the derivative from the right of a point, then the derivative does
not exist at that point. For example, if/(x) = \x\, then the derivative from the left at j: = 0 is - 1 and the derivative from the
right at a: = 0 is 1 . At j£ = 0, the derivative does not exist.

93. f(x) =

xsin(l/jc), X i^ 0

0,

x = 0

Using the Squeeze Theorem, we have -\x\ < x sin(l/x) < |a:|, a: =?^ 0. Thus, lim x sia(\/x) = 0 = /(O) and/is continuous at
X = Q. Using the alternative form of the derivative we have

,. fix) -/(O) ,. A:sin(lA) - 0 ,. , .
lim ''-^^ — 77^ = Imi ^"-^ = lim I sm

x-iO X — \J x^O X — \J J:->0

"i)-

Since this limit does not exist (it oscillates between - 1 and 1), the function is not differentiable at a: = 0.

g{x)-

X^ SV!\i\/x), X T^ 0

0,

0

Using the Squeeze Theorem again v.'e have -x'- < x'- sia(l/x) < x^,x i= 0. Thus, lim :? sin(l/j:) = 0 = /(O) and/is continu-
ous at X = 0. Using the alternative form of the derivative again we have

,. /W-/(0) ,. x2sin(lA)-0 ,. . 1 „
lim -''^-^ — r^ = lim —'—T = hm x sin - = 0.

jr^O X — 0 j:->0 X — 0 x-^0 X

Therefore, g is differentiable at x = 0, ^ '(0) = 0.

Section 2.2 Basic Differentiation Rules and Rates of Change

1. (a)

y = x'/2

yd) =5

3. y = 8
y' = Q

(b) y = x3/2

„l/2

y'W = I

5. >' = x6
y'= 6x?

(c)

7. y

y = x^

(d) y = ^

y' = 2x

y'=3x^

y'il) = 2

y\\) = 3

x' ""

9. y = ^ = xi/5

= -7x-8 =

-7

y 5 5x^/5

11. fix) = X + 1
fix) = 1

13. fit) = -2;2 + 3r - 6 IS. g(x) = x^ + 4x3

/'(;)= -4, + 3 g'W = 2x + 12x2

17. sit) = t^-lt + A
s'ii) = 'hi^ -1

77

19. y = ^ sin 6 - cos e

21. y = x^ — X cos X

23. V = 3 sin X

x

y' = —cos e + sin 0

y' = 2x + — sinx

x2

— 3 cos X

Section 2.2 Basic Differentiation Rules and Rates of Change 61

Function Rewrite

25. V

27. y =

29. y

3

(2x)3

>" = ?

>' = 7-^

>> = x

-1/2

Derivative
y'= -5;c-3

y' = -5^-^"''

= --r-3/2

Simplify

y =

^

2x^/2

31. fix) = ^ = 3x-2, (1, 3)

fix) = -6x-3 = -J-

/'(l) = -6

33. fix) = -| + |x3, (o, -^

21
Z'U) = yx2

/'(O) = 0

35. V = (2x + 1)2, (0, 1)

y'= 8;c + 4
j'(0)=4

37. /(e) = 4 sin 0 - e, (0, 0)

fie) = 4 COS e - 1

/'(O) = 4(1) -1=3

39.

fix) = x'- + 5 - 3^-2
/V) = 2x + 6x-3 = 2i +

41. ^(r) = ^2 - -^ = f2 - 4r3

g'(f) = 2f + 12r-* = 2r +

12

43. fix) = ^^ ^^^^^ = ;c - 3 + 4.t-2

/'W = l-3 ""

X3 ^3

47. /(x) = v^ - 6 ^ = xi/2 - 6x'/3

1 1 2

51. fix) = 6Vx + 5 cos x = 6.x:'^2 + 5 cosjc '

/'(x) = 3x ''2 — 5 sin JT = — = - 5 sin jc

45. y = x(x2 + 1) = x3 + X
y'= 3x2 + 1

49. his) = i-*/^ - ^2/3

4 9 4 7

t Y^l = -5-4/5 _ £c-i/3 = —Z f_

53. (a) >- = X* - 3x2 + 2
y' = 4x3 - ^x

At (1.0): y' = 4(1)3 _ 5(1) = _2.
Tangent line: y - 0 = -2(x - 1)

2x + y - 2 = 0
(b)

v;

V

\/^

(1.0?^

55. (a) fix) = ^ = ^1^"'^"

-3 —3

f'(x) = ^x"''/" = — —

At (1, 2), /'(I) = -^

Tangent line;

V - 2 = -Hx - 1)

3 . 7
3x + 2v - 7 = 0

y = --X + -

(b)

\

\^_

\

62 Chapter 2 Differentiation

57. y = X* - Sx^ + 2
>>' = 4j:^ - 16jr

= 4x{x^ - 4)

= 4x{x - 2){x + 2)
y' = 0 => j: = 0, ±2
Horizontal tangents: (0, 2), (2, - 14), (-2, - 14)

59. y

x2

2x ' = — ^ cannot equal zero.

Therefore, there are no horizontal tangents.

61. y = X + sin X, 0 < X < Itt

y ' = 1 + cos X = 0

cos JC=— 1 => X = TT

At X = TT, y = TT.

Horizontal tangent: (n, n)

63. AT^ — fcc = 4x - 9 Equate functions

2x - k = 4 Equate derivatives

Hence, k = 2x — 4 and
x^ - {2x - 4)x = 4x - 9-

For jc = 3, <: = 2 and for x

-x^ = -9=:>j: = ±3.
-3, /t= -10.

A; 3

65. - = — -X + 3 Equate functions

X 4

k^

X-

Equate derivatives

3 2

3 4^^
Hence, k = —x^ and —

4 a:

-3^1 3
4 4

3.3.,
■ — X + 3 =» -X = 3 :

. X = 2 =» A: = 3.

67. (a) The slope appears to be steepest between A and B.

(b) The average rate of change between A and B is
greater than the instantaneous rate of change at B.

(c)

69. g(x)=/(x) + 6^g'W=/'(x)

71,

ff/is linear then its derivative is a constant function.
fix) = ax + b
fix) = a

Section 2.2 Basic Differentiation Rules and Rates of Change 63

73. Let (x^, y^ and (xj, y^^ be the points of tangency ony = x^ and y = —x~ + 6x - 5, respectively. The derivatives of
these functions are

y' = 2x =^ m = 2x^ and y'= — 2x + 6 => m= — Ixj + 6.

m = 2x, = — Ivj + 6

JCl = -^2 + 3

Since y, = .r,^ and jj = -x^^ + 60:3 ~ 5,

V2 - yi (-x,^ + 6x2 - 5) - (-x,^)

ra = = = — 2x, + 6.

{-x.} + 6X-, - 5) - (-x, + 3)2

= -2x, + 6

X2 — {~X2 + 3)

{-x^ + 6x2 - 5) - Uj- - 6x2 + 9) = (-2x2 + 6)(2x2 - 3)
-Ixj' + 12t2 - 14 = -4x2- + 18x2 - 18
2x2' - 6x2 + 4 = 0
2(x2 - 2)(x2 - 1) = 0

1 or 2

1 => y2 = 0, X, = 2 and Vj = 4

Thus, the tangent line through (1,0) and (2, 4) is

y-0 =

(x - 1) => y = 4x - 4.

\2 - \)

Xj = 2 => y, = 3, X, = 1 and Ji = 1
Thus, the tangent line through (2, 3) and (1, 1) is

(i - r

V - 1 =

2 - 1

(x - 1) => y = 2x - 1.

75. /(x) = J~x, (-4,0)

'■"'-r"'-in

1

0 - V

ij'x -4-x
4 + X = 2Vxy
4 + X = 2VxVx
4 + X = 2x

X = 4, y = 2
The point (4, 2) is on the graph of/.
0-2

Tangent line: y - 2

-4-4
4y - 8 = X - 4

0 = X - 4v + 4

(.t - 4)

77. /'(I) = - 1

64 Chapter 2 Differentiation

79. (a) One possible secant is between (3.9, 7.7019) and (4, 8):
8 - 7.7019,

y-i

<^-4)

4 - 3.9
y - 8 = 2.981U - 4)
y = 5W = 2.981;c- 3.924

(b) fix) = \x^l^ ^ /'(4) = f(2) = 3

r(.r) = 3(;c - 4) + 8 = 3jc - 4

5(x) is an approximation of the tangent line Tt^:).

(c) As you move further away from (4, 8), the accuracy of the approximation T gets worse.

(d)

Ax

-3

-2

-1

-0.5

-0.1

0

0.1

0.5

1

2

3

/(4 + Ax)

1

2.828

5.196

6.548

7.702

8

8.302

9.546

11.180

14.697

18.520

7t4 + Ax)

-1

2

5

6.5

7.7

8

8.3

9.5

11

14

17

81. False. Let/(x) = x- and g{x) = x^ + 4. Then
/'(x) = g'W = 2x,but/(x)^g(x).

83. False. If y = ir^, then dy/dx = 0. (tt^ is a constant.)

85. True. If g(x) = 3/(.x), then g'(x) = 3/'(x).

87. f(t) = 2t + 7, [1,2]

fit) = 2

Instantaneous rate of change is the constant 2.
Average rate of change:

/(2)-/(l) [2(2) + 7] - [2(1) + 7]

= 2

2-1 1

(These are the same because/is a line of slope 2.)

89. f(x) = --. [1, 2]

fix)

x'

Instantaneous rate of change:
(1,-1) ^/'(1) = 1

2.-l)^rm-\

Average rate of change:

/(2)-/(l) (-1/2) -(-1) ^ j_
2-1 2-1 2

Section 2.2 Basic Differentiation Rules and Rates of Change 65

91. (a) s{t) = - 16r= + 1362

v(r) = -32r

sir) - s{\)
^"> 2-1 " ^^^^ - 1346 = -48 ft/sec

(c) v(t) = sXt)= -32/

Whenf = 1: v(l) = -32 ft/sec.
When t = 2: v(2) = -64 ft/sec.

(d) - 16/2 + 1362 = 0
1362

f2 =

16

/ = ^^» 9.226 sec

(e)v(^) = -32(^)

= -871362 « -295.242 ft/ sec

93. s(t) = -A.9fi + Vot + So
= -4.9f2 + 120r
v(r) = -9.8; + 120
v(5) = -9.8(5) + 120 = 71 m/sec
v(10) = -9.8(10) + 120 = 22 m/sec

95.

■S. 50--
.S '»

-t— I-

2 4 6 8 10
Time (in minutes)

(The velocity has been converted to miles per hour)

99. (a) Using a graphing utility, you obtain
« = 0.167V - 0.02.
(c) T= R + B = 0.00586V- + 0.1431v + 0.44

(e) -- = 0.01172V + 0,1431
flv

Forv = 40, r'(40) « 0.612.
Forv = 80, r'(80) « 1.081.
Forv = 100, r'(lOO) =« 1.315.

97. V = 40 mph = f mi/min
(f mi/min)(6 min) = 4 mi

V = 0 mph = 0 mi/min
(0 mi/min)(2 min) = 0 mi

V = 60 mph = 1 mi/min
(1 mi/min)(2 min) = 2 mi

(b) Using a graphing utility, you obtain

B = 0.00586v2 - 0.0239V + 0.46.
(d) 60

Time (in minules)

(f) For increasing speeds, the total stopping distance
increases.

101. A

, dA

' ds

When .r = 4 m,
dA

ds

= 8 square meters per meter change in i.

103.

^ 1,008,000 ,,^
C = :; + 6.3(2

dC_
dQ

Q
1,008.000

+ 6.3

C(351) - C(350) = 5083.095 - 5085 = -$1.91
d€

When Q = 350,

dQ

$1.93.

105. (a) /'(1 .47) is the rate of change of the amount of gasoline sold when the price is $1.47 per gallon,
(b) /'(1. 47) is usually negative. As prices go up, sales go down.

66 Chapter 2 Differentiation

107. y = ax- + bx + c

Since the parabola passes through (0, 1) and (1, 0), we have
(0, 1): 1 = a(0)2 + b{Q) + c ^ c = 1
(1, 0): 0 = a{\Y + b(\) + 1 =* fc = -a - 1.
Thus, y = ax^ + {—a — \)x + \. From the tangent line y = x - 1, we know that the derivative is 1 at the point (1, 0).
y' = lax + {-a — 1)
1 = 2a(l) + (-«-!)
1 =a - 1
a = 2

6= -a - 1 = -3
Therefore, y = 2x^ - l,x + \.

109. y = ]^ -9x

y' = 3x2 - 9 ■ ,. .

Tangent lines through (1, —9):

. y + 9 = (3x2 _ 9)(^ _ 1)
• (x3 - 9x) + 9 = 3x3 - 3x2 - 9x + 9

0 = 2x3 - 3x2 = ^2(2;c - 3)

X = 0 or X = 2
The points of tangency are (0, 0) and (f, -y). At (0, 0) the slope is ^'(0) = -9. At (f, -y) the slope is y '(2) = "I •
Tangent lines:

y-Q= -9(;c - 0) and y + f = "K^ "2)

n 9 27

y = -9x y = -4X - -^ ■ ■

9x + J = 0 9x + 4y + 27 = 0

111. /(x) = 2 . J, ^ o

[x2 + i), X > 2

/must be continuous at x = 2 to be differentiable at x = 2.
lim fix) = lim ox^ = 8a 1 8a = 4 + fo

lim /(x) = lim (x2 + /,) = 4 + Z7 J ^"^ ~ ■* " ^

x->2* x-»2*

fix)

3ax2, X < 2
2x, X > 2

For/to be diiferentiable at x = 2, the left derivative must equal the right derivative.

3a(2)2 = 2(2)
12a = 4

a = 3

i = 8a - 4 = -I

Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 67

113. Let/W = cosx

f(x + Ax)-f{x)

f'(x) = lim

Ax->0

= lim

Ax-»0

Ax

cos X COS Ajc — sin j: sin Ajc — cos x
Ajc

,. cosx(coszix- 1) ,. /sinAx

= liin : lim smx — : —

Ax->o Ajc /u->o \ Ax

= 0 - sinx(l) = -sinx

Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives

1. g(x) = (x- + DU^ - It)

g'(x) = {x'- + l)(2.r - 2) + ix^ - 2x)(2x)
= Ix^ - 2^2 + 2r - 2 + 2x3 - 4x2
= 4x^ - 6x- + lx - 2

S. f(x) = x^cosx

f'(x) = x3(— sinx) + cosx(3x-)
= 3x- cosx — x^ sinx

3. hit) = yiifi + 4) = fi/3(f2 + 4)

h'(t) = f'/3(2f) + (f2+4)l -2/3

= 2f4/3 +

t- + 4

3f2/3

7f2 + 4
3/2/3

7. /(x)

X2 + 1

fix) =

(x2 + 1)(1) - x(1t) 1 -x2

(x2 + 1)2 (x2 + 1)2

9. h(x)

rx

rl/3

X3 + 1 .x3 + 1

h\x) =

(X3 + l)jX-2/3 - X'''3(3x2)
U^Tl)^

(x3 + 1) - X(9x2)
3x2/3(x3 + 1)2

1 - 8x3

3x2/3(x3 + 1)2

11. g(x

gXx) =

x2(cosx) - sin.t(2t) _ x cos x - 2 sin x

(x2)2 - x3

13. fix) = (x3 - 3x)(2t2 + 3x + 5)

fix) = (x3 - 3x)(4x + 3) + (2x2 + 3;c + 5)(3;c2 _ 3)

= lOx^ + 12x3 - 3^2 _ 18^ _ 15
/'(0) = -15

15. fix) =

/'W =

/'(I) =

x2 - 4

X - 3

ix - 3)(2x) - (x2 -

-4)(1)

2r2 - 6.r - x^

U - 3)2

* 4

U - 3)2

.r2 - dx + 4
U - 3)2

1-6+4 1
(1 - 3)2 4

17. fix) = X cos X

fix) = (x)(-sinx) + (cosx)(l) = cosx - xsinx

Hf)=#-f(#)=f'-'»

68 Chapter 2 Differentiation

Function Rewrite

Of

nvafive

19. v^^^;^ .^.^^f.

y'

2 2
= 3^^3

21.. = 3^3 . = 1-^

y'

= -7;c-4

23. y = — y = 4v^, ;c> 0

y'

= 2;c-'/2

-/« = ^^^

.. . U^ - l){-2 - 2;c) - (3 - 2;c - x^)(2x)

/W- (^2_l)2

2^2 - 4;c + 2 2(x - 1)^

"U+D-^-'^i

/ 4 \ 4x
27./W=.(l-^^3)=.-^^3

29.

(x + 3)4-4x(l) U2 + 6;c + 9)-
-^^""^ U + 3)2 " (;c + 3)2

12

^2 + 6;c - 3

Simplih

, 2;c + 2

2a: + 5

2;c'/2 + 5;c-i

fix) =X-'/2-|;C-3/2 = X-3/2

^~i

U + 3)2

2;c - 5 ^ 2x- 5
2^Vx " 2x3/2

31. ftW = (53 - 2)2 = 56 - 4^ + 4
/i'(i) = 6^ - 12*2 = 6^2(^3 - 2)

33. fix)

2x- \ 2x- \

X — 3 xix — 3) ^2 — 3x

, , ^ U2 - 3a:)2 - (2x - l)(2x - 3) ^ 2^2 - 6x - 4^2 + 8a - 3
^ ^''' U2 - 3a)2 ^2 - 3x)2

-2^2 + 2x - 3
(x2 - 3x)2

2^2 - 2x + 3

x2(x-3)2

35. /W = (3jt3 + 4x)ix - 5)(x + 1)

fix) = (9x2 + 4)(^ _ 5)(^ + 1) + (3x3 + 4x)i\)ix + 1) + (3x3 + 4^)(^ _ 5)(i)

= (9x2 + 4)(^2 _ 4^ _ 5) + 3^4 + 3^3 + 4^2 + 4;c + 3;c4 _ i5_^ + 4j^2 _ 20x
= 9x* - 36x3 _ 4i;f2 _ 15^ - 20 + 6x^ - 12x3 + 8x2 - j^^
= ISx'* - 48x3 - 33x2 _ 2,2x - 20

37. /(x) =

/'(^) =

X2 + C2

X2-C2

(x2 - c2)(2x) - (x2 + c2)(2x)

(x2 - c2)2

39. fix) = fsmt

fit) = P cost + 2t sin t
= titcost + 2 sin t)

—4x^2

(x2 - c2)2

Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 69

41. fit) = ^ 43. fix) = -X + tan ;t

„/ V _ -tsinf - cosf _ f sin t + cos t

fix) = - 1 + sec^j: = laiv^x

45. g(f) = 4/^ + 8 sec f = f'/" + 8 sec f 47. y = ^''\ ^'"''^ = |(sec.t - tanx)

-^ 2cos.r 2

g'W =7^3'"' + 8secftanf = — J7T + Ssecrtanf ,3, , , 3 ,

^ ^r"' ;y = -(sec j: tan .t - sec- x) = Tsec .r(tan x - sec x)

3
= -(sec X tanx — tan^x - 1)

49. y — — CSC X — sin X 51. fix) = x^ tan x

y ' = CSC X cot X — cos x fix) = x- sec- x -H 2x tan x

COS X = x(x sec^ X -H 2 tan x)

= —r^ cos X

sin-x

= cosx(csc-x — 1)

= cos X cot'^ X

53. y = 2t sin X + x^ cos x 55. g(;^) = |1±_1 W _ 5)

>> ' = 2x cos X + 2 sin X -I- x-(— sin x) -t- 2x cos x

= 4x cos X -^ 2 sin x - x^ sin x S 'W = ^, , .^2 tfo™ of answer may vary)

(x + 2)2

57. g(e) =

1 — sin

,, . 1 - sin 9 -I- e cos e

ff (0) = r^ — Tz:; (form of answer may vary)

(sm 9-1)-

1 + CSC X "

59. y =

1 — CSC X

- , _ (1 - cscx)(-cscxcotx) - (1 -I- CSC x)(csc X cot x) _ -2 CSC X cot X
-*' ~ (1 - cscx)= ~ (1 - cscx)-

M ^ -2(2)(^) ^ _^
y[6) (1-2)2 4V3

61.

ftW =

sec t

f

h'it) =

f(sec r tan f) -

(sec r)(l)

•

t-

=

sec r(f tan r —

;2

i)

/iV) =

sec Triirtan tt

—

il

1

70 Chapter 2 Differentiation

63. (a) f(x) = {x^ -3x+ l){x + 2), (1, -3)

fix) = (;c3 - 3.t + 1)(1) + {x + 2)(3;c2 - 3)

= 4;c3 + 6;c2 - 6x - 5
/'(I) = -1 = slope at (1,-3).
Tangent line: y + 3 = —l(x- 1) => y = —x — 2

(b)

65. (a) fix) =
fix) =
/'(2) =

(2, 2)

;c - 1

(x - 1)(1) - xil) -1

ix - 1)2 ix - 1)2

1

= - 1 = slope at (2, 2).

(2 - 1)2
Tangent line: >> - 2 = - lU — 2) => 3? = -x + 4

(b)

-i^

\i \

67. (a) /W = tanx, (^ j, 1

/'(jc) = sec2x

/'(f) = 2 = slopeat Jl

Tangent line:

y-i=2[x

(b)

jm

J(

if

Y

y-i = 2x--

4j:-2y-Tr+2 = 0

69

/(-t) =

JC -

1

fix) =

ix-

- 1)(2x) - ;.

^(1)

ix - 1)2

^2

- 2x x(jc

-2)

ix - 1)2 (x - 1)2

fix) = 0 when x = 0 or x = 2.
Horizontal tangents are at (0, 0) and (2, 4).

71. fix) =
g'ix) =

(x + 2)3 - 3x(l) _ 6
ix + 2)2 (x + 2)2

(x + 2)5 - (5x + 4)(1) 6

ix + 2)2

ix + 2)2

, , 5x + 4 3x 2x + 4 ,, , ^

/ and g differ by a constant.

73. /(x) =x"sinx

fix) = x" cos X + nx" " ' sin X
= x"" ' (x cos X + n sin x)

When n = 1
When n = 2
When n = 3
When n = 4

/'(x) = X cos X + sin x.
/'(x) = x{x cos X + 2 sin x).
fix) = x2(x cos X + 3 sin x).
fix) = x^ix cos X + 4 sin x).

75. Area = Ait) = (2f + l)Vf = 2/3/2 + ^1/2
A'W = 2(|f'/2)+|r'/2

= 3?i''2 + 1,-1/2

6t + 1

2Vf

cm2/sec

For general n,f'ix) = x" ' (x cos x + n sin x).

Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 71

77. C= 100(202 +

X

""^ ioo(-^ +

■2 X + 30
30

1 < X

dx

(a) When ;c = 10

:^ {x + 30)2
dC

dx
dC

$38.13.

(b) Whenx = 15: -— = -$10.37.

dx

/"ft"'

(c) When a: = 20: -— = -$3.80.

dx

As the order size increases, the cost per item
decreases.

79. Pit) = 500
P'it) = 500
= 500
= 2000

11 '*' "

L 50 + f2j
(50 + r2)(4) -

- (4r)(2r)

(50 + Pjz
200 - 4t-

(50 + t^y_

" 50 = r

.(50 + a^J
P'(2) = 31.55 bacteria per hour

81. (a) sec x =

d. -. d

(cos;c)(0) - (1)(— sinx) _ sin.r _ 1 sin;c

(cos xy

cos X cos x cos X cos X

= sec X tan x

(b) CSC j:

sinx

-[csc^] =

d
dx

1
, sin X

(c) cot X =

cosx
sin.v

-[cot..] =

d
dx

"cos-t

.sin;c

83. fix) = 4x^/2

fix) = 6;ci/2

/'W = 3;c-/2

■N

3

fx

(sin.v)(0) - (1)(cosj:) cosjt

1 cos j:

(sin jc)-

sin X sm X sm x sin x

- CSC X cot x

cos-t] sin .t( — sin.r) — (cos. r)(cosj:) sin^jc + cos^x 1

(sin .r)-

sm''j:
85. fix) =

• ■ /W =

sin-.v

X -

- 1

ix

- 1)(1) -

.t(l)

-1

(-r - D-

ix

-D-

2

ix - D'

87. fix) = 3 sin x
fix) = 3 cos X
f'\x) = - 3 sin X

89. fix) = x^
fix) = 2r

= -) /:

91. /"(x) = 2v'x

/'•'>(.v)=^(2).r-'- = -Jp

93.

/(2) = 0

One such function is/(.v) = (x — 2)-.

95. fix) = 2gix) + hix)
fix) = 2g'ix)+h'ix)
fi2) = 2g'i2) + h'il)
= 2(-2) + 4
= 0

97. fix) =
fix) =
/'(2) =

hix)

hix)g'ix) - gix)h'ix)
[hix)y

h(2)g'i2) - g(2)h\2)
[h(2)Y

(-l)(-2)-(3)(4)

(-1)-

= -10

72 Chapter 2 Differentiation

99.

It appears that/ is cubic; so/' would be quadratic and
/"would be linear.

103. v(r)
, a(t)

lOOr

2t+ 15

(It + 15)(100) - (100?)(2)

{2t + 15)2

1500

(2r + 15P

(a) a(5) =

101. v(t) = 36 - r^, 0<r<6
a(r) = -2t
v(3) = 27 m/sec
a{3) = - 6 m/sec
The speed of the object is decreasing.

1500

(b) a(10) =

(c) a(20) =

[2(5) + 15?
1500

[2(10) + 15?

1500

[2(20) + 15]^

= 2.4 ft/sec2
= 1.2ft/sec2
» 0.5 ft/sec2

105. f(x) = g(x)h{x)

(a) fix) = gix}h'{x) + h(x)g'(x)

fix) = g{x)li"{x) + g'(x)h'{x) + h(x)g"ix) + h'{x)g'ix)

= g(x)hXx) + 2g'{x)h Xx) + h{x)g\x)
rix) = g(x)h"'(x) + g'{x)h"(x) + 2g'{x)h'\x) + 2g"ix)h'(x) + h(x)g"'(x) + h'{x)g%x)

= g{x)h '"{x) + 3g 'ix)h'{x) + 3g%x)h Xx) + g "'ix)hix}
f^%x) = g(xW'Kx) + gXx)h"Xx) + 3gXx)h"Xx) + 3g'\x)h'U) + Sg'UMx) + 3g"'{x)hXx)
+ g"Xx)hXx) + g^'KxMx)
= gixW'Kx) + 4gXx)h"Xx) + 6g'{x)h'{x) + 4g"'ix)hXx) + g^'Kx)h{x)

(X>) f w gwn W + j^(^ _ j)(^ _ 2) . . . (2)(1)^ ^'''^ ^''' ^ (2)(l)[(n - 2)(n - 3) ■ ■ • (2)(1)]^ ^' ^'

+ ,.J:[-:^^z'^-::^^'L.,g"'ix)h^-^^(^ + ■ ■ ■

(3)(2)(1)[(« - 3)in - 4) ■ ■ • (2)(1)
n{n- \)(n-2)- ■ ■ (2)(1)

[{n-l)(n-2)- ■ •(2)(l)](i;

g^"-'Kx)hXx)+g^"\x)h{x)

= gix)hMix) + !,(„": i),gW^"'-"(^) + 2!(/- 2)!g'^^)^^"""(^) + ■ ■ ■

-g(«-'>(x)A'(;t) + gW(x);i(;c)

(n- 1)!1!'
Note: «! = «(« — 1) ... 3 • 2 • 1 (read "n factorial.")

Section 2.4 The Chain Rule 73

107. f(x) = cos X

f'(x) = -sinjr

f"(x) = -cos a;

(a) Pi(x) = f'(a)ix - a) + fia
1

^1 7r\ 77 1

/ — = COS — = -
-'* 3/ 3 2

r/| ■''■\ • "■ v/3

/ljj = -s.n-=-—
/(- =-cos-=--

^f. - ^] .

2 V'

= -4l"-3J -T"l"-3J + 2
(c) P, 's a better approximation.

(b)

/■jX

<.

>

(d) The accuracy worsens as you move farther away
fromx = a = (ir/S).

109. False. If y = f{x)gix), then
dy

dx

- f(x)g'{x) + g{x)f'{x).

111. True

h'(c) = f(c)g'(c) + g(c)f\c)
= /(c)(0) + g(c)(0)
= 0

113. True

115. fix) = x\x\ =

X-, iix>Q
-x^, \ix < 0

fix)

fix) =

2x, ifjc > 0
-2x, \ix < 0

2, if a: > 0
-2, if.)c < 0

/"(O) does not exist since the left and right derivatives
are not equal.

Section 2.4 The Chain Rule

y=Mx))

1. y = {6x - 5)"

" = g(x)
u = bx — 5

■ y=f{u)

y = W

3. y = Ti^^nr

u = X- — 1

= ^

5. V = csc'j:

M = cscx

7. .V = (2x - 7)'

y'=3(2j<:- 7)-(2) = 6(2x- 7)^

9. ^(.v) = 3(4 - 9.V)-'

g'(x) = 12(4 - 9Af(-9) = - 108(4 - 9x)^

11. fix) = (9 - ;c=)=/'

13. /(f) = (1 - f)"-

/'(f)=^l-r)-'-(-l)

2vl -/

74 Chapter 2 Differentiation

15. y = (9;(2 + 4)1/3

17. y = 2(4 - j;2)i/'»

^'=2^ (4-;c2)-3'V2;c)

4^(4 - ;c2

19. >> = (;c- 2)-'

y'=-l(2-;c)-2(l) =

1

(^ - 2?

21. /(t) = (r - 3)-2
fit) = -2(r - 3)-3 =

a - 3)3

23. y = (x + 2)-'/2

^ = _i(^ + 2)-3/2 = \

25. f{x) = x\x - If

fix) = x\A{x - 2)3(1)] + (;c - 2)^(2;t)
= 2x{x - 2)3[2x + (x- 2)]
= 2x{x - 2)3(3x - 2)

27. y = xj\ - x^ = x{\ - x^y^

y =x

\(\ - x^)-'/^(-Xx)

+ (1 - x2)'/2(l)

= -xW - x^Y"^ + (1 - x2)'/2

= (1 - x')-''\-x^ + (1 - ;c2)]
1 -2x2

yn^

/ < ,' x + 5

31. g{x)= —

gW = 21

;c2 + 2,

;c + 5 \/(x2 + 2) - (x + 5)(2;i;)

35. y

;c2 + 2/\ (a;2 + 2)2

_ l{x + 5)(2 - 10;c - x^)
(x2 + 2)3

J~x + 1
;c2+ 1

1 - 3;c2 - 4;(3/2
275(;c2 + 1)2

The zero of y ' corresponds to the point on the graph of
y where the tangent line is horizontal.

29. y =

= X(x2 + l)-l/2

y' = /-^(x2 + l)-3''2(2x)l + (x2 + l)-'/2(i)

= -;c2(x2 + I)-3/2 + (^2 + l)-l/2
= (;c2 + l)-3/2[-;c2 + {x" + 1)]
1

(;c2 + 1)3/2

33. /(v) =

/'(v) = 3

1 - 2v
1 + V

1 - 2v\V(l + v)(-2) - (1 - 2v)

1 + V / V (1 + v)2

9(1 - 2v)2

37. «(f) =

(1 + v)*

3f2
Jfi + 2t- \

^ Itjfi + 3f - 2)
^ ^ ^ (r2 + 2f - 1)3/2

The zeros of g ' correspond to the points on the graph of
g where the tangent lines are horizontal.

Section 2.4 The Chain Rule 75

39. y

= V^

,^ J(x + l)/x
^ 2x{x + 1)

y ' has no zeros.

. . .^^

L.

v

(

41. s(t)
s\t)

-1{1 - t)J\ + t

JVT

The zero of s '{t) corresponds to the point on the graph of
s{t) where the tangent line is horizontal.

s

43. y

cos TTJC + 1

dy _ — vx sin vx — cos ttx - 1
clx~ x^

TTX sin TTX + cos TTX + 1

M

The zeros of y ' correspond to the points on the graph of y where the tangent lines are horizontal.

45. (a) y = sin X
y' = cosjc

y'io) = 1

1 cycle in [0, Itt]

(b) y = sin 2x
y ' = 2 cos 2x
y'iO) = 2
2 cycles in [0, 2Tr]
The slope of sin ax at the origin is a.

47. y = cos 3jc
dy

dx

= - 3 sin 3a:

51. y = sin (irx)- = sin (ir-x^)

y' = cos (tt x-)[27r\x] = 27r-.vcos(iT-j:-)

53. h(x) = sin 2jc cos 2x

h '(x) = sin 2t(- 2 sin Iv) + cos 2j:(2 cos 2jc)
= 2 cos- 2x — 2 sin- 2t
= 2 cos 4.1:.

Alternate solution: h{x) = — sin 4.r

h '(x) = - cos 4j:(4) = 2 cos 4.r

49. g{x) = 3 tan 4x
g'{x) = 12 sec-4jt

sc f, ^ cot.t cosj:

55. f(x) = = ^^;—

Sin X sin- X

fix) =

sin- .rl— sin .r) — cos xjl sin x cos .t)
sin'* .t

— sin^A: — 2 cos-.r — 1 — cos-.t

sin^ .v

sin- .V

76 Chapter 2 Differentiation

57. y = 4 sec^ X

>> ' = 8 sec a: • sec j: tan ^ = 8 sec^ xtanx

61. fix) = 3sec2(7rf- 0

/'W = 6sec(Trr - l)sec(Trr- l)tan(7rr- IKtt)

6Trsin(Trr — 1)

= 6irsec2(irr - 1) tan(irf - 1) =

cos^ (irt - 1)

59. /(fl) = J sin2 20 = |(sin 26)^

fie) = 2(3)(sin 20)(cos 2e)(2)
= sin2ecos2e = 5 sin 40

63. y = ^ + - sin(2;c)2
= -/x + - sin(4;c2)
£ = |;c-'/2 + icos(4x2)(8x)

1

2v^

+ 2x cos(2x)2

65. y = sin(cosx)

— = cos(cosar) • (— sinx)
ax

= — sin j: cos(cos x)

67. s(t) = it^ + 2t+ 8)'/2, (2,4)
J'W =|(f2 + 2f+ 8)-'/2(2f +2)
t+ 1

^'(2) =

Vr^ + 2r +
3

69- /W = ^ = 3(^3 _ 4)-i, / i,_3

/'W = -3(;c3 - 4)-2(3x2:

9;r2

ix' - 4y-

/'(-1) =

25

71. /(r) = i^, (0,-2)

/'(O

r- 1

(t - 1)(3) - (3f + 2)(1)

-5

(f - 1)^

/'(O) = -5

it - 1)'

73. y = 37 - sec3(2;c), (0, 36)

y' = -3 sec2(2x)[2 sec(2;c) tan(2x)]

= -6sec3(2x)tan(2x)
y'iO) = 0

75. (a) fix) = J3x^ - 2, (3, 5)
/'(x) = ^3;c^ - 2)->/2(6x)
3x

V3]t

/'(3) =

Tangent line:

>< - 5 = -(a: - 3) ^ 9;c - 5y - 2 = 0

(b)

\j

/"

/

77. (a) /(x) = sin 2x, (tt, 0)
fix) = 2 cos 2x
/V) = 2
Tangent line:

> = 2(;c - tt) ^> 2x - y - 27r = 0
(b)

Section 2.4 The Chain Rule 77

79. fix) = 2(;c2 - \f
fix) = 6(;c^ - mix)

= llxU" -2x^+1)
= llr^ - 24A-5 + Xlx

fix) = eOjc" - 72jc2 + 12
= 12(5x2 - l)(jc2 - 1)

81. /(x) = sin x2
/'(x) = 2xcosx2

/"(x) = 2x[2x(-smx=)] + 2cosx2
= 2[cos x^ — Ix^ sin x-]

83.

85.

The zeros of/' correspond to the points where the graph
of/has horizontal tangents.

The zeros of/' correspond to the points where the graph
of/ has horizontal tangents.

87. g(x)=/(3x)

g'(x) =/'(3.r)(3) ^ g'(x) = 3/'(3x)

89. (a) fix) = gix)hix)

fix) = g{x)h'ix) + g'ix)hix)
f'iS) = (-3)(-2) + (6)(3) = 24

('^^^w-fS

fix)
/'(5)

Mx)g'(x) - g(x);it^)

(3)(6) - (-3)(-2) 12 4
(3)2 9 3

(b) /(.r) = gihix))

fix) = g'ihix))h'ix)

/'(5) = g'(3)(-2) = -2g'(3)

Need g'(3) to find /'(5).

(d) fix) = [g(x)]^

/'(x) = 3[g(.r)]2g'(.t)
/'(5) = 3(- 3)2(6) = 162

91. (a) /= 132.400(331 - v)"'

/' = (-1)(132,400)(331 - v)--i-\)

132,400

" (331 - v) =

Whenv = 30,/'= 1.461.

(b) /= 132,400(331 + v)"'

/' = (-1)(132,400)(331 + v)-2(l)

-132.400
(331 + v)-

Whenv = 30,/'= -1.016.

93. e = 0.2 cos 8f

The maximum angular displacement is 9 = 0.2 (since
- 1 < cos 8f < 1).

M
dt

0.2[-8sin8t] = -1.6 sin 8r

When t = 3. dd/dt = - 1.6 sin 24 = 1.4489 radians per
second.

95. S = CiR- - r-)

f=c(2Rf-2r'-f
dt \ dt dt

Since r is constant, we have dr/dt = 0 and

^ = (1.76 X 105)(2)(1.2 X 10--)(10-5)
dt

= 4.224 X 10"- = 0.04224.

78

Chapter 2 Differentiation

97. (a) x= - 1.6372f3 + 19.3120f2 - 0.5082r - 0.6161
(b) C= 60x+ 1350

= 60(-1.6372r3 + 19.3120r2 - 0.5082« - 0.6161) + 1350
dC

dt

= 60(- 4.9 116(2 + 38.624r - 0.5082)

= -294.696r2 + 2317.44f - 30.492
The function — is quadratic, not linear. The cost function levels off at the end of the day, perhaps due to fatigue.

99. f{x) = sin ^x

(a) fix) = )3 cos fix
f"{x) = -0'smpx
f"'(x) = -p^cosfix

/(^' = /S" sin lix

(b) fix) + p^fix) = -/32 sin fix + fi^ism fix) = 0

(c) f ^^\x) = {- ly fi^'' sin fix
/(^--DW = (-!)*+ 1/32*-' cos /Sjc

101. (a) r'ix) = f'{g{x))g'ix)
r'(l)=/'(g(l))g'(l)

Note that ^( 1 ) = 4 and /'(4)

6-2

Also, g'{l) = 0. Thus, r'(l) = 0
(b) s'ix) = g'{fix))f'{x)
s'{4) = g'(f(4))/'(4)

Notethat/(4) = |g'(|

6-4 1 ^

T = -and

6-2 2

/'(4)

Thus, 5'(4)= 1(1

103. g = Vx(x + n)
= ~Jx- + nx

^ = i(;c2 + nx}-'/2{2x + n)
ax 2

2x + n i

Ijx^ + >u
(2;c + «)/2
VaCx + w)

[x + (x + «)]/2

VjcU + «)
£

105. g(x) = |2jc - 3|
2a:- 3

g'{x)

2JC-3

X9t

107. /j(x) = |x|cosx

h '(x) = — \x\ sinx + -r~\ cos x, x ^ 0

Section 2.5 Implicit Differentiation 79

109. (a) f(x) = tan -

/(I) = 1

/W = -sec2 —

.,„ s TT -, TTX TTX TT

/ W=-sec2— -tan— (-

/'(I) = J(2) = f

/"(l)=|(2j(l) = J

/',W=/'(l)(;c-l)+/(l) = |(;c-l)+l.

P.U) = |(f)(x - IF +/'(1)U - 1) +/(!) = |{:c - IP + |(x - 1) + 1

(b)

(c) Pj '5 ^ better approximation than P,

(d) The accuracy worsens as you move away from x = c = 1 .

111. False. Ify = (1 - xY'-, theny' = jd - x)-'/-(-l).

113. True

Section 2.5 Implicit Differentiation

1. x^ + f = 36
2x + 2yy' = 0

^1/2 + y/2 = 9

-1/2

,-1/2

5. .r^ - XV + y' = 4

3x' - jty ' — y + 2yy ' = 0

(2y — x)y ' = y — 3x^

Zy — X

7. x^y - y - X = 0

3.t3y2y'+ 3.A-' - y' - 1 =0

(3;c3>'= - l)v'= 1 - 3xn^

, _ 1 - 3.rK-^
•'■ 3.r3%- - 1

9. x^ - ?>x- + Zxy~ = 12

3x2 _ 3^2y- _ 6^^^, + 4;tyy' + 2r = 0

(4xv - 3x2)y' = 6xy - 3x- - 2y-

j _ 6xy - 3x- — 2y2
^ ~ 4x7 - 3x2

11. sin.v + 2cos 2y = 1
cos X - 4(sin 2y)y ' = 0

4 sin 2v

80 Chapter 2 Differentiation

13. sin.r = ;>:(1 + tany)

cos.r = x{sec'^ y)y' + (1 + tany){\)
, cos X — tan V — 1

y = T"

X sec- y

17. (a) ;c2 + y2 = 16

/ = 16 - a:^

±Vl6 - x^

15. y = sin(xy)

y' = [xy' + >']cos(xy)
y' - X cos{xy)y' = y cos(xy)

y cos(xy)

y =

I — X cos(xy)

(b)

y= Vl6-jc2

y = -V16-x2

/

(c) Explicitly
dy _l

dx

(16 - x?)-'"-(-2x)

(d) Implicitly:

2x + lyy' = 0

Vl6 - ;c2

±716"

19. (a) \(iy-= 144 - 9^;^

^ = 1^044 - 9x2) = ^(,6 _ ^2)

b^yie"

(b)

(c) Explicitly:

J = ±|(16-jc2)-/2(-2x)

3x

-3x -^x

4716 - X- 4(4/3)y 16y

(d) Implicitly:

18x + 32xv' = 0

-9x
16)-

21. xy = 4

xy' + y(l) = 0
xy'= -y

'■ = ?

At (-4,-1): y'=-\

2'- ^"-x= + 4

, , (x2 + 4)(2x) - (x2 -

- 4)(2x)

^•^^ " (x2 + 4)2

^^^

^^^ (x^ + AY

8x

-^ y(x2 + AY

At (2, 0), >>' is undefined.

Section 2.5 Implicit Differentiation 81

25. x2/3 + y2/3 = 5

r-1/3

-1/3

At (8,1): y'= --.

3/^

27. tan(a: + y) = jc

(1 +y')sec2(j( + >) = 1

1 - secHx + y)

sec^U + y)

— tan^U + y)

tan^U + y) + 1

— cin^l

sin^U + y)

x2+ 1

At (0,0): y' = 0.

29.

0

c= + 4)y =

= 8

(^^'

+ 4)y'

+ y(2x) =
y' =

= 0

-2xy
x~ + 4

=

-Zx[S/(x^ +
x^ + 4

-16;c

4)]

(x- + 4)-

. ,^ ,, , -32 1

At(2.1):y' = — =--

Or, you could just solve for y: y = — -

31. (x^ + y2)2 = 4x^

2(x^ + y-){2x + lyy') = 4.r=y' + y(8x)
4j:^ + 4x^ ' + 4xy + 4v^y ' = 4.t^' ' + 8xy

4j:n7 ' + 4y^y ' — 4x^ ' = 8xv — 4.r^ — 4xy^
4y '(j:^ + >'^ ~ X-) = 4(2xv' — .r' — xy~)

At(l, 1): y'= 0.

, _ 2xy — x^ — xy-

x?y + y^ — X-

33.

tany =

y'sec'v =

X

1

.v' =

1

= cos-

•y, -

n

sec- y

2

sec^v =

1 + tan

2y =

1 +

X-

y' =

1

1 +x-

37.

x^

- y^ =

16

Ix -

2yy ' =

y' =

0
x

y

X

- yy' =

0

1 - yy" -

(y'f =

0

\-yy"-

(;)= =

0

T -

- y'y" =

.X-

v"-

1

r -

1
X' _

77

< V < y

35. x^ + y2 = 36
2x + 2vv' = 0

„ y(-l)-^■n■'

y = ; ■■

y-

-V -l--r -

(-f)

-36

^

-^'^ 39. r = .x'
^ 2yv' ' = 3.r-

>\^i

.V"^

<^'

-<

fi

, _ 3£2 _ 3£2 ^ _ 3y .r^ 3y
2y 2y xy 2v \- 2t

„ 2T(3yO - 3y(2)

y =

, ,c

.^^

•^

^

r<

4x^
lx[i • (3y/lv)] - 6y

16

f y3

•1

K^

^ . M

^ 4.1- 4v

4.r=

1>

..^^

•.^>

V\^ ...c

'>

A

.■^'X

82 Chapter 2 Differentiation

41. v^ + v9 = 4

|;c-'/2 + ir'/^' = o

V^

At(9, l),:y'= -J

Tangent line: y - 1 = ~:;(x — 9)

y=-^x + 4
X + Sy - 12 = 0

43. x^ + f = 25
—X

At (4, 3):

Tangent line: y - 3 = —(x - 4) =^ 4;c + 3^ - 25 = 0

Normal line: .v - 3 = -{x - 4) => 3j: - 4>' = 0.

At (-3, 4):

Tangent line: y — 4 = -(x + 3) => 3j: - 4y + 25 = 0

Normal line: y - 4 = -r-ix + 3) => 4;c + 3y = 0.

45. x^ + y^ = r^

2x + 2yy' = 0

y ' = — = slope of tangent line

- slope of normal line

Let (xq, Jq) be a point on the circle. If Xq = 0, then the tangent line is horizontal, the normal line is vertical and, hence, passes
through the origin. H Xg i^ 0, then the equation of the normal line is

yo

y = —x

which passes through the origin.

Section 2.5 Implicit Differentiation 83

47. 25j;2 + I6f + 200;c - \60y + 400 = 0

50.x + 32yy' + 200 - \60y' = 0

, ^ 200 + 50.t
^ ~ 160 - 32.V

Horizontal tangents occur when x = -4:

25(16) + 163'2 + 200(-4) - 160y + 400 = 0

>'(>'- 10) = 0=*>' = 0, 10
Horizontal tangents: (-4, 0), (-4, 10).

Vertical tangents occur when >> = 5:

25jc2 + 400 + 200.t - 800 + 400 = 0

25;c(.x: + 8) = 0 => .v = 0, - 8

Vertical tangents: (0, 5), (- 8. 5).

(-8.5)

49. Find the points of intersection by letting y^ = 4x in the equation 2x^ + y^ = 6.
2^2 + 4x = 6 and (x + 3)ix - 1) = 0
The curves intersect at (1, ±2).

Ellipse:

Parabola:

4x + lyy' = 0

2yy ' = 4

Ix

, 2

y=-j

' ^y

At (1,2), the slopes are:

y'=-l

y'= 1-

At(l, -2), the slopes are:

y'= 1

y'=-l.

Tangents are perpendicular.

51. y = —X and x = sin v

Point of intersection: (0, 0)

y = -x: X = sin y:

y' = -I 1 = y ' cos y

y' = sec y

At (0, 0), the slopes are:

y'= -1 y'= 1.

Tangents are perpendicular.

53. xy= C x^ -f = K

xy' + y = Q 2x-2yy'=Q

y , X

y=-- y=-

. b^ = 4x|

f

f

k

^=.

|lr2 + y2 = 6

C= I

\

'"7-

J-

\

At any point of intersection [x. y) the product of the
slopes is (-y/.ir)(.t/y) = - 1. The curves are orthogonal.

84

Chapter 2 Differentiation

55. 2/ - 3jt* = 0

(a) Ayy' - llr^ = 0

Ayy' = \2x^

' = 12^= 3^

57. cos Try - 3 sin ttj: = 1

(a) — IT sin (Try)^' — 3t7 cos ttat = 0

(b) — 77 sin(T7>')-p - Btt cos(irx)— = 0
at at

, - 3 cos TTX

y = — ■

sin try

59. A function is in explicit form if y is written as a function
of x: y = f(x). For example, y = x^. An implicit equation
is not in tlie form y = f{x). For example, x^ + y^ = 5.

■ / \dy . , Jbi
-smlTT-y)— = 3 cos(Tr;c)— ■

61. (a) x^ = 4(4;c2 _ ^2)
4/ = 16;c2 - y^

f = 4x^ - -x*

4

4x^ - -y^

(b)

v = 3=*9

4;c2 - 7A^
4

36 = 16x' - x^
y^ - ]6x^ + 36 = 0

^,^16± 7256^144^^^^

Note that jc^ = 8 ± 728 = 8 + 2^7 = (l ± Vv)".
Hence, there are four values of x:

-1-V^, 1-V7, -1 + V7, 1 + 7?

x(8 - x^)

To find the slope, 2yy ' = 8x — x^

■y =

2(3)

For a: = - 1 - Jl,y' = jIV? + 7), and the line is

>-, = \[Jl + 7)(x + I + 77) + 3 = ^[(77 + i)x + 877 + 23)].
For X = \ — 77, y' = jl 77 - 7), and the line is

yi = Kv^ - 7)(x - 1 + 77) + 3 = ^[(77 - l)x + 23 - 877].
For j: = - 1 + Jl,y' = -^iVl - l), and the line is

Vi = -1(77 - 7)(;c + 1 - 77) + 3 = -5[(77 - l)x - (23 - 87?)].
Forx = 1 + 77,^' = -5(7^ + 7), and the line is

y, = -}(77 + 7)(;c - 1 - 77) + 3 = -^[{^ + l)x - (877 + 23)].
—CONTINUED—

V

X

-ff

^

Section 2.6 Related Rates 85

61. —CONTINUED—

(c) Equating y^ and y^,

-\{ji - i){x + 1 - v?) + 3 = -|(y7 + i){x - 1 - y?) + 3

(V7 - i){x + 1 - V?) = (77 + 7)(jc - 1 - Vv)
V7;c +V7-7-7JC-7 + 7^7 = Jlx -Jl-l + lx-1- 1 Jl

16V7 = \Ax
877

nx = _ , then v = 5 and the hnes intersect at | „ . 5

63. Let /(a:) = .t^ = ;t^''*, where p and q are nonzero integers and g > 0. First consider the case where p = 1. The derivative of
/(jc) = x'/* is given by

:f [.'/.] = lim f^^^^)-f^-^ = nn, /W^M

dJC :^r^0 ^x t-^x t — X

where t = x + Ajc. Observe that

fit) - f(x) _ f'/" - x^^i _ t'^l" - x^li

t - X

t - X {t"")" - {x^li)"

(fl/9 _ y\lq^(f\-(\lq) + jl-(2/,)^l/<, _!_... + jl/«^I-(2/9) + yA-illqf)

_ 1

~ fl-(l/9) + fl- (2/9)^1/? _|. . . . + jl/9^1-(2/?) + ^1 -(!/«)■

As f — >.i:, the denominator approaches g.x'~"^*'. That is.

Now consider/(.t) = ji^/' = (.x^)'/«. From the Chain Rule,

g iit g q q \ q

Section 2.6 Related Rates

1. y= ^x

dy_ ^ I 1 \dx
dt XlJ'x] dt

dt dt

(a) When .v = 4 and (iv/dr = 3,

dt 2v^^^^~4-

(b) When .r = 25 and dyjdt = 2,

^ = 2725(2) = 20.
dt

3.

.rv' =

4

''1t =

0

dy _
dt

vWv
xjdt

dt

(-

x\dy
\ldt

(a) When x =

= 8.

V = 1

y = 1/2, anddxM= 10.

(b) When .r = 1. v = 4. and dy/dt = -6.

dt

-i,-« = i

86 Chapter 2 Differentiation

5. y = x2 + 1
dx

dt

= 2

dt dt

(a) When ;c = - 1
dy

dt

= 2(-l)(2) = -4 cm/sec.

7.

J =

tan j:

ate _
dt

2

dy
dt

sec^j:

^
A

(a) When x= -it/3,
dy

-j- = (2)2(2) = 8 cm/sec.
dt

(b) When x = 0,
dy

, = 2(0)(2) = 0 cm/sec.
dt

(c) When x = \,

^ = 2(1)(2) = 4 cm/sec.
af

dx dy . .

9. (a) — negative => — positive

(b) — positive => — negative
ar dt

(b) When x = - tt/4,
dy

-j-= (72)2(2) = 4 cm/sec.

(c) When x = 0,

^ = (1)2(2) = 2 cm/sec.

dy
11. Yes, >> changes at a constant rate: ~r = a

No, the rate -p is a multiple of —r-
dt *^ dt

dx
dt'

13. D = Jx-+f = 7x2 + (^2 + 1)2 = 7x^ + 3x2+1

— = 7

dt

dD 1, . , , ,x—i /-,/,,. , \dx 2x^ + 3x

dt 2' 'dt Vx^ + 3x2 +

dx

4x3 + 6x

1 dt Jxf + 3x2+1

15.

A =

77r2

dr _
df

3

dt

2Trr —
dt

(a) When r =

6,

dA _
dt

27r(

(b) When r =

24

dA

dt

= 27r(6)(3) = 36-77 cm2/min.

= 2Tr(24)(3) = 1447rcm2/min.

2 s 2

eh, e

cos X = ~=^ h = s cos -

2 s 2

. 1,, \ L . e\ 6

A = -bh = -\2s sin -II 5 cos -

= yl2sm-cos-l=-sine

dA s2

,de

de 1

(b) -— = — cos fl-— where ^- = - rad/min.
d/ 2 df dt 2

77 dA

^'^^"^=6'd7 = y- 2 8

V3\/l\ 735

^^"^=3'dr=2(2A2

52

(c) If dS/df is constant, dA/dt is proportional to cosft

Section 2.6 Related Rates 87

4 dV

19, V = -7rr3, ^ = 800
3 dt

dV _ , dr

-J- = Airr- — -
dt dt

1 /rfV

i"47^r2UJ"47^r^^^°^^

(a) When r = 30,

"7 = :; — r^Tv^? (800) = — - cm/min.
£?r 47r(30r 9tt

(b) When r = 60, ^

1

dt 47r(60)'

(800)

1
18-n-

cm/min.

21. s = 6;c2

= 3

= 12x

dx
dt

(a)

When

x= \,

ds

— = 12(1)(3) = 36cmVsec.

dt

(b) Whenx = 10,
ds

dt

12(10)(3) = 360cm7sec.

23. V = -irr-h =-Tr(-hAh [since 2r = 3h]

4

— = 10

dt

^ = ^^2 ^ ^ _ 4(rfV/^f)
dr ~ 4 dt"^ dt~ 9-771%-

When/! = 15

d/i _ 4(10)

dt 977(15)- 40577

ft/min.

25.

(a) Total volume of pool = -(2)(12)(6) + (1)(6)(12) = 144 m^

Volume of Im. of water = -(1)(6)(6) = 18 m^

(see similar triangle diagram)

% pool filled = Ts(100%) = 12.5%

(b) Since for Q < h < 2, b = 6/i, you have

V = \bh(6) = 3bh = 3(6h)h = Wr

1 1

dV

... dh

1 dh

36/!— =

= -=>--r

dt

dt

4 dt

1

144/1 144(1) 144

m/min.

88 Chapter 2 Differentiation

27. x2 + / = 252

^ dx dy

Ix — + 2y — = 0
dt dt

dy _ j2£
dt y

dx
dt

Ix . dx
— since —- - 2.
y dt

(a) When x = l,y= 7576 = 24, -f

dt

dy -2(7) -7

24

12

ft/sec.

(b) A = -xy

When;f = IS,)' = 7400 = 20.

rfy

-2(15) _ -3

When;c = 24,j = 7,^ =
dt

dt
2(24)

20
-48

ft/ sec.

ft/sec.

dy dx

iVJt^'Jt.

dx

From part (a) we have a: = 7, 3; = 24, — - = 2,

dt

dy 1

(c)

tan 0 = -

y

^ „ d6 1 dx X dy
sec^ e — = ; r--f

dt y dt

dt

de ^ \,\ dx X dy^

— = cos- 6\- ■ ^ • -r

dt V,y dt y~ dt\

Thus,

dA
dt

)_

2

527

24

12

+ 24(2)

21.96 ftVsec.

Usingx= 7,^ = 24,^ = 2,^
^ dt dt

24

de

12

and cos 8 = —, we have -r = \t7

25

dt

24 \2

25

24^ ^ (24)- 1 12

12

rad/sec.

29. When y = 6, ;t = 712^ - 6- = 6 V3, and

s= Vx- + (12 - y)2
= VIO8 -I- 36 = 12.

;c2 + (12 - y)2 = 52

12 -V

^^

^^

■

n...

X

^^te

y

\

^

y<

n

x-r + {y- \2)-f = s —
dt dt dt

Also. ;c2 -K /= 122

^ dx , ^ dy „ dy —x dx
dt dt dt y dt

™, dx , , ,^J—xdx

Thus, X— -I- [y- 12 — —

dt \ y dt

dx
dt

X +

\2x

ds_
dt'

ds
= 'Jt

dx _ sy ds

(12)(6)

dt \2x ' dt (12)(673)

(-0.2)

5V3

15

m/sec (horizontal)

(iy ^ -;cflDc ^ -6v^ (-VI) ^ J_

rff

y rfr

15

m/sec (vertical).

Section 2.6 Related Rates 89

31.

(a)

s^ =
dx

dt

dy _
dt

-450
-600

dt

dt

^ dt

ds

x{dx/dt)

+ y{dyldt)

100 200

Distance (in miles)

When x= 150 and >> = 200, s = 250 and
ds _ 150(-450) + 200(-600)

dt

250

= -750mph.

250 1
(b) r = — = J hr = 20 min

33.

1
s~ =

X —

dx

dt

90- + x-
30

-28

. ds

2s— =
dt

dx ds X
dt dt s

dx

' dt

Whenj:

s =

= 30,

5oyio

= V90- + 30= =

ds
dt "

^13^'-'^

-28

yio

5 —

8.85 ft/sec

35.

«f

-'' • llv

- 15.x =

6y

y - X

dt

dy 5 dx 5 , _, 25 . ,

^., d{y-x) dv dx 25 ^ 10,.,

90 Chapter 2 Differentiation

37. x(t) = ^sm^,x^- + y^= 1
2 6

277

(a) Period: — 77 = 12 seconds

77/6

(b)When. = i.^7l-(l)^^f,

73'

Lowest point: I 0,

(c) When x = -,y

VMI

and r = 1

39. Since the evaporation rate is proportional to the surface
area, dV/dt = kiAirr-^). However, since V — (4/3)Trr',
we have

dV . ^dr

dt dt

Therefore,

dt dt

dx l/7r\ TTt TT TTt

~r = T 7" COS -T = TX COS —r

dt 2\6/ 6 12 6

X2 + / = 1

^ dx „ dy „ dy -x dx

2x— + 2vT^ = 0=>-f = — .

dt ■ dt dt y dt

Thus,

di
dt

Speed

1/4 TT / 77

-77 / 1 \^ ^ -77 1

■VSt

yi5V12/ 2 24 75 120 •

-7577

120

7577

120

m/sec

41.

py\.^

1.3^V0.3^+V'3^ = 0

dt dt

^ dt dt

..3pf=-v4

dt

dt

43.

tan e =

30

3 m/sec.

sec- d

dd_ ]_di

dt ~ 30 dt

de 1 ^dy

— - = TT COS' e • — -

rfr 30 rff

When >- = 30, e = 77/4 and cos 6 = v z/2. Thus,

rfe 1 /l\,,v 1 .,
^=30l2r^ = M'^/^^'=-

Section 2.6 Related Rates 91

45.

tan e = -, y = 5

X

^= -600mi/hr
dt

.^li

,-T

,=5

, ^^.dO 5 dx

(sec^e)— = — r • —

do

dt ''''\-7^)-d-t = U[-7^)jt

= (-§ )(|)f = (-sin2 0)(|)(-6OO) = 120sin'9

(a) When 6 = 30°, ^ = ^ = 30 rad/hr = ^ rad/min.
at 4 2

(b) When e = 60°, y = 12o(|) = 90 rad/hr = | rad/min.

(c) When 6 = 75°, ^ = 120 sin^ 75° « 1 1 1.96 rad/hr = 1.87 rad/min.
dt

47, — = (10rev/sec)(2irrad/rev) = 20'n-rad/sec

(a) cos 9 = —

. „de I dx

-'"^^Yt^ToTt

^=-30sin.^ =
dt dt

(b) 2

wo

A A

0

vAT

-30 sin e(207r) = -60077 sin e

(c) \dx/dt\ = I -60077 sin 0| is greatest when sin e = 1 => 6= (tt/I) + mr(oT90° + n ■ 180°)
\dx/dt\ is least when 6 = mriorn • 180°).

(d) For e = 30°,

dx

dt

-600T7sin(30°) = -6OO77- = - 3OO77 cm/sec.

For e = 60°, ^ = -60077sin(60°) = -60077^ = -300V^ 77 cm/ sec
dt 2

49. tan e = — => j: = 50 tan e

— - = 50 sec- 6 —-
dt dt

2 = 50 sec-

1^

— - = rrcos^ e, -— < e < —

dt 25 4 4

92 Chapter 2 Differentiation

51. x~ + )r = 25; acceleration of the top of the ladder = -jj

dx dy

First derivative: lx—- + 2y— = Q
dt ■ dt

dx , dy
X— + y— = 0
dt dt

.... d-x , dx dx , dhi dy dy .

Second denvative: x — -r + —r ' —r ~^ TT^ + ~; T = 0

dt- dt dt dt^ dt dt

d^ (I
dt^ '

dy^ _1_
dt 12'

dx
dt

dh _ fdx\2 _ fdyy
'^ \dt) [dtj .

dx . . d^x

When X = 7, y = 24, — - = — — -, and — - = 2 (see Exercise 27). Since -— is constant, -rr = 0.
■^ -'* '^ -'• dt dt^

d^ ^ 1
dt'^ " 24

-7(0) - {ly

24

:] =

1441 24

625
144

= -0.1808 ft/sec^

53. (a) Using a graphing utility, you obtain mis) = -0.8815^ + 29.105 - 206.2

(b)^ = ^j;=(- 1.762. + 29.10)f
dt ds dt dt

ds
(c) Ifr = s(1995J, then. = 15.5 and— = 1.2.

Thus. ^ = (- 1.762(15.5) + 29.10)(1.2) = 2.15 million.
dt

Review Exercises for Chapter 2

1. fix)

= X- -Ix + Z

f'(x)

f(x + Ax)-f{x)

= lim

Aj->o Ax

,. [{x + Axf - 2{x + Ax) + 3]- [x^ - 2x + 3]

Ax-»0 Ax

•

,. ix" + 2x(Ax) + {Ax)- -2x- 2(Ax) + 3) - {x^ -

- 2x + 3)

iit->o Ax

,. 2x(Ax) + (Ax)- - 2(Ax) ,. ,^
= hm — ^^ — ^--~ 5^ — - = lim (2a: + A;c - 2) =

AJ-.0 Ax _ Ax-^O

= 2x - 2

3. f(x)

= Vx + I

5. /isd

fix)

fix + Ax) -fix)
— lim

^x-M Ax

,. (Vx + Ax + 1) - (Vx + 1)

5. / is differentiable for all .t =?^ — 1 .

= lim

Ax->0

lim

Ax

^x + Ax - yx Vx + Ax + Vx
Ax Vx + Ax+ 7x

(x + Ax) — X

Ax-^o Ax(Vx + Ax + Vx)

r 1 1

^-0 Vx + Ax + Vx 2Vx

Review Exercises for Chapter 2 93

7. fix) = 4 - |;c - 2|

(a) Continuous at x = 2.

(b) Not differentiable at jr = 2 because of the sharp turn
in the graph.

4 1
9. Using the limit definition, you obtain g '{x) = -x - -

4 1 -3
Atx=-l,g'(-l)=---- = —

11. (a) Using the limit defintion, f\x) = "ix^.

At X = - !,/'(- 1) = 3. The tangent line is

>-- (-2) = 3(;c -(-!))
.V = 3x + 1

(b) -4f

13. g'(2) = lim.

gW - g(2)

2 X-1

= lim-^(--V'

jr->2 JT — 2

,. A^ - ;c2 - 4
= lim

x-*^ X — 2

,. (x - 2)U2 + x + 2)
= lim r

lim U^ + ;c + 2) = 8

a:-»2

IS.

17. >- = 25
y' = 0

19. /W = x»

f'(x) = ar'

21. /i(r) = 3r*
A '(f) = 12f'

23. /U) = .r3 - 3x'

f'(x) = 3x- - 6x = 3.x(.r - 2)

25. A(x) = 6V^ + 3^ = dr'/2 + Sx'/'
/I'W = 3;t-'/2 + X-V3 = 3 _^ 1

29. fid) = 20 - 3 sin 9
/'(0) = 2 - 3 cos e

27. g(r) = |r-

-4 -4

31. fie) = 3 cos e

sin e

fie) = -3 sine -

cos e

94 Chapter 2 Dijferentiation

33. F = 200yr
100

F'{t) =

Vr

(a) When T = 4, F'(4) = 50 vibrations/sec/lb.

(b) When 7=9, F'(9) = 335 vibrations/sec/lb.

35. s(t) = - 16f2 + 5o

^(9.2) = - 16(9.2)2 + So = 0

5o = 1354.24

The building is approximately 1354 feet high (or 415 m).

37. (a)

20 40 60

Total horizontal distance: 50
(b) 0 = X - 0.02x2

0=x 1

50

implies x = 50.

39. x{t) = f - 2t + 2 = it- 2){t - 1)
(a) vit) = x'it) = 2t - 3

ait) = v'(r) = 2
(c) v(f) = Oforr = i

^ = (l-2)(|-i) = (4)(l) = -i

(c) Ball reaches maximum height when x = 25.

(d) y = x- 0.02x2
y'= 1 - 0.04x

y'(0) = 1
y'ilO) = 0.6
^'(25) = 0
y'i30) = -0.2
y'(50)=-l

(e) y'(25) = 0

(b) v(f) < 0 for f < 5.
(d) x(f) = Oforf= 1,2.

lv(l)| = |2(l)-3| = l

|v(2)| = |2(2) - 3| = 1

The speed is 1 when the position is 0.

41. fix) = (3.x2 + 7)(x2 - 2.x + 3)

fix) = (3x2 + 7)(2;^ - 2) + (x2 - 2x + 3)(6x:)
= 2(6^3 - 9x2 + 16^ _ 7)

45. fix) = 2x - x-2

fix) = 2 + 2x-^ = 2( 1 + -J
. _ 2(x3 + 1)

X3

43. hix) = Jlc sin ;t = x'/2 sin x
1

h'ix) =

47. /(:c) =
fix) =

2V5"

X2 +

+ J~XI.

1

Jt2- 1

(x2 - l)(2x + 1) - (^2 + X - l)(2x:)

ix" - 1)2

-(;c2+l)
(x2 - 1)2

49. fix) = (4 - 3x2)-'

6x

/'(x) = -(4 - 3x2)-2(-6x) ^4 _ 3^,^,
53. >> = 3^2 sec X

y'

ji. y —

cosx

cos X ilx) - x\- sin x)
cos2 X

2x cos X + x2 sin x

cos2x

55. y = -jctanjc

3; ' = — x: sec2 X — tan X

Review Exercises for Chapter 2 95

57. y = X cos j: — sin j:

y' = — j: sin ;c + cos X - cos j: = -xsin;c

59. gU) = r^ - 3r + 2
g'it) = 3^2-3
g"{t) = 6r

61. f(0) = 3 tan 9
f'(d) = 3 sec2 6
f'(e) = 6 sec e (sec 6 tan 0) = 6 sec^ 0 tan

63. >> = 2 stn j: + 3 cos x

y' = 2 cos x — 3 sin x
y" = — 2 sin ;t — 3 cos x
y" + y = — (2 sin ;c + 3 cos x) + (2 sin jc + 3 cos x)
= 0

65. f(x) = (1 - .r3)'/2

1,

/'W=^l-x5)->/2(-3x2)
3;c2

2Vl -:t3

67. /zW = (^ ^^'

h'(x) = 2

X^ + ly

;c-3\/y-+ 1)(1) - (;c - 3)(2x)\

a:= + 1

+ 1)^ ;

2U - 3)(-.t^ + fa + 1)

69. /(i) = {s- - 1)5/2(53 + 5)

f'(s) = (52 - 1)V2(3^2) + (^ + 5)(|)(^2 _ 1)3/2(2^)

= sis' - l)3/2[35(i2 - 1) + Sis' + 5)]
= sis' - 1)3/2(8^3 - 3i + 25)

71. y = 3 cos(3.r + 1)
y' = -9sin(3;c + 1)

73. y = "z CSC 2x

y ' = r{ — CSC 2x cot 2i:)(2)

- CSC 2x cot 2t

.r sin 2x

= r{l — cos 2x) = sin^.r

2 2

77. y = - sin3/2 x - - sin'^/^x

= sinl/2

inV2 ,

sm'''^ JTCOS j: — sm^'-xcos j:
(cos j:)Vsin j:(1 — sin^x)

c3

= ( cos= xl V sin a:

79. V =

.T + 2

, _ ix + 2)Trcos TT.r — sin tt.v
•'' " (-v + 2)2

81. /(/) = fit - ly

fit) = t(t- DVt - 2)

The zeros of/' correspond to the points on the graph off
where the tangent line is horizontal.

83. gix) = 2x(.r + D-'/^

X + 2

g'U) =

ix + 1)3/2

g ' does not equal zero for any value of t in the domain.
The graph of g has no horizontal tangent lines.

•■\

^

" J

96 Chapter 2 Differentiation

85. fit) = {t+ l)i/2(r + l)'/3 = (r + l)V6
5

/'(f)

6(f + !)'/«

/' does not equal zero for any x in the domain. The
graph of/has no horizontal tangent lines.

^

^

/■

87. y = tanVl - x

c=yr

2Vl -.«

y ' does not equal zero for any x in the domain. The graph
has no horizontal tangent lines.

r^

89. y = 2;t2 + sin 2jc
>-' = 4;c + 2cos2i
y" = 4 — 4s'm2x

91. /(a:) = cot X

93. /(r) =

fix) = -csc^a;

/"= -2cscjf(-cscj:

cotjc)

/'(f) =

= 2 csc^ X cot ;c

/"(f) =

t

(1-

ty

f +

1

(1-

f)'

2(f4

-2)

(1 - f)^

95. gie) = tan 30 - sin(e - 1)

g'ie) = Ssec^se- cos(e- i)

g"(e) = 18 sec2 3etan 361 + sin(e - 1)

97. r= 700(f2 + 4r+ 10)-'

- 1400(f + 2)
" (f2 + 4r + 10)2

(a) When f = 1,

^, -1400(1 + 2)

(1 + 4 + 10)2
(c) When ; = 5,

- 1400(5 + 2)

== - 18.667 deg/hr.

r

(25 + 30 + 10)

5= -3.240 deg/hr.

(b) When r = 3,

^, _ - 1400(3 + 2)

(9 + 12 + 10)2
(d) When t = W,

-1400(10 + 2)

7"

(100 + 40 + 10)2

-7.284 deg/hr.

= -0.747 deg/hr.

99. x2 + 3xy -f- y = 10

2x -I- 3xy' -I- 3y -I- 3>'2y'= 0

3(a: + /)>>'= -(Ix -(- 3>')

-(2A: + 3y)
^ Hx + f)

101.

yv^t ~ x^ = 16

2V^ — X , __ 2vxy — y
iVy ^ " 2Jx

, ^ ijxy - y _ ijy ^ lyjx - yVy
2jx iVxy — X Ixjy — x^x

Review Exercises for Chapter 2 97

103. xsxny = y cos x

{xcosy)y' + smy = ->> sinj: + >''cos;c

>''(^cos>' - cos x) = — ysinjt — siny

, _ y sin a: + sin y
cos X — X cos y

105. Jt^ + y2 = 20
2x + lyy' =Q

At (2, 4): y'= --

i4

\,

Tangent line: y — 4 = — xU - 2)

;c + 2y - 10 = 0
Normal line; y - 4 = l{x - 2)
2x-y = Q

107. y= ^x
dy

dt

= 2 units/sec

dy 1 ^v ^^2^^ = 4^

dt 2jx dt

dt

dt

1 dx

(a) When x = — , — = 2 V2 units/sec.

dx

(b) When j: = 1, — = 4 units/sec.

dt

dx

(c) When j: = 4, — = 8 units/sec.

dt

109.

5

h ~

1/2
2

S =

^'

dt

1

Width of water at depth h:

w = 2 + 2s = 2 + 2\hx

A + h

''-f(^-^> = j«*'*

dV 5,, ,.dh

dt

dh ^ 2{dV/dt)
dt ~ 5(4 + h)

dh

When h = 1. ^ = rr m/min.
dt 25

111. s{t) = 60 - 4.9r
s\t) = -9.8r
5 = 35 = 60 - 4.9f-
4.9r= = 25
5

tan 30°

74.9
.r(f) = Jls(t)

^=73^=V3(-9.8h^
= - 38.34 m/sec

98 Chapter 2 Differentiation

Problem Solving for Chapter 2

1. (a) x~ + (y - r)^ = r^ Circle

X- = y Parabola
Substituting,

(■y _ ;.)2 = ^ _ y

y'^ — 2ry + r' = r^ — >>
y^ — 2ry + >> = 0
y{y - 2r + 1) = 0

Since you want only one solution, let 1 — 2r = 0 => r
Graph y = x'^ and x'^ + [y - 5)' = 4

(b) Let (x, y) be a point of tangency: x^ + (y - b)^ = 1 => 2j: + llj" - b)y' = 0 => y '
y = ;c'^ => >; ' = 2x (parabola). Equating,

r

2jc^

Z.-y

(circle).

b-y

2{b - y) = 1

Also, a;^ + (y - Z))^ = 1 and y = x^ imply
y + (y - fc)2 = 1 =* y + l^y - (y + ^j j = 1

Center: 0

Graph y = x- and x-

■ y--= l=>y = - and/7 = -.

3. (a) f(x) = cos;c
/(O) = 1
/'(0) = 0

P,{x) = 1

(c)

P\ix) = % + OiX

P,(0) = ao=»ao= 1
P'i(O) = a, =* a, = 0

(b) f(x) = cosx
/(O) = 1
/'(O) = 0
/"(0)=-l

P^ix) = 1 - I;c2

j:

-1.0

-0.1

-0.001

0

0.001

0.1

1.0

cosx

0.5403

0.9950

« 1

1

« 1

0.9950

0.5403

i'aW

0.5

0.9950

= 1

1

" 1

0.9950

0.5

PjW is a good approximation of f{x) = cos x when x is near 0.

(d) fix) = sin;c

/(O) = 0

/'(O) = 1

/"(O) = 0

/"'(0)= -1

P^ix) =x-\x^

Oq + a,j; + flj^^ + ^s-"^

P,m= ao^ao = 0
P\{0) = a, =^ a, = 1
^"3(0) = 2^2 => 02 = 0
^"'3(0) = 603 => 03 = -

PjW = Oq + OiX + a^K-

P^iO) = flo =5> ao = 1
/"2(0) = ai=»a, = 0
P"2(0) = 2a2^a2= -i

Problem Solving for Chapter 2

99

5. Letp(jc) = Aj^ + Bx^ + Cx + D
p'{x) = 3Ax- + 2Bx + C
At (1, 1); A + B + C + D = 1 Equation 1

3A + 2B + C =14 Equation 2

At (-1, -3): -A + B-C + D=-3 Equations
3A - 2B + C = -2 Equation 4

Adding Equations 1 and 3: 25 + 2D = -2
Subtracting Equations 1 and 3: 2A + 2C = 4

Adding Equations 2 and 4: 6A + 2C = 12
Subtracting Equations 2 and 4: 45 = 16

Hence, S = 4 and £> = k(-2 - 2B) = -5

Subtracting 2A + 2C = 4 and 6A + 2C = 12, you obtain 4A

Thus, pW = 2^3 + 4x- - 5.

■ A = 2. Finally, C = 5(4 - 2A) = 0

7. (a) .x^ = a-x^ - a^y^
a^y- = a-x — x^

±Va-.v--.V>

„ . Va-.r- - -r^

Graph: >>, = and Vj =

^cP-x- — .r*

(b)

frb

^

a= 1

vj.,

(±a, 0) are the Ar-intercepts, along with (0, 0).
(c) Differentiating implicitly,
Ax^ = 2a- .1: — 2a^yy'

y = — r-^; = r = Q^2x- = a-=^x = —7=.

2ay a^ V2

4

->
a-

V = ±7

^ . I a a\ I a a\ ( a a\ l-a a\

Four poutts: ^-^, -j. ^^, --j. ^-^. ^ j. (yj' "^J

100 Chapter 2 Differentiation

9. (a)

(b)

90 100
jVo/ drawn to scale

(0, 30)

1.6)
no. 3)

60 70
!^ot drawn to scale

Line determined by (0, 30) and (90, 6):

y-30

30-6
0-90

U-0)

24

'90-" 15^^

?

+ 30

-4 10

When X = 100, y = -j-r(100) + 30 = — > 3 => Shadow determined by man.

Line determined by (0, 30) and (60, 6):

-2
When x = 70,y = -z-(70) + 30 = 2 < 3 => Shadow determined by child.

(c) Need (0, 30), {d, 6), {d + 10, 3) collinear.

30-6

6-3 24 3

0' d

d- id+ 10)^ d 10

d = 80 feet

dx

(d) Let V be the length of the street light to the tip of the shadow. We know that — =

dt

For X > 80, the shadow is determined by the man.

y y ~ X 5 , dy 5 dx —25

5^ = -^=^^ = ?^'^^ = 4^ = ~-

For X < 80, the shadow is determined by the child.

y_ ^ y - X
30 3

Therefore,

10

10 100 _,rfy 10^ -50

dt

-25

4
-50

9

;c > 80
0 < .x < 80

dy

—r is not continuous at j: = 80.

dt

11. L'(x) = lim

Ajt->0

= lim

Aj:->0

L{x + ^x) - L{x)
Ax

Ljx) + L(Ax) - L(x)

Ax

lim

L{Ax)

<ii^o Ajc

Also, Z.'(0) = lim

UAx) - L(0)

Ax->0 Ax

But, L(0) = 0 because Z,(0) = L(0 -I- 0) = L(0) -f- L(0) => L(0) = 0.

Thus, I, W = Z.'(0), forall;c.

The graph of L is a line through the origin of slope L'{0).

Problem Solving for Chapter 2 101

13. (a)

z (degrees)

0.1

0.01

0.0001

sinz

0.0174524

0.0174533

0.0174533

(b) lim^-^- 0.0174533

z->0 z

, . ^ ,. smz 77
In fact, lim —

J->0 z

(c) —(sin z) = lim

dz A;-»0

= lim

180

sin (z + Az) - sin z

Az

sin z • cos Az + sin Az • cos z - sin z

Az

I sin z r + lim

oL V Az /J Aj:-»oL

= lim I sin z

= sinz(0) + cosz(j^ =-f^cosz

cos z

sin Az

Az

(d) 5(90) = sin(^ 9o) = sin ^ = 1; C(180) = cos(^ 180 ) = - 1

—S{z) = — sin(cz) = c • cos(cz) = 757:C(z)
az az ISO

(e) The formulas for the derivatives are more complicated in degrees.

15. jit) = a 'it)

(a) j(t) is the rate of change of the acceleration.

(b) From Exercise 102 in Section 2.3,
s{t) = -8.25f2 + 66;

v(r) = - 16.5f + 66
a(t) = - 16.5
a'(t)=j(t) = 0

CHAPTER 3
Applications of Differentiation

Section 3.1 Extrema on an Interval 103

Section 3.2 Rolle's Theorem and the Mean Value Theorem . 107

Section 3.3 Increasing and Decreasing Functions and

the First Derivative Test 113

Section 3.4 Concavity and the Second Derivative Test .... 121

Section 3.5 Limits at Infinity 129

Section 3.6 A Summary of Curve Sketching 136

Section 3.7 Optimization Problems 145

Section 3.8 Newton's Method 155

Section 3.9 Differentials 160

Review Exercises 163

Problem Solving 172

CHAPTER 3
Applications of Differentiation

Section 3.1 Extrema on an Interval

Solutions to Odd-Numbered Exercises

1. fix)

fix)

x^ + 4

(^ + 4)(2;c) - U=)(2t)

&x

U2 + 4)-

tr- + 4y

/'(O) = 0

3. fix) =x +

27

2x2

27
X + —x

fix) = 1 - 27x-3 = 1

27

/'(3) = 1 - |J = 1 - 1 = 0

5. /(x) = U + 2)2/3
f'i—2) is undefined.

7. Critical numbers: x = 2
X = 2: absolute maximum

9. Critical numbers: x = 1, 2, 3
X = 1,3: absolute maximum
X = 2: absolute minimum

11. f(x) = xKx -3) = x^ -3x-
f'(x) = 3;c2 - 6x = 3x(.t - 2)
Critical numbers: x = 0, x = 2

13. ^(f) = fv/4 - t, t < 3

gV) = r

i(4_ f)-l/2(_l)| + (4-,)l/2

= ^4-r)-'/2[-f + 2(4-r)]
8 - 3r

2V4 - /
Critical number is t

15. hix) = sin^x + cosx, 0 < .t < 27r

h'(x) = 2 sin .t cos AT — sinx = sinx(2 cos.r — 1)

TT 5 TT

On (0, 27r), critical numbers: x = — , x = tt, x = —

17. fix) = 2(3 - x), [- 1. 2]

fix) = - 2 => No critical numbers
Left endpwint: (- 1, 8) Maximum
Right endpoint: (2, 2) Minimum

19. fix) = -X- + 3x. [0. 3]
/'(x)= -2r + 3
Left endpoint: (0. 0) Minimum
Critical number: (2.4) Maximum
Right endpoint: (3. 0) Minimum

103

104 Chapter 3 Applications of Differentiation

21. f{x)=x^-jx', [-1,2]
f'(x) = 3x2 - 3x = 2x(x - 1)

Left endpoint: \~'^'~2J '^™'""'"
Right endpoint: (2, 2) Maximum
Critical number: (0, 0)

V

Critical number: 1,

23. fix) = 3a;2/3 - 2x, [- 1, 1]

/'W = 2;c-'/3 - 2 = 2(LzJ^

Left endpoint: (- 1, 5) Maximum
Critical number; (0, 0) Minimum
Right endpoint: (1, 1)

25. ,W = -^,[-l,l]

1

6t

(t^ + 3)2

g'it)

Left endpoint: I ~ 1. T 1 Maximum
Critical number: (0, 0) Minimum
Right endpoint: ( 1. T ) Maximum

27. his) = y-h;, [0, 1]
h'is) =

s-2'

-1

(^ - 2)2

Left endpoint: I 0, - — 1 Maximum
Right endpoint: (1, - 1) Minimum

29. fix) = cos TTX, 0, -
fix) = - TT sin TTX

Left endpoint: (0,

1) Maximum

/l V3^
Right endpoint: I -, — r-

Minimum

4 7131: r, «n

31. y = - + tan—, [1, 2]

X O

, -4 77 ^ TTX _

TT , ira 4

On the interval [1,2], this equation has no solutions.
Thus, there are no critical numbers.

Left endpoint: (l, v^ + 3) = (1, 4.4142) Maximum

Right endpoint: (2, 3) Minimum

33. (a) Minimum: (0, —3)
Maximum: (2, 1)

(b) Minimum: (0, -3)

(c) Maximum: (2, 1)

(d) No extrema

35. fix) = x^-2x

(a) Minimum: (1, — 1)
Maximum: (—1,3)

(b) Maximum: (3, 3)

(c) Minimum: (1, —1)

(d) Minimum: (1, — 1)

Section 3.1 Extrema on an Interval 105

37. f(x)

2x + 2, 0 < .r < 1
4x-, 1 < X < 3

Left endpoint: (0, 2) Minimum
Right endpoint: (3, 36) Maximum

39. fix) = J— ^A\ A]

Right endpoint: (4, 1) Minimum

41. (a) 5

(1.4.7)1

(0.4398.-1.0613)

Maximum: (1, 4.7) (endpoint)
Minimum: (0.4398, - 1.0613)

(b)

fix) = 3.2^5 + 5jc3 - 3.5x, [0, I]

fix) = 16.r^ + 15^2 - 3.5

16x^ + \5x- - 3.5 = 0

, ^ -15 ± Jil5)- - 4i\6)i-3S)
2(16)

-15 ± 7449

32

/- 15 + .
V 32

'449

- 0.4398

/(O) = 0

/( 1 ) = 4.7 Maximum (endpoint)

<V^

/449

= -1.0613

Minimum: (0.4398, - 1.0613)

43. fix) = (1 + x3)'/2, [0, 2]

fix) = ^Ki + x^)-''-

f%x) = J(X* + 4x)(l + .t5)-3/2

f"\x) = -^(.t« + 20x^ - 8)(1 + x3)-V2

O

Setting/'" = 0, we have .r« + 20.r3 -8 = 0.

jc3 =

-20 ± 7400 - 4(1)F8)

x= i/- 10 ± yiM = 73 - 1

In the interval [0, 2], choose

= V- 10 ± v-^ = 73 - 1 = 0.732.
/"( {/-\0 + 7i08) = 1.47 is the maximum value.

45. fix) = ix+ l)-^\ [0, 2]
/'(x)=|(x+l)-'/3

fix) = -^(.T + i)-*/3

f"{x) = ^(.v + 1)-'/^

/(4)(.^) = _|^(^ + 1

-10/3

/^%) = f§u + l)-"/^

|/*^*(0)| = — is the maximum value.
81

106 Chapters Applications of Differentiation

47. f(x) = tan X

/is continuous on [0, 7r/4] but not on [0, tt]. iim tan j: = oo.

X — »7r/2

49.

51. (a) Yes

(b) No

53. (a) No

(b) Yes

55. P = VI - RI^ = 12/ - 0.5/2, 0 < / < 15

P = 0 when / = 0.

P= 67.5 when /= 15.

P'= 12 - /= 0

Critical number: / = 12 amps

When / = 12 amps, P = 72, the maximum output.

No, a 20-amp fuse would not increase the power output.
P is decreasing for / > 12.

57.

3s^l^-C0Sd\ TT „ IT

s = 6hs + — ■ - — I- < e<-

2 \ smO / 6 2

dS^3fi
dd 2^

• Vscsc e cot 0 + csc^ e)

= -^csc g{- Vscot e + esc e) = 0

CSC 6 = V^cot 0
sec S = Vs

e = arcsecVs = 0.9553 radians

5|'f) = 6fo + ^V3)

5(arcseCv/3) = 6fc + -~-{j2)

S is minimum when d = arcsecVS = 0.9553 radians.

59. (a) y = ax^ + bx + c

y' = 2ax + b •• ^

The coordinates of S are (500, 30), and those of A are (- 500, 45).
From the slopes at A and B,

-1000a + b= -0.09

1000a + b = 0.06.

Solving these two equations, you obtain a = 3/40000 and fc = -3/200. From the points (500, 30) and (-500, 45),
you obtain

30 =

45 =

40000
3

50°' + '"^[wo^ ^ '^

40000 50°' -^ii)^^-
75

In both cases, c = 18.75 = — -. Thus,
4

3 2__3_ 75
^ 40000"^ 200^ 4'

—CONTINUED—

Section 3.2 Rolle's Theorem and the Mean Value Theorem 107

59. —CONTINUED—

(b)

X

-500

-400

-300

-200

-100

0

100

200

300

400

500

d

0

.75

3

6.75

12

18.75

12

6.75

3

.75

0

For -500 < X <Q,d={a:^ + bx + c)- (-0.09^).
For 0 < ;c < 500, rf = (or^ + to + c) - (0.06.r).
(c) The lowest point on the highway is (100, 18), which is not directly over the point where the two hillsides come together.

61. True. See Exercise 25.

63. True.

Section 3.2 Rolle's Theorem and the Mean Value Theorem

1. Rolle's Theorem does not apply to /(j:) = 1 — |x - 1|
over [0, 2] since/ is not differentiable at .r = 1.

5. fix) = xJITa

j:-intercepts: (-4. 0), (0, 0)

f'{x) =x^x + 4)-i/2 + {x + 4)1/2

)-'/^(f

= {x + A)-"A^+ (x + A)

fix) = I 2^ + 4 ]{x + 4)-'/2 = 0 at;c =

3. fix) =x^ - x-2 = ix-2)ix+ I)
jc-intercepts: (-1,0), (2,0)

fix) = 2x-l = 0atx = -.

7. fix) = x--2x, [0, 2]

/(0)=/(2) = 0

/is continuous on [0, 2]. /is differentiable on (0, 2).
Rolle's Theorem applies.

fix) = 2x - 2

2x-2 = 0^.t=l

c value: 1

9./(;c) = (:c- l);.r-2)(x-3),[l,3]

/(l)=/(3) = 0

/is continuous on [1, 3]. /is differentiable on (1, 3).
Rolle's Theorem applies.

fix) =x^ - 6x- + ll.r - 6

fix) = 3x- - \2x+ 11

3a- - 12x + 11 = 0=>x= ~

11. /(.r)=x2/3- l,[-8,8]

/(-8)=/(8) = 3

/is continuous on [—8. 8]. /is not differentiable on
(—8, 8) since /'(O) does not exist. Rolle's Theorem does
not apply.

6-73 6 + v^

108 Chapters Applications of Differentiation

/(-l)=/(3) = 0

/is continuous on [- 1, 3]. (Note: The discontinuity, x = —2, is not in the interval.)/is differentiable on (— 1, 3). Rolle's
Theorem applies.

f,, X (x + 2)(2;c - 2) - U^ - 2x - 3)(1) .
/ U) = 7 — -^:^ = 0

ix + 2)2

c value: — 2+V5

x^ + 4;c - 1

ix + ly

= zi4M = _2±V5

15. /W = sinx, [0,2it]

/(0)=/(27r) = 0

/is continuous on [0, 2Tr]./is differentiable on (0, lir).
Rolle's Theorem applies.

f'{x) = cos X

77 377

c values: — , — —

2 2 V

17. fix) = — - 4 sin^jc,

■ [»f]

/(o)-/m.o

/is continuous on [0, 77/6]. /is differentiable on (0, 77/6).
Rolle's Theorem applies.

fix) = 8 sin a: cos x = 0

77

19. fix) = tan X, [0, 77] ,

/(0)=/(77) = 0

/is not continuous on [0, 77] since/(T7/2) does not exist.
Rolle's Theorem does not apply.

77

8 sm X cos X

477

1 • 0

-smlx

3

277

sin2x

1

— arcsm

.217/

x

X =»

0.2489

c value:

0.2489

.1. fix) =

= w-

1, [-1,1]

/{-1) =

= /(!) =

0

/is continuous on [- 1, l]./is not differentiable on
(— 1, 1) since/'(0) does not exist. Rolle's Theorem does
not apply.

Section 3.2 Rolle's Theorem and the Mean Value Theorem 109

23. fix) = 4a: - tan ttx.

'4' 4

A-4J^A4J-o

/is continuous on [-1/4, 1/4]. /is differentiable on
(- 1/4, 1/4). Rolle's Theorem applies.

f'{x) = 4 - V sec- TTX = 0

25. fit) = -16f2 + 48r + 32
(a) /(I) =/(2) = 64

Q^) V = f'(t) must be 0 at some time in (1 , 2).
/'(f) = -32r + 48 = 0

t = - seconds

sec* TTX = —
77

sec TTX = ±-

'77

^1 1 \ Jtt

X = ±— arcsec — ■;= = +— arccos -^^ —

""■ V77 •"■ 2

« ±0.1533 radian
c values: ±0.1533 radian

27.

tangent line secant line

29. fix) = ^^, [0, 6]

/has a discontinuity at jr = 3.

31. fix) = x~ is continuous on [-2, 1] and differentiable on

(-2,1).

/(l)-/(-2) 1-4
1 - (-2) 3

= -1

fix) = It = - 1 when x = --. Therefore.

'=-2-

33. fix) = .r-'^ is continuous on [0, 1] and differentiable on
(0, 1).

/(l)-/(0)

1 - 0

= 1

fix) = j.v-'/5 = 1

27

110 Chapters Applications of Differentiation

35. fix) = V2 - X is continuous on [— 7, 2] and
differentiable on (—7, 2).

/(2)-/(-7)_0-3 _ _1
2 - (-7)

f'(x) =

9

3

-1

1

272-

X

3

272-

X =

= 3

72-

X -

3
" 2

2 -

X -

_ 9
" 4

X =

1

4

37. /(x) = sin X is continuous on [0, -ir] and differentiable on
(0, tt).

/(7r)-/(0) 0-0

= 0

77-0 V

f'(x) = cos X = 0

77

/w = -

X

yon

1

2'

2

X +

(a)

1

:^

^,,v^^secant

-1

(b) Secant line:

Slope =^(^^^

-/(-1/2)
-(-1/2)

2/3 -(-1)
5/2

2
3

2

= f(;c-2)

3.y - 2 =

= 2x - 4 •

,

Sy-

- Ix

f 2 =

= 0

(c) f'{x) =

(x + 1)^ 3

{x + 1)^ = 1

j: = -1 ± ^

V 2

-..f

In the interval [- 1/2, 2], c = - 1 + (76/2).

-1 + (76/2) _ -2 + 76 -2

f{c)

[-1 +(76/2)] + 1
2

76 76

+ 1

1l /f\

Tangent line; j - 1 + -^ = -Ax + 1

, ^ 76 2 76_^2
y — 1 -I — :::— = -x — - — i- -
^ 3 3 3 3

33- - 2;c - 5 + 276 = 0

Section 3.2 Rolle's Theorem and the Mean Value Theorem

111

41. /W = v^, [1,9]
(1,1), (9. 3)
3-1 I

1

(a)

(b) Secant line: >> — 1 = -(x - 1)

1 ^3
y = ~x + —

4 4
0 = X - 4^ + 3

(c)

fix)

ijk

/(9)-/(l) 1
9-1 4

1 _ 1

2Vc"4

7? = 2
c = 4
{c,/(c)) = (4, 2)

'^ = /'(4) = 4

Tangent line: y - 2 = -(j: - 4)

V = -.X + I
4

0 = .t - 4>' + 4

43. 5(f) = -4.9f2 + 500

5(3) - 5(0) _ 455.9 - 500

(a) K,

3-0

14.7 m/sec

(b) 5(f) is continuous on [0, 3] and differentiable on (0, 3).
Therefore, the Mean Value Theorem applies.

v(f) = s\i) = -9.8r = - 14.7 m/sec
-14.7

-9.8

1 .5 seconds

45. No. Let/(.r) = .r= on [- 1, 2].

/'(.t) = Ix

/'(O) = 0 and zero is in the inter\al (—1,2) but
/(-l)^/(2).

47. Let 5(t) be the position function of the plane. If f = 0 corresponds to 2 p.m., 5(0) = 0, 5(5.5) = 2500 and the Mean Value
Theorem says that there exists a time fg, 0 < fg < 5.5, such that

S%) = v(ro) = ^ffy » 454.54.

Applying the Intermediate Value Theorem to the velocity function on the intervals [0. fg] and \t^. 5.5], you see that there are at
least two times during the flight when the speed was 400 miles per hour. (0 < 400 < 454.54)

112 Chapters Applications of Differentiation

49. (a) /is continuous on [— 10, 4] and changes sign,

(/(-8) > 0, /(3) < 0). By the Intermediate Value
Theorem, there exists at least one value of x in
[- 10, 4] satisfying/U) = 0.

(c)

(b) There exist real numbers a and b such that

-\0 < a < b < A and /(a) = f(b) = 2. Therefore,
by Rolle's Theorem there exists at least one number c
in (- 10, 4) such that /'(c) = 0. This is called a criti-
cal number.

(d)

8--

iX

(e) No,/' did not have to be continuous on [—10, 4].

51. /is continuous on [-5, 5] and does not satisfy
the conditions of the Mean Value Theorem.
=>/is not differentiable on (-5, 5).
Example: f{x) = |;c|

M*-i

53. False. /U) = l/j:has a discontinuity zlx = 0.

55. True. A polynomial is continuous and differentiable everywhere.

57. Suppose that p{x) = x^+ 1 + ax + b has two real roots Xj and X2. Then by Rolle's Theorem, since p{xy) = p{x-^ = 0, there
exists c in (x,, x^ such that p '(c) = 0. But p '{x) = {In + l)x^ + a t^ 0, since n > 0, a > 0. Therefore, p(x) cannot have two
real roots.

59. \fp{x) = Ax- + Bx + C, then

fib) - f{a) _ (Ab- + Bb + C) - (Aa^ + Ba + C)

p '(x) = 2Ax + B

b — a b — a

A{b^ - a2) + B{b - a)

b- a

{b - a)[A(b + a) + B]
b- a

= A(b + a) + B.
Thus, 2Ax = A{b + a) and x = (b + a)/2 which is the midpoint of [a, b].

61. fix) = 2 cos X differentiable on (— oo, oo).
f'(x) = — 2 sin X

-5 ^ /'(x) < 2 =>f'(x) < 1 for all real numbers.
Thus, from Exercise 60, /has, at most, one fixed point, (x == 0.4502)

Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 113

Section 3.3 Increasing and Decreasing Functions and the First Derivative Test

1. fix) = X-- 6x + i
Increasing on: (3, oo)
Decreasing on: (—00, 3)

3. y =

}?

'ix

Increasing on: ( - 00, - 2), (2, 00)
Decreasing on: (-2,2)

5- fix) = ^

fix)

x'

Discontinuity: x = 0

Test intervals:

-00 < JT < 0

0 < .t < 00

Sign of fix):

/'>o

/'< 0

Conclusion:

Increasing

Decreasing

Increasing on (- 00, 0)
Decreasing on (0, 00)

7. six)

2x- S

g'ix) = Ix - 2
Critical number: x = 1

Test intervals:

-00 < j: < 1

1 < .r < cso

Signof g'W:

g' <Q

g' >o

Conclusion:

Decreasing

Increasing

Increasing on: (l,oo)
Decreasing on: (— 00, 1)

9. .v =

y' =

xVl6 — x^
-2(;c2 - 8)

Domain: [-4, 4]
-2

V16 - x"- yi6

Critical numbers: x = ±2^2

=ix - 2V2)(.v + 272)

Test intervals:

-4 < X < -2V2

-2v^ < .t < 2v^

2V2 < X < 4

Sign of _v ':

>-' < 0

v' > 0

y' < 0

Conclusion;

Decreasing

Increasing

Decreasing

Increasing on ( - 2 VI, 2 V2)
Decreasing on (-4, -2V5), (2^2, 4)

11- fix) = x^

6x

fix) = 2x - 6 = 0
Critical number: x = 3

13. fix) = -2x^ + 4x + ?,
fix) = -4.V + 4 = 0
Critical number: .v = 1

Test intervals:

-00 < j: < 3

3 < j: < 00

Sign of/'U):

/'< 0

/'>o

Conclusion:

Decreasing

Increasing

Increasing on: (3, 00)
Decreasing on: (-00, 3)
Relative minimum: (3, - 9)

Test intervals:

-00 < x < 1

1 < .V < 00

Sign of/'t.v)-

/'> 0

/' < 0

Conclusion:

Increasing

Decreasing

Increasing on: {-00,1)
Decreasing on: (1, 00)
Relative maximum: (1.5)

114 Chapters Applications of Differentiation

15. fix) = 2x3 + 3x2 - 12a:

fix) = 6x2 + 6x - 12 = 6(;^ + 2)(x - 1) = 0

Critical numbers: x = —2, 1

Test intervals:

-oo < X < -2

-2 < X < 1

1 < X < oo

Sign of/'(x):

/'>o

/' < 0

/'>0

Conclusion:

Increasing

Decreasing

Increasing

Increasing on: (— oo, —2), (1, oo)
Decreasing on: (-2,1)
Relative maximum: (—2,20)
Relative minimum: (1,-7)

17. fix) = x2(3 - x)

3x2

fix) = 6x — 3x2 = 3^('2 _ J,)
Critical numbers: x = 0, 2

Test intervals:

— oo < X < 0

0 < X < 2

2 < X < oo

Sign of /'(x):

/'<o

/'>0

/'<o

Conclusion:

Decreasing

Increasing

Decreasing

Increasing on: (0, 2)
Decreasing on: (-oo, 0), (2, oo)
Relative maximum: (2, 4)
Relative minimum: (0, 0)

19. /(^) =

x^ - 5x

fix) = x^ - 1

Critical numbers: x = —1,1

Test intervals:

-oo < X < -1

- 1 < X < 1

1 < X < oo

Sign of/'(x):

/'>0

f <Q

f> 0

Conclusion:

Increasing

Decreasing

Increasing

Increasing on: (— oo, — 1), (1, oo)
Decreasing on: (-1,1)
Relative maximum: (—1,5)
Relative minimum: (l, -5)

Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 115

21. fix) = .r'/3 + 1

fix) = -x'^/^ = — ^
Critical number: x = 0

Test intervals:

-oo < j: < 0

0 < j: < oo

Sign of/'(-t):

r > 0

/'>0

Conclusion:

Increasing

Increasing

Increasing on: (-00,00)
No relative extrema

23. /W = (x - 1)2/3
2

/'W

3U - 1)'/'
Critical number: x = 1

Test intervals:

-co < X < 1

1 < .V < CO

Sign of/'U):

/'< 0

/'>0

Conclusion:

Decreasing

Increasing

Increasing on: (l,oo)
Decreasing on: (-00, 1)
Relative minimum: (1,0)

25. fix) = 5 - |x - 51

fix) =

x

|x- 5
Critical number: x = 5

l-l, X >

Test intervals:

-00 < X < 5

5 < jc < 00

Sign off'ix):

f'>0

/'<o

Conclusion:

Increasing

Decreasing

Increasing on: (—00, 5)
Decreasing on: (5, 00)
Relative maximum: (5. 5)

27

fix) =x + ^

fix) = 1 - A =

X-- 1
X-

Critical numbers:

x= -1,

1

Discontinuity: x

= 0

Test intervals:

- 00 < v < - 1

- 1 < .r < 0

0 < .t < 1

1 < JT < oo

Sign off'ix):

/'>0

/'<0

/'<o

/'> 0

Conclusion:

Increasing

Decreasing

Decreasing

Increasing

Increasing on: (-00, - 1). (1, 00)
Decreasing on: (-1, 0), (0, 1)
Relative maximum: (- 1, -2)
Relative minimum: (1,2)

116 Chapter 3 Applications of Dijferentiation

29. f{x)

fix)

{x^ - 9)(2.r) - (x'){2x) _ -\8x

ix^- - 9r

Critical number: x = 0
Discontinuities: x = —3, 3

(x^ - 9)2

Test intervals:

-oo < jc < -3

-3 < JC < 0

0 < X < 3

3 < X < oo

Sign of/'U):

/' > 0

/' > 0

/' < 0

/'<0

Conclusion:

Increasing

Increasing

Decreasing

Decreasing

Increasing on: (-oo, -3), (-3, 0)
Decreasing on: (0, 3), (3, oo)
Relative maximum: (0, 0)

31. fix) =

fix) =

X' - 2a: + 1
X + \

(x + l)(2x - 2) - (x^ -2x+ 1)(1) _ x^ + 2jc - 3 (x + 3)(x - 1)

ix + ir
Critical numbers: x = —3, 1
Discontinuity: ;<: = — 1

ix + IT'

ix + ir

Test intervals:

-oo < X < -3

-3 < X < -1

- 1 < .1 < 1

1 < ;c < oo

Sign of/'U):

/'> 0

/'< 0

/'<o

/'> 0

Conclusion:

Increasing

Decreasing

Decreasing

Increasing

Increasing on: (— oo, -3), (1, oo)
Decreasing on: (—3, — 1), (— 1, 1)
Relative maximum: (-3,-8)
Relative minimum: (1,0)

33. fix) = - + cosjc, 0 < X < Itt

fix) = - - sin ;c = 0

Critical numbers: x = —,—r

0 0

Test intervals:

0<x<f

0

77 577

Sir

-^ < X < 277
0

Sign of fix):

f >o

/'<o

/'> 0

Conclusion:

Increasing

Decreasing

Increasing

Increasing on: ( 0, — j, ( — , 277

Decreasing on:

77 577

6'T

Relative maximum:

Relative minimum:

77 77 + 6v3

6' 12

577 577 — 6V3
T' 12

Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 117

35. fix) = sin^ X + sin X, 0 < X < Itt

fix) = 2 sin j: cos x + cos x = cos x(2 sin a: + 1 )

_ . . , , IT 7tt 37T IItt

Cntical numbers: x = —, — -, -r-. — r—
2 6 2 o

Test intervals:

0<.<f

TT 777

2<^<T

777 3t7

377 1177

2 < •'^ < 6

1177

•— - < .r < 277
6

Sign of/'(:t):

/'>0

/'< 0

/'>0

/'<0

/'> 0

Conclusion:

Increasing

Decreasing

Increasing

Decreasing

Increasing

I ^ '''■\ /777 37r\ (IItt ^
Increasing on: ( 0, -I, I — , — I, I — , 277

Decreasing on:

Relative minima:

77 7t7\ /377 II77

2' 6 y V 2

777 1\ /11t7 1

6 ' 4

Relative maxima: I — , 2 I, ( — , 0

37. f(x) = 2xV9 - x\ [-3, 3]
2(9 - 2x^-)

, 2(9 - 2^2)

Critical numbers: x = +— ^ = +

J2 2

(d) Intervals:

-3,

3^2

fix) < 0
Decreasing

3v^ 372
" 2 ' 2

/'(.x) > 0
Increasing

/'(.r) < 0
Decreasina

/is increasing when/' is positive and decreasing
when/' is negative.

39. /(r) = t^ sin f, [0, 277]

(a) fit) = t^cost + 2t sin t
= t(t cos t + 2 sin f)

(b)

(c) t(t cos / + 2 sin t) = 0

t = OoT t = - 2 tan r

f cot r = - 2

t = 2.2889. 5.0870 (graphing utility)
Critical numbers: r = 2.2889. t = 5.0870

(d) Intervals:

(0.2.2889) (2.2889,5.0870) (5.0870.277)

/'(f) > 0 /'(f) < 0 /'(f) > 0

Increasing Decreasing Increasing

/is increasing when/' is positive and decreasing when
/' is negative.

118 Chapters Applications of Differentiation

41. fix) = 5 ; = -^^ :r^ = x^ -7,x, x + ±1

fix) = gix) =x^ - 3xf0Tal\xit±i_

fix) = 3a:2 - 3 = 3(JC2 - 1), X 7t ±1 /'(j;) 7t 0

/symmetric about origin
zeros of/: (0, 0), (± 73, o)
No relative extrema

Holes at (-1,2) and (1,-2)

43. fix) = c is constant =*/'W = 0

45. /is quadratic =>/' is a line.

I I I I I

-2--
-4--

I I I I I > J

47. /has positive, but decreasing slope

I iTT

-2-
-4-

In Exercises 49-53,/'W > 0 on (-oo, -4),f'(x) < 0 on (-4, 6) and/'W > 0 on (6, oo).

51. gix) = -fix) S3, gix) = fix- W)

49. gix) = fix) + 5
8'ix)=f'ix)
g'(0)=/'(0) < 0

g'ix) = -fix)
g'i-6) = -f'i-6) < 0

g'ix) = fix- 10)
g'(0)=/'(-10) > 0

I> 0, x < 4 =>/ is increasing on (-00, 4).

undefined, x = 4
< 0, X > 4 =»/is decreasing on (4, oo).

Two possibilities for fix) are given below.

(a)

(b)

Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 119

57. The critical numbers are in intervals (-0.50, -0.25) and
(0.25, 0.50) since the sign of/' changes in these intervals,
/is decreasing on approximately (-1, -0.40), (0.48, 1),
and increasing on (-0.40, 0.48).

Relative minimum when x ~ ~ 0.40.

Relative maximum when x = 0.48.

59. fix) = X, g{x) = sin x. 0 < .v < tt
(a)

fix) seems greater than gix) on (0, tt).
(b)

x > sin j: on (0, it)

X

0.5

1

1.5

T

2.5

3

fix)

0.5

1

1.5

2

2.5

3

six)

0.479

0.841

0.997

0.909

0.598

0.141

(c) Let hix) = fix) — gix) = X - sin x

h 'ix) = 1 — cos ;c > 0 on (0, tt).

Therefore, hix) is increasing on (0, it). Since
hiQ) = 0, hix) > 0 on (0. tt). Thus.

jr — sin^: > 0

X > sin.r

fi.x) > gi.x) on (0, tt).

61. V = kiR - r)r^ = kiRr^ - r^)
v' = kilRr- 3r=)

= krilR - 3r) = 0
r = 0 or f /?
Maxmium when r = f /?.

v/? R

63. P = 7— — ' p ..,, V and /?, are constant

(a, + Kj)"

rfP _ (/?, + R.JHvR,) - vR,R.[2iR, + j?,)(l)]
d/fj (^, + /?,)*'

^ v/;,(/?, - ^,) ^ ^

Maximum when ^j = /?; .

65. (a) S = 0.1198t^ - 4.4879r^ + 56.9909F - 223.0222r + 579.9541
(b) 1500

(c) S' = 0 for f = 2.78. or 1983, (311.1 thousand bankruptcies)
Actual minimum; 1984 (344.3 thousand bankruptcies)

67. (a) Use a cubic polynomial

fix) = fljx' + a2.x- + a,.v + a^.

(b) fix) = 3aj.x'- + la^x + a,

(0, 0): 0 = ao (/(O) = 0)

0 = a, (/'(O) = 0)

(2,2): 2 = 8a3 + 4fl, (/(2) = 2)

0 = 12a3 + 4a, (/'(2) = 0)

(c) The solution is Op = a, = 0. O; = -. a,

3 ,

fix) = -|.v-' + |.v

(d)

\

(2. 2)

(0. 0)

\

120 Chapters Applications of Differentiation

69. (a) Use a fourth degree polynomial /W = a^x'^ + a^x^ + a,^^ + "i-* + "^o-
(b) fix) = 4a^x^ + 3^3x2 + 2^2^ + a,

(0, 0): 0 = Qo (/(O) = 0)

0 = a, (/'(O) = 0)

(4, 0): 0 = 256^4 + 640, + IGa^ (/(4) = 0)

0 = 256a^ + 48a3 + 8^2 (/'(4) = 0)

(2, 4): 4 = I6a^ + Sa, + 4^2

0 = 32a. + 12a, + 4a,

(/(2) = 4)
(/'(2) = 0)

(c) The solution is Oq = a, = 0, a, = 4, aj = — 2, 04 = —

fix) = -x" - 2^ + Ax-
id)

\

(2.4) J

A/

(0,0)

(4.0)

71. True

Let h{x) = f{x) + g{x) where /and g are increasing. Then
h'ix) = fix) + g'ix) > 0 since /'(x) > Oandg'U) > 0.

73. False

Let/(x) = x^, then/'(x) = 3x^ and/only has one
critical number. Or. let/U) = x^ + 3x + 1, then
fix) = 3(x' + 1) has no critical numbers.

75. False. For example, fix) = x^ does not have a relative extrema
at the critical number x = 0.

77. Assume that/'(x) < 0 for all x in the interval (a, b) and let x^ < x-^ be any two points in the interval. By the Mean Value
ve know there e

fix.) -fix,)

Theorem, we know there exists a number c such that x^ < c < x^, and

. /'(c)

Since/'(c) < 0 and-nr, - x, > 0, then/(x2) — /(x,) < 0, which implies that/Cxj) < fix). Thus,/is decreasing on the

interval.

79. Let/U) = i\ + x)" - ivn - 1. Then
fix) = n(l + x)"-^ - n

= «[(! +j:)"~' - 1] > Osincex > Oandn >1.
Thus./W is increasing on (0, 00). Since /(O) = 0 =^fix) > 0 on (0, 00)
i\ + x)" - nx - I > 0 => (1 + jc)" > I + nx.

Section 3.4 Concavity and the Second Derivative Test 111

Section 3.4 Concavity and the Second Derivative Test

1. y = x'- - X — 1, y" — 2
Concave upward: (-00,00)

3. fix) =

24

^ ^ - 144(4 - x'^)
x~+ 12'-' ~ (;c2 + 12)3

Concave upward: (-00, -2), (2, 00)

Concave downward: (—2,2)

5. fix) = J^, y

4(3x^ + 1)

r ■' (x2 - 1)3

Concave upward: (-00, - 1), (1, 00)
Concave downward: (-1,1)

7. fix) = 3x^ - x^
fix) = 6x- 3x2
fix) = 6 - 6x
Concave upward: (-00. 1)
Concave downward: (1, 00)

9. v = 2x- tanx, (-^.7

y' = 2 — sec-x
>'"= —2 sec-x tanx

Concave upward: ( - — , 0

Concave downward: 0,

11. fix) = x> - 6x- + \2x

fix) = 3x- - 12x + 12

fix) = 6(x - 2) = 0 when x = 2.

The concavity changes at x = 2. (2, 8) is a f)oint of
inflection.

Concave upward: (2, cc)
Concave downward: (—00, 2)

13.

fix)

1

2x2

fix) = x3 - 4x
fix) = 3x2-4

fix) = 3x2 _ 4 = Owhenx

2

^73-

Test interval:

2

CO < X < r-

J3

2 "^

~~r^ < X < -7=

J3 V3

2
^ < X < 00

73

Sign of/"(x):

fix) > 0

/"(-v) < 0

fix) > 0

Conclusion:

Concave upward

Concave downward

Concave upward

Points of inflection:

20

'■V3'

122 Chapter 3 Applications of Differentiation

15. /(;c) = x(x - 4)3

fXx) = x[3{x - 4)2] + U - 4)3

= U - 4)^(4x - 4)
f"(x) = 4{x - \)[2{x - 4)] + 4(x - 4)2

= 4(a: - 4)[2U - 1) + (x - 4)]
= 4(x - 4)(3x - 6) = \2{x - 4){x - 2)
f"{x) = 12(jc - 4)(jc - 2) = 0 when j; = 2, 4.

Test interval:

-oo < X <1

2 < j: < 4

4 < jt < oo

Sign of /"W:

fix) > 0

fix) < 0

/"(x) > 0

Conclusion:

Concave upward

Concave downward

Concave upward

Points of inflection: (2, - 16), (4, 0)

17. f(x) = xjx + 3, Domain: [-3, oo)

/'W=.(tj(-.3)-./2.vm^|^

/"W

6v<m - 3U- + l){x + 3)-'/2 _ 3(x + 4)

4U + 3) 4{x + 3)3/2

/"(jc) > 0 on the entire domain of/ (except for x = - 3, for which/"(x) is undefined). There are no points of inflection.
Concave upward on (-3, oo)

19. /W = -

f\x) =
/"W =

;c2+ 1

1 -;c2
(;c2 + 1)2

2x(x2 - 3)

(x2 + 1)3

= 0 when a: = 0, ±V3

Test intervals:

— oo < X < — V3

-73 <;c < 0

0 < ;c < 73

73 < X < oo

Sign of/'W:

/"<o

/">0

/"<0

/">0

Conclusion:

Concave downward

Concave upward

Concave downward

Concave upward

Points of inflection:

■73,-^1 (0,0), (73,^

21. fix) = sin( - 1, 0 < a: < 47r

fix) = icos(f

/"W = -^sini 2

/"(;c) = 0 when x = Q, Itt, 4t7.
Point of inflection: (27r, 0)

Test interval:

0 < .^r < 277

277 < j: < 477

Sign of/"W:

/" < 0

/">0

Conclusion:

Concave downward

Concave upward

Section 3.4 Concavity and the Second Derivative Test 123

23. f(x) = sec(;<: - -\,Q < x < Att

fix) = secfjr - -j tanix - -

f"(x) = sec^l X - —\ + seel x - —\ tan^j x - —\ i= Q for any x in the domain of/.

Concave upward: (0, tt), {Itt, Ztt)
Concave downward: (tt, 2tt), (Stt, 47r)
No points of inflection

25. f{x) = 2 sin;<: + sin 2;c, 0 < j: < 277
fix) = 2 cos j: + 2 cos 2x
f"{x) = — 2 sin j: - 4 sin 2x = — 2 sin ;c(l +4 cos x)

f"{x) = 0 when x = 0, 1 .823, tt, 4.460.

Test interval:

0 < A- < 1.823

1.823 < .X < TT

77 < ,x: < 4.460

4.460 < .V < 277

Sign of /"(a):

/"< 0

/"> 0

/"< 0

/">0

Conclusion:

Concave downward

Concave upward

Concave downward

Concave upward

Points of inflection: (1.823, 1.452), {77, 0), (4.46, - 1.452)

27. /(a) = A^ - 4x3 + 2

/'(a) = 4a3 - 12a' = Ax-{x - 3)

/"(a) = 12a' - 24a = 12a(a - 2)

Critical numbers: a = 0, a = 3

However, /"(O) = 0, so we must use the First Derivative
Test. /'(a) < 0 on the intervals (-00, 0) and (0, 3); hence,
(0, 2) is not an extremum./"(3) > 0 so (3, -25) is a
relative minimum.

29. fix) = (a - 5)2
/'(a) = 2(a-5)
f"{x) = 2
Critical number: a = 5

/"(5) > 0
Therefore, (5, 0) is a relative minimum.

31. /(a) = a3 - 3a2 + 3

fix) = 3a2 - 6a = 3a(a - 2)
/"(a) = 6a - 6 = 6(a - 1)

Critical numbers: a = 0, a = 2

/"(O) = -6 < 0
Therefore, (0, 3) is a relative maximum.

/"(2) = 6 > 0
Therefore, (2, — 1) is a relative minimum.

33. g{x)=x\6-xy

g\x) = x{x - 6)2(12 - 5.r)

g"(x) = 4(6 - x)(5x~ - 24a + 18)

Critical numbers: x = 0, -f , 6

^"(0) = 432 > 0
Therefore, (0, 0) is a relative minimum.

g"{fj = - 155.52 < 0
Therefore, \-f, 268.7) is a relative minimum.

g"(6) = 0

Test fails by the First Derivative Test. (6. 0) is not
an extremum.

124 Chapters Applications of Differentiation

35. f(x) = .x2/3 - 3
2

fix) =
fix) =

-2

Critical number: a; = 0

However, /"(O) is undefined, so we must use the First
Derivative Test. Since /'(x) < 0 on (— oo, 0) and
f'(x) > 0 on (0, oo), (0, -3) is a relative minimum.

37. f(x) =x + -
■' X

fix) = 1 -

Critical numbers: x = ±2

f'i-2) < 0
Therefore, (—2, —4) is a relative maximum.

/"(2) > 0
Therefore, (2, 4) is a relative minimum.

39. fix) = cosx - X, 0 < x < 417
fix) = - sin ;c - 1 < 0
Therefore, / is non-increasing and there are no relative extrema.

41. fix) = O.lx^ix - 3)3, [- 1, 4]

(a) fix) = 0.2;c(5;c - 6)U - 3)^
/"W = U - 3)(4;c2 - 9.6a: + 3.6)

= OAix - 3)(10.x2 _ 24.x + 9)

(b) /"(O) < 0 =» (0, 0) is a relative maximum.

/"(f) > 0 =*• (1.2, - 1.6796) is a relative minimum.

Points of inflection:

(3, 0), (0.4652, -0.7049), (1.9348, -0.9049)

/is increasing when/' > 0 and decreasing when
/' < O./is concave upward when/" > 0 and con-
cave downward when/" < 0.

43. fix) = sin jc - -sin 3x + -sin 5x, [0, n]

(a) fix) = cos X - cos 3x + cos 5x

fix) = 0 when x = T- ^ = T' ^ = "g"-

fix) = — sin X + 3 sin 3a: — 5 sin 5x

fix) = 0 when x = ^, x = ^, x = 1.1731. .t == 1.9685
6 6

(b) /"( — 1 < 0 => ( — , 1 .53333 1 is a relative maximum.

Points of inflection: I -, 0.2667], (1.1731, 0.9638),

(1.9685,0.9637), [y"' 0-2667

Note: (0, 0) and (ir, 0) are not points of inflection
since they are endpoints.

(c)

The graph of/ is increasing when/' > 0 and decreas-
ing when/' < O./is concave upward when/" > 0
and concave downward when/" < 0.

Section 3.4 Concavity and the Second Derivative Test 125

45. (a)

/' < 0 means/decreasing

/' increasing means
concave upward

(b)

/' > 0 means/ increasing

/' increasing means
concave upward

47. Let/U) = ;d.
f"(x) = Ux^
/"(O) = 0, but (0, 0) is not a point of inflection.

49.

51.

53. >

55. y

57.

/"is linear.

/' is quadratic.

/is cubic.

/concave upwards on (— oo. 3), downward on (3. oo).

126 Chapters Applications of Differentiation

59. (a) n = 1:

fix) = 1
fix) = 0
No inflection points

n = 2:

/W = (;c - 2)2

/'W = 2U - 2)

/"W = 2

No inflection points

Relative minimum:
(2,0)

n = 3:

fix) = (x- ly

fix) = 3{x - If
fix) = 6{x - 2)
Inflection point: (2, 0)

,/

Point of
inflection

n = 4:

/W = {x- ly

fix) = 4(;c - 2)3

/W = 12(;c - 2)2

No inflection points:

Relative minimum:
(2,0)

J.

Conclusion: If n > 3 and n is odd, then (2, 0) is an inflection point. If n > 2 and n is even, then (2, 0) is a relative minimum.
(b) Let fix) = ix- 2)", fix) = nix - 2)"~\ fix) = n(« - \)ix - 2)"-\
For n > 3 and odd, « — 2 is also odd and the concavity changes at x = 2.
For n > 4 and even, n — 2 is also even and the concavity does not change at j: = 2.
Thus, j: = 2 is an inflection point if and only if « > 3 is odd.

61. fix) = ax^ + bx^ + ex + d
Relative maximum: (3, 3)
Relative minimum: (5, 1)
Point of inflection: (4, 2)

fix) = 3ax^ + 2bx + cf'ix) = 6ax + 2b

/(3) = 27. + % + 3c + rf=3 ,„._,..^..__._49^^8,^^^_1

/(5) = 125a + 25b + 5c + d= 1
/'(3) = 27a + 6fc + c = 0, /"(4) = 24a + 2fc = 0
- 49a + 8fc + c = - 1 24a + 2* = 0
27a + 6fc + c = 0 22a + 2fc = - 1

22a + 2b = - 1 2a
a = ii)= -6,c = f,rf= -24
fix) = \x^ - 6x^ + f ;c - 24

= 1

Section 3.4 Concavity and the Second Derivative Test 127

63. f(x) = ax^ + bx^ + ex + d

Maximum: (-4, 1)

Minimum: (0, 0)

(a) f'(x) = 3ax- + 2bx + c, f"{x) = 6ax + 2b
m = 0=^d = 0

/(-4) = 1 =
/'(-4) = 0 =

/'(0) = 0 =

Solving this system yields a = j^ 3"^ i> = 6a = ^.
fix) = j,x^ + ^,x^-

-64a + 16fc - 4c = 1
48a - 8b + c = 0

c = 0

(b) The plane would be descending at the greatest rate at
the point of inflection.

fix) = 6ax + 2b = j^x + ^
Two miles from touchdown.

0

x = -2.

65. D = 2x' - 5Lx^ + 3L-x^

D' = 8.r3 - \5Lx^ + 6L~x = .r(8.x- - 151.x + 6L-) = 0

15L± 733Z. /l5 ± 733 V
, = 0or.. = =(^_^^JZ.

By the Second Derivative Test, the deflection is
maximum when

15

16

L = 0.578L.

67. C = 0.5.V- + 15.r + 5000
C^^ = 0.5..-.15^^

X X

C = average cost per unit

^.0.5-^ = 0when.v=100
dx X-

By the First Derivative Test, C is minimized when
X = 100 units.

69.

sXt) =

S"(t) =

5000P

8 + ^2

80,000f

(8 + t^r
80,000(8 - it-)

(8 + a'
S"(t) = 0 for r = ^/873 « 1.633.
Sales are increasing at the greatest rate at f = 1.633 years.

71. /(;c) = 2(sin.r + cos;c),
f'(x) = 2(cos X — sin .r),
f"{x) = 2{-sin.v - cos.t),

P,(jc) = 2v^ + o(x - Jj = 2V^

/( - I = 272

0

/"(f) = -^

Pi'{x) = 0
P,(x) = 272 + o(.v - fl + ^(-272)(.v - f )^ = 2 V2 - .^(.v - ^

PAx) = -272(.r - fl

P^x) =-lJl

The values of/, P,, Z'^, and their first derivatives are equal at .r = ttJA. The values of the second derivatives of/ and P, are
equal at x = Tr/4. The approximations worsen as you move away from x = 7r/4.

128 Chapters Applications of Differentiation

73. /U) = Vl -X,
1

/'W = -

fix) = -

2Vl - X
1

4(1 - .t)3/2'

/(O) = 1

/'(o) = 4

/"(O)

p,W = i + (-|)u-o) = i-|

PAx) = -i-

^

s^. .

/.

N

/>.'(.) = 4- J • .

P.'U) = -|

The values of/, Pj, P,' ^°d their first derivatives are equal at x = 0. The values of the second derivatives of/ and P2 are equal
at j: = 0. The approximations worsen as you move away from x = 0.

75. fix) = X sin

fix)

r COSi

x^ \x

1 n\ . n

--cos - + sm -
X \xl \x

fix) = --
x

i ^'"(i

+ sin

+ -^ cos ^ cos - = — T sm -

X \xl X- \xj x' \x.

\\ 1

A

./^

V

^(J..)

Point of inflection: | — , 0

When X > I/tt,/' < 0, so the graph is concave downward.

77. Assume the zeros of/ are all real. Then express the function as/W = a(x — r^)ix — r-^ix — r^) where r^, r^, and rj are the
distinct zeros of/. From the Product Rule for a function involving three factors, we have

fix) = a[ix - r,)(x - r^ + ix - r,)(jf - r^) + (x - r^ix - r^)]

fix) = a[ix - r,) + ix-r,J + ix- r,) + (x - r^) + (x - r^) + (x - r^)]

■ = a[6x — 2(ri + rj + rj)].

Consequently, fix) = 0 if

2(''i + ''2 + '"3) '"1 + '"•' + '"3 /
jc = 7 = f = (Average of r,, r,, a«d rj).

79. True. Lety = ax^ + bx^ + ex + d. a ¥= 0. Then y" = 6(Xt + 2fe = 0 when x= - ib/3a)
and the concavity changes at this point.

Section 3.5 Limits at Infinity 129

81. False.

fix) = 3 sin Jt + 2 cos x
fix) = 3 cos j: - 2 sin X
3 cos .V — 2 sin X = 0

3 cos j: = 2 sin x
2 = tan.r
Critical number: x = tan '(2)

/(tan"' 2) ~ 3.60555 is the maximum value of .v.

Section 3.5 Limits at Infinity

83. False. Concavity is determined by /".

1. f(x) =

3x2

.t2 + 2
No vertical asymptotes
Horizontal asymptote: v = 3
Matches (f)

3. fix.

X- + 2
No vertical asymptotes
Horizontal asymptote: y = 0
Matches (d)

5. fix)

4sin j:

X- + 1
No vertical asymptotes
Horizontal asymptotes: y = 0
Matches (b)

7. f(x) =

4x + 3

/ \-^l

2x- 1

X

10°

10'

102

103

lO-*

10^

10^

fix)

7

2.26

2.025

2.0025

2.0003

2

2

lim fix) = 2

9. fix)

-6x

V4? + 5

X

10"

10'

10-

103

10"

10-^

10«

fix)

-2

-2.98

-2.9998

-3

-3

-3

-3

lim/W = -3

11. fix) = 5 - -

1

.V2+ 1

x

10"

10'

10-

103

10*

10^

10^

fix)

4.5

4.99

4.9999

4.999999

5

5

5

lim fix) = 5

130 Chapters Applications of Dijferentiation

13. (a) Mx) = — -r- = ; = 5a - 3 H — 7

x-^ x^ x^

lim h(x) — 00 (Limit does not exist)

-r— ♦00

lim h(x) = 5

f{x) ^ 5x3 - 3;c^+ 10 _ 5 3 ^ iQ

lim /lU) = 0

. ^2 + 2

Jr^ooX" — 1
^2 + 2

15. (a) lim ,

... I v^

(b) lim

J:~*o:

(c) lim

Jr'-»oo X^ — \

x^ + 2

X - 1

= 0
= 1
= 00 (Limit does not exist)

5 - 2x3^2
17. (a) lim - , , = 0

(b) lim

(c) lim

5 - 2t3/2 _ _2
3x3/2 - 4 3

5 -2x^/2
3x-4

= — 00 (Limit does not exist)

10 r ^- 1 ,• 2 - (1/x) 2-0 2
19. lim = lim -^ , . = r r = r

i-^00 3x + 2 x->c^ 3 + (2/x) 3+0 3

2L lim

X — >oD X~ ~ 1

lim

lA _o ^

1 - (1A2) 1

23. lim

5x2

lim

5x

x + 3 x'-X-a= 1 + (3/x)
Limit does not exist.

25. lim —r^=

j:->-oo ^x'^ — X

lim ^

( forx < 0 we have x = — J]?)

= lim

-1

x^-x^i -d/x)

= -1

2x + 1
27. Iim — , .. = lim

2 +

*-=<= Vx2 — X Jr->-oo / .Jx^

(for X < 0, X = - 7?)

lim — , ,= = — 2

^^°o Vx + O/x)

29. Since (- 1/x) < (sin(2x))/x < (1/x) for all x =5^ 0, we
have by the Squeeze Theorem,

,. 1 ,. sin(2x) ^ ,. 1
lim — < lim < lim -

X -*co X X -*oo X X -*oo X

0 < lim ^^^^ < 0.

Jr-»oo X

sin(2x)
Therefore, lim = 0.

X -*oc X

3L lim

1

X ->oo 2x + sin X

Section 3.5 Limits at Infinity 131

33. (a) fix)

x+ 1

lim — V—

x->aoX + 1

lim

= -1

>x+ 1
Therefore, y = 1 and y =

1 are both horizontal asymptotes.

.^M

,, ,. .1 ,. smt
35. lim jr sm — = lim = 1

x^ac X (-»0* t

(Let a: = 1/r.)

37. lim (a: + V;c= + s) = lim

= lim

^-»-°=a: - Va:2 + 3

39. lim {x - Vat + x)

lim

- v? +

+ vVT:

= lim

X + Vx^ + X
-1

t ^°° a: + V-t- + X X-.00 1 + Vl + (lA)

41.

JC

10"

10'

102

103

10^

10^

10«

/w

1

0.513

0.501

0.500

0.500

0.500

0.500

lim (x — ^x(x - 1)) = lim

— Vj^ — X X + V-t- — x

= lim

x+ v^

^=»x + Vx^ - X

1

'™ — / 7—r

■< ^=° 1 + Vl - (1/x

2

^

43.

X

10«

10'

10-

10-'

10^

10^

10*

/(x)

0.479

0.500

0.500

0.500

0.500

0.500

0.500

Letx = 1/f.
lim X sin r-

,. sin(f/2) 1 sin(f/2) 1

= lim = hm — 7- — = -

, -,0* f r ^0* 2 f/2 2

ri^/::

132 Chapters Applications of Differentiation

45. (a)

(b) lim fix) = 3 lim f'(x) = 0

(c) Since lim f{x) = 3, the graph approaches that of a
horizontal line, lim f'{x) = 0.

47. Yes. For example, let/(x) =

6U - 21

j(x - 2)2 + r

49. >- =

2 +x

1 -;c

Intercepts: (-2,0), (0,2)
Symmetry: none
Horizontal asymptote: y = — 1 since

lim

2+ X

1 = lim .

X ->oo \ — X

»-oo 1 — X

Discontinuity: x = 1 (Vertical asymptote)

x^ - 4
Intercept: (0,0)
Symmetry: origin
Horizontal asymptote: y = 0
Vertical asymptote: x = +2

53. y

x^ + 9
Intercept: (0,0)
Symmetry: y-axis
Horizontal asymptote: y = 1 since

lim

= 1 = lim

Relative minimum: (0, 0)

4-
3--
2
1

1 2 3

Section 3.5 Limits at Infinity

133

55. y =

2x2

Intercept: (0,0)
Symmetry: y-axis
Horizontal asymptote: y = 2
Vertical asymptote: x = ±2

57. xy- = 4

Domain: a: > 0

Intercepts: none

Symmetry: x-axis

Horizontal asymptote: y = 0 since

lim -^ = 0

lim —

Vx-

Discontinuity: x = 0 (Vertical asymptote)

59. V =

2x

1 - X
Intercept: (0, 0)
Symmetry: none
Horizontal asymptote: y = —2 since

1- 2r . ,. 2x

urn = — 2 = lim .

j:-»-oo I — X .r->oo i — X

Discontinuity: x = 1 (Vertical asymptote)

H — I — I — l-»-i
2 3 4 5

61. V

Intercepts: (±v'^. O)

Symmetry: y-axis

Horizontal asymptote: y = 2 since

lim (2 - 4j ) = 2 = lim (2 - V\-

x->-ooV X-/ x^oo\ X-j

Discontinuity: .r = 0 (Vertical asymptote)

63. V = 3 + -

X

2 2
Intercept: v = 0 = 3 H — ^- =

XX

Symmetry: none
Horizontal asymptote: y = 3
Vertical asymptote: x = 0

.,.-|-|o

134 Chapters Applications of Differentiation

65. y =

Domain: (—00, —2), (2, 00)

Intercepts: none

Symmetry: origin

Horizontal asymptote: none

Vertical asymptotes: x = ±2 (discontinuities)

I -°

I 16
1 12

11/

I I I

-5-4-3-2-1

A

-12

-16--

-20--

67. /W = 5 - A = ^^^
X- x'-

Domain: (— 00, 0), (0, 00)

f'(x) = 3 => No relative extrema

/"W

''"'"^ = ---J =i. No points of inflection

Vertical asymptote: ;c = 0
Horizontal asymptote: y = S

|y = 5|

V = o|

6

\

r~

69. f{x) =
/'W =

/"U) =

x^ - ^

U^ - 4) - ;t(2jc)
{x^ - 4)2

-U2-H4)

=i^ 0 for any x in the domain of/.

+ (;c2 + 4]
(x^ - 4)2

{x^ - 4)2
(;c2 - 4)2(-2;c) + (x^ + 4)(2)(x2 - 4)(2;c)

lx{x^ + 12)

= 0 when ;t = 0.

L

r^-'?!-^ ^"

(a:2 - 4)3

Since/"(;c) > 0 on (-2, 0) and/"(;«:) < 0 on (0, 2), then (0, 0) is a point of inflection.
Vertical asymptotes: x = ±2
Horizontal asymptote: y = Q

71. /U)

.X - 2

x-2

;c2 - 4;c + 3 U - 1)(a: - 3)

^ (x2 - 4x + 3) - (;c - 2)(2;c - 4) ^ -x^ -h 4j: - 5
■^ ^""^ (x2 - 4x + 3)2 (x2 - 4x + 3)2 "^ "

/"W

(x2 - 4jc -t- 3)2(-2x + 4) - (-^2 + 4x - 5)(2)(j:2 - 4x + 3)(2x - 4)

i^^

"^

^Mi^

2(^3 - 6^2 -F 15x - 14)

(jc2 - Ax + 3)*

= 0 when x = 1.

(x2 - 4x + 3)3

Since/"U) > 0 on (1, 2) and/"U) < 0 on (2, 3), then (2, 0) is a point of inflection.
Vertical asymptote: x = l,x = 3
Horizontal asymptote: >> = 0

Section 3.5 Limits at Infinity 135

73. /W =
fix) =

fix) =

3;c

jAx'^ + 1
3

(4x2 + 1)3/2

-36x

=> No relative extrema

(4^2 + 1)V2
Point of inflection: (0,0)
Horizontal asymptotes: y —
No vertical asymptotes

0 when jc = 0.

'^ I

nS

75. g(x) = sin

X- 2

3 < X < oo

-2 cos

g\x)

X- 2

(^ - 2?
Horizontal asymptote: y = \

„ , . . X 77

Relative maximum: = —

X - 2 2

No vertical asymptotes

277
77- 2

5.5039

77. /(x)
(a)

x^ - 3x- + 2
x(x — 3)

^(x) = X +

x(x - 3)

■M

/=

(b) /(x) =

t^ - 3x- + 2
x(x - 3)

xKx - 3) ^ 2
x(x — 3) x(x — 3)

9

= X +

x(x - 3)

g{x)

(c)

fT?T-')

/

/ r^^ 1

■J 1 V = sm( 1 ) 1
■ — ' — ' — ■ — ' — ■ — ■ — ' — ' — ■

The graph appears as the slant asymptote v = x.

79. C = 0.5x + 500
C = ^

X

C = 0.5 +
lim 10.5 +

500

X

500

= 0.5

136 Chapters Applications of Differentiation

81. line: inx - y + 4 - 0

12 3 4

(a) d =

|Ati + B,vi + C\ _ \m{3) - 1(1) + 4|

|3m + 3 1

Vm2 + 1

V"r + 1

(b)

(c) lim (f(m) = 3 = lim dim)

The line approaches the vertical Une x = 0. Hence, the
distance approaches 3.

83. (a) TiU) = -0.003/2 + o.677; + 26.564
(b)

(c)

r, =

1451 + 86r

58 + /

(d) r,(0) = 26.6

TjCO) » 25.0
86

(e) lim r.

1

86

(f) The limiting temperature is 86.
r, has no horizontal asymptote.

85. Answers wiU vary. See page 195.

87. False. Let/U) =

Vx^ + 2

(See Exercise 2.)

Section 3.6 A Summary of Curve Sketching

1. /has constant negative slof>e. Matches (D)

3. The slope is periodic, and zero at j: = 0. Matches (A)

5. (a) fix) = OfoTx = -2 and^; = 2

/' is negative for — 2 < ,t < 2 (decreasing function).

/' is positive for x > 2 and x < — 2
(increasing function).

n?) fix) = 0 at X = 0 (Inflection point).

/"is positive for x > 0 (Concave upwards).
/" is negative for X < 0 (Concave downward).

(c) /' is increasing on (0, oo). (/" > 0)

(d) /'(x) is minimum at x = 0. The rate of change of/ at
X = 0 is less than the rate of change of /for all other
values of x.

Section 3.6 A Summary of Curve Sketching 137

7. y =

y =

X- + 3

6x

{x^ + 3)2

18(1 - .Q

= 0 when x = 0.

= 0 when x = ± 1 .

Horizontal asymptote: y = 1

i—h-x

V

v'

y"

Conclusion

-oo < X < -\

-

-

Decreasing, concave down

x= -I

1

4

-

0

Point of inflection

-1 < X < 0

-

+

Decreasing, concave up

x = 0

0

0

+

Relative minimum

0 < X < \

+

+

Increasing, concave up

x= 1

1

4

+

0

Point of inflection

1 < X < oo

+

-

Increasing, concave down

9. y =

- 3

1

:j < 0 when x ^ 2.

U - 2)

^ U - 2)3

No relative extrema, no points of inflection

Intercepts: (-, o), (o,-y

Vertical asymptote: x = 1
Horizontal asymptote: v = — 3

11. y =

Ix

< Oifxifc ±1.

X- - 1

, -2ix- + 1)

„ 4x(x- + 3)

Inflection point: (0, 0)
Intercept: (0, 0)
Vertical asymptote: a = ± 1
Horizontal asymptote: y = 0
Symmetry with respect to the origin

138 Chapters Applications of Differentiation

13. g(;c)=., + -

g'(x) = 1 -

.V2+ 1

{x^ + 1)2

{X^ + 1)2

V3

Owhenx = 0.1292, 1.6085

„, , 8(3x2 _ 1)
^ W = (^ + 1)3 = 0 when X = ± 2

g"(0.1292) < 0, therefore, (0.1292, 4.064) is relative maximum.
^"(1.6085) > 0, therefore, (1.6085, 2.724) is a relative minimum.

Points of inflection:

2 ,2.423j,(^^, 3.577

Intercepts: (0,4), (- 1.3788, 0)
Slant asymptote: y = x

(0.1292. 4.064)

(-f . 2,423)

y

(1.6085. 2.724)

(-1.3788.
H — f-<4

15. /(x) = ^^ = X + -
.r X

/'(x) = 1 - -r = 0 when x = ± 1.

x^

fix) =4^0
x'

Relative maximum; (—1, —2)

Relative minimum: (1,2)

Vertical asymptote: x = 0

Slant asymptote: y = x

-4 -2 '

(1,2)

2
(-1.-2)

x2 - 6x + 12 4

17. y = : = X — 2 +

y'= 1

x-4

4

x-4

{x - 4)2

(x - 2)(x - 6)
ix - 4)2

0 when x = 2, 6.

y

ix - 4)3
j" < 0 when x = 2.

Therefore, (2, —2) is a relative maximum.
y" > 0 when x = 6.
Therefore, (6, 6) is a relative minimum.
Vertical asymptote: x = 4
Slant asymptote: y = x - 2

(0,-3) J,

A' y = ^

(6.6)

■7^ — I 1 1 h-^'

6 8 10

(2,-2)

Section 3.6 A Summary of Curve Sketching 139

19. .y = xjx - 4,
Domain; (-oo, 4]

y ' = — , = 0 when x = - and undefined when x = 4.

Bx - 16

16

J - .,. M ;-> = 0 when x = ^- and undefined when x = 4.
4(4 — x)^'^ 3

Note: X = y is not in the domain.

y

y '

y"

Conclusion

8
-oo < X < -

+

-

Increasing, concave down

8
"=3

16

3V3

0

-

Relative maximum

8

- < X < 4

-

-

Decreasing, concave down

x = 4

0

Undefined

Undefined

Endpoint

21. hix) = xV9 - x= Domain:

9 - 2x-
/! '(x) = , = 0 when x = ±

V9 - X'

3 < X < 3

3 372

72

h"{x) = 1^' ..^? = 0 when x = 0

(9 - x2)3/:

Relative maximum:

3V^ 9

Relative minimum: —

2 '2
372 9

2 ' 2;
Intercepts: (0, 0), (±3, 0)
Symmetric with respect to the origin
Point of inflection: (0, 0)

23. y = 3x2/3 _ 2x
y' = 2x-'/3 - 2 =

2(1 - x'/3)

_l/3

= 0 when x = 1 and undefined when x = 0.

-2

y =

3x^/3

< 0 when x ^ 0.

y

y'

y"

Conclusion

-oo < X < 0

-

-

Decreasing, concave down

X = 0

0

Undefined

Undefined

Relative minimum

0 < X < 1

+

-

Increasing, concave down

X = 1

1

0

-

Relative maximum

1 < X < oo

-

-

Decreasing, concave down

140 Chapters Applications of Differentiation

25. y = x^-3x- + 3

y ' = 3x- — 6x = 3x{x - 2) = 0 when x = Q, x = 1
y" = 6j: - 6 = 6(a- - 1) = 0 when x = 1

y

y'

y"

Conclusion

-CXD < .T < 0

+

-

Increasing, concave down

;c = 0

3

0

-

Relative maximum

0 < X < \

-

-

Decreasing, concave down

x= 1

1

-

0

Point of inflection

1 < ;c < 2

-

+

Decreasing, concave up

x = 2

-1

0

+

Relative minimum

1 < X < CO

+

+

Increasing, concave up

(-0.879, 0)

(2, -1)
-(1.347.0)

27. y = l- X- x^

y'= -\-3x'^

No critical numbers

y" = — 6x = 0 when x = Q.

y

y '

y"

Conclusion

-oo < X < Q

-

+

Decreasing, concave up

x = Q

2

-

0

Point of inflection

0< j: < oo

-

-

Decreasing, concave down

29. f{x) = 3.r3 - 9;c + 1

fix) = 9;c2 - 9 = 9(;c2 - i) = o whenx = ±1
f"{x) = \%x = Owhenjc = 0

fix)

fXx)

/"(-t)

Conclusion

-oo < Jf < — 1

+

-

Increasing, concave down

.X = -1

7

0

-

Relative maximum

-1 < ;t < 0

-

-

Decreasing, concave down

x = Q

1

-

0

Point of inflection

0 < X < 1

-

+

Decreasing, concave up

X = 1

-5

0

+

Relative minimum

1 < .)C < oo

+

+

Increasing, concave up

(0.112,0)

Section 3.6 A Summary of Curve Sketching 141

31. y = 3y + 4^3

y' = 12r^ + llx' = \2x-(x + 1) = 0 when.r = 0, jc = - 1.
y"= 36x2 + 24.t = 12.t(3.r + 2) = 0 whenx = 0, jr = -f.

y

y '

y"

Conclusion

- OO < .V < - 1

-

+

Decreasing, concave up

x= -1

-1

0

+

Relative minimum

-1 < X < -f

+

+

Increasing, concave up

1

X = -f

16

21

+

0

Point of inflection

-| < X < 0

+

-

Increasing, concave down

x = 0

0

0

0

Point of inflection

0 < X < OO

+

+

Increasing, concave up

33. fix) = x^ - 4x3 + 16x

fix) = 4.r3 - 12t2 + 16 = 4(x + l)(x - 2)= = 0 whenx = - l,x = 2.
fix) = \Zx- - 24x = 12r{x - 2) = 0 when x = 0, x = 2.

/(-v)

/'W

fix)

Conclusion

- OO < X < - 1

-

+

Decreasing, concave up

x= -1

-11

0

+

Relative minimum

- 1 < X < 0

+

+

Increasing, concave up

x = 0

0

+

0

Point of inflection

0 < X < 2

-

-

Increasing, concave down

X = 2

16

0

0

Point of inflection

2 < X < OO

+

+

Increasing, concave up

(-1.679. 0)

35. y = x5 - 5x

y'= 5x* - 5 = Six* - 1) =
y" = 20x3 = 0 when x = 0.

0 whenx = ±1.

y

y'

y"

Conclusion

-OO < X < -1

+

-

Increasing, concave down

x= -1

4

0

-

Relative ma.ximum

- 1 < X < 0

-

-

Decreasing, concave down

x = 0

0

-

0

Point of inflection

0 < X < 1

-

+

Decreasing, concave up

x= 1

-4

0

+

Relative minimum

1 < X < OO

+

+

Increasing, concave up

142 Chapters Applications of Differentiation

37. y = \lx - 3|

, 2(2x - 3) . . . ^ 3
y — t; TT undefined aXx = -.

^ \lx - 3| 2

)>"= 0

y

y'

Conclusion

-OO < X < 2

-

Decreasing

3
X = 2

0

Undefined

Relative minimum

5 < .V < OO

+

Increasing

39. y = sin X — — sin 'ix, Q < x < 2tt
18

1 , „ , 77 3-77

>> = COS x — - COS 3j: = 0 when x = —, -^.
0 2 i

,, . 1 . - . , ^ TT Stt Itt IItt

>> = - sm j: + - sin 3x = 0 when a: = 0, — , —-, tt, — -, — — .
2 6 6 6 6

Relative maximum

\2'18/

Relative minimum:

377 _29
2 ' 18

Inflection points: ( -^, - j, ( ^, ^ ), (tt, 0),

TT 4\ /Stt 4
6'9/'V 6'9

7tt _4\ /JJjr _4
6' 9M 6 ■ 9

_,^ _ TT TT

41. y = 2x — tan .r, — — < j: < —

^ 2 2

43. y — 2(cscjc + sect), 0 < jr <

y ' = 2 - sec- x = 0 when x = ±— .

4

y " = — 2sec^ j: tan :«: = 0 when j: = 0.

Relative maximum: \—,— — 1
\4' 2

Relative minimum: ——,1 — x
\ 4' 2

Inflection point: (0, 0)
Vertical asymptotes: x = +—

y ' = 2(sec j: tan j: — esc x cot a:) = 0 => jc = 7r/4

Relative minimum: I — , A^Jl

Vertical asymptotes: x = 0, x = —

Section 3.6 A Summary of Curve Sketching 143

__ , . 377 377

45. g{x) = x\.znx, — — < ^ < ^

„ . X + smx cos X ^ ,
g W = ; = 0 when x = 0

g"W

cos^x
2(cosa: + jr sin;c)

Vertical asymptotes: x = -

BtT it TT 377

2 ' 2' 2' 2
Intercepts: (- 77, 0), (0, 0), (77, 0)
Symmetric with respect to y-axis.

Increasing on ( 0, — j and ( — , —
Points of inflection: (±2.80, 0)

47. f(x)

2Qx 1 \9x^ - 1

;c2 + 1 X x{x- + 1)

X = 0 vertical asymptote

y = 0 horizontal asymptote
Minimum: (-1.10, -9.05)

Maximum: (1.10,9.05)

Points of inflection: (-1.84, -7.86). (1.84, 7.86)

49. y

Vjc2 + 7
2

^^^^

(0, 0) point of inflection

y = ±1 horizontal asymptotes

51. /is cubic.
/' is quadratic,
/"is linear.

S3.

(any vertical translate of/ will do)

144 Chapters Applications of Differentiation

55.

(any vertical translate of/ will do)

57. Since the slope is negative, the function is decreasing on
(2, 8), and hence/(3) > /(5).

59. fix

4{x - 1)2

jc2 - 4jt + 5
Vertical asymptote: none
Horizontal asymptote: y = 4

^

The graph crosses the horizontal asymptote >» = 4. If a
function has a vertical asymptote six = c, the graph
would not cross it since /(c) is undefined.

61. h(x) =

6 -

2x

3 -

■ X

2(3

-x)

2, if jc ^ 3

Undefined, if a: = 3

3 -X
The rational function is not reduced to lowest terms.

j;2 - 3r - 1 3

63. fix) = ^^^ = -X+1 +

x-2

x-2

\

\

\

i \

The graph appears to approach the slant asymptote
y = -.V+ 1.

hole at (3, 2)

65. fix) = . ■ ■ (0, 4)

vGc^+T'

(a)

\.

'J\/x

^^

On (0, 4) there seem to be 7 critical numbers:
0.5, 1.0, 1.5,2.0,2.5,3.0,3.5

,, > _ —cos Tr<:(xcos TTX + lirix^ + l)sin ttx) _

(t>) / W - (^2 _^ jp/2 - "

Critical numbers = ^, 0.97, |, 1.98. |, 2.98, |.

The critical numbers where maxima occur appear to
be integers in part (a), but approximating them using
/' shows that they are not integers.

Section 3.7 Optimization Problems 145

67. Vertical asymptote: x = 5
Horizontal asymptote: v = 0
1

y = z — ?

69. Vertical asymptote: x = 5
Slant asymptote: y = 3jc + 2

1 3x~ - 13x - 9

y = 3x + 2 +

x-5

X- 5

71. fix) =

(.V - bV

(a) The graph has a vertical asymptote at x = b. If
a > 0, the graph approaches oo as x-^b.\i a < 0,
the graph approaches — oo as x — > Z?. The graph
approaches its vertical asymptote faster as \a\ — >0.

(b) As b varies, the position of the vertical asymptote
changes: x = b. Also, the coordinates of the
minimum [a > 0) or maximum [a < 0) are changed.

73. fix) =

3y
.r^ + 1

(a) For n even, /is symmetric about the y-axis. For n odd,
/is symmetric about the origin.

(b) The A-axis will be the horizontal asymptote if the
degree of the numerator is less than 4. That is,

n = 0, 1,2,3.

(c) « = 4 gives y = 3 as the horizontal asymptote.

(d) There is a slant asymptote y = 3j: if n = 5:

3;c^
X* + 1

= 3.r

3.r

(e)

X* +

n

0

1

2

3

4

5

M

1

O

3

2

1

0

N

2

3

4

5

2

3

75. (a) 2750

(b) When t= 10, MlO) = 2434 bacteria.

(c) N is a maximum when t ~ 7.2 (seventh day).

(d) N%t) = 0 for f - 3.2 .

13 250

(e) lim Nit) = -^ — = 1892.86

(-►oo 7

Section 3.7 Optimization Problems

1. (a)

First Number, x

Second Number

Product, P

10

110 - 10

10(110- 10) = 1000

20

110-20

20(110 - 20) = 1800

30

110- 30

30(110 - 30) = 2400

40

1 10 - 40

40(110 - 40) = 2800

50

1 10 - 50

50(110 - 50) = 3000

60

110-60

60(110 - 60) = 3000

—CONTINUED—

146 Chapters Applications of Differentiation

1. — CONTEWED—

(b)

First Number, x

Second Number

Product, P

10

110 - 10

10(110 - 10) = 1000

20

110 - 20

20(110- 20) = 1800

30

110- 30

30(110 - 30) = 2400

40

110-40

40(110-40) = 2800

50

110- 50

50(110- 50) = 3000

60

110- 60

60(110 - 60) = 3000

70

110- 70

70(110 - 70) = 2800

80

1 10 - 80

80(110 - 80) = 2400

90

1 10 - 90

90(110- 90) = 1800

100

110 - 100

100(110 - 100) = 1000

The maximum is attained near jc = 50 and 60.
(c) P = 41 10 - x) = llOx - ;c^

(d) MOO

(e) ^ = 1 10 - 2;c = 0 when x = 55.
ax

dx'-

= -2 < 0

The solution appears to be x = 55.

P is a maximum when x = 1 10 — jf = 55.
The two numbers are 55 and 55.

3. Let X and y be two positive numbers such that xy = 192.

S = x+y = x-\

■ X

dx

d^S 384

192

= Owhen.1 = V192.

> 0 when x = V192.

dx~ x^
5 is a minimum when x = y = Vl92.

5. Let X be a positive number.
S = x + -

-r- = I ? = 0 whenx = 1.

dx x'-

-pr = -J > 0 when x= \.
dx- j^

The sum is a minimum when jc = 1 and l/;t = 1.

7. Let X be the length and y the width of the rectaiigle.

lx + ly= 100

y = 50 - X

A = xy = .)c(50 — x)

dA

^- = 50 - 2a: = 0 when x = 25.

dx

d^-A
dx^

= -2 < Owhen^; = 25.

A is maximum when j: = _y = 25 meters.

9. Let x be the length and y the width of the rectangle,
xy = 64
64

P = Zx + 2y = 2x + 2(^) = 2x + ^

£/P , 128 „ ^

-— = 2 — — 0 when x = 8.

dx X-

d^P 256

> 0 when x = 8.

dx'^ x3
P is minimum when x = y = 2> feet.

Section 3.7 Optimization Problems 147

11. d = J{X - 4)2 + (v^ - 0)2

^

7x+ 16

Since d is smallest when the expression inside the radical
is smallest, you need only find the critical numbers of

fix) = 3^ -lx+ \6.
fix) = 2x-l = 0

By the First Derivative Test, the point nearest to (4, 0) is

(7/2, 7772).

y

4

3-

te^/x)

13. d = V(x - If + [x-^ - (1/2)P
= Jx* - Ax + (17/4)

Since d is smallest when the expression inside the radical
is smallest, you need only find the critical numbers of

f{x) =x^-4x + T-

f'(x) = 4;c3 - 4 = 0

X = 1

By the First Derivative Test, the point nearest to (2. ^) is
(1, 1).

3 (4,0)

15. ^ = fcr((2o -x) = kQoK - kx^

d^Q

-^ = fcgo - 2kx

= k{Qo - 2x) = 0 when ;c = %

^=-2k < Owhen.r = %
dx> 2

dQ/dx is maximum when x = Qq/2.

17. xy = 180,000 (see figure)

360.000\

S = X + 2y = \x + ■
of fence needed.

where 5 is the length

dS

= 1

360,000

dx X-

d-S 720.000

0 when x = 600.

dx-

x^

> 0 when x = 600.

5 is a minimum when .r = 600 meters and v = 300
meters.

19. (a) A = 4(area of side) + 2(areaofTop)

(a) A = 4(3)(11) + 2(3)(3) = 150 square inches

(b) A = 4(5)(5) + 2(5)(5) = 150 square inches

(c) A = 4(3.25)(6) + 2(6)(6) = 150 square inches

150 - 2x2

(c) S = Axy + Zx' = 150:

4x

,^ , 4\5Q-2x^\ 75 1 ,

V = x-y = x'\ = ^r A' ~ :t •'^

V 4.x: / 2 2

r = y-|.r2 = 0^.r = ±5

(b) V = (length)(width)(height)

(a) V = (3)(3)(I1) = 99 cubic inches

(b) V = (5)(5)(5) = 125 cubic inches

(c) V = (6)(6)(3.25) = 117 cubic inches

\^\

^

,'

y^

By the First Derivative Test, .r = 5 yields the maximum volume. Dimensions: 5 x 5 x 5. (A cube!)

148 Chapters Applications of Differentiation

21. (a) V = x{s - Ixf, Q < X < -

dV

dx '

ItU - 2x)(-l) + {s- 2xY

= (s — 2x)(s — 6x) = Q when x = —,- {s/2 is not in the domain).

2 6

d'^V

^ = 24x - 85

dir

—-^ < 0 when x = —.
dx- 6

2s^ 5

V = — — is maximum when x = — .

27 6

(b) If the length is doubled, V = 27 (25)' = 8(575'). Volume is increased by a factor of 8.

23. 16 = 2y + ;c +

'(f)

12 = Ay + 2x + TTX

Zl -' Ix ~ TTX

A=^+2U

t^l x\2 (32 — 2x — TTX

\x +

( )

r

r

^r*

;;:::;::;:;::;::;:;:;;f:dii

2 4 8

^ = 8-x-f;c + ^;c = 8-
ctc 2 4

0 when x

8

xll.f
32

1 + (77/4) 4 + 77'

_=_^l+_j<Owhenx = ^^

_ 32 - 2[32/(4 + tt)] - 77(32/(4 + tt)]

16

4 + TT

16 32

The area is maximum when y = -:; — ; — feet and x = - — ; — feet.

4+77

4+77

25. (a)

y - 2 _^ 0 - 2
0 - 1 ~ .r - 1

y = 2 +

x- 1

L= Jx^ + y2

= /^

+ 2 +

;c - 1

;c^ + 4 +

X - 1 (x - 1)

;, X > 1

Z, is minimum when x = 2.587 and L == 4.162.

—CONTINUED-

Section 3. 7 Optimization Problems 149

25. —CONTINUED—

(c) Area = A{x) = -xy = -x\l + ^— -^

X + ■

X - 1

U - 1)2 = 1
X- 1 = ±1

;c = 0, 2 (select x = 2)
Then y = 4 and A = 4.
Vertices: (0, 0), (2, 0), (0, 4)

27. A = 2x>' = 2xV25 - x^ (see figure)

J 25 - 2r- \ „ , 572 ...

= 21 , _ 1 = 0 whenx = y = ^- = 3.54.

By the First Derivative Test, the inscribed rectangle of maximum area has vertices

,572 \/ 572 5^
~ 2 ' /'V~ 2 2

5 /5
Width: ^^; Length: 572

29. xy = 30 =^ y = —

A = {x + 2)[— + l] (see figure)

dA
dx

= (x + 2)1-^1 + I— + 2) = ^^^^^-T-^ = 0whenx = 730.

-W\ , /30

v2

. = ^=V30

730

By the First Derivative Test, the dimensions (x + 2) by {y + 2) are (2 +
7.411). These dimensions yield a minimum area

30) by (2 + 730) (approximately 7.477 by

31. V = Trr^h = 22 cubic inches or h =

22

(a)

Radius, r

Height

Surface Area

0.2

22

2-7HO.2)

.^•^ " .Ha2) J

= 220.3

77(0.2)2

0.4

22
tHO.4)-

2t7(0.4)

01
f\ 1 1

« 111.0

_°-' ' M0.4)=J

0.6

22

277<0.6)

->0

= 75.6

7t{0.6)-

L°-^ ' 77<0.6)2.

0.8

22

7T<0.8)2

27t(0.8)

11

= 59.0

L°-^ ' 7710.8) =

—CONTINUED—

150 Chapters Applications of Differentiation

31. —CONTINUED—

(b)

Radius, r

Height

Surface Area

0.2

22

277(0.2)[0.2 + J^^ J

== 220.3

0.4

22

277(0.4)

n^ 1 22 1

- 111.0

7T<0.4)2

°-^ ' 77(0.4)2]

0.6

22

277(0.6)[0.6 + ^2^2^^,

-75.6

77{0.6)2

0.8

22

77(0.8)2

277(0.8)[0.8 . ^^^^,

= 59.0

1.0

22

277(1.0)

'lO 1 22 "

= 50.3

77(1.0)2

.'•° ' 77(1.0)2_

1.2

22

277(1.2)

['■2 ^77(0)2] -45.7

41.2)2

1.4

22

277(1.4)

['■4 ^77(14)2] -43.7

77(1.4)2

1.6

22

277(1.6)

[^■^^77(iy-43.6

77(1.6)2

1.8

22

277(1.8)

b'uf.^K

= 44.8

77(1.8)2

2.0

22

77(2.0)2

27K2.0)

b-' ^ 4?0)2_

= 47.1

The minimum seems to be about 43.6 for r = 1.6.

33. Let X be the sides of the square ends and y the length of the package.

P = 4;c + y = 108 ^. .V = 108 - 4x
V = ;c2y = ;c2(108 - 4a:) = 108^2 - 4x^
dV

dx

= 216.x - 12a:2

12a:(18 - x) = Owhenx = 18.

22] ^ , 44
r + — r = 277r2 + —

The minimum seems to be 43.46 for r ~ 1.52.

(K AA

(e) — = 47rr - ^ = 0 when r = yil/77 «= 1.52 in.
dr r^

22
h= — r = 3.04 in.
7rr2

Note: Notice that

22

22

77r2 77(11/77)2/3 Vir'

= ^|i^)-^'.

d^V

dx"-

= 216 - 2Ax = -216 < Owhenx = 18.

The volume is maximum when x = 18 inches and y = 108 — 4(18) = 36 inches.

Section 3.7 Optimization Problems 151

35. V = -TTX^h = -Trx^{r + Vr^ - x^} (see figure)

3
cix~ 3''

2r^ + 2rVr2 - x^ - 3;c2 = 0

2rV/-2 - ;c2 = 3x^ - 2r^

0 = 9x^ - 8x-r2 = x\9x^ - Sr^)
2V2r

■ ^ = 0,^-

By the First Derivative Test, the volume is a maximum when

-. 4/-

X = — - — and h = r

(O.r)

■..-V7^)

+ Vr2 - x~

Thus, the maximum volume is

3277r3

-Hm

81

cubic units.

37. No, there is no minimum area. If the sides are x and y, then Ix -¥ 2y = 20 => y = 10 — .x.
The area is A{x) = jc(10 — x) = \Qx — x^. This can be made arbitrarily small by selecting x ~ 0.

39. V= 12 =-T7-7-3 + nr-h

12 - {4/3)T7r3 12 4

h = ^ = — J - -r

7rr- Trr'^ 3

12 4
S = Atrr- + lirrh = Airr^ + IttA — n - z;r

\Trr~ 3

24 8

-TTr~ +

24

— - = -irr T = 0 when r = IJ^Itt = 1.42 cm.

dr 3 r-^

— -r = -TT H — r > 0 when r = </9/Trcm.
dr- i r"

The surface area is minimum when r = 1/9/tt cm
and /! = 0. The resulting solid is a sphere of radius
r « 1.42 cm.

41. Let X be the length of a side of the square and y the length
of a side of the triangle.

4x + 3>' = 10

A 2^1 (^
A = x'- + -y\—y

(10 - 3v)' , V3 ,

• ■■ .•■; = 16 +— ^"

■•■ ■ | = i(io_3v)(-3)+^y^O

-30 + 9y + 4y3y = 0

30

■^ " 9 + 4v^

d-A 9 + 4^3

dy~

> 0

A is minimum when
30

y = 7= and .r = ~.

9 + 473 9 + 4V3

152 Chapters Applications of Differentiation

43. Let 5 be the strength and k the constant of proportionality.
Given /z^ + w- = 24^ h^ = 24^ - w^,

S = kwh^

S = kw(576 - w^) = k(576w - w^)

dS

dw
d'-S

= k(576 - 3w^) = 0 when w = 8^3, /i = 8V6.
= -6kw < Owhenw = SVS.

dw^
These values yield a maximum.

45. R = — sin 2d
8

dR 2v^ tt TiTT

—- = — ^cos 29 = 0 when 0 = — , —r-
dd g 4 4

^ = -^sin 29 < 0 when 6 = ^.
dd^ g 4

By the Second Derivative Test, R is maximum
when 8 = n/A.

47. sin a = - => i

s

, 0 < a < -

sm a 2

h , ^ 2 tan a ^

tano = -=>« = 2tana=>i = — : = 2 sec a

2 sm a

k sin a ksma k .

I = ;:; — = ; — = - sm a cos- a

s- 4 sec - a 4

-r- = T[sin a(—2 sin a cos a) + cos- a(cos a)]
da 4

= -rcos afcos^a - 2 sin- a]
4

= - cos a[l - 3 sin^ a]

„ , IT 377 , . 1

= 0 when a = —, -r-, or when sm a = ±—7= .
2' 2' V3

Since a is acute, we have
1

sm a

A = 2 tan a = 2 -7= = v^ feet.

73 \^l

Since (cm)/{doP) = {k/4) sin a(9 sin^a - 7) < 0 when sin a = 1/V3, this yields a maximum.

49.

s =

Jx'^ + 4,L

= 7i

+ (3 - .v)2

ne = ?- =
dT

y.x-2 + 4 Vx'~ -
2

- 6x -1- 10
4

;c- 3

dx

x^

2Vx2 + 4
9 - 6;c -^

4Vx-
x^

■ - 6a- + 10

ii:;t^

jc2 + 4 4(^2 - 6a -I- 10)
A^ - 6^3 -t- 9x2 + 8a - 12 = 0

You need to find the roots of this equation in the interval [0, 3]. By using a computer or graphics calculator, you can determine
that this equation has only one root in this interval (a = 1). Testing at this value and at the endpoints, you see that a = 1 yields
the minimum time. Thus, the man should row to a point 1 mile from the nearest point on the coast.

Section 3.7 Optimization Problems 153

51 7- = v^tM^ ^ V.r" - 6;c + 10

0^ ViV;c2 + 4 v.Va:^ - 6j: + 10
Since

X X — 3

= sin 0, and . ^ = — sin

V^^ + 4
we have

Vj:2 - 6x + 10

sin 9^ sin 62 sin 6, sin 62

Vi V2

Since

d^T 4

= 0 =

a[i2 v,U2 + 4)3/2 ^,^(^2 _ 5^ + 10)3/:
this condition yields a minimum time.

> 0

I

t ^
* £^

53. /W = 2 - 2sinx

3
2
I--

(a) Distance from origin to y-intercept is 2.

Distance from origin to j:-intercept is tt/2 ~ 1.57.

(b) d:

ix- + r
3

'x- + (2

: sin x)-

^

(0.7967. 0.9795)

Minimum distance = 0.9795 at.v = 0.7967.

(c) Let/U) = d\x) = ;c= + (2 - 2 sin.r)-.

f\x) = 2a: + 2(2 - 2sin;c)(-2cos.r)

Setting/'W = 0, you obtain .v == 0.7967, which
corresponds to d = 0.9795.

55. Fcose= k(W- Fsine)

^ kW

cos 6 + k sin 6

dF _ -kW{k cose- sine)

0

de (cos e + k sin 6)-

k cos e=sinfl=>A: = tane=>0 = arctan ^■
Since

1 k-

cos fl + ^- sin 0

the minimum force is
kW

Jk^+ 1 v//t= + 1

ftW

= Jk~ + 1,

cos e + /t sine Vit^ + 1'

1 «-J

154 Chapters Applications of Differentiation

57. (a)

Base 1

Base 2

Altitude

Area

8

8+16 cos 10°

8 sin 10°

-22.1

8

8+16 cos 20°

8 sin 20°

= 42.5

8

8 + 16 cos 30°

8 sin 30°

= 59.7

8

8 + 16 cos 40°

8 sin 40°

= 72.7

8

8 + 16 cos 50°

8 sin 50°

= 80.5

8

8+16 cos 60°

8 sin 60°

= 83.1

(b)

(c) A = {a + b)-

, 8 sin e

[8 + (8 + 16 cos e)]- ^

64(1 + cos e)sin e, 0° < 9 < 90°

(e) 100

Base 1

Base 2

Altitude

Area

8

8 + 16 cos 10°

8 sin 10°

= 22.1

8

8 + 16 cos 20°

8 sin 20°

= 42.5

8

8 + 16 cos 30°

8 sin 30°

= 59.7

8

8 + 16 cos 40°

8 sin 40°

= 72.7

8

8 + 16 cos 50°

8 sin 50°

= 80.5

8

8 + 16 cos 60°

8 sin 60°

= 83.1

8

8+16 cos 70°

8 sin 70°

= 80.7

8

8 + 16 cos 80°

8 sin 80°

= 74.0

8

8 + 16 cos 90°

8 sin 90°

= 64.0

The maximum cross-sectional area is approximately
83.1 square feet.

dA
(d) — = 64(1 + cos e)cos e + (-64 sin e)sin 6
du

= 64(cos e + cos^ d - sitP- 6)
= 64(2cos2e + cos e- 1)
= 64(2 cose- l)(cose+ 1)
= 0 when e = 60M80°, 300°.
The maximum occurs when 6 = 60°.

59. C = 1001

C'= 100'

200

x + 30

1 < X

400

30

(x + 30)2

Approximation: x ~ 40.45 units, or 4045 units

61. 5, = {Am - 1)2 + (5m - 6)^ + (10m - 3)^

- = 2(4m - 1)(4) + 2(5m - 6)(5) + 2(10m - 3)(10) = 282m - 128 = 0 whenm = -^.

dS__
dm

Line: y
S =

64

i4r

/^

\141
256

141

- 1

+

320
141

m

- 6

- 6

640
141

10

64
141

858
141

6.1 mi

^, o |4m - 11 |5m - 61 1 10m - 3
63. 5, = I , , ' + ' , _ ' +

Vm2+ 1 Vm2+ 1

m2+ 1

Using a graphing utility, you can see that the minimum occurs when x = 0.3.
Line: y = 0.3jc

|4(0.3) - 1| + |5(0.3) - 6| + 1 10(0.3) - 3|

^3 =

7(0.3)2 + 1

= 4.5 mi.

Section 3.8 Newton's Method

155

Section 3.8 Newton's Method

1. fix) =x^-3
fix) = 2x
X, = 1.7

n

x„

fiXn)

f'ixj

f'W)

X f^'^y

" nx„)

1

1.7000

-0.1100

3.4000

-0.0324

1.7324

2

1.7324

0.0012

3.4648

0.0003

1.7321

3. f(x) = sinj;

f'(x) = COS.X

;cj = 3

n

Xn

n^n)

/'UJ

■" /'UJ

1

3.0000

0.1411

-0.9900

-0.1425

3.1425

2

3.1425

-0.0009

-1.0000

0.0009

3.1416

5. f{x)=x^ +x- 1
fix) = 3x2+1

Approximation of the zero of/ is 0.682.

n

Xn

fi^n)

n\)

,fUJ

X ^^'"^
" f'(-\)

1

0.5000

-0.3750

1.7500

-0.2143

0.7143

2

0.7143

0.0788

2.5307

0.0311

0.6832

3

0.6832

0.0021

2.4003

0.0009

0.6823

/U) =
/'U) =

3Vx — 1 — X
3

2v^

Approximation of the zero of/ is 1.146.
Similarly, the other zero is approximately 7.854.
, /W = A^ + 3
fix) = 3.t2
Approximation of the zero of/ is - 1.442.

11. fix) = x' - 3.9^2 + 4.79x - 1.881
/'W = 3x' - 7.8x + 4.79

n

.r„

/UJ

n\)

/UJ

X ^^-'"^
" /'UJ

1

1.2000

0.1416

2.3541

0.0602

1.1398

2

1.1398

-0.0181

3.0118

-0.0060

1.1458

3

1.1458

-0.0003

2.9284

-0.0001

1.1459

n

•^„

fi^J

n^n)

/UJ
/'UJ

'" /'UJ

1

-1.5000

-0.3750

6.7500

-0.0556

-1.4444

2

-1.4444

-0.0134

6.2589

-0.0021

-1.4423

3

-1.4423

-0.0003

6.2407

-0.0001

-1.4422

n

■^n

/UJ

/'UJ

/UJ
/'UJ

, /UJ
" /'UJ

1

0.5000

-0.3360

1.6400

-0.2049

0.7049

2

0.7049

-0.0921

0.7824

-0.1177

0.8226

3

0.8226

-0.0231

0.4037

-0.0573

0.8799

4

0.8799

-0.0045

0.2495

-0.0181

0.8980

5

0.8980

-0.0004

0.2048

-0.0020

0.9000

6

0.9000

0.0000

0.2000

0.0000

0.9000

Approximation of the zero of/ is 0.900.
-CONTINUED—

156 Chapters Applications of Differentiation

11. —CONTINUED—

n

x„

no

/'UJ

nx„)

/'UJ

X f^'"^

" nx„)

1

1.1

0.0000

-0.1600

-0.0000

1.1000

Approximation of the zero of/ is 1.100.

n

Xn

nxj

/'UJ

/'UJ

" /'UJ

1

1.9

0.0000

0.8000

0.0000

1.9000

Approximation of the zero of/ is 1.900.

13. /W = .r + sin(.x + 1)
f'(x) = 1 + cos(;c + 1)
Approximation of the zero of/ is -0.489.

n

^.

nxj

/'UJ

nx,}

fixj

X ^^'"^

" fixj

1

-0.5000

-0.0206

1.8776

-0.0110

-0.4890

2

-0.4890

0.0000

1.8723

0.0000

-0.4890

15. h(x) = fix) - g(x) = 2;c + 1 - ^x + 4
1

h'(x) = 2

Ijx + 4

Point of intersection of the graphs of /and g occurs
when X = 0.569.

«

^„

h{x„)

hXx^)

X ^^'"^

1

0.6000

0.0552

1.7669

0.0313

0.5687

2

0.5687

-0.0001

1.7661

0.0000

0.5687

17. /zU) = /(j:) - g(x) = jc - tan X ,

h'{x) = 1 — sec-^jc

Point of intersection of the graphs of/ and g occurs
when X = 4.493.

n

Xr,

h(xj

k'(x„)

hix,}
h'ixj

X ^^'"^
" h'ixj

1

4.5000

-0.1373

-21.5048

0.0064

4.4936

2

4.4936

-0.0039

-20.2271

0.0002

4.4934

19. f(x) = x^ - a = 0
fix) = 2x

v2 _ „

2x,.^ - x,^ + a _ Xj^ + a _ x,- a

2x, 2Xi 2 2x:

i

1

2

3

4

5

X!

2.0000

2.7500

2.6477

2.6458

2.6458

77 = 2.646

23 , ^ 3.,^ + 6

i

1

2

3

4

Xl

1.5000

1.5694

1.5651

1.5651

76" « 1.565

Section 3.8 Newton's Method 157

25. fix) = 1 + cos X
f'(x) = -sinx

Approximation of the zero: 3.141

n

^n

/UJ

/'UJ

" fix J

1

3.0000

0.0100

-0.1411

-0.0709

3.0709

2

3.0709

0.0025

-0.0706

-0.0354

3.1063

3

3.1063

0.0006

-0.0353

-0.0176

3.1239

4

3.1239

0.0002

-0.0177

-0.0088

3.1327

5

3.1327

0.0000

-0.0089

-0.0044

3.1371

6

3.1371

0.0000

-0.0045

-0.0022

3.1393

7

3.1393

0.0000

-0.0023

-0.0011

3.1404

8

3.1404

0.0000

-0.0012

-0.0006

3.1410

27. y = 2jc3 - 6x2 + 6;c - 1 = f(^^)
y' = 6x^ - 12x + 6 =f'{x)
x, = l
fix) = 0; tlierefore, the method fails.

n

Xn

/UJ

/'UJ

1

1

1

0

29. y = -x^ + 6x2 - 10^ + 6 = /(x)
y'= -3x2 + lit - 10 =/'(x)

^1

= 2

Xj

= 1

X3

= 2

X4

= 1

and so on

Fails to

converge

31. Answers will vary. See page 222.

Newton's Method uses tangent lines to approximate c such that /(c) = 0.
First, estimate an initial x, close to c (see graph).

Then determine Xj by Xj = x.

/u.

fW

Calculate a third estimate by Xj = Xj

fix,)

Continue this process until \x„ - x„ + 1 1 is within the desired accuracy.
Let x„ + 1 be the final approximation of c.

W)

33. Let g(x) = fix) — X = cos x — x
g 'ix) = — sin X — 1 .
The fixed point is approximately 0.74.

n

x„

8ix„)

sXx„)

S^x„)
g\x„)

" S\x„)

1

1.0000

-0.4597

-1.8415

0.2496

0.7504

2

0.7504

-0.0190

-1.6819

0.0113

0.7391

3

0.7391

0.0000

-1.6736

0.0000

0.7391

158 Chapters Applications of Differentiation

35. fix) = jc3 - 3;t2 + 3, f'{x) = 3;c2 - 6x
(a)

(c) .^i =

1

Continuing, the zero is 2.532.

(e) If the initial guess jc, is not "close to" the desired zero
of the function, the x-intercept of the tangent line may
approximate another zero of the function.

(b) X, = 1

#4 « 1.333

f(Xl)

Continuing, the zero is 1.347.

(d)

The j:-intercepts correspond to the values resulting
from the first iteration of Newton's Method.

37. /U) = - - a = 0

X

fix) = -3

_^l"/^ 2" = x„+ x„'^[y - a]=x„+x„' x/a = 2x„- x^a = x„{2 - oxj

39. fix) = X cos X, (0, tt)

fix) = —X sin x + cos X = 0

Letting F{x) = fix), we can use Newton's Method as follows.

[F'{x) = — 2 sin X + X cos x]

n

Xn

F(x„)

nx„)

Fix„)
Fix,)

" F\x„)

1

0.9000

-0.0834

-2.1261

0.0392

0.8608

2

0.8608

-0.0010

-2.0778

0.0005

0.8603

Approximation to the critical number: 0.860

Section 3.8 Newton's Method

159

41. >'=/U) = 4-^2, (1,0)

d^- 7U -iy + (y- 0)2 = VU - D' + (4 - x-r- = ^x*-lx--2x + 17

li is minimized when D = x* — 7x- - Zx + 17 is a minimum.
gix) = D' = 4x^ - Hx - 2
g'ix) = 12r= - 14

n

•'^n

gix„)

sK)

1

2.0000

2.0000

34.0000

0.0588

1.9412

2
3

1.9412
1.9385

0.0830
-0.0012

31.2191
31.0934

0.0027
0.0000

1.9385
1.9385

x- 1.939
Point closest to (1. 0) is = (1.939, 0.240).

43.

, ,. . . ^ Distance rowed Distance walked
Minimize: T = — z ^ 1-

Rate rowed

Rate walked

_ vU2 + 4 V.T- - 6.x:

10

4
x-3

1.939.0.240)

3jx- + 4 4jx- - 6x + 10

4xjx- - 6x + 10 = -3U - 3)7.^2 + 4

16x2(^2 - 6x + 10) = 9U - 3Hx^ + 4)

7.r^ - 421^ + 43a" + 216a- - 324 = 0

Let /(a) = 7x* - 41v3 + 43^2 + 216x - 324 and /'(a) = 28.r3 - 126^2 + 86.1; + 216. Since /(I) = - 100 and/(2) = 56, the
solution is in the interval (1, 2).

n

Xn

fi^n)

n^)

/(■vj

, /K)
" /'(-^J

I

1.7000

19.5887

135.6240

0.1444

1.5556

2

1.5556

-1.0480

150.2780

-0.0070

1.5626

3

1.5626

0.0014

49.5591

0.0000

1.5626

Approximation: x == 1.563 miles

45. 2,500,000 = -76.x^ + 4830.x~ - 320,000

76x^ - 4S30x- + 2,820,000 = 0
Let/(x) = 16x^ - 4830.r2 + 2,820,000

f'(x) = ll%x~ - 966aT.
From the graph, choose .t, = 40.

n

^n

/(.vj

/'(•v„ )

/UJ

, /UJ
" /'UJ

1

40.0000

-44000.0000

-21600.0000

2.0370

37.9630

2

37.9630

17157.6209

-38131.4039

-0.4500

38.4130

3

38.4130

780.0914

-34642.2263

-0.0225

38.4355

4

38.4355

2.6308

-34465.3435

-0.0001

38.4356

The zero occurs when x == 38.4356 which corresponds to $384,356.

160 Chapters Applications of Differentiation

47. False. Let/W = (jr - \)/{x — 1). .r = 1 is a discontinuity. It is not a zero oif{x). This statement would be true if
fix) = pix)/qix) is given in reduced form.

49. True

51. fix) =\^ - 3.t2 + \x

fix)

3 2
1= — v^

6x + :

Let .»:[ = 12.

Approximation: a; = 11.803

Section 3.9 Differentials

n

-^.

/UJ

fK)

/'UJ

" /'UJ

1

12.0000

7.0000

36.7500

0.1905

11.8095

2

11.8095

0.2151

34.4912

0.0062

11.8033

3

11.8033

0.0015

34.4186

0.0000

11.8033

1. fix) =X'
fix) = 2x

Tangent line at (2, 4): >- - /(2) = /'(2)(x - 2)
y - 4 = 4ix- 2)
y = 4x — 4

X

1.9

1.99

2

2.01

2.1

fix) = X-

3.6100

3.9601

4

4.0401

4.4100

rU) = 4a: - 4

3.6000

3.9600

4

4.0400

4.4000

3. fix) = ;c= -. .

fix) = 5x^

Tangent line at (2, 32): y - /(2) = f'il)ix - 2)
y - 32 = 80U - 2)

>> = 80.x - 128

.r

1.9

1.99

2

2.01

2.1

/U) = ^

24.7610

31.2080

32

32.8080

40.8410

rU) = 80.x: - 128

24.0000

31.2000

32

32.8000

40.0000

5. fix) = s\nx
fix) = cos X

Tangent line at (2, sin 2):

y-/(2)=/'(2)(;c-2)
3; - sin 2 = (cos 2)ix - 2)

y = (cos 2)(x - 2) + sin 2

X

1.9

1.99

2

2.01

2.1

fix) = sin a:

0.9463

0.9134

0.9093

0.9051

0.8632

Tix) = (cos 2)(.x -

- 2) + sin 2

0.9509

0.9135

0.9093

0.9051

0.8677

3 2

1 j: , X

l.y=fix) = Wj'ix

Ay=fix + ^x) -fix,
= /(2.1)-/(2)
= 0.6305

2,Ax = dx = 0.1

flfy=/'(.x)a!x
= /'(2)(0.1)
= 6(0.1) = 0.6

Section 3.9 Differentials 161

9. y = fix) = x^ + \,f'{x) = 4x\ x= - 1, Ax = A = 0.01

Ay = f(x + Ax)- f(x) dy = f'(x) dx
= /(-0.99)-/(-l) =/'(-l)(0.01)

= [(-0.99)" + 1] - [(- ir + 1] « -0.0394 = (-4)(0.01) = -0.04

11. y = 3x^ - 4

dy = 6.t dx

15. y = xj\ - .v2

dy = \x

vr

+ Vl - -v- dx

Zx-

vT^^

iv

.r + 1

13.

>' =

2x-

1

dy--

-

3 .

(2;c-

-l)^'^

17.

y --

= 2x~

• cot-x

dy--

--(2 +

2 cot X csc-

.t)di:

=

= (2 + 2 cot ;c + 2

cot^.

x)dx

,o 1 /677nr- 1

19. y = -C0s(-^-

dy = — IT sin — ^7: | tic

\ 2

21. (a)/(1.9) =/(2 - 0.1) -/(2) +/'(2)(-0.1)

= 1 + (1)(-0.1) = 0.9
(b) /(2.04) =/(2 + 0.04) «/(2) +/'(2)(0.04)

== 1 + (1)(0.04) = 1.04

23. (a) /(1.9) =/(2 - 0.1) «/(2) +/'(2)(-0.1)

« 1 + (-jK-O.l) = 1.05
(b) /(2.04) =/(2 + 0.04) «/(2) +/'(2)(0.04)

= 1 + (-5)(0.04) = 0.98

25. (a) g(2.93) = ^(3 - 0.07) « ^(3) + ^'(3)(-0.07) 27. (a) ^(2.93) = ^(3 - 0.07) « ^(3) + ^'(3)(-0.07)

- 8 + (-5)(-0.07) = 8.035 . = 8 + 0{-0.07) = 8

(b) g(3.1) = ^(3 + 0.1) « g(3) + g'{3){0.\) . (b) g{2.\) = g{3 + 0.1) » ^(3) + g'(3)(0.1)

= 8 + (-5)(0.1) = 7.95 = 8 + 0(0.1) = 8

29. A=x-

x=\l

Ax = dx = ±^
dA = Ixdx

AA - (M = 2(12)(±^)
= ±f square inches

31. A = 7rr2 :

r= 14
Ar = dr = ±\

AA -^ dA = iTTr dr = 7r(28)(±5)
= ±777 square inches

162 Chapters Applications of Differentiation

33. (a) X = 15 centimeter

Ax = dx = ±0.05 centimeters

A=x^
dA = 2xdx = 2{15)(±0.05) .

= ± 1.5 square centimeters
Percentage error:

T = Bi = »■"«'■ ■ = ;'

W'^-^--^ 0.025
A x^ X

dx , 0.025

<

X ' 1

= 0.0125 = 1.25%

35. r = 6 inches

Ar = dr = ±0.02 inclies

4

dV = Anr'^dr = 477<6)2(±0.02) = ±2.887r cubic inches

(b) 5 = 4irr2

dS = 87rrrfr= 8i7<6)(±0.02) = ±0.967r square inches

dV ^ Airr^ dr ^ 3dr

(c) Relative error: y ~ (4/T,)Trr^ ~ r

= 7(0.02) = 0.01 = 1%
0

Relative error:

dS Sirrdr 2dr

Airr^
2(0.02)

= 0.000666 .

37. V = TTrVi = 407rr2, r = 5 cm, /z = 40 cm, dr = 0.2 cm
AV « rfV = SOTrrrfr = 80tt<5)(0.2) = SOwcm^

39. (a) T = Ittjm

ai —

/Ug

Relative

error:

dT

(ndD/i

gjL/g)

: T

27rv

'L/g

dL

IL

= — (relative error in L)

= -(0.005) = 0.0025

jnn 1

Percentage error: —(100) = 0.25% = -%

(b) (0.0025)(3600)(24) = 216 seconds
= 3.6 minutes

41. e =

-- 26°45'= 26.75°

de = ±15' = ±0.25°

(a) h = 9.5 CSC 6

dh = -9.5 CSC e cot Odd

^= -cotede .

h

dh

h

= (cot 26.75°)(0.25°)

Converting to radians, (cot 0.4669)(0.0044)
== 0.0087 = 0.87% (in radians).

(b)

= cotede < 0.02

0.02 tan e

de ^ 0.02

d ~ e(cot0)

de _^ 0.02 tan 26.75°
e ~ 26.75°

0

_ 0.02 tan 0.4669
0.4669

= 0.0216 = 2.16% (in radians)

Review Exercises for Chapter 3 163

43. r = ^(sin2e)

Vo = 2200 ft/sec
0 changes from 10° to 11°
(2200)2

dr =

»='«(il

16

Tso

{cos 28) d6

de={\\ - 10)

180

_ (2200)2 /'20irV it

16 '^°Tl80/Vl80

4961 feet

= 4961 feet
47. Let/(x) = ifx,x = 625, dx= -\.

fix + At) -/(x) + f'(x) dx= i/i +

A-'J?

dx

fix + A.r) = t/624 = ^625 +

4(4/625}

j(-l)

= 5-- = 4.998
Using a calculator, ^624 = 4.9980.

51. In general, when Ax -^ 0, dy approaches Ay.

45. Let/(;c) = ^,x= I00,dx= -0.6.
fix + Ax)^fix)+f'{x)dx

= Vx H -pdx

l-fx

fix + Ax) = 7995

= VTOO + — ^(-0.6) = 9.97
2V100

Using a calculator: 7994 = 9.96995

49. Let/(x) = Vx,x = 4.ir = 0.02,/'(x) = l/(2v^).
Thffli

/(4.02)-/(4)+/'(4)di

74^02 - V4 + ^^(0.02) = 2 + ^(0.02).

53. True

55. True

Review Exercises for Chapter 3

1. A number c in the domain of/ is a critical number if /'(c) = 0 or/'
is undefined at c.

/'Wis 3.. /V)=o

undefined. '

3. g(x) = 2x + 5 cos X. [0, 2-77]
^'(.r) = 2 - 5 sin X

= 0 when sinx = |.

Critical numbers: x == 0.41, x = 2.73
Left endpoint: (0, 5)
Critical number: (0.41,5.41)
Critical number: (2.73, 0.88) Minimum
Right endpoint: (277,17.57) Maximum

18

V

/

16.28. 17j7y
_^.73, 0.881/

4

164 Chapters Applications of Differentiation

5. Yes. /(- 3) = /(2) = 0. /is continuous on [- 3, 2],
differentiable on (— 3, 2).

f'(x) = (x + 3)0x - 1) = 0 for X = |.
c = 3 satisfies /'(c) — 0.

/(l)=/(7) = 0
(b) /is not differentiable atx = 4.

9. /(x) = x2/3, 1 < X < 8

/'W = |x-'/3

/(fc)-/(a) 4- 1 _3
b- a 8-1 7

/'(c) = |c-/3 = I

'14\3 2744 ^^^,

11. /(x) — X — COS X, — — < X < —

J V / '2 2

/'(x) = 1 + sinx

/(fc)-/(a)_ (77/2) -(-77/2)

fc-a (77/2) - (-77/2)

/'(c) = 1 + sin c = 1
c = 0

= 1

13. /(x) = Ar2 + Sx + C

/'(x) = 2Ax: + B
/(xj) - /(x,) /l(x/ - x,2) + S(x3 - X,)

= A(xi + Xj) + S
/'(c) = 2Ac + B = A(xi + Xj) + B
2Ac = A(xj + Xj)

Xj + X2

c = - — r — = Midpoint of [x,, XjJ

15. /(x) = (x - l)2(x - 3)

/'(x) = (x - 1)2(1) + (x - 3)(2)(x - 1)

= (x - l)(3x - 7)
Critical numbers: x = 1 and x = t

Interval:

-00 ~< X < 1

1 < X < 1

3 < X < 00

Sign oif'(x):

fix) > 0

fix) < 0

fix) > 0

Conclusion:

Increasing

Decreasing

Increasing

17. h{x) = v^(x - 3) = x3/2 - Sx'/^
Domain: (0, 00)

;!'(x)=|xl/2-|x-'/2

= 3 V2(^ _ . 3(x - 1)
2"^ ^ ^ 2v^

Interval:

0 < X < 1

1 < X < 00

Sign of h '(x):

h'{x) < 0

h'{x) > 0

Conclusion:

Decreasing

Increasing

Critical number: x = 1

Review Exercises for Chapter 3 165

19. h{t)

h'[t)

i,4

- 8f
8 = 0 when t = 2.

Relative minimum: (2, - 12)

Test Interval : -

oo < f < 2

2 < t < oo

Signof/j'W:

h'(t) < 0

h '(t) > 0

Conclusion:

Decreasing

Increasing

21. y = - cos(12r) - - sin(12/)

V = y'= -4sin(12f) - 3cos(120

77 1

(a) When t = '^,y = 'z inch and v = >> ' = 4 inches/second.

(b) y' = -4sin(12f) - 3cos(12r) = Owhen

sm(12r) 3 ,,^, 3

— TTTTT = -7 => tan(12r) = --.
cos(12r) 4 4

3 4

Therefore, sin(12r) = -— and cos(12i) = -. The maximum displacement is

I I)-K-|)=^'"^*^■

(c) Period: ^ = f

Frequency:

1 _ 6

77/6 77

23. f{x) = X + cosx, G < X < 2tt
f\x) = 1 - sin.r

77 377

f"{x) = - cos .r = 0 when x

r 2

^ . ^. „ ■ (tT 77\ /377 377

Pomts of mflection: I — , — I, I — , —

Test Interval:

0 < .V < ^

77 377

3- < -t < Y

377

— < .r < 277

Sign of /"(.v):

f"{x) < 0

f"(x) > 0

fix) < 0

Conclusion:

Concave downward

Concave upward

Concave downward

V2

25. g(x) = 2x2(1 - x^)

g'ix) = -4x{2x^ — 1) Critical numbers: x = 0, ±

g"(x) = 4 - 24x2

g"(0) = 4 > 0 Relative minimum at (0. 0)

g'i +—/= = — 8 < 0 Relative maximums at ±^7=, —

V V2; V V^ 2

27.

(5.^5))

29. The first derivative is positive and the second derivative is
negative. The graph is increasing and is concave down.

166 Chapter 3 Applications of Differentiation

31. (a) D = 0.0034r* - 0.2352/3 + 4.9423/2 - 20.8641/ + 94.4025
(b) 3«9

(c) Maximum at (21.9, 319.5) (=1992)
Minimum at (2.6, 69.6) (« 1972)

(d) Outlays increasing at greatest rate at the point of inflection (9.8, 173.7) (= 1979)

33. lim

2r2

= lim

X ->oo 3jc2 + 5 JT -»o=i 3 + 5/x^ 3

5 cos X
35. lim = 0, since |5 cos j-l < 5.

JT^OO X

37. /!(x) =

2x + 3

X- 4

Discontinuity: x = 4

, 2x + 3
lim — —

.X -»oo X — 4

nml±m = 2

.CO 1 - {4/x

Vertical asymptote: x = 4
Horizontal asymptote: y = 2

39. f(x)

- 2

Discontinuity: x = 0

lim I 2| = -2

J— »oo \X

Vertical asymptote: x = 0
Horizontal asymptote: 3; = — 2

41. fix) =x^ +

243

Relative minimum: (3, 108)
Relative maximum: (—3, — 108)
200

Vertical asymptote: x = 0

43. fix) =

1

1 + 3x^

Relative minimum: (-0.155,-1.077)
Relative maximum: (2.155, 0.077)

Horizontal asymptote: y = 0

45. fix) = 4x-x'^ = xi4 - x)

Domain: (-00, 00); Range: (-00,4)
fix) = 4 - 2x = 0when;<: = 2.
fix) = -2

Therefore, (2, 4) is a relative maximum.
Intercepts: (0, 0), (4, 0)

Review Exercises for Chapter 3 167

47. f(x) = xVl6 - x2. Domain: [-4, 4], Range: [-8,8]
Domain: [-4,4]; Range: [-8,8]

f'(x) = 1^^ ^ = 0 when x = ±2 V2 and undefined when x

f"i

Vie -X-

2x{x^ - 24)

±4.

(16 - x=p/2

/'(-2V2) > 0

Therefore, (-2^2, —8) is a relative minimum.

/"(272) < 0
Therefore, [2^, 8) is a relative maximum.

Point of inflection: (0, 0)
Intercepts: (-4, 0), (0, 0), (4, 0)
Symmetry with respect to origin

. hVz i)

2V1, -8

(-4.0) J 1(4.0)

I I t I f I * I — H^-t
i -6 I -2 i \ 2 4 6 8

^ (0. 0)

49. fix) = U - D^U - 3)2

Domain: (—00, 00); Range: ( — 00,00)

11

f'(x) = (x - iy{x - 3)(5x - 11) = 0 whenjt = 1, y, 3.

ii+ye

f"{x) = 4(x - l)(5x2 - 22.x + 23) = Owhen.r = 1,

/"(3) > 0
Therefore, (3, 0) is a relative minimum.

4t) <"

Therefore, ( — , 1 is a relative maximum.

Pomts of inflection: (1, 0), ( ' ^ ~ ^^, 0.60 ), ( " t 0.46

Intercepts: (0, -9), (1, 0), (3, 0)

51. fix) = .r'/5(.v + 3)2/3

Domain: (—00, 00); Range: (—00, 00
x+ 1

/W =

(X + 3)'/3;c2/3

-2
x^'Hx + 3)^/3

= 0 when x = —I and undefined when x = —3,0.
is undefined when .x = 0,-3.

By the First Derivative Test (-3, 0) is a relative maximum and (- 1. - v'^) is
a relative minimum. (0, 0) is a point of inflection.

Intercepts: (-3,0), (0,0)

168 Chapters Applications of Differentiation

53. fix)

x+ 1

X - 1

Domain: (-oo, 1), (1, oo); Range: (-oo, 1), (1, oo)
fix) = -r—^- < 0 if j: ^ 1.

f'Xx) =

ix - ly

Horizontal asymptote: y = I
Vertical asymptote: x = I
Intercepts: (- 1, 0), (0, - 1)

55. fix)

1 + x^

Domain: (-oo, oo); Range: (0, 4]

fix) =

-8x

(1 + x^-y-

= 0 when jc = 0.

/W= (i+,2)3 =Owhen. = ±— .

/"(O) < 0

Tlierefore, (0, 4) is a relative maximum.

Points of inflection: (±V3/3, 3)
Intercept: (0, 4)
Symmetric to the y-axis
Horizontal asymptote: >> = 0

57. fix) = x^ + x + -
■' X

Domain: (— oo, 0), (0, oo); Range: (— oo, — 6], [6, oo)

.„ ^ ^ n . 4 3xf* + x^ - 4 ^ ,
/ ix) = ix- +1 2 = ^ = 0 when x = ±\.

8 6;c^ +

fix) = 6x + ^ 3

7t 0

/"(- 1) < 0
Therefore, (- 1, -6) is a relative maximum.

/"(I) > 0

Therefore, (I, 6) is a relative minimum.

Vertical asymptote: x = 0
Symmetric with respect to origin

(-I.-6)

r\\

(1.6)

Review Exercises for Chapter 3

169

59. fix) =\x^-9\

Domain: (-00, oo); Range: [0, oo)

2x(jip- - 9)
f'(x) = — p^ — -j— = 0 when jc = 0 and is undefined when x =

2(x^ - 9)
fix) = -r^ — TTT is undefined at x = ±3.

F - 9|

/"(O) < 0
Therefore, (0, 9) is a relative maximum.

Relative minima: (+3, 0)
Points of inflection: (±3,0)
Intercepts: (±3, 0), (0, 9)
Symmetric to the )'-axis

61. fix) = X + cos JC

Domain: [0, lir]; Range: [1, \ + 2tt]
fix) = 1 — sin.T > 0, /is increasing.

fix) = - cos j: = 0 when x = —, -r-.
J \ J 2' 2

Points of inflection: — > TT 1 1 ^;~' ~;~

\2 2/ \ 2 2

Intercept: (0, 1)

±3.

(2ir, 2ir + l). ^

2ir-

Z''

. (f.f)/

"

v

</

(0, t)'

^•f)

1 ' i ' 21 "

63. x- + 4r - 2v - 16v + 13 = 0

(a) (.x:= - 2jc + 1) + 4(v= - 4>' + 4) = - 13 + 1 + 16

ix - 1)2 + 4Cy - 2)2 = 4
ix - 1)2 , (y-2)2

1

The graph is an ellipse:
Maximum: (1,3)
Minimum: (1, 1)
(b) X' + 4f -Ix- 16y + 13

0

2;c + 8v

dx

i4 = o

dx

^i%y - 16) = 2 - 2x

2 - 2jc

1 -X

dl_

dx 8y - 16 4v - 8

The critical numbers are .r = 1 and y = 2. These correspond to the points (1, 1), (1. 3), (2, - 1), and (2, 3).
Hence, the maximum is (1, 3) and the minimum is (1, 1).

170 Chapters Applications of Differentiation

65. Let r = 0 at noon.

L = rf- = (100 - 12/)2 + (- lOt)- = 10,000 - 2400t + 244?^

^ = -2400 + 488f = 0 when r = ^ - 4.92 hr.
dt 61

Ship A at (40.98, 0); Ship B at (0, -49.18)

d^ = 10,000 - 2400f + 244t^

=» 4098.36 when t « 4.92 « 4:55 p.m..
d » 64 km

(0,0)

(100-12(.0)

;

^'-4 (100,0)

(0, -100

67. We have points (0, v), {x, 0), and (1, 8). Thus,

y - 8 0-8 Sx

m = = ■ 7 or y

0-1 X - 1 ' X- V

8.t V

Let/(j:) = L^ = x- +

X- 1

f\x) = 2x+ 128'

X - 1

U-1)

U - 1)2 J

= 0

64x .

Jf ~ 7 TTT = 0

x\(x — 1)' — 64] = 0 when jc = 0, 5 (minimum).
Vertices of triangle: (0, 0), (5, 0), (0, 10)

2 4 6 8 10

69. A = (Average of bases)(Height)

v

- + ^^35^ + 2sx -x\ ^ ,

j (.see iigurej

dA_}_
dx 4

; , '''^' ""' , + 73.2 + 2sx - x^
L V3s2 + 2sx - x^

l{ls-x)(s + x) ^ ,
= — , , = 0 when x = 2s.
4j3s^ + 2sx - X-

A is a m

aximum when x = 2s.

|V3i^ + 2j:

71. You can form a right triangle with vertices (0, 0), {x, 0) and (0, y).
Assume that the hypotenuse of length L passes through (4, 6).

y - 6 6-0 6x

m = 7 = -: or J

0-4 A- x
Let/(x) = l2 = Jc2 + >>2 = ;c2 +

-x - 4
6.t

f'(x) = 2x + 12

,x- 4

X

ix - 4)2

= 0

\x - 4/

jr[(j: - 4)3 - 144] = 0 when .t = 0or;c = 4+ ^144.
L == 14.05 feet

Review Exercises for Chapter 3 171

73.

i-,

CSC e = — or L, =6 CSC 6 (see figure)

6

csci ^-0j = -^otU = 9 csc(y - e

L = L, + L, = 6 CSC e + 9 CSC

e = 6 CSC e + 9 sec e

cf0

-6csc ecot e + 9 sec dlan 6 = 0

tan' 0 = - => tan

j/2
1/2

seed = Vl + tan= 6 = , / 1 +

/

2\2/3 V32/3 + 22/3

CSC 6 =

: e V3-/3 + 2-'

/3

Z. = 6

tan e 2'/3

(32/3 + 22^^ „ (32/3 + 22/3)'/^

L y

X 6

UyC

Xe '

\

/{e _

(!-«)

9

21/3

+ 9-

31/3

3'/3

= 3(32/3 + 22/3)3/2 ft == 21.07 ft (Compare to Exercise 72 using a = 9 and fc = 6.;

75. Total cost = (Cost f)er hour) (Number of hours)
^ /v2 ^VllO\ llv , 550

^=ii_ 550 ^ llv- - 33,000
dv ~ 60 V' 60v-

= 0 when v = ^3000 = 10^30 = 54.8 mph.

dv- ~ v3

> 0 when v = lOv^ so this value yields a minimum.

77. fix) = .v3 - 3x - 1

From the graph you can see that/(.r) has three real zeros.
f\x) = 3.x2 - 3

n

K

/(^„)

/'(^J

1

-1.5000

0.1250

3.7500

0.0333

-1.5333

2

-1.5333

-0.0049

4.0530

-0.0012

-1.5321

n

^n

/(■^J

/'(-^J

/(-T„)

X ^^'"^
" f'ix„)

1

-0.5000

0.3750

-2.2500

-0.1667

-0.3333

2

-0.3333

-0.0371

-2.6667

0.0139

-0.3472

3

-0.3472

-0.0003

-2.6384

0.0001

-0.3473

n

Xn

fix„)

/'(■vj

1

-1.9000

0.1590

7.8300

0,0203

1.8797

2

1.8797

0.0024

7.5998

0.0003

1.8794

The three real zeros of/(j:) are .r = - 1.532, x = -0.347, and x = 1.879.

172 Chapters Applications of Differentiation

79. Find the zeros of/(;c) = X* — x — 3.
fix) = 4;t^ - 1
From the graph you can see that/(j:) has two real zeros.
/changes sign in [—2, — l].

n

Xn

/UJ

f'iXn)

1

- 1.2000

0.2736

-7.9120

-0.0346

-1.1654

2

-1.1654

0.0100

-7.3312

-0.0014

-1.1640

On the interval [-2, -l]:x~ -1.164.
/changes sign in [1, 2].

n

x„

/UJ

/'UJ

1

1.5000

0.5625

12.5000

0.0450

1.4550

2

1.4550

0.0268

11.3211

0.0024

1.4526

3

1.4526

-0.0003

11.2602

0.0000

1.4526

On the interval [1, 2]; x = 1.453.

81. y = x{\ - cos j:) = x - j: cos j:

die

1 + j: sin^: — cos;c

dy = (1 + ;c sin j: — cos x) dx

83. 5 = 47rrl rfr = Ar = ±0.025

dS = S'nrdr= 877(9)(±0.025)
= ± 1. 8 TT square cm

f(100) = ^(100) = ^(100)

47rr
2(±0.025)

(100) = ±0.56%

4

dV = 4Trr2 dr = 47r(9)2(+0.025)
= ±8.1 77 cubic cm

3dr

f'><»> = iJ^«

3(±0.025)

(100)

(100)

rO.83%

Problem Solving for Chapter 3

1. Assume y^ < d < y^. Let g{x) = f{x) — (/(jc — a), g is continuous on [a, b] and therefore
has a minimum (c, g(c)) on [a, b]. The point c cannot be an endpoint of [a, b] because

g'{a)=f'(a} -d = y^-d<Q

g'ib) =f'{b) -d = y^~d>Q

Hence, a < c < b and g'{c) = 0 => f'{c) = d.

Problem Solving for Chapter 3 173

3. (a) For a = —3, —2, —\,0, p has a relative maximum at (0, 0).

For a = 1, 2, 3, p has a relative maximum at (0, 0) and 2 relative minima.

(b) p'{x) = Aax' - \lx = Ax{ax^ - 3) = 0

0.±

p"{x) = \2ax- - 12 = 12(a.v2 - 1)

For x = 0, p"(Q) = -12<0=>phasa relative maximum at (0, 0).

(c) If a > 0..r = ±

P ±

- are the remaining critical numbers.

= 12( — I — 12 = 24 > 0 => p has relative minima for a > 0.

a I \a

(d) (0, 0) lies on y = -Ix-.

^3

Let.t = ±

a

. Then

^(^)="(!r-<i . .

9 / /3\

Thus,>'= — = -3 ±^/- = -3j:- is satisfied by all the relative extrema of p.
a \ \ a I

5. p{x) = x"^ + ax^ + \

(a) p '(x) = 4:^ + lax = 2x(2.c- + a)
p"(x) = I6;c= + 2a

For a > 0, there is one relative minimum at (0, 0).

(b) For a < 0, there is a relative maximum at (0, 1).

(c) For a < 0, there are two relative minima at .v = ±

(d) There are either 1 or 3 critical points. The above analysis
shows that there cannot be exactly two relative extrema.

7. f(x} = - + .r2

X

fix) = — T + 2j: = 0 => -^ = 2v
X- x-

2 \ X

/"W = | + 2

If c = 0,f{x) = x-'^ has a relative minimum, but no relative maximum.

If c > 0, X = 3/ - is a relative minimum, because/"! -V - J > 0.

c .

If c < 0, .V = ..^V r is a relative minimum too.

Answer: all c.

174 Chapters Applications of Differentiation

9. s./^'^-f^f-f'lf'-^^ = k.
(b - ay

Define F{x) = f{x) - f(a) - f'{a)ix - a) - kix - af.

F{a) = 0, F{b) = fib) - f(a) - f'{a){b -a)- kib - a)^ = 0

F is continuous on [a, b] and differentiable on (a, b).
There exists c,, a < c, < fc, satisfying F'(c,) = 0.

f W = fix) — f'ia) — 2kix - a) satisfies the hypothesis of RoUe's Theorem on [a, c,]:
F'ia) = 0, F'(ci) = 0.

There exists Cj, a < C2 < c, satisfying F"ic2) = 0.

Finally, F"ix) = fix) - Ik and F'Xcj) = 0 implies that

Ti,„o ; /(fe) - fia) - fia)ib - a) f'jc^ _^ffu\ ft \ j_ t'( \(u ^ _■_ ^ ft \(u \2
Thus, k = _ .^ = —7^ =^fib) = fia) +fia)ib - a) + -f (c,)!^ - a)^.

, , tan 4>i\ - 0.1 tan (/)) 10 tan </> - tan- 0
■ ^'^' ~ 0.1 + tan(/) ~ 1 + 10tan</)

^,, J, (1 + lOtan (^)(10sec2<^ - 2tan<^sec2<^) - (lOtan (^ - tan2(^)lOsec-0 „
^^"^^^ (1 + 10 tan 4.) = °

=> (1 + 10 tan (^)(10 sec- 0 - 2 tan (^ sec^ <^) = ( 1 0 tan <^ - tan- <^)10 sec^ 0

=* 10 sec- (/) - 2 tan (^ sec-^ <^ + 100 tan <^ sec- (^ - 20 tan^ </> sec- <^

= 100 tan (^ sec* </> - 10 tan^ <^ sec- <^

=> 10 - 2 tan <^= 10 tan- 0

=> 10 tan^ (^ + 2 tan (^ - 10 = 0

— 2 -I- /4 + 400
tan (^ = =^^^- = 0.90499, - 1.10499

Using the positive value, <f> = 0.7356, or 42.14°.

13. V = -2400irsine
v'= -2400iTCOse = 0

t* = — + 2/jTr, — -(- 2«ir, n an mteger

Problem Solving for Chapter 3 175

X y 4

15. The line has equation T + T = 1 or y = --x + 4.

Rectangle:

Area = A = xy = x[ --x + 4J = ~:;x- + 4x.

A '{x) = -|.r + 4 = 0=>|x = 4

Dimensions: r x 2 Calculus was helpful.

X V

Circle: The distance from the center (r, r) to the line T "*" J ~ ' ~ ^ must be r.

fn-

- 1

12

7r- 12

/l

1

5

12

|7r- 12

5r = |7r - 12| =^ r = 1 or r = 6.

Clearly, r =1.

X y
Semicircle: The center lies on the line T + T = ^ ^^ satisfies x = y = r.

Thusf + ^=l=>^r=l

r = —. No calculus necessary.

17. >> = (1 + j:')-'

^ (1 + x-y-

" 2(3a-^ - 1) . _ ^1 ^V3

/': 1 1 h-

3 3

The tangent line has greatest slope at ( — — , - I and least slope at . ,

V3 3

73 3

19. (a)

.V

0.1

0.2

0.3

0.4

0.5

1.0

sinj:

0.09983

0.19867

0.29552

0.38942

0.47943

0.84147

sinj: < .r

(b) Let f(x) = sin x. Then f'(x) - cos x and on [0, x] you have
by the Mean Value Theorem,

f,, . fix) -no) „

f(c) = ^_o ' 0 < c <.v

cos(c)

Hence,

X

sin.r

|cos(c)| < 1

|sinA:| < |jc|
sin.v < .V

CHAPTER 4
Integration

Section 4.1 Antiderivatives and Indefinite Integration 177

Section 4.2 Area 182

Section 4.3 Riemann Sums and Definite Integrals 188

Section 4.4 The Fundamental Theorem of Calculus 192

Section 4.5 Integration by Substitution 197

Section 4.6 Numerical Integration 204

Review Exercises 209

Problem Solving 214

CHAPTER 4
Integration

Section 4.1 Antiderivatives and Indefinite Integration

Solutions to Odd-Numbered Exercises

1. Ml ^A= Aj-x^-i ^r\^ _Q.-4 -9

M.^^^j^^^^"^^^ = -^'"

d n

3. ^1 -x3 - 4x + C] = X- - 4 = {x - 2)ix + 2)

5.f = 3r^
dt

7.$; = x3/2

y = fi + C

3' = \x?n + C

Check: -^^^
at

+ C] = 3/2

Check: £ Ir^/^ + C

Given

Rewrite

Inteerate

Simplify

9. 3/^aL^:

x"^ dx

_^4/3

4

11. -^dx
J xwx

x-^'^-dx

^-

~i*"

13./i<i.

- ip"^

^S)-

-i-

15. {x + 3)dx =

y + 3;t + C

17. (Zr - 3.r-)ciT: = x' -

Check:

dx

- + 3. + C

= ;^/2

= .v + 3

Check: —[;<:-- .r^ + C] = 2x - 3.v=

/<

19. I (x' + 2) (ir = -r'' + 2jr + C

Check: jijx* + 2.x + C| = .v^ + 2

/'

21. I (x3/2 + 2x + 1) a[r = t^v^''^ + ,.; + ^ + ^

Check: M^'' + x- + x + c\ = x^'- + Iv + 1

23. \l/^dx= j.r

jc^''' 3

Check: 4(|^-'^' + C) = .x^/s = l/^-

25. 3 ^^ = -^^^ ^^ = 3T + C =

Check: ^(--i^ + C=^
dx\ Ix^ j .r"

-^ + c

177

178 Chapter 4 Integration

27. {^^^j^dx = I (;c3/2 + x'/'- + x-'/2) dx = '^x?'^ + |jc3/2 + 2x'/2 + C = -^;c'/2(3x2 + 5x + 15) + C

15

Check: -f fl-^'' + 1^^' + 2x"' + c) = i^'^ + x'/^ + ;c-'/2 = ""' "^ ^^"^ '
dx\5 3 / y^

'■/

/<

29. U + l)(3x - 2) dlx = (3x2 + ;c - 2) ^^

= ^3 + -x2 - 2x + C
2

Check: 4-\^ + ~:^ - 2x + C] = 3x^ + x - 2

dx\

= ix+ l)(3x - 2)

31. J/vS^y = J;

//2 dy = -yin + C

Check:

dj2
dy

±yin + cj = //2 = y2^

33. Uf =

\dx = x + C

\

35. I (2 sin X + 3 cos x) dx = — 2 cos x + 3 sin x + C

Check: — (x + C) = 1

OK

Check: -7-(— 2 cos x + 3 sin x + C) = 2 sin x + 3 cos x
ox

/

37. (1 - CSC t cot i)dt = t -\- esc r + C

\

39. (sec^ e - sin 0) dS = tan e + cos e + C

Check: —it + esc r + C) = 1 - esc r cot r

Check: —(tan e + cos 6 + C) = sec^ 0 - sin 0
dv

1. (tan-y + \)dy = \.

41. (tan- y + \)dy = sec' >> dy = tan y + C

Check: —(tan y + C) = sec- y = tan- y + \
dy

43. /(x) = cos X

C = -2

45. /'(x) = 2

/(x) = 2x + C

f(x) = It + 2

/W = Tx

Answers will vary.

49. ^ = 2x - 1, (1, 1)
dx

-\

(Ix - \)dx = x^-x + C

1 = (1)2- (1) -t- C => C= 1
>> = x^ — X -1- 1

Answers will vary.

Section 4.] Antiderivatives and Indefinite Integration 179

51. V- = cos X, (0, 4)
dx

53. (a) Answers will vary.

^h

y = I cos X dx = s'mx + C

4 = smO + C=> C = 4

y = sin.ic + 4

(b)J = 4 -1,(4,2)
dx 2

y=--x+C

42
2 = — -4 + C
4

2 = C

55. fix) = 4x,f(0) = 6

f(x) = 4r ^ = 2i2 + C

y{0) = 6 = 2(0)2 + c => C = 6
/(;c) = 2;c2 + 6

57. h'{t) = 8f3 + 5, /!(]) = -4

;i(f) = (8r3 + 5)dt = 2r* + 5f + C

/!(1) =-4 = 2 + 5 + C^C = -11
/!(?) = 2f^ + 5f - 11

59. f"{x) = 2
/'(2) = 5
/(2) = 10

/W = I 2 dx = 2j: + Ci

/'(2) = 4 + C, =5=>C, = 1
/'(;t) = 2x + 1

fix)= \{2x+ l)dx = x- + x+ Cj

/(2) = 6 + Q = 10 =^ Q = 4
/(x) = jr + .r + 4

= /

63. (a) h(t) = {l.5t + 5)dt = 0.75r + 5t + C

h{0) = 0 + 0 + C=12=i.C=12
/!(;) = 0.75r + 5f + 12
(b) ;!(6) = 0.75(6)2 + 5(6) + 12 = 69 cm

61. f"(x) = x-3/2
/'(4) = 2
/(O) = 0

fix

\x~^/-dx = -2r-i/2 + C, = -- %r + C,

/'(4) = -- + C, = 2 ^ C, = 3

/'(x) = -^ + 3

/(;c)

(-2r-'/2 + 3) iv = -4.v"- + 3.r + C,

/(O) = 0 + 0 + C, = 0=* C, = 0
fix) = -4x^'- + 3.T = -4^^ + 3.V

180 Chapter 4 Integration

65. /(O) = - 4. Graph of/' is given.
(a)/'(4)«-1.0

(b) No. The slopes of the tangent lines are greater than 2
on [0, 2]. Therefore, /must increase more than 4 units
on [0, 4].

(c) No,/(5) < /(4) because /is decreasing on [4, 5].

(d) /is an maximum at x = 3.5 because /'(3. 5) = 0 and
the first derivative test.

(e) /is concave upward when/' is increasing on (-oo, 1)
and (5, oo)./is concave downward on (1, 5). Points
of inflection at j: = 1,5.

67. ait) = -32ft/sec2

v(f) = \ -32 dt = -32t+ Ci

v(0) = 60 = C,

s{t) = (-32f + 60)dt = - \6t- + 60f + C^

5(0) = 6 = C.

5(r) = -16r^ + 60f + 6 Position function

The ball reaches its maximim height when

v(t) = -32r + 60 = 0

32? = 60 ■ ' ■

15
t = — seconds

o ' - .

15

161 -^ I +601

(f

+ 6 = 62.26 feet

asy

71. a(f) = -9.8

v(r) = -9.8 ^f= -9.8<+ C, .
v(0) = Vq = C, ^ v(f) = -9.8r + vo
/(/) = (-9.8r + Vo) dt = -4.9t^ + v^t + C^
m = 5o = Q ^ fit) = -4.9f2 + v^t + So

(f) /"is a minimum at x = 3.
(g)

69. From Exercise 68, we have:

sit) = -16r2 + V(,f

s '(f) = - 32f + Vji = 0 when r = — = time to reach
maximum height.

it2)—''[f2r'it2)^'''

'°' + ^ = 550

64 32

35,200
187.617 ft/sec

73. From Exercise 71, /(f) = -4.9t^ + lOf + 2.

V (f) = -9.8f + 10 = 0 (Maximum height when v = 0.)
9.8f = 10
10

t =

9.8

/I

©=■

7.1m

75. a = -1.6

- v(f) = — 1.6 (if = — 1 .6f + Vq = — 1.6f, since the stone was dropped, Vq = 0.

sit) = (-1.6f)rff = -0.8f2 + 5o

j(20) = 0 => -0.8(20)2 + So = 0

5o = 320

Thus, the height of the cliff is 320 meters.
v(f)=-1.6f
v(20) = -32 m/sec

Section 4.1 Antiderivatives and Indefinite Integration

181

77. x{t) = r3 - 6f2 + 9f - 2 0 < r < 5

(a) vW = x'(t) = 3f2- 12r + 9

= 3(r= - 4f + 3) = 3(f - l)(f - 3)
a(t) = v'W = 6f- 12 = 6(r- 2)

(b) v(f) > 0 when 0 < r < 1 or 3 < r < 5.

(c) a(t) = 6(r - 2) = 0 when t = 2.
v(2) = 3(l)(-l) = -3

79. v{t) = -^ = r'/2

f > 0

xit)

= /.

(f) dt = 2f'/= + C

x{\) = 4 = 2(1) + C => C= 2
jc(r) = 2t'^' + 2 position function

a(r) = V '(r) = - r' ^^^ = TI72 acceleration

81. (a) v(0) = 25 km/hr = 25 • j^ = — m/sec

v(13) = 80km/hr = 80-|^ = ^m/sec

a(f) = (3 (constant acceleration)
v{t) =at+C

v(0) =
v(13) =

250
36

800
36

550
36

250

v(r) = «r + —

13a +

13a

250
36

550 275 , ,^^ , ,
'' = 468=l3i'='-''^'"/^^'^"

(b) ^W = aY + ^f Wo) = o)

275 (13)- 250

„ (lmi/hr)(5280 ft/mi) 22

**^- (3600sec/hr) " 15 "'^^'^

(a)

83. Truck: v{t) = 30

. s{t) = 30f (Let 5(0) = 0.)
Automobile: a{t) = 6

v(/) = 6f (Let v(0) = 0.)
5(f) = 3/2 (Let s(<S) = 0.)
At the point where the automobile overtakes the truck:
30f = 3r=
0 = 3r - 30r
0 = 3r(r - 10) when t = 10 sec.

(a) 410) = 3(10)- = 300 ft

(b) v(10) = 6(10) = 60 ft/sec = 41 mph

;

0

5

10

15

20

25

30

V, (ft/sec)

0

3.67

10.27

23.47

42.53

66

95.33

V,(ft/sec)

0

30.8

55.73

74.8

88

93.87

95.33

,c)5,(,)./v,(,).,.5Jp^-!if^^ + 0,3679,

= /

5,(f) = V,(f) dt

0.1208r3 6.7991/2

(b) V(f) = 0.1068/2 - 0.0416/ + 0.3679
VjCr) = -0.1208/2 4. 6.7991/ - 0.0707

0.0707/

3 2

[In both cases, the constant of integration is 0 because 5,(0) = Sjfi) = 0]
5,(30) = 953.5 feet
52(30) «= 1970.3 feet
The second car was gomg faster than the first until the end.

182 Chapter 4 Integration

87. a(t) = k
v{t) = kt

s(t) = ^P since v(0) = s(0) = 0.

At the time of lift-off, kt = 160 and {k/2)f = 0.7. Since {k/l)fi = 0.7,

t =

¥-V¥=«

\Ak = 1602

1602

1.4

^ 18,285.714 mi/hr2
= 7.45 ft/sec^.

89. True

91. True

93. False. For example, Ix-xdxi' Ixdx- I.

X dx because — + C ^t ( y + C, jfy + Cj

95. fix)

1, 0 < .r < 2
3x, 2 < .x: < 5

I .t + Ci, 0 < .r < 2

ll" "^ ''2'
/(I) = 3 =^ 1 + Ci = 3 =i> C, = 2
/ is continuous; Values must agree ?& x = 2:
4 = 6 + C2=>C2 = -2

fjc + 2, 0 < ;r < 2

fix) = 3a:2 ^ ^

— -2, 2<x<5

The left and right hand derivatives at .t = 2 do not agree. Hence / is not differentiable at x = 2.

Section 4.2 Area

5 5 5

1. ^(2/ + 1) = 2^/ +^1= 2(1 +2 + 3 + 4 + 5) + 5 = 35

i=\ i=\

3 Y '

-1 + 1 + 1+1 + 1 158

9 1

" ' ^ 2 ^ 5 ^ 10 + 17

8

'■2.

85

15.f2, = 2f,^2M=420
1=1 /=i L -i J

5. ^c = c + c + c + c = 4c

20 19

17. ;^(i-i)2=2r-
1=1 >=i

'19(2(

))(39)

-iiK-fn

L 6 J

2470

Section 4.2 Area 183

15

2

;=1

15

19. 2 '■(/- 1)^ = !:'■'- 21 '' + E

= 1 i=l

^ \5\\6Y _ .15(16)(31) 15(16)
4 6 2

= 14,400 - 2.480 + 120

= 12,040

21. sum seqUB 2 + 3, ;c, 1, 20, 1) = 2930 (n-S2)

2(r- + 3) = 20i2o±iM20LLi)^3,2o,

20
i=l

(20)(21)(41)

60 = 2930

23. 5 = [3 + 4 + f + 5](1) = f = 16.5
i = [1 + 3 + 4 + f](l) = f = 12.5

25. 5 = [3 + 3 + 5](1) = 11
.y = [2 + 2 + 3](1) = 7

- «^) = J^ - Vf (^ M) ^ 4^ = '^-^;-^^^ = «-™

'<^. - G)* VKa - VK^

3/l\ 1 + 72+^/3

4V4

« 0.518

M-M4—

i.l.i.i.i»o.«.

31. lim

8lV^(«+lFl 81.

~r : = ~r lim

/j''/ 4 J 4n-»oo

n"* + 2n3 + n-

>-7

33. lim

n— >oo

18VH" + 1)

18,.
— hm

n- + n

> = 9

35. ;^ii4^ = -L2(2< + i) = A

5(10) = ^ = 1.2

- 5(100) = 1.02
5(1000) = 1.002
5(10,000) = 1.0002

.n{n + 1)

2 1- n

n + 2

■-Sin)

37-1

^6k(k - 1) 6 ^

^^(k^-k)

n{n + 1)(2« + 1) n{n + 1)

2«- + 3n + 1 - 3« - 3

[2n- - 2] = 5(n)

5(10) = 1.98

5(100) = 1.9998

5(1000) = 1.999998

5(10,000) = 1.99999998

,„ ,• v^/16'\ ,. 16 ^. ,. 16/r7(« + D'
39. hm y -T = lim — V/ = lim — P-r — -\ = Urn
n-><x> l^^\ n'- 1 n->oo n- fr'^ n->Qo n-\ 2

m

8 lim 1 +- = 8

n^^ooV /I/

184 Chapter 4 Integration

41. lim > -t(i - 1)^ = lim ^ > i^ = lim -^\

n->co OL « J n-'ooL OV 1

43. lim y{\+ -\{-\ = 2 lim -[y 1 + i y il = 2 lim -

l/n(n+l)

= 2 lim [1+4:^

i-»oo L 2n^

211+2

45. (a)

(b) Ax

2-0 2

n n

Endpoints:

o<,g)<.g)<.^.<„-„g)<„g) = .

(c) Since >> = j: is increasing, /(m,) = /U,._ J on [;c,-_ ,, x,].
^(n) = J/U,_,)Ax

(2\ '

(d)/(W,)=/(j:,-)on[A:,-_„x,]

(<■ - 1)

*.=|/«-=,|/(!M,[.M!)

47. >> = -2x + 3 on [0, 1]. (Note: \x = ^-^^ = -

V n n

= 3

2 ^ . , 2(n + l)n „ 1

lE' = 3

2«2

= 2

Area = lim s{n) = 2

49. 3^ = ^2 + 2 on [0, 1]. (Note: Ax = -

-"Hm-t

+ 2

[il'

(e)

X

5

10

50

100

s{n)

1.6

1.8

1.96

1.98

S{n)

2,4

2.2

2.04

2.02

(f) lim 2

(<• - 1)

^^= iim^|;a-i)

«/ n^oo n^

.. 4r«(«+_i) ]

= hm — n

n->OG n\ 2 J

= lim r^(^l±il - i] = 2

/!->00 L " ''J

4 \«(n + 1)

= lim

= lim2i^±i) = 2

3 1

^ n(n + 1)(2« +1) ^ ,^ _ . , ,

+ 2 = -5^ 7^\ ^ + 2 = -2+- + — +2

6n^ 6\ n n-

H 1 — -^i

Area = lim S{n) = —
n-»oo 3

Section 4.2 Area 185

51. y = 16 - A-2 on [l, 3]. ( Note: Ax = -

«(.)^|/(. -!)© = ,!

9 "

1

r

\t>n

n

,

15

4r 4i'

16 - I 1 + —

4 n{n + l)(2;i + 1) 4 «(>; + 1)

«"

30 - -^(/! + l)(2n + 1) - -(/I + 1)
6/r /!

53. y = 64 - .x^ on [1, 4]. (Note: Ax

^(«) = |/ii+7

n /\n

64 - 1 +

3(

'' i= 1

63

27i-' 27r 9i

63n

27 n^n + lY 27 >;(>; + 1)(2k + 1) 9 n(n + 1)'

= 189-fi(.+ l)-fi,(n.l)(2..1)-?^i^
4n- 6/!- 2 /!

81 27 513

Area = lim s{n) = 189 - ^ - 27 - -^ = — - = 128.25

n ->ao 4 2 4

55. y = -t^ - x3 on [- 1. 1]. Note: A.^

l-(-l)

Again, T{n) is neither an upper nor a lower sum,

2i\/2
1 = 1

1=1
n

= 1

1+^

■1 +

7;\3'

, 4( 4/-

1 -- + —
n n~

, 6i 12/2 8,-3

■1 + - + —

n n- n-

= 2

i=l

20/ 26i: _ sii

/I H- n'

4^

20A . 32,^

16^

nl n

-Si-^I'^^I^^-^Z''

n

n ;

4, , 20 n[n +1) 32 n(" + lH2'! +1) 16 ^!-(/i + 1)-

-(") — - ■ — ~ + -J • 7 T ■ ;

n n- 2 If' 6 ;;* 4

4-10|l+iUi^2 + ^ + A

41 1 + - + -
;; II-

32 2

Area = lim T{ii) = 4- 10 + ^-4 = -

n— >cc 3 3

186 Chapter 4 Integration

2-0 2

57. /( v) = 3y, 0 < .V < 2 Note: ^y =

«.=i/w.,=|/(f)e)=|3(f)e

12 A. /12\ n(n + 1) 6(n + 1) ^ 6
= 711'= -I- -^— = = 6 +

Area = lim S{n) = lim ( 6 + - 1 = 6

59. /(y) =r, 0 <y < 3 (Note: Ay

3-0 3

«")vS/(!B = S(!)B=5i.-

nj \ni n

^ Tl n(n + \){2n + 1) ^ 9 l2n- + ^in + \\ ^27 9
" «3 ■ 6 " nA 2 / 2n 2n2

Area = lim S{n) = lim ( 9 + ^ + -^ ) = 9

n-»oo 'i-»oo V 2« 2«^/

61. gW

/ 3-12

= 4y2 - y3, 1 < V < 3. Notc: Ay = = -

V n n

sw = 2^ 1 +

2i\/2

= E

4 1 +

n /\n

2[
n

1 +

-)1-

2 " r, 4i API r, 6( 12r 8/3
-y41+ — + — r- -1+ — + —z- + —r
1 1^1 L n n^ ] L n n- n _

2^r3,M^4._8.1

10 n{n + 1) ^ 4 n{n + l)(2n +1) 8 n\n + IP]

44

Area = lim 5(n) = 6+ 10 + --4

n — ♦cc 3 3

63. /W = a:2 + 3, 0 < X < 2, « = 4
Let c,. = ' ' '■

65. /W = tan ;c, 0 < ;c < -. n = 4

Letc,

X,. + JC,._

Ax = -, c, = - c

2 /f<^3 4-^4 4

4' ^ 4

4

Area«X/('^,)Ax=2t,' + 3]^

2

69
8

^')-(^-)-(i-)-(^-

.77 77 377 577 777

Ax = — , Ci = — , C2 = TT, c, = T:r, c^

Area - 2/(0 ^=E(tanc,)h^

4

I

1=1 /=1

77/ 77

377

32

—I tan— + tan :7:r + tan XT + tan-:;:^! « 0.345

577

32

777

32

67. /(xj = v^ on [0, 4].

n

4

8

12

16

20

Approximate area

5.3838

5.3523

5.3439

5.3403

5.3384

(Exact value is 16/3)

Section 4.2 Area 187

69./(;c) = tan(^)on[l,3].

n

4

8

12

16

20

Approximate area

2.2223

2.2387

2.2418

2.2430

2.2435

71. We can use the line y = x bounded by x = a and x = b.
The sum of the areas of these inscribed rectangles is the
lower sum.

The sum of the areas of these circumscribed rectangles is the
upper sum.

a b

We can see that the rectangles do not contain all of the area in
the first graph and the rectangles in the second graph cover
more than the area of the region.

The exact value of the area lies between these two sums.

73. (a)

8--

6- ^^^

4- /^^ H

' i >

_ [ }... .If. .i

i ' ■ -

Lower sum:

5(4) = o + 4 + 55+6=15|=f« 15.333

(c)

(b)

Upper sum:

S(A) = 4 + 5| + 6 + 6f = 21I7 = ^ « 21.733

(d) In each case, A.x = A/n. The lower sum uses left end-
points, (i — l)(4//i). The upper sum uses right endpoints,
{i)(A/n). The Midpoint Rule uses midpoints, (/ - ;)(4/n).

Midpoint Rule:

M(4) = 2f + 45 + 5f + 6f = fij - 19.403

(e)

n

A

8

20

100

200

s{n)

15.333

17.368

18.459

18.995

19.06

S{n)

21.733

20.568

19.739

19.251

19.188

M{n)

19.403

19.201

19.137

19.125

19.125

(f) s(n) increases because the lower sum approaches the exact value as n increases. S(n) decreases because the upper sum
approaches the exact value as n increases. Because of the shape of the graph, the lower sum is always smaller than the
exact value, whereas the upper sum is always larger.

188 Chapter 4 Differentiation

75.

77. True. (Theorem 4.2 (2))

I ■! J J 4i >

b. A = 6 square units

79. f(x) = sin X

M

Let A, = area bounded by f(x) = sin x, the x-axis, x = 0 and x = tt/I. Let Aj = area of the rect-
angle bounded by .v = L y = 0, .«: = 0, and x = Tr/2. Thus, Aj = (7r/2)(l) = L570796.
In this program, the computer is generating N^ pairs of random points in the rectangle whose
area is represented by Aj. It is keeping track of how many of these points, N,, lie in the region
whose area is represented by Aj. Since the points are randomly generated, we assume that

N.,

A,

N.

■A,.

1

fW =

sm(xy^ (f'l)

0.75-

/^/, :,'

0.5-

/

/./,:/ "

025-

/

'

Z. JL

The larger Afj is the better the approximation to A,.

81. Suppose there are n rows in the figure. The stars on the left total 1 + 2 + •
n{n + I) stars in total, hence

+ «, as do the stars on the right. There are

2[1 + 2 + •
1 + 2 +

+ «] = n{n + 1)
• + n = \(ri){n + I).

83. (a) y = (-4.09 x \Q-'^)x' + 0.016.x2 _ 2.67x + 452.9

(c) Using the integration capability of a graphing utility,
you obtain

A « 76,897.5 ft^.

(b)

Section 4.3 Riemann Sums and Definite Integrals

-5 .->

1. f{x) = ^x.y = 0,a: = 0,x = 3, c, = ^

n^

3^ _ 3(i - 1)^ _ 3 ,

lim J;/(c,)Ax, = lim % JK\{2i " D
n ->oc /s'l 1 ->oc ,^| v n~ n^

= lim ^2(2'^-')

= lim

3V3

n{n + l)(2n + 1) n{n + 1)

]

-i'^j^K^^'^^-'-m

= 373

;i-o].

2v^ = 3.464

I '„ I.. ' ..,1 > '

3 3(2)' 3(>.-l)^^

Section 4.3 Riemann Sums and Definite Integrals 189

3. V = 6 on [4, 10]. (Note: Ax = — - — = -. ||A|| -* 0 as n ^ oo

n n

|/w--|/(-!)(M<M:?-'

f

6dx= lim 36 = 36

5. y = jc3on[-l, 1]. ( Note: A.T = —^ = - ||A||^0 as h-»oo

n^^, n^ .4', n^ .4',

, 6/ 12/- 8/3"

■1 + ~ + ^

n n- n'

+ 6(l+iU4(2+^ + AU4fl+^ + ^^ 2

n n^ n

r

)^ dx = lim - = 0

7. >' = x2 + 1 on [1,2]. Note: At

2-1 1

>0 as n—^oo

2/(c,)A^. = i:/ 1 +

/\/i

1 + -I + 1

n

, 2/ /2 ,

1 + — + — + 1

n n-

n-,-^, «^e'l \ "/ 6\ /! n-l 3 2/! 6??-

I

10 3 1

^^'^'^'^ = }T^[-j^2;t^^-)= ,

10

9. lim y (3c,. + 10) At, = (3;c + 10) dx
on the interval [— 1, S].

11. lim y 7c,- + 4 At, = Jx^ + 4
ti!-.o ,-e,

-dr

on the interval [0, 3].

13. -idx
Jo

-f

(4 - Lxl) dx

17. (4 - r-) dx

I

19. smxdx

Jo

v^ dy

23. Rectangle

A = W; = 3(4)

Jo

Adx = 12

^^vsxx -.N sN ^^ Rectangle

^^f^

190 Chapter 4 Integration

25. Triangle

A = ^bh = ^4)(4)

-i

xdx=%

27. Trapezoid

A =

b, + b. /5 + 9

r

^h

{2x + 5)dx= 14

^4M

''yj&i Trapezoid

^

^ — h

1 2 3

29. Triangle

A = ^M = ^(2)(1)

/:

(1 - \x\)dx= 1

31. Semicircle

A = ^■TTr^ = |77<3)2

£

V9^^A = ^

4- - Semicircle

r4 ri /•4

In Exercises 33-39, x^dx = 60, Lcrfjc = 6, dx = 2

jfiix = —6

33. L<:dc= -

37. U - 8) iic = ;c ate - 8 d:^ = 6

8(2) = - 10

41. (a) /U) dx = fix) dx+ /(;c) etc = 10 + 3 = 13

(b) I /(;c)dx= -j /(;c)<ic= -10 •

(c) \f{x)dx = Q

(d) 3/(x) otc = 3 /(jc) (it = 3(10) = 30
Jo Jo

35,

39.

.[4.^ = 4/;

xdx = A{6) = 24

I (ix3 - 3x + 2J^ = I x^dc - 3 xdr + 2 dx

= -(60) - 3(6) + 2(2) = 16

43. (a) I [fix) + gix)'\ dx= \fix)dx+ \ gix) dx
= 10 + (-2) = 8
(b) J [gix) - fix)] dx=\ gix) dx - yix) dx

= -2 - 10 = -12

J Igix) dx = 2\

(c) Igix) dx = 2\ gix)dx=^ 2(-2) = -4

(d)

I 3/(x) dx^sl fix)dx = 3(10) = 30

45. (a) Quarter circle below X-axis: — jitt^ = —-^tKX)^ = —tt

(b) Triangle: ^_bh = 5(4)(2) = 4

(c) Triangle + Semicircle below x-axis: -2(2)(1) - 5^2)^ = -(1 + 27r)

(d) Sum of parts (b) and (c): 4 - (1 + 27r) = 3 - 27r

(e) Sum of absolute values of (b) and (c): 4 + (1 + 27r) = 5 + 2ir

(f) Answer to (d) plus 2(10) = 20: (3 - 2ir) + 20 = 23 - 2ir

47, The left endpoint approximation will be greater than the
actual area: >

49. Because the curve is concave upward, the midpoint
approximation will be less than the actual area: <

Section 4.3 Riemann Sums and Definite Integrals 191

51. /W

x-4

is not integrable on the interval [3, 5] and /has a
discontinuity at ;c = 4.

^— :^ — ^'

a. A = 5 square units

55. y

r-t-T'

Jo

d. I IsmiTxdx'- 2^1)(2) « 1

I

57. xjl - xdx

n

4

8

12

16

20

L{n)

3.6830

3.9956

4.0707

4.1016

4.1177

M(n)

4.3082

4.2076

4.1838

4.1740

4.1690

R{n)

3.6830

3.9956

4.0707

4.1016

4.1177

59.

Jo

sin^ X dx

n

4

8

12

16

20

L(n)

0.5890

0.6872

0.7199

0.7363

0.7461

M{n)

0.7854

0.7854

0.7854

0.7854

0.7854

R{n)

0.9817

0.8836

0.8508

0.8345

0.8247

61. True

63. True

67. fix) = ,v' + 3,T, [0, 8]

j^o = 0, ,i:, = 1, X2 = 3, -Xj = 7, .r4 = 8
Ax, = 1, Ar, = 2. Axj = 4, zLv4 = 1
c, = 1, Cj = 2, Cj = 5, c^ = 8

X/(c,) A.t = /(I) A.V, + /(2) Axj + /(5) ^x, + /(8) !^x^
1 = 1

= (4)(1) + (10)(2) + (40)(4) + (88)(1) = 272

65. False

\ (-x)±x =
Jo

192 Chapter 4 Differentiation

69. f(x)

1, jc is rational
0, X is irrational

is not integrable on the interval [0, l]. As ||A|| — > 0, /(c,) = 1 or/(c,) = 0 in each subinterval since there
are an infinite number of both rational and irrational numbers in any interval, no matter how small.

71. Let/(x) = x~,0 < X < 1, and Ajc, = 1/n. The appropriate Riemann Sum is

n n / : \ 2 1 in

1 r,-, r,^ 07 -,1 ,• 1 n{2n + l)(n + 1)

n->co rj' n->oc n' 6

,. 2«2 + 3w + l ,. /I 1 1

= hm T-^ = hm {- + — + —2

)4

Section 4.4 The Fundamental Theorem of Calculus

1. fix) =

3. fix) = xjx~ + 1

xVjc^ + 1 cic = 0

J

1-0=1

f,

7. U - 2) (ic =

>(i-)-(4-

1 \ -

11. ilt- \Ydt= (4f--4f
Jo Jo

Ix

« /I

+ 1) rf/

rr^ - 2f2 + f

1°
3

4-2+1=1
0 3 3

" '-l-aUM-D-l

"■/>'

2 rf/ =

l-"-^ir(l

,3/2

+ 2 I = -4

' = if 1 - 2
0 3\2 3

J_
18

21. I (r'/3-f2/3)rf,=

1,4/3 _ lf5/3

4 5

]>»

|2j:- 3| A:= (3 - 2x) d:x + (:

Jo Jo J3/2

3 3\ _27

4 5) 20

23. |2j: - 3| (ir = I (3 - 2x) lic + (2j; - 3) a!^; split up the integral at the zero x =

= [

3x- x^

3/2

_3.];^.f?-2)-0 + (9-9)-f^-?U2f|-^Ul

3/2 \2 4

4 2/ \2 4/ 2

Section 4.4 The Fundamental Theorem of Calculus 193

25. I |.r2 - 4| <ir = I (4 - x'-) dx + I (.t^ - 4) dx

0

-I

f-1

-|8-;U(9-12)- ^3

23
3

Jo

27. (1 + sinx}dx =

X — cos X

(tT + 1) - (0 - 1) = 2 + TT

29.

-77/6

sec- .t dx =

.-77/6

tan.x

77/6
-,7/6

73 / v^\ 273

r'7/3 r T^/^

31.

4 sec 6 tan 0

. -77/3

de =

4 sec e

= 4(2) - 4(2) = 0

-V3

10.

Jo

33. 10.000(r - 6)di = 10,000

- 6t

5135,000 35. A = (.v - x-) dx

' ^1
0 6

37. A = (3 - .v)7^a[t = (3.t'/2 - .rV2) j^. =
Jo Jo

2^-3/2 _ 2,.5/2

-(10 - 2v)

1273

I

77/2

39. A = I cos X dx =

sin.x

77/2

41. Since y > 0 on [0. 2],

A = {3x- + 1) dx =

= 8 + 2 = 10.

43. Since y > 0 on [0, 2],

A = (.t^ + x) dx
Jo

x" x'-

4^2

4 + 2 = 6.

45. (x - 27^)rf.v =

X^ 4^3/2-

/(c)(2 - 0) =
c - 27c =

.2 3

6 - 8^2
3

3 - 4.-2

872

c-2V^+ 1 = - .^" - + 1

{Vc-l

,2 6 - 4^'2

./^-1 =

/6 - 4x 2

*V^]

c« 0.4380 ore = 1.7908

194 Chapter 4 Integration

Tr/4

47. I 2 sec^ xdx =

J—TTj

■/4

fic]

J-7r/4

2tanx = 2(1) - 2(-l) = 4

f)]

2 sec- c = —

2 4

sec c = ±-

c = ±arcsec(— ^

= ±arccos -— « ±0.4817

49.

1

2 - (-2)
Average value

£'"-"

)& =

4;t - -j:^

273

4 — jc- = r when jc- = 4--orx = + ^
3 3 3

+ 1.155.

sin x ate

Average value = —

TT

sinx = —

77

= 1

0 77

(-¥•!) 3 (^-f)

53. If/ is continuous on [a, b] and F'U) = f{x) on [a, fc],

(0.690, 1)

then fix) dx = F(b) - F{a).

x« 0.690, 2.451

Jo

55. I f{x)dx = - (area of region A) = -1.5

Jo

/•6 (-6

59. I [2 + /(jc)] dx= 2dx+ \ f(x)dx
= 12 + 3.5 = 15.5

61. (a) F{x) = ksec-^x
F(0) = A: = 500
Fix) = 500sec2x

57.

f \fix)\ dx = -Ifix) dx + Ifix) dx =
Jo Jo Jl

(b)

1^/3 -oJo

500 sec- xdx =

1500r l'^/^

1500,

[tanx]^^'

(v^-O)

77

= 826.99 newtons
~ 827 newtons

63.

-oJo

(0.1729r + 0. 1552^2 - 0.0374^3) dt '^ ^

0.08645r2 + 0.05073?^ - 0.00935?^

0.5318 liter

Section 4.4 The Fundamental Theorem of Calculus 195

65. (a)

The area above the j:-axis equals the area below the
jf-axis. Thus, the average value is zero.

The average value of S appears to be g.

67. (a) V = -8.61 x 10"^ + 0.0782^ - 0.208r + 0.0952
(b) ^

(c)

69. F(x)

Jo

-8.61 X lO-V 0.0782r3 0.208^2

+ 0.0952f

5r

0 2

F(2)=--5(2)= -8

f(5)=^-5(5)=-^

F(8)=y-5(8)=-8

73. F{.x) = I

cos ddd = sin 6

sin X - sin 1

F{2) = sin 2 - sin 1 = 0.0678
F(5) = sin 5 - sin 1 -= - 1.8004
f(8) = sin 8 - sinl =0.1479

= 2476 meters

71.FW = 0^v = £lOv-.V=^];

i°+ 10=10 1-1

X \ X

F{2) = 10(^-j = 5

f(5) = 10^-j = 8

F(8) = 101

7\ 35

75. (a) \ {t + 2)dt= [l

+ 2t

= —X- + Zx

77. (a)

3

-,4/3

4

' = |(.^4/3 _ 16) = 3 ,/3 _ j2

8 4 4

(b) 4f^/3 - 12

= rl/3 = 3/

79. (a) sec^ r A = tan r

J.r/4 L Ax'A

tan .t — 1

(b) — [tan.t - IJ = sec-.r

81. F(x) =1 (f- - 2t) dt
F'(x) = x^ -2x

83.

. F{x) = Vr» + 1 df
F'(.v) = Vx^+ 1

= 1'

85. FU) = I t cos tdt

F'{x) = .TCOSX

196 Chapter 4 Integration

87. Fix) = {4t + \)dt

I

= [2(.r + 2)2 + (x + 2)] - [2x2 + x]
= 8.V + 10
F'(x) = 8

rsi

89. f U) =

Jo

Jtdt

23/2

3

= f(sinx)V2
0 3

F'W = (sin j:)'/2cosx = cos-tv/sinx
Alternate solution

Fix)

Jo

^tdt

F'ix) = Vsin JT -p (sin x) = Vsin j:(cos x)
ax

Jo

93. g{x)= \fit)dt

giO) = 0, gH) = ^, g(2) - 1, gi3) » ^, g(4) = 0

Alternate solution:

Fix) = (4f+ l)rff

ro rx

= (4r + 1) rff +

(4f+ 1)A +
0 Jo

(4r + \)dt+ i4t + 1) rff

(4f+ 1)A + (4f + 1)A

I Jo

F'ix) = -i4x + 1) + 4(x + 2) + 1 = 8

Jo

91. Fix) = I sin f 2 rff
Jo

F'ix) = sin(x3)2 • 3^2 = 3x^ sin x^

95. (a) Cix) = 50001 25 + 3 | f''"' dt

[4

3(25 . 3/;,

= 5000 25 + 3

Ij5/4

= 5000(25 + y^'") = 1000(125 + 12x5/")

(b) C(l) = 1000(125 + 12(1)) = $137,000
C(5) = 1000(125 + 12(5)5/4) ^ $214,721
C(10) = 1000(125 + 12(10)5/") ^ $338,394

g has a relative maximum at x = 2.

97. True

99. False

; I x-^dx = J x-^dx + I x-2atc

Each of these integrals is infinite. /(x) = x ^
has a nonremovable discontinuity at x = 0.

By the Second Fundamental Theorem of Calculus, we have
1

f'ix) =

lU '

(l/x)2 + 1\ xV X2 + 1

' 4-^^ = 0.

1 + X2 x2 + 1

Since fix) = 0,/(x) must be constant.

Section 4.5 Integration by Substitution 197

103. x{t) = t^ - 6t^ + 9t - 2
x'{t) = 3f2 - I2t + 9
= 3(f2 - 4/ + 3)
= 3(r - 3)(t - 1)

Total distance = |j:'(')|rf'
Jo

= \'3\{t-3){t-l)\dt
Jo

= 3 J (i- - 4t + 3)dt - 3 (f2 - 4r + 3)A + 3 (t^ - 4t + 3)dr

= 4 + 4 + 20
= 28 units

105. Total distance = \x'{t)\dt
= I Ht)\dt

= 2(2 - 1) = 2 units

Section 4.5 Integration by Substitution

f{g{x))g'{x)dx

u =g{x) du = g'{x)dx

1. I {5x^ + \Y{\Qx) dx 5x- + \

x'+ 1

I tan- X sec- x

dx

tanjc

XQx dx

2x dx

sec^ X dx

/'

7. 1(1 +2.t)-'2d:t= ^' \^^' + C

Check: ^

(1

^ + C] = 2(1 +

ixy

». I (9 - x2)'/2(-lv) dx = ^^ , J'^' + c = ^(9 - .v-)3/- + C

Check: -^[|(9 - .r^)^/: + c
dxi3

3/2 ■ ^ 3
2 3

3 2

(9_^)i/:(_2^)= y9-.r=(-lT)

198 Chapter 4 Integration

11. f.r3(.x^ + 3P<ic = ^f(

.r3(.x^ + 3Ydx^^\(x' + 3)2(4x3) ir = - ^"^ t ^^' + C = ^"^ .t ^^' + C

4 3

12

Check:

£[^^ + c]=^^^(4x3) = (^ + 3)V)

13. |.rV-l)^A = |J(.

.r2(x3 - 1)" A = :^ I (x^ - Ij-'CSx^) dx = ^

w-m^c = ^^^^^^c

-]

15

Check:

^

Cv^ - 1)
15

+ C

5(.r3 - 1)^(3x2)
15

= x2(x3 - I)'*

rVFT2 dt = ^ I (r^ + lY'^lt) dt = \^^^^ + C= ^^±^ + c

15. I rVFT2 * = I r

"(r2 + 2)3^

Check:

dt

+ C

3/2(f2 + 2)'/'(2f)

(?2 + 2)'/2t

17. [5x(l-x2)'/3d^c=-|J(l-x^;
Check: 4[-^l - x^)''/^ + c

5 (1 _ y2W3 15

)>/3(-2x) ^ = -^ • ^ ^^3^ + C = -yd - x2)V3 + c
15 4,

--^ • ^1 - x2)i/3(-2x) = 5x(l - x2)i/3 = 5xyr^

O 3

19- J(]-f]^^ = -||(1 - x')-\-2x)dx= -

1 (1 - x^

+ C^

4(1 - x-:

+ C

Check:

^

.4(1 - x^Y

^+ c

= i(-2)(l-xr'(-2x) = ^^-:^

-k

1

+ ^3)2<i^ = jjd + ^')-'(3^-) 0^^ =

(1 + x^.

-1

+ c

1

3(1 + x3)

+ C

Check:

dx

1

L 3(1 + x3)

+ C

-i(-i)(i+x3)-w) = ^Y:^

23.

/^%-=-i/<'-

,<i,--jJ(l-x")-'»(-2«)i<.-iiL_i!

2-1 1/2

+ C= - Vl - x2 + C

Check: —[-(1 - x^)'/^ + C] = --(1 - x2)-'/2(-2x) = -y^==

dx 2 yn^

25.

[1 + (lA)?

[1 + iUt)V

4

Check:

dr

+ C

+ C

l.-J. . IV/ 1\ 1/. , 03

4(4)i+7jl-^j = ?l'^7

(2x)

^^•/;fc^=iK"^^4[

Check: ^V2i + C] = ^(2x)-'/2(2)

1/2-

1/2 J

+ C= V2x + C

dj:'-

vs

Section 4.5 Integration by Substitution 199

29.

: ^W^r- + 2;c3/2 + 14.'/= + cl = :n±^±I

+ 35) + C

Check;

31.

j r^ff - 7) * = f ('' - 2;) * = |f4 - r- + C

Check:

dt

-t" - t^ + C

= f3 - 2f = f-U - y

33. [(9 - y)^dy = [(Qy'/^ - y/z) rfy = 9^|y3/2

|v5/2 + c = |v3/2(15 - >') + C

Check: — 7^/^(15 - y) + c

6^/2 _ 1^/2 + c

^^•^=/h^7ife]

tic

v/l6~- x^
= 4lxdx- lliie- x-)-^'H-2x)dx

-(f)-

(16 - x2)i/2

1/2

+ C

2x2 _ 4716 - X- + C

= 9v'/2 - y/2 = (9 - y)v^
^^•^ = J(.v2 + lx-3P^

- + 2.x - 3)-2(2x + 2) dx

{x- + 2.x - 3)-

2(.x2 + 2x - 3)

+ C

c

39. (a)

dv

(b) ^ = .xV4^^, (2, 2)
dx

-/■

^/'

xv^4^^dx = -^ (4 - .x2)i''-(-2xdx)

= -i • ^(4 - .r2)V2 + c = -{(4 - x-)^'- + c

(2, 2): 2 = -^(4 - l-y- + C => C = 2

V = -^(4 - .x2)3/2 + 2

41. J 77 si

sin TTxdx = — cos ttx + C

45.j^_coslde=-jcos\[-j,)de

-sin- + C

6

43.

I sin Iv ix = - I

(sin 2jc)(2;t) dx = -- cos Ix + C

200 Chapter 4 Integration

47. \smlxcos.2xdx-=^\ (sin 2x)(2 cos 2x) dx = - i^iH^L + C = - sin^ 2x + C OR

\^^^ + C, = -\cos^lx + C, OR

sin 2x cos 2xdx = rr I 2 sin 2x cos Ixdx = ir I sin 4j: A — - cos 4j: + C-,

49. I tan"* .v sec- xdx = ^^ + C = - tan^jc + C

. I sin 2x cos 2j; A; = — I (si

I sin 2x cos 2x dx = — - I (cos 2j:)(-2 sin 2x) dx = — -
/si.2.cos2,A = i|2™2xc.s2.<*:.i|:

I

I CSC^ X I

51. — —dx = - (cot j:)"3(-csc2.r) otc

= -i£5L^ + c = —^ + C = )-tm^x + C = i(sec2;c - 1) + C = |sec2;c + C,
— 2 2 cot^ X 2 2 2

53. cot' xdx = \ (esc- j: - 1) A = - cot .r - jc + C 55. /(x) = cos ^ (it = 2 sin ^ + C

Since /(O) = 3 = 2 sin 0 + C, C = 3. Thus,
/(;c) = 2sin| + 3.

51. u = X + 2, X = u ~ 2,dx = du

\xjx + 2dx = \(u - 2)^du

/"

3/2 _ 2„l/2) J„

= |«V2 _ l„3/2 + c

= ^(3« - 10) + C

= ^x + 2)3/2[3(x + 2) - 10] + C

= ^jc + 2)3/2(3;c - 4) + C

59. u = 1 — X, X = I — u, dx = — du

I du

jjcVl -Xtic = - 1(1 - Mfv^
= - J(«>/2 -

2m3/2 + „5/2) ^j,

= -(|m3/2-1«V2 + |„7/2) + c
2«3/2^

105

■(35 - 42m + 15m2) + C

-r^(\ - xy/^[35 - 42(1 - x) + 15(1 - x)^] + C
-j|^(l - x)3/2(15a:- + 12x + 8) + C

Section 4.5 Integration by Substitution 201

61. M = 2x- l,x = r{« + l),dx = -zdu

J J2x - 1 J V" 2

!<3/2 + 2ttl/2 _ 3„-l/2) ^„

= |(|m5/2 + |„3/2 _ 6„1/2J + c

1/2

= ^(3m= + 10« - 45) + C
60

V2x- 1

60

{?>(2x - 1)' + \Q{2x - 1) - 45] + C

-^Jlx - \{\2x- + at - 52) + C
60

^V2jc - 1(3^2 + 2;^ _ 13) + c

63. « = a: + 1, .X = M — I. dx = du

/

: oLv = ;=^ du

(X + 1) - V.V + 1 J M - V«

(y;i + 1)(7^- 1)

7ii(v«- l)

(1 +«-i/2)rfu

rf«

= -(« + 2«'/2) + C
= -M - 2V« + C

= -(.r + 1) - 2V-r + 1 + C
= -.r - iVxTl - 1 + C

= -(.v + iJTTl) + c,

where Cj = — 1 + C

65. Let u = x~ + I, du = 2v dx.

J x(.r2 + 1)3 dr = ^1 (;c2 + m2x) dx

U- + 1)^

67. Let H = .v' + 1, rf" = 3.r- dx

Ix-J^P^Tldx = 2 • I Cr' + l)»/-(3.r2)a^t

Ls 3/2 J,
= ^[27-2./2]=12-§V2

202 Chapter 4 Integration

69. Let M = 2jc + 1, Jm = 2 dx.

r4

Jo ji^n 2jo

^ dx = ]-\ {lx+ \)-"\2)dx =

Jl.

;x + i[ =V9-VT = 2

71. Let tt = 1 + v^, du = — -p dx.

Ijx

73. H = 2 — j:, j: = 2 — M, (ic= —du

Whenx = \,u= \. Whenx = 2, w = 0.

1

2v^/^=rTT7^Ji-~2+'-2

['(;C - 1)72"^ A = [" - [(2 - m) - iJv^dM = f («3/2 - „l/2) ^„ = |,,5/2 _ |„3/2l° = _ r| _ |1 =

_4_
15

I

r/2

75. I cosl-jf)(ic

3 . 1
— sin -X
2 V3

V2

3/73\ 3v^

77. u=x+\,x = u— \,dx = du

When .t = 0, M = \. When x = 7, m = 8.

Area

^M

xi/x + ldx= {u - \)l/u

= I {u"'^ - m'/^) du

-uy^ - -u^l^

384

12

3 _ 3\ 1209
7 4/ ~ 28

79. A = I (2 sin jc + sin 2;c) ate = - 1 2 cos * + r cos 2a;

'. A = I (2 sin jc + sin It) ate = — 2

^£:Mf)--£>(f)©-[--(c:-(^-')

81. Area

83.

Jo

Jc , , ,,, 10

, dx^ 3.333 = —

Jlx + 1 3

r

85. xjx - 3dx'= 28.8

144

f

Jo

87. I |0 + cos^lrfe = 7.377

r^

J

89. \{2x- \y-dx = ^\{2x- \yidx = h2x - 1)^ + C, = |x3 - 2^:^ + .r - ^ + C,

». j(2x- \r~dx = ]A(

\{2x- \)-dx = (4;c--4x+ 1)

They differ by a constant: Cj = C,

i& = —x^ — 2jc + X + C2

1
6'

Section 4.5 Integration by Substitution 203

91. fix) = x\x^ + 1) is even.

x\x^ + 1) ir = 2 [x" + x'-)dx = 2

5 "^ 3

32 8
T + 3

272
15

93. fix) = x{x'^ + 1)3 is odd.

xix- + 1)3 iv = 0

Jo

95. x'±x
Jo

(a)
(c)

the function x^ is an even function.

(b) I x2iT = 2j>dLr = y

(d)

3^:^ ^- = 3 x'-

dx= 8

97.

U' + 6x- -2x-2)dx= ix^ - 2x) dx + (6.r- - 3) rf.t = 0 + 2 i6x- -

3)dx = 2

2x' - 3.x

232

99. Answers will vary. See "Guidelines for Making a Change of Variables" on page 292.

101. fix) = xix- + 1)2 is odd. Hence

, j^xix^ + D-

dx = 0.

dV
103. ^

dt it + 1)2

Vit

( k , k ^

V(0) = -A: + C = 500,000
V'(l) = -i't + C = 400,000

Solving this system yields A' = —200.000 and
C = 300,000. Thus,

V(r) = ^ + 300.000.

When t = 4, V'(4) = $340,000.

105.

b - a

f

74.50 + 43.75 sin ^
6

dt =

b - a

^^ ^„ 262.5 tTt
74.50/ cos — -

77 6

(a)-

(b)

,, ,„ 262.5 Trt
74.50/ cos —

77 6

262.5 T7T
74.50/ cos —

77 6

3 1/ ''62 5 \

= - 223.5 + - — - I = 102.352 thousand units

0 3\ 77

^'447 + ^^^ - 223.5 I = 102.352 thousand units
77

'''T2

262.5 777

74.50/ cos —

77 6

'- 1 / 262.5 262. 5\

= — 894 + 1 = 74.5 thousand units

0 12\ 77 77

204 Chapter 4 Integration

b -
(a)

(b)

(c)

-S:

1 c 1 r 1 1 1*

107. I [2 sm(607rf) + cos(1207r;) ] dt = -ttt- cos(60iTf) + -— — sin(120Trt)

b — a\_ jOtt 12U TT

^^

1

(1/60) - 0
1

-^cos(60..) + Tisi„(12O.0]7 = 60[(^ + o) - [~^^] = ^ » 1.273 amps

(1/240) - 0

1 1 i'/^'"' r

cos(6077-f) + -— - sin(120-7rf) = 240

12077 Jo L

3077

' + '

30V2ir 12077/ V 3077,

-(5 - 272) ^ 1.382 amps

1

(1/30) - OL 3077

-^7— cos(6077t) +

I2O77

sin(12077.)];^° - '{[£) - (-i)] - 0

I amps

109. False

|(2x+ \f-dx = ]A{2x+ \f2dx = hlx+ If + C

111. True

rio rio rio

(ax' + bx- + ex + d) dx = \ {ax^ + ex) dc + {bx^ + d) dx = Q + 1
J-10 J-10 J-10

Odd Even

J-10
(to- + d)dx
0

113. True

4 sin a: cos xdx = l\%\n2xdx = — cos 2jc + C

4- sin a: cos j: lic = 2 [ s

115. Let « = j: + /i, then du = <ic. When j: = a, « = a + ft. When x = b, u = b + h. Thus,

J-fc rb + h rb + h

f(x + h)dx=- f{u) du = fix) dx.

a Ja + h Ja + h

Section 4.6 Numerical Integration

;.6667

I. Exact: JJ,. ^ = [i^]^ | » 2.,

Trapezoidal: j x^ ^ - ^[o + 2^)' + 2(1)= + ll^J + (2)A = ~- = 2.7500
Simpson's: J jc^otc - ifo + 4^^)^ + 2(1)^ + 4(|j' + (2)^1 = | - 2.6667

3. Exact:
Trapezoidal
Simpson's

4.000

+ 2(^)\ 2(1)3 + 2(1)' + (2)3

17

4.2500

x^dx^

0 + 4(|y + 2(1)3 + 4(1]' + (2)3

^ = 4.0000

Section 4.6 Numerical Integration 205

5. Exact: \ }? dx = \KA = 4.0000

Trapezoidal: I x' oLr ~ -

o-iy-(!r-er-»)-er-(!r-er^

Si™p».V [^ * . ^[O + 4(i)' . 2g)' . 4g)' . 2(,). + 4(5)' + 2(f)' + 4(^)' + 8

4.0625

4.0000

7. Exact:

/:

-Jxdx

:,j/2

^18_ 1^ = ^.12.6667
4 3 3

Trapezoidal: I ^dx =
Simpson's: I Vx dx

,37 ^ /2 ^ /47 ^ /26 ^ /57 ^ /31 ^ /67 ,
4 V8 \'4 VS V4

- 12.6640

2 + 4^/? + V2T + 4,/^ + V26 + 4^/" + VJT + 4^/? + 3

= 12.6667

9. Exact:

1U +

;(ix

X + 1

- Ill

= -T + :;■ = 7 = 0.1667
1 3 2 6

Trapezoidal

Simpson's:

i

(x + 1

' ^«^

' . 2f.....l-^U 2( '

4 ^((5/4) + \r-j A((3/2) + l)-y n((7/4)

•) + 1)-] ^ 9J

KMf-l^f^^--

J, U + 1)=

ix

^.41

1

+ 2

1

4 "V((5/4) + If) -V((3/2) + 1)V \((7/4) + 1)^

+ 41

1

p)4]

^i^lMi^i^f^i)--

11. Trapezoidal: Vl + a^ dx - ^[l + 2^1 + (1/8) + ijl + 2j\ + (27/8) + 3] = 3.283

Jo 4

f

Simpson's: | Vl + x" dx « -[1 + 4Vl + (1/8) + ijl + AJ\ + (27/8) + 3] = 3.240
Graphing utility: 3.241

13. JxJ\ - xdx= Jxi\ - x) dx
Jo Jo

Trapezoidal

Jo

x) dx ~ —

I

Simpson's: I ^x(l — x) dx ~

0+4^

■-!

i._i„, ,.

vf4)-VlH)

« 0.342

= 0.372

Graphing utility: 0.393

206 Chapter 4 Integration

15. Trapezoidal:

/:

t/2 J'ttFI

cos(x^) dx = • — - —

== 0.957

cos 0 + 2 cosi " . I +2 cos

^V2V , ^ ..Y3VV2V

+ 2 cosI

+ cos .. / —

r

Simpson's: | cos(a:^) dx = — rr —

== 0.978

cos 0 + 4 cosI " / 1 + 2cos( r | + 4 cost V—\ + cosI

Graphing utility: 0.977

\:

Yl. Trapezoidal: sin jc^ ^ ^ -sTrCsinll) + 2 sin(1.025)2 + 2 sin(1.05)2 + 2 sin(1.075P + sin(l.l)2] -= 0.089

80

r

Simpson's: | ^mx^dx'^ Tio'-^'"^^^ "^ 4sin(1.025)2 + 2 sin(1.05)2 + 4sin(1.075)2 + sin(l.l)2] == 0.089

Graphing utility: 0.089

19. Trapezoidal:

/:

r/4

X tan X dx ■

32

0 + 21

16

^^"(5)^2(7!)^

^U2f^ltani

16

'16

f)n]--

194

Simpson's: \^\ tanxdx^ ^[o + 4(^) tan(:^) + 2(ff ) tan(ff ) + 4(^] tan(ff ) + j] » 0.186

Graphing utility: 0.186

21. (a)

The Trapezoidal Rule overestimates the area if the graph
of the integrand is concave up.

23. fix) = x^

fix) = 3x^

fix) = 6x

fix) = 6

f'Kx) = 0

(2 - 0)3
(a) Trapezoidal: Error < ,^. (12) = 0.5 since

/"W is maximum in [0, 2] when jc = 2.
(2 - 0)^

(b) Simpson's: Error

/^W = 0.

180(4")

{0) = 0 since

25./"(x) = ^in[l,3].

(a) \f"{x)\ is maximum whence = 1 and |/"(1)| = 2.

2'
Trapezoidal: Error < 7^(2) < 0.00001, n^ > 133,333.33, n > 365.15; let« = 366.

/W'(;c)=^in[l,3]

(b) \p'*>{x)\ is maximum when jc = 1 and [/"'(l)! = 24.

2'
Simpson's: Error < 75777(24) < 0.00001, «■* > 426,666.67, « > 25.56; letn = 26.

Section 4.6 Numerical Integration IQTI

27. f(x) = ym

(a) f"(x)

1

4(1 + x)V2

in [0, 2].

\f"{x)\ is maximum when.t = 0 and |/'(0)| = -.

Trapezoidal; Error < Ty^lj) < 0.00001, n- > 16,666.67, « > 129.10; let « = 130.

(b) /(*>W =

-15

16(1 + xy/^

in [0, 2]

|/('*>(jc)| is maximum when a: = 0 and \f'^\Qi)\ =

16'

Simpson's: Error < 7^77(77 I < 0.00001,/?'' > 16,666.67, n > 11.36; let « = 12.
I0O/2 \16

29. fix) = tan(jc2)

(a) f"(x) = 2 stc\x-)[\ + Ax- tan(x2)] in [0, 1].

\f"(x)\ is maximum when .r = 1 and |/"(1)| = 49.5305.

Trapezoidal: Error < .7 ^ (49.5305) < 0.00001, n- > 412.754.17. n > 642.46; let n = 643.

(b) /'"'(x) = 8 sec2(jt2)[12T:2 + (3 + 32^) tan(x2) + 36x- tan2(.:c2) + A^x^" tan^i^?)] in [0, 1]

\f''\x)\ is maximum whenx = 1 and |/'"(1)| " 9184.4734.
0-0)5,

Simpson's: Error <

180W'

-(9184.4734) < 0.00001, «'' > 5,102,485.22, n > 47.53: let n = 48.

31. Let/(x) = Ar* + Bx- + Cx + D. Then/'^H^) = 0.

Simpson's: Error < . (0) = 0

180/2

Therefore, Simpson's Rule is exact when approximating the integral of a cubic polynomial.
1

Example:

I

x^dx =

r /iv 1

1

0+4^ +1

L 12^ J

4

This is the exact value of the integral.

33. f(x) = 72 + 3.t- on [0. 4].

n

LM

M{n)

R(n)

Jin)

Sin)

4

12.7771

15.3965

18.4340

15.6055

15.4845

8

14.0868

15.4480

16.9152

15.5010

15.4662

10

14.3569

15.4544

16.6197

15.4883

15.4658

12

14.5386

15.4578

16.4242

15.4814

15.4657

16

14.7674

15.4613

16.1816

15.4745

15.4657

20

14.9056

15.4628

16.0370

15.4713

15.4657

208 Chapter 4 Integration

35. fix) = sinv^on[0,4].

n

Lin)

Min)

Rin)

Tin)

Sin)

4

2.8163

3.5456

3.7256

3.2709

3.3996

8

3.1809

3.5053

3.6356

3.4083

3.4541

10

3.2478

3.4990

3.6115

3.4296

3.4624

12

3.2909

3.4952

3.5940

3.4425

3.4674

16

3.3431

3.4910

3.5704

3.4568

3.4730

20

3.3734

3.4888

3.5552

3.4643

3.4759

37. A

Jo

ix COS X dx

Simpson's Rule: « = 14

r,r/2

^ V,cos.:.fa==-|^70cos0 + 4^-cos- + 2^-cos- + 4^-cos-

0.701

' TT ■7T\

TT cos —

2 2

Jo

39. W= 100xVl25 - x^dx

Simpson's Rule: « = 12

100xVl25 -x^dx'^ -^
Jo 3(12)

-«'^)v--(^r-ov— (f

'^^(u)-\^~^ ~ [nf + • • • + O] - 10,233.58 ft • lb

41.

-0

:dx Simpson's Rule, n = 6

■Jr "^ ^,^. [6 + 4(6.0209) + 2(6.0851) + 4(6.1968) + 2(6.3640) + 4(6.6002) + 6.9282]

= :^[113.098]« 3.1416
36

1000
43. Area - ^(J^^25 + 2(125) + 2(120) + 2(112) + 2(90) + 2(90) + 2(95) + 2(88) + 2(75) + 2(35)] = 89,250 sqm

45. I s\n^dx = 2, « = 10
Jo

By trial and error, we obtain t ~ 2.477.

Review Exercises for Chapter 4 209

Review Exercises for Chapter 4

1. y

./,

3. \{2x'^ + X - \) dx = ^}? + ^rx^ - X + C

/^-/

X H — r;\dx = —x- h C

X-) 2 x

J(4x-

3 sin x) dx = 2x- + 3 cos x + C

9./'U)= -2x, (-1,1)

/W = -Ixdx = -x^ + C
Whenx = -1:

>-= -1 + C= 1

C = 2

y = 2 - ^2

11. a(r) = a

v(f) = a rfr =

ar + C,

v(0) = 0 + Ci = 0 when C, = 0.
v(f) = at

-\'

s{t) = I ar rfr = -r + C;

s(G) = 0 + C, = 0 when C. = 0.

i(f) = 2''

5(30) = -(30)- = 3600 or

. = « = 8ft/sec^.
v(30) = 8(30) = 240 ft/sec

13. a(t) = -32

v(r) = -32f + 96
s(r) = - 16r= + 96f

(a) v(f) = - 32f + 96 = 0 when t = 3 sec.

(b) 5(3) = - 144 + 288 = 144 ft

96 3

(c) v(t) = - 32f + 96 = — when t = - sec.

(d) 5I

(1) =

= -16(|j + 96(|j= 108 ft

15. (a) 2(2'- 1)

(b) S/'

i=l
10

(c) 2(4' + 2)

210 Chapter 4 Integration

17. y = ^r—T, Aat = -, /J = 4
X- + 1 2

S{n) = 5(4) = ^r^ +

10

+ .J^ +

10

2\_ 1 (1/2)2 + 1 (1)2 + 1 (3/2)2 + i_
13.0385

s{n) = s{A)

^h

10

+ -^+ 10

10

_(l/2)2 +1 1 + 1 (3/2)2 +1 22+1
« 9.0385
9.0385 < Area of Region < 13.0385

4
19. y = 6 — jc, Ax = -, right endpoints

Area = lim V f{ci) Ax

lim -

n-*oo n

6n

4 n{n + 1)1
n 2 J

= lim

24-8 ^^-^ I = 24 - 8 = 16

21. y = 5 - x^, Ax = -
n

Area = lim V /(«') Ajc
= lim 2

-2 +■

= lim^2ri+i^-^l

,. 3r 12«(n +1) 9 «(n + 1)(2m + 1)

= hm - n + — ^

n->oo n\_ n 2 n- 6

lim[3 + 18^^-?(^l±il%^^l

n^-oo L n 2 «' J

= 3 + 18 - 9 = 12

23. X = Sy - r, 2 < y < 5, Ay = -

n

Area = lim V

5

(-f)-(-f]

©

= lim - 2

No+*^' 4 12'-
n n

9(^"

= lim - 1

1

->
>
i

6+^-^1

= lim -

^ , 3 n(n + 1) 9 «(« + l)(2n + 1)

"" +

n 2 n 6

18 + --9

27
2

Review Exercises for Chapter 4 211

25. lim y {lei - 3) ^xi = (2x-'i)dx

-f

$5-\x-5\dx= I (5-{5-x))dx= \ xdx = ^-

(triangle)

29. (a) J [/W + gW] d.x= \ fix) dx + \ g{x) dx = 10 + 3 = 13
(b) f [/W - gW] ^ = J /W dx-\ g{x)dx=\Q-3 = l

I

(c) [2/(;c) - SgW] ^ = 2

J/Wd:x-3[

gW ^ = 2(10) - 3(3) = 11

(d)

5/(^) dx = 5\ f{x)dx = 5(10) = 50

31. j (^ + l)

dx

4

4(16) + 8

]-[l

+ 1

73

(0

Jo

33. {1 + x)dx =

2x +

.^-u

'•/:

35. (4r3 - 2t) dt

r - t-

= 0

37. f ;cv^a^r = f ;t3'2^ = Hr^^'T = Jlv^)' - (74)'] = ^(243 - 32) =
J4 J4 l5 J4 5 5

P'r/4 r

39. sindde = -cos

3 17/4
0

f)+l = l+^ = ^^

2 / 2 2

41. (2v- l)at( = \x

\(x--9)dx= J -9x1'

= (y-36)-(9-27)

64 _ 54 ^ W
3 3 3

212 Chapter 4 Integration

Jo

45. {x- x^)dx =

2 4

' =1-1 = 1

0 2 4 4

47. Area

=r

:dx =

^x L(l/2)Ji

4;c'/2

= 8(3 - 1) = 16

1 f 1 ri p 2 2

49. —pdx = -ijx = t(3 - 2) = - Average value

2 _ 1

5 Vx

25

2 4 6 8 10

51. F'W = ;cVl + ;c3

53. F'W = ;c2 + 3x + 2

f (;c2 + 1)3 die = JC

;c^ 3

55. I (;c2 + \f dx = \{x^ + ■ix^ + l,x^ + $ dx = J + -x^ + x^ + X + C

57. M = a:^ + 3^ ^^ = 3;t:2 ^

I /' att = |(x3 + 3)-'/2;c2^ = tI(-^ + 3)-'/2 3jc2d:^ = |(;c3 + Zf- + C
J Jx" + 3 J 3J 3

59. M = 1 — 3;c2, £/m = — 6j; (&

/

41 - ix-'Ydx= -| (1 - Ix-'Yi-exdx) = -^(1 - 3;c2)5 + C = ^(3;c2 - 1)^ + C

1. /si

61. I sin' X cos xdx = - sin'' jt + C

63. . "" ^ dx = (1 - cose)-'/2sin0de = 2(1 - cos e)'/^ + C = 2^/1 - cos e + C
J J\ - cos 0 J

tan";

tan" ^ ' X

65. I tan" x sec^ xdx ^^ — + C,ni^ -\

n + \

'■/'

i/<

67. [ (1 + sec TTxf- sec Tr;c tan 7rx<36c = — | (1 + sec Trj:)-(7rsec Tr;ctan nx) dx = —(1 + sec ttxY + C

69.

j'xix^ -4)dx = ^j y- 4)(2.) ^ = I^^^Y^]' _ = ^[0 - 9]

Review Exercises for Chapter 4 213

4-2 = 2

71. I , ' dx = I (1 + x)-"^dx= [2(1 + xY'^
Jo VI + ->c Jo L

Ti. u = \ — y, y = \ — u, dy = — du

Whenv = 0, « = 1. Wheny = 1, « = 0.

2tt\ {y + l)Vl -ydy = 2tt\ -[(1 - u) + \]V^du

= 27r| (m3/2 - 2M'/2)d„ = 2T7r|«^''2 - |«

75. [ cos(f ) dx = 2|^ cos(f ) I ^ = [2 sin(f

°_ 28t7
1 15

77. u = \ — x,x = 1 — «, a[r= —du

When x = a, M = 1 - a. When x = fe, « = 1 — /).

Pa,b= \ ^x^/V^~xdx = —\ -i\-u)Judu

3/2

l-a 4

2„3/2

15

(3m - 5)

i-fc

l-a

(1 - xy/^

{3x + 2)

"^3/2

■~—{3x + 2) I = 0.353 = 35.3%

2 Jo.50

(b) P,

'0.50. 0.75 ~ 2 (3j: + 2) I

*_ (1 - fc)^/^

(3Zj + 2) + 1 = 0.5

(1 - fc)3/2(3fc + 2) = 1
fe = 0.586 = 58.6%

79. p = 1.20 + 0.04r

„ 15.000 f'"' ,
c = —ri-\ Pds

(a) 2000 corresponds to f = 10.

C = -^f"[l.20 + 0.04r]dr
^ Jio

1.20r + 0.02f'

^ i5,ooor

~ M [

" 24,300

10 M

(b) 2005 corresponds to f = 15.

C =

15,000

M

1.20r + 0.02r^

"^ _ 27,300
15 ~ M

81. Trapezoidal Rule (« = 4): J_ -p^^ »= |[y^ + l + (',25)3 + TtW ^ 1 + (^75)3 + TT2^
Simpson's Rule (n = 4): I , dx " ~r7:\~r

+ -4^ + - '^

+ 1' 1 + (1.25)3 1 + (1 5)3 1 + (J 75)3 1 + 2

= 0.257

0.254

Graphing utility: 0.254

83. Trapezoidal Rule (,

« = 4):

Jo

^ COS xdx ~ 0.637

Simpson's Rule (n = 4): 0.685
Graphing Utility: 0.704

SS. (a.) R < 1 < T < L

(b) S(4) = ^^[/(O) + 4/(1) + 2/(2) + 4/(3) +/(4)]

4 + 4(2) + 2(1) + 4i^] + i

5.417

214 Chapter 4 Integration

Problem Solving for Chapter 4

1. (a) L(l)

=IV-

1

(b) L '{x) = - by the Second Fundamental Theorem of Calculus.

X

L\\) = 1 .

(c) L{x) = 1 = I -rfrfor;c « 2.718

-^n^

dt = 0.999896 (Note: The exact value of x is e, the base of the natural logarithm function.)

f

P'l f 1

(d) We first show that \ -dt = \ -dt.

To see this, let m = — and du = — dt.

Xy X^

P'l f 1 f 1 f 1

Then -dt= — (x, du) = - du = \ -dt.

h t JiA.^i JiA," JlA.f

/■jTAj Pm / f\

Now, i-(x,x-,) =1 -dt = \ —du\ using « = —

Ji t JiA,« V Xi/

f 1 Pn

Ji/x,« Ji "

P'l P^l

= -du+ -du

Ji " Ji «

= LUi) + LUj).

= IH't

-2--

3. 5(;c) = I sini ^ Uf
(a)

(b)

V2 V5 2 V? vS-y7 2Vl3

The zeros of y = sin ^r— correspond to the relative
extrema of S{x).

(c) 5'W = sin-^ = 0

= mr =^ X' = 2n => X = V2»2. n integer.

Relative maximum at jc = ~Jl = 1.4142 and x = V6 «= 2.4495
Relative minimum at x = 2 and x = ,/8 = 2.8284

.2

(d)5"(x) = cos(^)(7rx) = 0

= — + AjTT => x^ = 1 + 2n => X = Vl + 2n, « integer

Points of inflection at x = 1, V3, Js, and ^7.

Problem Solving for Chapter 4 215

5. (a)

(8,3)

(b)

x

0

1

2

3

4

5

6

7

8

1

7

7

1

Hx)

0

"2

-2

~2

-4

~2

-2

4

3

(c) f(x) =

-X, 0 < .r < 2

X - 4, 2<x<6

ij: - 1, 6 < ;c < 8

0 < -r < 2

F(x]

., [(-^72), ......

= /(f) dt = I (.x:V2) - 4x + 4, 2<A<6
■''' L(l/4):r2_ ^ _ 5 ^ < x < "

F'(x) = f(x). F is decreasing on (0, 4) and increasing on
(4, 8). Therefore, the minimum is — 4 at .r = 4, and the
maximum is 3 at .r = 8.

(d) F"{x)=f\x) =

f-l, 0 < Jt < 2
1, 2 < .r < 6
J.,

2

6 < j: <

At = 2 is a point of inflection, whereas x = 6 is not.
(/is not continuous at j: = 6.)

cos.ri±c « cosi ^1 + cosi — ?= j = 2 cosi — p 1 = 1.6758

7. (a) cos.ri±c « cosi — -i= \ + cos'

I cos a: dx: = sin X = 2 sin(l) = 1.6829

Error. 11.6829 - 1.6758 = 0.0071

(b)

r.i

1

; dx ~

1

1

1 +A-2 1 + (1/3) 1 + (1/3) 2

(Note: exact answer is 7r/2 == 1 .5708)

(c) Let p(x) = ax^ + bx- + ex + d.

b.^ cx-
+ —- + —- + <ir
3 2

1

lb
3

= =r + Id

'^--ir'\ji

.,|..).g..).|..

9. Consider Hv) = [fix)]- => F\x) = 2f{x)f\x).'\\ms,

\f{x)f\x)dx= \\

Ja Jo ^

F'(x)dx
Fix)

= ^[F(b) - F(a)]

= pibT- - f(ar-]

11. Consider x^ dx =

Jo

The corresponding Riemann Sum using right endpoints is

Si

"' = e

' - 1^ ^

= \[l^ + 25 + ■ ■ ■ + n5]

Thus, lim S{n) = lim

1^ + 25 +

~ 6'

216 Chapter 4 Integration

13. By Theorem 4.8, 0 < /W < M => \ f{x) dx < I M dx = M(b - a).

Ja Ja

rb cb

Similarly, m < f(x) => m{b — a) = mdx < f(x) dx.

Ja Ja

Thus, m{b - a) < \ f{x) dx < M{b - a). On the interval [0, 1], 1 < JT^H? < V2 and fo - a = 1.

Ja

Thus, 1 < J\ + x^dx < 72. (Note: I yr+~?c&= 1.0894J

15. Since-|/W| < f{x) < \f{x)\,

J'b rb rb rb rb

\f{x)\dx< \f{x)dx< \f{x)\dx ^ \ f{x)dx <\ \f{x)\dx.

a Ja Jo Ja Jo

1 r

100,000

. l-nit - 60)1 _, 100,000
1 + sm r— 7 dt =

36.5

'\

365

365 l-nit -
t - ^r-COS -

2lT

t - 60)T' ^
365 Jo

100,000 lbs.

CHAPTER 5

Logarithmic, Exponential,

and Other Transcendental Functions

Section 5.1 The Natural Logarithmic Function: Differentiation .... 218

Section 5.2 The Natural Logarithmic Function: Integration 223

Section 5.3 Inverse Functions 227

Section 5.4 Exponential Functions: Differentiation and Integration . . 233

Section 5.5 Bases Other than e and Applications 240

Section 5.6 Differential Equations: Growth and Decay 246

Sections.? Differential Equations: Separation of Variables 251

Section 5.8 Inverse Trigonometric Functions: Differentiation 259

Section 5.9 Inverse Trigonometric Functions: Integration 263

Section 5.10 Hyperbolic Functions 267

Review Exercises 272

Problem Solving 278

CHAPTER 5

Logarithmic, Exponential, and Other Transcendental Functions

Section 5.1 The Natural Logarithmic Function: Differentiation

Solutions to Odd-Numbered Exercises

1. Simpson's Rule: /j = 10

X

0.5

1.5

2

2.5

3

3.5

4

Jl '

-0.6932

0.4055

0.6932

0.9163

1.0987

1.2529

1.3865

Note:

Jl t Jos t

dt

3. (a) In 45 » 3.8067

ri

(b) I -dt == 3.8067

\y

5. (a) In 0.8 = -0.2231
ro.8
(b) I -dt == -0.2231

ro.8

-dt == -0.:
Jl t

7. f{x) = \nx + 2

Vertical shift 2 units upward
Matches (b)

11. f{x) = 3 in ;c
Domain: x > 0

13. fix) = In 2x
Domain: jc > 0

9. fix) = In U - 1)

Horizontal shift 1 unit to the right
Matches (a)

15. fix) = In(x - 1)
Domain: x > \

17. (a) In 6 = In 2 + In 3 = 1.7917

(b) Inf = In2 - ln3 = -0.4055

(c) In 81 = In 3'' = 4 in 3 « 4.3944

(d) In 73 = in 3'''2 = i |n 3 « 0.5493

19. Inf = In 2 - In 3

21. In — = In X + In y - In z

23. In 7a2+ 1 = ln(a2 + i)i/3 = - in(a2 + 1)

218

Sections.] The Natural Logarithmic Function: Differentiation 219

25. Inf^^-T-^V = 3[ln(;c2 - 1) - In.r^]

x'

= 3[lnU- + 1) + InU - 1) - 3 In.x]

27. In z(z - 1)2 = In z + ln(z - 1)^
= In z + 2 ln(z - 1)

29. ln(;c - 2) - InU + 2) = In

x-2

x + 2

— 1 r» , / ,x , , / 1 .\T 1 , -tU + 3)- , , /x(x + 3)2

31. -[21ii(a: + 3) + Inx - InU^ - 1)] = - In ^^, _ ^ = In ^ '^^ _ ^

33 . 2 In 3 - - \n(x- + 1) = In 9 - lnVx= + 1 = In , ^

2 Vj;' + 1

35. 3

37. lim ln(x - 3) = -oo

j-»3*

39. lim ln[x2(3 - x)] = In 4 - 1.3863

X — »2

41. y = In j:^ = 3 In .r

, 3

At(l,0),v' = 3.

43. >' = In JT' = 2 In ;c
At(l,0),v'= 2.

45. g(x) = Inx- = 21nA;

47. V = (ln;c)''

^ = 4(lnx)3(i)=^(ill^
ax \x/ X

49. V = In xjx^ - 1 = In .X + - In(x2 - 1)

_ 2.r2 - 1
dc X 2\x2 — 1/ xU' — 1)

^ = i + V ^

51. /W = In^— - = In.T - In(.r- + 1)

„, . 1 2x l-x-

/ w = r

x x^ + 1 x(x^ + 1)

53. g{t) =

Int

g'it)

t-(l/t) - 2t\nt 1 - 2Inr

55. .v = In(Inx-)

^ = -!-— (In.r2) = (^^'/-^^ = - = 1
djc In x^ a[r ' In x- x In x- x In x

57. y = \n

yj^ = f[ln(x+l)-ln(x-l)]

dy ^ J_r_l 1_1 ^ 1

dx 2Lx + 1 X - ij ~ 1 -x2

59. f(x) = In

J 4 + x^ 1

-ln(4 + x^) - Inx

fix)

4 + x^ X x(x~ + 4)

220 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

61.

y =

■v^c^+T

+ ln(;c + Jx- + 1 )

dl^ -x[x/Jx- + l) + y/jC^ + 1

dx~ £-

1

1

1 +

x + -Jx- + 1 / V V;c2 + 1
1

J^fT\+ .

x^J^^^Tx \x + JWT\j \ J^T\ I x^JlFTl 7]?TT ^-J^FTx

+ ■

1 + x- v^2~rT

63. y = In I sin .r|

dV _ cos.T
dx sinjc

cotx

65. y = In

COSJC

COS j: - 1
= Inlcosxl - Inlcosj: — ll

-sin a:

dy _ — sinx _

dx cos X cos X — I

— tan.r +

sin X
cosj: - 1

67. >> = In

■ 1 + sin x

2 + sinx

= ln| — 1 + sin xl — ln|2 + sinx|

dy _ cos X cos x

dx — 1 + sin .t 2 + sin X

_ 3 cos X

(sinx - l)(sinx + 2)

69. fix) = sin 2x In x^ = 2 sin 2x In x
fix) = (2 sin 2x)(-J + 4 cos 2x in x

= - (sin 2x + 2x cos 2x In x)

X

2
= ^sin 2x + X cos 2x In x^)

71. (a) >- = 3x2- In X, (13)

dy I

— = 6x

dx x

. When X = 1, v" = 5.
dx

Tangent line: y - 3 = 5(x - 1)

y = 5x - 2

0 = 5x - 3^ - 2

(b)

73. x2 - 31n>' -l-y^ = 10
y dx dx

-le-'

dy _ 2x 2x3;

dx " i-i/y) -ly~ l-ly^

75. y = 2(lnx) + 3

77. y = j-Inx

Domain: x > 0

1 (x + l)(x - 1)

>> =x

0 when x = 1 .

1

/'=l+^>0
Relative minimum: ( 1 , 2)

^

Section 5.] The Natural Logarithmic Function: Differentiation 221

79. y = xlnx

Domain: ;c > 0

y' = xl-] + \nx = 1 + In .r = 0 when x = e~K
y"=- > 0

Relative minimum: (e ', — e ')

(f-'.-f-')

81. V =

Injc
Domain: 0 < jc < l,.x > I

, (in;c)(l) - (x){l/x) _ In-r- 1

y =

= 0 when x = e.

(ln;c)2 (Inx)'

(ln;c)^(lA) - (Inx - l)(2/;c) In a: 2 - \nx

{\nxr
Relative minimum: (e, e)
Point of inflection: {e~, e^/2)

x(\nx)^

0 whenx = e^.

83. fix) = ln;c, /(I) = 0

/'W=^, /'(1)=1

/"W = -A, /"(i) = -i

PiW =/(l) +/'(1)U - 1) = X - 1, P,(l) = 0
/'2W =/(l) +f'Wix - 1) + |/"(1)U - IP

= U-l)-^(x- 1)% P,(1) = 0

P,'(.r) = 1, P/(l)= 1

P/W = 1 - (x - 1) = 2 - .r, P.'(l} = 1

M;c)= -1, P/i;i)= -1

85. Find.x such that In.r = — a:.
fix) = ilnx)+x = 0
1

fix) = - + 1

" fix J -''■

1 - ln,T„

1 +.v„

M

1

2

3

^.

0.5

0.5644

0.5671

/UJ

-0.1931

-0.0076

-0.0001

The values of/, Pj, Pj, and their first derivatives agree at
X = 1. The values of the second derivatives of/ and P,
agree at j: = 1 .

87. y = x^x^ - 1

Inr = In.v + |ln(.t= - 1)

y\dxj X ;c2 - 1

dt \xix^ - 1)J ^/p^H"

Approximate root: x = 0.567

222 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

89.

y =

1

]ny = 2 ln;c + -ln(3;c - 2) - 2 In^ - 1)

lf^U2^ 3

y\dx) X 2(3a- — 2) x — I

dy
dx

y

3x^ - \5x +

2xOx - 2){x - 1).

3x!^ - 15;>:^ + 8jc
2{x - iyj3x - 2

91.

1)3/2

V7+T

3 1

Iny = \nx + - InU - 1) - - ln(;t + 1)

3/ 1

U 1

y\dxj X 2\x — 1/ 2\x +

dy
dx

2Lx

X + 1

y^4x^ + \x- 2
21 xix''- 1)

{2x^ + 2x- \)Jx- 1
{x + 1)3/2

93. Answers will vary. See Theorems 5.1 and 5.2.

95. In e' = ;c because f{x) = In x and g(x) = e^
are inverse functions.

97. (a) /(l) 9^/(3)

(b) fix) = 1 - - = 0 forA: = 2.

X

99.

/3 = 10 log,,

10\ 10-16

l«lnf ^

In 10 VIO

10
In 10

{In/ + 161n 10] = 160 + lOlogig/

j8(10"'°) = 7^[ln 10-'° + 16 In 10] = 7^[- 10 In 10 + 16 In 10] = 7^[6 In 10] = 60 decibels
1 In 10 m 10 In 10

101. (a) You get an error message because In h does not exist
for /z = 0.

(b) Reversing the data, you obtain

h = 0.8627 - 6.4474 In p. ■ '

[Note: Fit a line to the data (x, y) = (Inp, h).]

(C) 25

(d) If p = 0.75, h == 2.72 km.

(e) If h = 13 km, p = 0.15 atmosphere.

(f) /! = 0.8627 - 6.4474 In/?

1 = -6.4474--^ (implicit differentiation)
p dh

dp ^ p

dh -6.4474

For A = 5, p = 0.5264 and dp/dh = -0.0816 atmos/km.

For h = 2Q),p = 0.0514 and dp/dh = -0.0080
atmos/km.

As the altitude increases, the rate of change of pressure
decreases.

103. (a) fix) = In x, gix) = J'x

{h) fix)^\nx,gix)=ifx

^■'"4-'«^i^

For X > 'X, g 'ix) > fix), g is increasing at a faster rate
than/for "large" values of z.

fix)

For X > 256, g Xx) > fix), g is increasing at a faster rate
than/for ' ;arge" values of .v./(.v) = In x increases very
slowly for "large" values of x

Section 5.2 The Natural Logarithmic Function: Integration 223

105. False

In j: + In 25 = ln(25x) # ln(;t + 25)

Section 5.2 The Natural Logarithmic Function: Integration

BH

-dx = 5\n \x\ + C

3. « = .r + 1, rf« = dx
1

J.r + 1

dx = InLc + 1 + C

5. u = 3 — 2x, du = —2dx

^ dJc=-l-\—^{-2)dx

Js-Zx'^ 2j3 -2x

1

In 3 - 2x\ + C

7. u = x^ + I, du = 2x dx

-ln{x2+ 1) + C

InV.r^ + 1 + C

/^-/h!

ott

- 4ln|;c| + C

11. M = .r^ + 3x- + 9x, du = 3(x~ + Zx + 3) dx

r .t- + 2x + 3 ^ 1 f3{x^ + 2j: + 3)
J x3 + 3jc- + 9.t: " 3J .r^ + 3a2 + 9.r

■ ln|x3 + 3x^ + 9x\+ C

..j^f^.-j{.-^^jh].

4x + 61nU + 1| + C

"■/^^:^*^/(''--?tiK

y-2;c + |ln(;c2 + 2) + C

21. H = a; + I, du = dx

= Ux+ l)-'/2(ic

= 2{x + 1)1/2 + C

lvlTT* = l<

= 2V.r + 1 + C

-/^4^-/('-7^)-

= Y + 51n|;c - 3| + C

19. M = in X, du = - dx

X

\

{\nxf , _ 1„_.,3

dx = -(ln.r)' + C
X 3

f 2r flv - 2 + 2

=/n--/u^-

= 21nU - 1|

(.V - 1)

+ C

224 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

1

25. M = 1 + Jlx, du = -^= dx =^ {u - \)du = dx
1

}\ + Jlx J

Jlx

■du

du

1 + V2jc / "

= « — ln|M| + C,

= (l + 72I) - Injl + JYx\ + Ci
= Jlx- ln(l + v^) + C
where C = Cj + 1 .

27. « = V^ - 3, rfw = -pdx => 2(m + 3) du = dx

ijx

\-^^dx = 2j^.« = l\^^^^^du = 2j(« . 6 + f).«

= 2

+ 6m + 91ni«|

Ci = M^ + 12« + 181n|M| + C,

= [Jx - 3f + 12(v^ - 3) + 18 \^Jx - 3| + C,
= ;»:+ 6V5 + IBlnlv^ - 3! + C where C = C, - 27.

f

29. P^rfe=ln|sine| +C

sin 6
(m = sin Q, du = cos 6 dS)

31. CSC 2j: ^ = - (

I CSC 2j: ^ = - I (esc 2.i;)(2) dx

■ Inlcsc 2j: + cot 2x\ + C

33. f-^^^

J 1 + sin r

37.. = |:

df = Inll + sinil + C

2- X

■dx

= -3

-dx

(1.0)^

i

:^

?V:^

x-2
= -31n|;f - 2| + C

(1,0): 0 = -31n|l - 2| + C ==* C = 0
y = -3 1n|;<;- 2\

,_ , secxtanx . , 1 , ,"

35. i r dx = Inlsec x - 1 1 + C

f

39. i = tan(2e) rf0
1

(0.2), 4

^1'

tan(2e)(2rf0)

A I I A ,

-ln|cos2e| + C

(0, 2): 2 = -- ln|cos (0)| + C =J. C = 2

i = --ln|cos2e| + 2

41,

dx X + 2

(a)

(0,1)

1=^*

(b) y

X + 2
>.(0) = 1 ^ 1 = In 2 + C

rfx = ln|x + 2| + C
C =

Hence, y = ln|x + 2| + I - In 2 = In

- in 2
X + 2

Section 5.2 The Natural Logarithmic Function: Integration 225

43.

fl^'^^tf'^l^^^'ll

= -In 13 « 4.275

45. M = 1 + \n X, du = — dx

X

i

(1 + \nxY

dx =

^1 +ln.r)3

-ff^-/>

X + 1

dx

-x^- X- \n\x+ 1|

-In 3

49.

f-

cos 6

sin 0

rfe

ln|e - sin e\

= In

2 - sin 2

1 — sin 1

« 1.929

51. -Inlcosjcl + C = in

cos.r

+ C = Inlsecr + C

53. Inlsec^c + tan.r + C = In

= In

(sec X + tan .r)(sec j: — tan jc)

(sec X — tan x)

+ C = In

sec'.t - tan-^.x:

secx - tanj:

+ C

1

sec X - tan x

+ C = — Inlsecr — tan j: + C

. I ^a!j; = 2(1 + v^) - 21n(l + v^x) + C,

= 2[Vx - ln(l + Jx)] + C where C = C, + 2.

57. cos(l - x)dx = -sin(l - .t) + C

p/2 r

59. I (cscr — sinjf)fl[x = — ln|cscjc + cot.i:| + cosj:

Jtt/A L

V2

72

ir/4

= ln(V2 + 1) - -^ « 0.174

Note: In Exercises 61 and 63, you can use the Second Fundamental Theorem of Calculus or integrate the function.

=B

61. F(x) = \ -dt
F'ix) = i

62.fix) = [\dt = ['\dt-[\dt

F'ix) = f - i = 0
3a' .V

65.

A = 1.25
Matches (d)

67. A = I '^^^-^dx = I (x + -]dx

'\^

+ 4 In .r

' = (8 + 41n4)-^

15

— + 8 In 2 ~ 13.045 square units

*^

226 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

69. r2sec^d^ = -r
Jo 6 ttJo

ttxXtt ,

12

In

;— + tan —
6 6 I Jo

11

77
12

In

sec— + tan— ln|l + 0|

3 3 77

ln(2 + 73) - 5.03041

71. Power Rule

73. Substitution: (m = j:^ + 4)
and Log Rule

75. Divide the polynomials:
= j: — 1 +

x+ 1

X + 1

J p g ("4

77. Average value = r- ~dx = 4 x'''^dx

4 - 2J2 X- J2

-4i

-M

79. Average value

^tI'

Inx , 1

dx =

X e — 1

\_nxir

1 /I

e - 1V2
1

2€ - 2

= 0.291

81. Pit) = r fo25/' " (3000)(4)J^ "p^-^g/? = 12,000 In |1 + 0.25r| + C

P(0) = 12,000 ln|l + 0.25(0)1 + C = 1000

C = 1000

P(t) = 12,000 ln|l + 0.25f| + 1000 = 1000[121n|l + 0.25f| + 1]

P{3) = 1000[12(ln 1.75) + 1] « 7715

83.

50

^r3Sfi-=HH~

+ 3x\

_ 40

« $168.27

85. (a) 2x-- y^ = S
• ■ ^2 = 2x2 - 8

y, = V2x2- 8
:y2 = - V2x2- 8

\

^

/

'\

Let/: = 4andgraphr.^ ^ I - if^)

(c) In part (a), 2^^ - / = 8
4jc - 2yy ' = 0

, 2x

In part (b), y^ = - = 4j: '
x

2yy' =

,'_-^ _ ~2y _ -2y _ -y
yx^ y'^ x'^ 4x 2x'

Using a graphing utility the graphs intersect at (2.214, 1.344). The slopes are 3.295 and -0.304 = (- l)/3.295, respectively.

Section 5.3 Inverse Functions 227

87. False
I

(lnx) = ln(;c'^2)^(,n^)i/2

Section 5.3 Inverse Functions

89. True
"l

il

dx = InU + C,

= ln|x| + ln|C| = ln|C;c|, C ^0

1. (a) f(x) = 5;c + 1

/(gW) =/(^) = 5(^1 + l=x

giifU)) = g(5x + 1) = ^^""^P ' = X

3. (a) fix) = x'
g(x) = ^x

f{g{x))=f[ifx) = {irxy = x

g(/W) = g(:^)= W = x

(b)

I _2--
-3

5. (a) fix) = vT^^

gU) = x= + 4, .V > 0
/(gW)=/U- + 4)

= VU^ + 4)-4=v^ = ;c
= (y;c -4)2 + 4 = x-4 + 4

(b)

12--
10-

4

2

-\ 1 1 l-»-J

2 4 6 8 10 12

7. (a) fix) =

1

g{x) = -

f(g{x)) = Y/^ = ^
g(/W) = tV = ^

l/x

(b)

H 1 1—*--'

9. Matches (c)

11. Matches (a)

228 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

13. fix) = ja: + 6

One-to-one; has an inverse

15. f{e) = sin d

Not one-to-one; does not have an

17. h(s) =

1

s-2
One-to-one; has an inverse

19. /W = \nx

One-to-one; has an inverse

21. g(x) = U + 5)3

One-to-one; has an inverse

-2

23. f(x) -- {x + af + b

f'(x) = 3(;c + a)2 > 0 for all x.

/is increasing on (— oo, oo). Therefore, /is strictly
monotonic and has an inverse.

25. /W = ^ - 2x^

f'ix)=x?-4x = 0 when ;c = 0, 2, -2.

/is not strictly monotonic on (— oo, oo). Therefore, /does
not have an inverse.

27. /(x) = 2 - X - ;c3 . -.

fXx) = - 1 - 3;c2 < 0 for all x.
/is decreasing on (— oo, oo). Therefore, / is strictly monotonic and has an inverse.

29. fix) = lx--b=y

x = '-^
2

X + 3

y =

/-'W =

2
X + 3

31. /W = x5 = y

x= 4/y

y= Vx

/-'(x) = 5/^ = x'/5

33. /(;c) = v^ = y

X = y^

y = x'-
f-\x) = jc2, jc > 0

Section 5.3

Inverse Functions

229

35. fix) = JA- x'^ = y, 0 <x <2

x= J A -y^

= JV^J^

f-\x) = V4 - x\ 0 < ;c < 2

37. fix) = Vx- 1 = y
jc = y3 + 1

y = x^ + \

/-'W =x3+ 1

r

^'r

4-

^

The graphs of /and/"' are
reflections of each other
across the line y = x.

39. fix) = x^l^ = .V, ;( > 0
x = y3/2

f'\x) =;c3/2, ;c > 0

The graphs of/and/"' are
reflections of each other
across the line y = x.

41. /(x)=^#= = y
77;c

-^-yr^

/-

2

f::^

-^z

Jly

1 < ;<: < 1

The graphs of/ and/ ' are
reflections of each other
across the line v = x.

43.

X

1

2

3

4

f-\x)

0

1

2

4

2 3 4

45. (a) Let x be the number of pounds of the commodity
costing 1.25 per pound. Since there are 50 pounds
total, the amount of the second commodity is 50 — x
The total cost is

V = 1.25.r + 1.60(50 - x)

= -0.35a: + 80 0 < .r < 50.

(b) We find the inverse of the original function:

V = -0.35;c + 80
0.35a: = 80 - >>

X = f (80 - y)
Inverse: y = -^(80 - x) = f (80 - .x).
.r represents cost and y represents poimds.

(c) Domain of inverse is 62.5 < .v < 80.

(d) If .t = 73 in the inverse function.

y = -^(80 - 73) = -x = 20 pounds.

230 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

47. fix) = (x - 4)2 on [4, oo)

f'(x) = 2U - 4) > 0 on (4, oo)

/is increasing on [4, oo). Therefore, /is strictly
monotonic and has an inverse.

49. f(x) = -^ on (0, oo)

fix) = — r < Oon(0, oo)
x'

/is decreasing on (0, oo). Therefore, /is strictly
monotonic and has an inverse.

51. f(x) = cos X on [0, tt]

f'(x) = — sin x < 0 on (0, it)

/is decreasing on [0, it]. Therefore, /is strictly monotonic and has an inverse.

53.

fix)

x^--4

X^y — 4y = X

x^y — X — 4>' = 0

a = y, b = — 1, c = —4y

yon (-2, 2)

^ 1 + Vl - 4(y)(-4v) ^ 1 ± Vl + 16>-2
""' _ 2y 2> .

f(l - WTj6?)/2x, ifx ¥= 0
0, if X = 0

y=r'ix)

Domain: all x
Range: -2 < y < 2

^

^

The graphs of /and/"' are
reflections of each other
across the line >> = x.

55. (a), (b)

/

/-'

-fl

-^

(c) Yes, /is one-to-one and has an inverse. The inverse
relation is an inverse function.

57. (a), (b)

(c) g is not one-to-one and does not have an inverse.
The inverse relation is not an inverse function.

59. /U) = Jx - 2, Domain: x>2

fix)

> OfoTx > 2.

2yr^

/is one-to-one; has an inverse

Jx - 2 = y
x-2=f
x = f + 2
y = x^ + 2
f'^ix) = x^ + 2,x > 0

61. fix) = |x- 2|,x < 2
= -(x-2)
= 2 -X
/is one-to-one; has an inverse
2 — X = >>
2 — y = x
f~\x) = 2 - X, X > 0

63. fix) = ix - 3)2 is one-to-one for x > 3.
(x - 3)2 = >-
X - 3 = v^
x= Vy + 3
y=Vx + 3
/-'(x) = ^ -I- 3, X > 0
(Answer is not unique)

65. /(x) = |x -I- 3| is one-to-one forx > -3.
X -I- 3 =y
X = y — 3
y = X — 2
/-H3c) = X - 3, X > 0
(Answer is not unique)

Section 5.3 Inverse Functions 231

67. Yes, the volume is an increasing function, and hence
one-to-one. The inverse function gives the time /
corresponding to the volume V.

69. No, C{t) is not one-to-one because long distance costs are
step functions. A call lasting 2.1 minutes costs the same as
one lasting 2.2 minutes.

71.

fix) = :c^ + 2x - 1, /(I) = 2 = a
fix) = 3.r2 + 2

if-r{2)-j;rj^

1

1

1

/'(/-'(2)) /'(I) 3(iP + 2 5

73. /W = sin;t,/(|)=| = a

(/-

fix) = cos .r

1

1

2/ /'(/-'(1/2)) /V/6) cos(7r/6)

1 _ 2V3
73/2 3

75. /W = ^ - -■ /(2) = 6 = a

fix) = 3.r= + ^

(J ) yo) f'( f~^((i\\ f'(i\

1

1

/'(/-H6)) /'(2) 3(2)2 + (4/22) 13

77. (a) Domain/ = Domain/ ' = (—00,00)
(b) Range/= Range/"' = (—00,00)
(c)

3-1-

//

(d)

/(.) = ^, (|,^

/I

'X 1^1

/u

/-H;c)

(/-')'(.r) =

(/-

</.t-

79. (a) Domain/ = [4, 00), Domain /^^ = [0. oc)
(b) Range/ = [0, 00), Range/"' = [4, 00)
(c)

4 6 8 10 12

(d)

fix) = TT^, (5, 1)

Ax) =

2jx - 4

/'(5) = ^
/"'(x) = X- + 4, (1,5)
(/"■)'(.r) = 2x
(/-')'(1) = 2

232 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

81. .r = y3 - V + 2

^ ^d\ , , dy

dv

' Mi-4A),^ '

1

dx liy- - 14v' ' "ate 3-14 11

Alternate solution; let /(a) = x^ - Ix^ + 2.
Then/'(jc) = Sjc^ - 14Aand/'(l) = -11.

In Exercises 83 and 85, use the following.
/(x) = |a: - 3 andg(x) = x^
r\x) = %{x + 3) and g-\x) = i^

83. (/-' "g-')(l) =/-'(g-Hl)) =/-'(!) = 32

85. (/-' »r')(6) =/-'(/-'(6)) =/-'(72) = 600

In Exercises 87 and 89, use the following.

f(x) = X + 4 and gix) = 2* — 5

r'W=x-4andg-'W = ^

87. (g-' ./-i)W = r'(/-'W)
= g-'U - 4)

^(x- 4) + 5
2

x+ \

89. (/°g)(x)=/(gW)

= /(2x-5)
. = (Zx - 5) + 4
= 2x- 1

Hence, (/"g)-'U) =

X + 1

(Note: (/-g)-' =g-'"/-')

91. Answers will vary. See page 335 and Example 3.

93. y = x-^ on {— CO, ca) does not have an inverse.

95. /is not one-to-one because many different x- values yield
the same >'-value.

Example: /(O) = /(tt) = 0

(2rt — IItt
Not contmuous at , where n is an mteger

97. Let if'gjix) = y then x = (/■= g)^'(y). Also,

{f'g)ix)=y
figix)) = y
g(x)=r'(y)

X = g-'{f-'(y))
= ig-' 'f-')(y)

Since/and g are one-to-one functions,

(/°5)-' = g^'°r'-

99. Suppose g(x) and h{x) are both inverses of/(x). Then the graph of/(x) contains the point {a, b) if and only if the
graphs of gix) and h{x) contain the point (b, a). Since the graphs of g{x) and h(x) are the same, g{x) = h{x).
Therefore, the inverse of/(x) is unique.

Section 5.4 Exponential Functions: Differentiation and Integration 233

101. False

Let/W = x\

105. Not true

X, 0 < ;c < 1

1 - jc, 1 < X < 2 ■

Let/W

/is one-to-one, but not strictly monotonic.

103. True

107.

fix)

(/-')'(o)

dt

71 + t
1

ym^

/'(2) i/yi?

;, /(2) = 0

= yi7

Section 5.4 Exponential Functions: Differentiation and Integration

1. e° = \
Inl = 0

3. In 2 = 0.6931

g0.6931. =2

5. e'"-' = 4
X = 4

7. e' = 12

JC = In 12 = 2.485

9. 9 - 2?* = 7
2^^= 7
e' = 1
x = 0

11. 50e-^ = 30

-X = In

-4

0.511

15. InU - 3) = 2
X — i = e-

.t = 3 + e2 == 10.389

13. In .r = 2

.r = e= - 7.3891

17. ln7x + 2 = 1

Jx + 2 = e' = e
X + 2 = e-

jt = e^ - 2 = 5.389

19. V = e

21. >• = e-"^

Symmetric with respect to the .%-axis
Horizontal asymptote: v = 0

234 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

23. (a)

^

V

Horizontal shift 2 units to the
right

(b)

/f

k

A reflection in the jc-axis and a
vertical shrink

(c)

^

Vertical shift 3 units upward
and a reflection in the y-axis

25. y = Ce'"

Horizontal asymptote: v = 0
Matches (c)

27. y = C(l - €-<")
Vertical shift C units
Reflection in both the x- and y-axes
Matches (a)

29. f(x) = e^"

g(x) = InV^ = -\nx

31. f(x) = e--l
g(x) = ln{x + 1)

33.

;

g

^

As j: — > oo, the graph of /approaches the graph of g.

lim (1 +

X — »oo

^r

35. 1 +

1

1,000,000

1.000,000

« 2.718280469

e «= 2.718281828

e > 1 +

1

1,000,000

37. (a) y = e^^

At(0, l),y'= 3.

39. f{x) = e^
f'(x) = 2e^

45. g(i) = (e-' + e')^

g'W = 3(e-' + e')V-e-')

(b) y = e-

41. /(jc) = e-2'+^'

f-"

47. >' = In<r''

Ik-

y' = -Ze-^"

At(0, l),.v' =

• -J

.

43.

y =

dy _
dx

Ij-x

49.

y =
di_

ln(l +
2e^

e^)

dx I + e^

Section 5.4 Exponential Functions: Differentiation and Integration 235

51. y =

e' + e-

2{e^ + e-")'^

dy
dx

= -lie' + e-'Y^e' - e'")
-2(e^ - e'')

{e' + e-^)2

53. y = x^e^ - Ixe' + le' = e^ix^ - 2x + 2)
^ = e^(2x - 2) + e'ix^ - 2x + 2) = .tV

55. /(;c) = e-' In j:

Z'U) = e'4-] - e-'\BX = e~4- - In^J

57. y = e'isin x + cos x)

—- = e^icQis x — sin x) + (sin x + cos xSie' )
dx

= e^(2 cos j:) = le' cos at

59. xey - mx + 3.V = 0

xey^- + ey - 10 + 3^ = 0
dx dx

dy
dx

(xey + 3) = \0- ey

dy _ 10 - gy
dx " xey + 3

61. fix) = (3 + 2x)e-^'

fix) = (3 + 2x)(-3e-3') + le-^"

= (-7 - 6A;)e-3^
/W = (-7 - 6x){-3e-^') - 6e-3x

= 3(6;c + 5)e-^'

63. J! = e^{cosV2jc + sinv/2Jc)

>>' = e^(- V2sin ^x + V2cos V2;>:) + e-^fcosVlv + sin v^)

= ej{l + 72)cosV2x + (l - >/5)sin>/2»:]
>>"= ej_-{j2 + 2)sin v/2x + (^2 - 2)cos J2x] + e^[(\ + V2)cos72Jc + (l - v^sin Vlt]
= e{(- 1 - 2V2)sin ^x + (- 1 + 2V2)cos V2a]
-2>'' + 3v = -2e-'f(l + V2)cos Vli: + (l - 72)sin Vlr] + 3e{cosV2.i: + sin Tlr]

= e{(l - 2v/2)cos y2.v + (l + 272)sin Vlx] = -y"
Therefore, -2y' + 3y = -y" => y" - 2.v' + 3.v = 0.

65. f(x) =

e^ + e-

f'(x) = ^ .^ " = 0 when x = 0.

fix) = ^-^-^ > 0

Relative minimum; (0, 1)

67. gW = -^,-U-2)V2

8'{x) = ^(.v - 2)g-(-2)V2
V2Tr

g"(x) = -^(.t - l)(.r - 3)g-(-2'V2

v/2^

1

Relative maximum: ( 2, — ;= ) ~ (2, 0.399)

'Z7T
1

Points of inflection: I 1,

^r- 1 3

,.-1/: ==

(1,0.242), (3,0.242)

236 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

69. fix) = x-e-'^

fix) = -jc^e--- + 2x6''^ = xe'^il - x) = 0 when jc = 0, 2.
fix) = -e''i2x - x^) + e-^(2 - 2x)

= e-^ix^ - 4x + 2) = 0 when;c = 2 ± V2.
Relative minimum: (0, 0)
Relative maximum: i2,4e~^)

x = l± Jl

y = (2 ± v^2^-(2±>^
Points of inflection: (3.414, 0.384), (0.586, 0.191)

°(2±V2,(6±4V2)c-"*^')

71. g{i) = 1 + (2 + t)e-'

. g'ii) = (1 + t)e-'
g"(i) = te''

Relative maximum: (—1,1 + e)
Point of inflection: (0, 3)

(-1,3.718)

{-\.\*e)

^0.3)

73. A = (base)(height) = Ixe'"'
dA

dx

= -Ax^e''^ + le-^

= 2e-^(l - Ix^) = Owhenx =

72

A = J2e~''^

75. y =

1 + ae-''"'

-x/b

.A-i'

,a>0,b>0,L>0
at

,-xlb

y =

(1 + ae-^l»Y (1 + ae-^I^Y

y"=

(1 + ae-''"'Y
(1 + ae-'/>')\~e-''^\ + 2(y <?--/* V^g-V*

(1 + ae-'''>'f

^ Lae-^"'[ae-''/'> - 1]
(1 + ae-'/^f fc2

y"= Olfae-"'" = 1
L

—r- = In - => j: = fo In a

b \aj

yib In a) =

1 +ae-(i'in«)/* 1 +a(l/a) 2
Therefore, the y-coordinate of the inflection point is L/2.

Section 5.4 Exponential Functions: Dijferentiation and Integration 237

77, e"^ = X =^ fix) = X- e'"
fix) = 1 + e"^

X, = 1

X2 = x,- |r4 = 0.5379

79. (a)

' f'(x2)

0.5670

^4 = *3 - TTfH " 0-5671
/U3)

We approximate the root of /to be .1: = 0.567.

(b) When x increases without bound, l/x approaches zero,
and e'''-' approaches 1. Therefore, f(x) approaches
2/(1 + 1) = 1. Thus, /(.r) has a horizontal asymptote
at y = 1. As j: approaches zero from the right, l/x
approaches co, e^^' approaches 00 and/(.r) approaches
zero. As x approaches zero from the left, l/x approach-
es — 00, e'/^ approaches zero, and/(.r) approaches 2.
The limit does not exist since the left limit does not
equal the right limit. Therefore, x = 0 is a
nonremovable discontinuity.

81.

h

0

5

10

15

20

P

10,332

5.583

2,376

1,240

517

]nP

9.243

8.627

7.773

7.123

6.248

(a) 12

y = -0.1499/! + 9.3018 is the regression
line for data (h. In P).

(c) 12.000

gOh

(b) In P = ah + b

P = pnh + b —

P = Ce^, C = e"
For our data, a = -0.1499 and C = e"0'8 = 10.957.7
P = 10,957.7e-o'''99''

(d) ^ = (10,957.71)(-0.1499)e-''"'9'"
all

= - 1642.56e-o '«»'■

For/i = 5,^ = -776.3. For /! = 18,^ « -110.6.
ah all

83. fix) = e'^fio) = 1
fix) = \e^'\fXQ) = i

fix) = iw^/"(0) = \

P: jf^'^

P,(;f) = 1 + 2(^-0)=:^+ l.PilO) = 1

P,'W=|,P/(0)=|
P2W = 1 + |(-v - 0) + |(.x- - 0)^- = I + ^ + 1, P,(0) = 1

P/W = \, Pz'tO) = ^

The values of/, f ,, Pj and their first derivatives agree at x = 0. The values of the second derivatives of/ and P;

agree at .v = 0.

238 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

85. (a) y = e^

/

(c) y = e'

x^ x^

/

>

Qo) y = e

^2 = 1 + ^ + ( y

^ ^

/

87. Let u = 'bx, du = 5 dx.

/

e ' 5dx = e^ + C

91.

Jxe-^dx=-|J,

e-'^{-2x)dx = -|e"^ + C

89. Let u = -2x,du = -2dx.

ju^=4|U(-2)^=[4.-

1 e^ _ 1

93. I ^dx = 2 I e^(-^ W = 2e-^ + C

95. Let H = 1 + e"-', du = -e""^dj:

I ^ "^ dx = - \-—^-^dx = -ln(l + e-^) + C = ln( ^ . ) + C = a: - ln(e^ + 1) + C
J 1 + e"^ J I + e-' ^ ' \e^+ 1/

3 3

97. Let u = -, du = — :;dx.
X X-

/:f-4H-i)-

-igSAl

'I

{e^ - 1)

99. Letu= I - e^,du = -e'dx.

le'JT^^dx = - (1 - e'y'-(-e')dx

= -^(1 - e'f'^ + C

101. Let « = e^ - e"-', <^u = (e^ + e-'^)dx.

/

e* — e

-iic = In e"^ - e"^ + C

103. r J" dx= \5e-^dx- \e-^dx

= -T^"^ + e"-* + C

Lsinm^Qs 7^^ = _Lsin,rx(^

COS T7;c) lie

— rtSin 7j:r

+ c

Y

107. I e-^Xw.(e-^)dx= - [tan(e~^)](-e-"^)att
= ln|cos(e-^)| + C

Section 5.4 Exponential Functions: Differentiation and Integration 239

109. Let u = ax^, du = lax dx. (Assume a i= 0)
y = I xe"^' dx

= \xe^(

— e'^(2ax) dx = —e^ + C
2a j 2a

/i

111. fix) = \^(e' + e-^)dx = ]-(e- - e'^) + C,

/'(O) = C, = 0
"l

/w

/:

^{e' - e-')dx = -(e^ + e"^) + C^

/(O) = 1 + Cj = 1 => Co = 0

f(x} = -(C + e-')

113. (a)

(0, d;

(b) ^ = 2e-^/% (0, 1)

= \le-''^dx= -4 e

1

>-= \2e-''^dx= -A\e-''^\--dx

= -4e-^/2 + c
(0, 1): 1 = -4e° + C= -4 + C => C = 5

t

115. I £^dA:=e' =e5-l = 147.413

76

117. I xe-'^^'^dx =

I

.lp-^/4

= -le-^l^- + 2 « 1.554

119. (a) /(m - v) = e"-" = (£")(£-") =
(b) /(fcjc) = e^- = (e')^ = [/W?.

e" _ /(«)

(-6

121. 0.0665 e-o.oi39Cr-48)^^,
J48

Graphing Utility: 0.4772 = 47.727c

123.

e'dt >
Jo Jo

\dt

e^-\>x^!-e^>\+ X for.v > 0

125. /(.v) = e". Domain is (—00. 00) and range is (0. oc).
/is continuous, increasing, one-to-one. and concave
upwards on its entire domain.

lim e'' = Q and lim e'' = 00.

a — » - 00 j:— >oo

127. Yes. fix) = Ce^.Ca constant.

129. e ' >

Jo

e'^dx > 0.

240 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

131. fix)

\nx

(a) f'(x) = ^ — = 0 when x = e.

On (0, e),f'{x) > 0 => /is increasing.
On {e, oo),/'(jc) < 0 => /is decreasing.

(b) For e < /4 < S, we have:

In A InS

B\t\A > A\nB
InA^ > InS''

A« > B^.

(c) Since e < tt, from part (b) we have e'^ > tt"".

Section 5.5 Bases Other than e and Applications

■-(r

' -^(r

5. log^l = log2 2-3 = -

At fo = 6, y =

(if

_ 1

~ 4

At fo = 10, y =

10/7

- 0.3715

7. log, 1 = 0

9. (a) 23 = 8

11. (a) logioO.Ol = -2

\ '

log28 = 3
(b) 3-'=^

10-2 = 0.01

(b) logo^8= -3

0.5-3 = 8

l0g3T=-l

%-' = 8

13. y = 3^

X

-2

-1

0

1

2

y

1

9

1
3

1

3

9

\ 1 — >-»

15. v =

X

-2

-1

0

1

2

y

9

3

1

1

3

1

9

17. h{x) = 5^-2

JC

-1

0

1

2

3

y

125

X

25

i
5

1

5

/

4-

/

3-

/

2-

1

\

1-

w.

1 2

3

A

*

Section 5.5 Bases Other than e and Applications 241

19. (a) logio 1000 = X

W = 1000

.t = 3

(b) logioO.l =x

i(y = 0.1

JC= -1

23. (a) x^- x = log; 25

X- - X = logj 5- =

2

;c2 - X - 2 = 0

(x + 1)U - 2) = 0

X = -10R;c = 2

25. 32^ = 75

2jcln 3 = In 75

. = 1'"^5^,Q..

2 In 3

21. (a) log3;c= -1

3-' =x

I
-f = 3

(b) log2J: = -4

2-" = X
1

(b) 3a; + 5 = log2 64
3jc + 5 = log2 2* = 6
3.r = 1

27. 23--' = 625

(3 - .r)ln 2 = In 625
In 625

3 - x =

In 2

3 -

In 625
In 2

-6.288

29. |l+«ff^3
12rln|l +~) = ln3

f =

1

In 3

12 /, ^ 0.09
T+I2-

== 12.253

31. logjU - 1) = 5

.r - I = 2^ = 32
.r = 33

33. logj x~ = 4.5

;c2 = 3^-5

.t = ±v^^ ±11.845

35. gU) = 6(2'--)-25
Zero: x= -1.059

^

(-1.059.0)

/

\

X.^

37. /lU) = 32 logioU - 2) + 15
Zero: 5 = 2.340

242 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

39. f(x) = 4^
g{x) = log4;c

x

-2

-1

0

i

2

1

fix)

1

16

1
4

1

2

4

x

1
16

1

4

1

2

4

gW

-2

-1

0

i

2

1

41. /U) = 4^

/'(x) = (ln4)4-

43. >- = 5^-2
ax

45. g{t) = t^2'

g'it) = tHln2)2' + (2t)2'
= r2'(rln2 + 2)
= 2'ti2 + t\n2)

47. /i(e) = 2-9cos ttO

h'ie) = 2-s(-7rsin ttO) - (ln2)2-»cos trd
. = -2-^(In 2) cos -770+ TTSin Trd]

49.

^
^

logs -'^

1
xln3

51. fix) = log.

/'W

55. git)

8 'it)

- 10S2

X - 1

= 2 log2 x - logj ix - 1)

2
xln

1

2 (;t-l)ln2

X- 2

(In 2)a:(jc - 1)

lOlog^r 10 /In A
r ln4V r /

10

rr(lA) - Inrl

In 4

L f' J

10

r, ,„ .1

5

/2ln4'-

;Mn2

(1 -Inf)

53. y = logj V^^^^ = I log; ix^ - 1)

^ = 1 2;c ^ X

dx 2' [x- - l)ln5 ~ U^- l)ln5

57.

y = X

'./x

In y = - In j:

X

^ = ^(1 - Inx) = 2x(2/-'-2(l - ln;c)

59. y=ix-2Y*^

Iny = (x + l)ln(x - 2)
1

Kl)=<-«u-.

+ InU - 2)

^
^

X + I

-^ + Hx - 2)

x - 2

= U - 2)--

i^

+ InU - 2)

61.j3^^ = |^

+ C

Section 5.5 Bases Other than e and Applications 243

In 2 L 2j

2 In 2 In 4

5./.5-=^=-l|5

65. \x5'''~ dx= -]r\5 '\-2x)dx

1\5-

\ll In 5
-1

+ C

2 In 5

(5-^'-) + C

r 3^

J 1 +3-

67. I . : d[»;, M = 1 + 3-', c/m = 2(ln 3)3^ dx

1

2 In 3)3-

2 In 3

m

d[x

^ ln(l+3^) + C

2 In 3

69. ^ = 0.4^/3^ (0, \
dx \ 2

Jo.4^^^^. = 3[o.4'/'(|

dx

In 0.4

0.4^/3 _,_ (; = 3(in 2.5)(0.4K3 + ^

.V = 3 In 2.5(0.4)^/3 + _ _ 3 ]„ 2.5

(a)

3(1 - 0.4^3) 1
In 2.5 2

71. Answers will vary. Example: Growth and decay problems.

(b)

7

73. y

(8,3)»

• (2.1)

(1,0)

-»-\ 1 1 1-

X

1

2

8

y

0

1

3

(a) y is an exponential function of -v: False

(b) y is a logarithmic function of .r: True; v = log^.v

(c) X is an exponential function of v: True, 2* = x

(d) y is a linear function of x: False

2 4 6

75./(.^) = log,x ^ f'(x) =
six.

1

xln2

x' =^ g'(x) = x^(l + In.x)

[Note: Let y = g(x). Then: In y = In a-"^ = x In .r

— y = A' • — h In .V
y .V

y' = y(l + In.v)

y' = x'^(l +lnA) = g'(j:).]

^(a) = .ir => ^ '(.t) = 2x

Ma) = 2' =» A''(.v) = (In 2)2^

From greatest to smallest rate of growth:
gix), k{x), h{x)J(x)

77. C{t) = P(1.05)'

(a) C(10) = 24.95(1.05)'°

-$40.64

(b) ^ = P{\n 1.05)(1.05)'
at

j^

Whenf = 1: — = 0.051P
dt

When f = 8: — = 0.072P
dt

(c) ^ = (In 1.05)[P(1.05)']

= (In 1.05)C(r)

The constant of proportionality
is hi 1.05.

244 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

79. P = $1000, r = 35% = 0.035, t = 10

/ 0 035\"'"
A= lOOofl +^^)

A = 1000e<oo35)(>o) = 1419.07

n

1

2

4

12

365

Continuous

A

1410.60

1414.78

1416.91

1418.34

1419.04

1419.07

81. P = $1000, r = 5% = 0.05, ( = 30
0.05 yo"

A = 10001 1 + 1

A = lOOOeCo^'^o = 4481.69

n

1

2

4

12

365

Continuous

A

4321.94

4399.79

4440.21

4467.74

4481.23

4481.69

83. 100,000 = P^-oo^i => p = lOO.OOOe^

t

1

10

20

30

40

50

p

95,122.94

60,653.07

36,787.94

22,313.02

13,583.53

8208.50

85. 100,000 = P 1 +

OOSV^'
12

P = 100,000 1 +

0.05 V'^
12

t

1

10

20

30

40

50

p

95,132.82

60,716.10

36,864.45

22,382.66

13,589.88

8251.24

0 06V365)(8)

87. (a) A = 20,000( 1 + ^j » $32,320.21

(b) A = $30,000

(c) A = 8000 1

0.06V

,(365)(8)

+ 20,0001

365/
$12,928.09 + 25,424.48 = $38,352.57

0.06Y36s)(4)
365/

= $34,985.11
Take option (c).

(365)(8)

+ 1 +

0.06V
365/

(365){4)

+ 1

89. (a) lim 6.7e<-*8'>/' = e.le" = 6.7 million ft^

r-»oo

322.27

(b)

V-

-(48.1)A

V"(20) == 0.073 million ftVyr
V'(60) = 0.040 million ftVyr

91. y

300

3 + ne'0.062Sjc
(a) '<»

(b) If ;c = 2 (2000 egg masses), y = 16.67 == 16.'

(c) If y = 66.67%, then x = 38.8 or 38,800 egg masses.

(d) y = 300(3 + ne-oos^^o:)-!

,_ 318.756-°°^^^
^ (3 + 17e-oo625x)2

„_ 19.921875e-'"^^^(17g-°'^^^ - 3)

^ (3 + 17e-0"625^)3

17^-0.0625^ - 3 = 0 => ;c = 27.8 or 27,800 egg masses.

Section 5.5 Bases Other than e and Applications 245

93. (a) B = 4.7539(6.7744)'' = 4.7539e''»32^

(b) '20

(c) SV) = 9.0952e'»'3M
5 '(0.8) == 42.03 tons/inch
5 '(1.5) = 160.38 tons/inch

95. (a) f(t) dt = 5.67
Jo

I g{t) dt - 5.67
Jo

h(t) dt - 5.67
Jo

(b)

(c) The functions appear to be equal: fit) = git) = hit)
Analytically,

/3\2r/3

/W=4| =4

4(^)' = ^(0

;i(t) = 4^-0.653886, = 4|-g-0.653886J, = 4(0.52002)'

^9i/3Y

«(f) = 41

4(0.52002)'

No. The definite integrals over a given interval may be
equal when the functions are not equal.

97.

=

IQQOe-"'^'

dt

Jo

=

\ 2000 „,1
.-0.06

10
0

=s;

$15,039.61

99.

101. False, e is an irrational number.

t

0

1

2

3

4

y

1200

720

432

259.20

155.52

y = Cik!)

When r = 0. y = 1200 => C = 1200.

y = 1200(/f )

720 432 259.20 155.52
1200 - °-^- 720 - °-^' 432 " "■^- 259.20

Let k = 0.6.

y = 1200(0.6)'

103. True.

105. True.

figix))

= 2 + e'-'f^-^

£M = .^and£[e-1=-

= 2 + .r - 2

= X

g{f{x))

= ln(2 + e- -

e' = e'" when .v = 0.

= In e^ = X

ie°){-e-°) = -1

= 0.6

246 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

f=^>(! -)-(«) =^

^'^^=idt^'~

y[{5/4) -y] 25 5

lny-ln(|-y)=|f+C

/(r(vi7^)'^^^B*

g(2/5)t + C = Q g(2/5)(

(5/4) - y
y{0) = 1 =^ C, = 4 =» 4e'^5>'

(5/4) - y

4g(2/5)/5 _ ^\ ^ ^ 2^(2/5), = 4g(2/5);y + y = (4e(2/* + l)y
5e(2/5>' 5 1.25

4^(2/5)t +1 4 + g-0.4/ J + 0.256-°"

Section 5.6 Differential Equations: Growth and Decay

l.^ = . + 2 3. ? = . + 2

abc dx

X?- dy

• >> = I (.X + 2)dx = y + 2x + C jj^ = A

■■ ■ ■•■ )jh'^y = j'^

Injy + 2| = jc + Ci
>- + 2 = e^**^' = Cr'
>> = Ce^ - 2

,_5jc 7. y' = V^ );

W = 5x y

\yy'dx= ISxdx ■- \—dx= iVxdx

lydy=\5xdx . j^^}^'^

i,2 = |,2^C, ln, = f.3/2 + C,

,2 _ 5,3 = c ^ = ^'^^^'"'"'^-

= gC, g(2/3);(5/^

Section 5.6 Differential Equations: Growth and Decay 247

9. (I + x^)y' - 2xy = 0

1 +.t2
2x

y \ + x^
jjdx = jj^^dx

- dx

(dv^ f 2x

J y Ji+x'

Inv = ln(l + X-) + Ci
\ny = ln(l + x^) + In C
Iny = lnC(l + x'^)
y = C(l + x^)

11.

dQ^k

dt t'

dt

dQ= -- + C

t

dN
13. =r- = k{250 - s)
ds

If-/

k{250 - s)ds

dN = -x(250 - 5)2 + C

A' = -t(250 - ip + C

15. (a)

(b)

di
dx

_dy_
y-6

x(6->), (0,0)

n

Inb - 6| = ^- + C

>- - 6 = e--""/2 + c = c,e-'^/2
.V = 6 + C.g-'^'/^
(0, 0): 0 = 6 + Ci =^ Ci = -6 ^ y = 6 - 6e-^/2

17. ^4,. (0.10,

/-/■

rrfr

y = -r^ + C

10 = -(0)2 + c

y = -r2+10

C= 10

19. |=_i,, (0,10)

/f^H

\dt

In V

f + Ci

y = e-(r/2) + C, == gC, g-,/2 = Cg-r/2

10 = Ce° :^ C = 10
y = lOe-'/^

y = Ce'-'' (Theorem 5.16)
(0, 4): 4 = Ce" = C

(3,10): 10 = 4e3* =4. /t = | ln[|

When x = (,,y = Ae^'^ ^^(^nm = 4^^(5/2)^

23. ^ = kV

dt

V=Ce*' (Theorem 5.16)
(0, 20,000): C = 20,000

(4, 12.500): 12,500 = 20.000e* ^ <•' = ^ ln(|
When r = 6, V = 20,000e '-""''■ *^'*' = 20,O0Oe^^5/8)'«
= 20,000(|j " = 9882.118

248 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

25. V = C<^\ ( 0, -1, (5, 5)

^ = 2

, = -^

5 = 2^*

In 10

k =

y = i g(ta 10/5), = l(10'/5} or 3^ = ieO.4605,

29. A differential equation in x and y is an equation that
involves x, y and derivatives oiy.

27.

y = Ce^, (1,1), (5,5)

1 = Ce*

5 = Ce5*

5Ce* = Ce5*

Set = g5«.

5 = 6^"

/t = ^ - 0.4024

4

■y = Ce°-*'2'"

1 = CfiO'W^i

C « 0.6687 (C = 5-1'"')

y - 0.6687e0-^24,

31.

rfy 1
-dx = -2^

dy

— > 0 when xy > 0. Quadrants I and in.

33. Since the initial quantity is 10 grams, y = iOeIi"(i/2)/i62o],_ ■^^hen / = 1000, y = iQ^^iUD/iiiojvxa) « 552 grams. When
/ = 10,000, >' = iOe['"<i/2Vi620](io,ooo) « 0.14 gram.

35. Since >- = c^^^U2)/\(,io],^ ^g ^ave 0.5 = Cet'"('/2Vi620]Cio,ooo) ==^ c = 36.07.
Initial quantity: 36.07 grams.
When t = 1000, we have y = Ce[i"(i/2)/i62o](iooo) ^ 23.51 grams.

37. Since the initial quantity is 5 grams, we have y = 5.0e['"*'''^'^^'^*.
When? = 1000,>' = 4.43 g.
Whenf= 10,000, v=» 1.49 g. ''

39. Since y = Ce^^iUD/i'i.im ^ ^e have 2.1 = Cet"'"/2»/24.360](iooo) ==> c = 2.16. Thus, the initial quantity is 2.16 grams. When
t = 10,000, y = 2.16e['"('/2'/2«*]«'°'0'») = 1.63 grams.

dy
41. Since— = ky. y = Ce'" or y = y^e'".

1

7% = yoe

1620*

-In 2
1620

y = y g-'ln2)//I620_

When t= 100, y = y^e''^^" 2)/i6.2 „ ^^(0.9581).
Therefore, 95.81 % of the present amount still exists.

43. Since A = 1000e°°^', the time to double is given by
2000 = 1000e°°'5' and we have

2 = gO.oer
In 2 = 0.06f

' = o:o6'"-^^y'''''-

Amount after 10 years: A = 1000e*°°«'<'°' = $1822.12

Section 5.6 Differential Equations: Growth and Decay 249

45. Since A = 750e" and A = 1500 when t = 7.75, we have
the following.

1500 = 750s''"'-

r = :^ = 0.0894 = 8.94%

7.75

Amount after 10 years: A = 750eO"89''(io) = $1833.67

47. Since A = 500e" and/4 = 1292.85 when t = 10, we have
the following.

1292.85 = SOOe'O'-

ln(1292.85/500)

'■ = lo

The time to double is given by
1000 = 500e'"»5o,
In 2

= 0.0950 = 9.50%

0.095

= 7.30 years.

0 075\<i2)(20)
49. 500,000 = P| 1 + ^^ j

0.075 Vz-w

P = 500,000 1 +

= $112,087.09

12

51. 500,000 = P 1 +

P = 500,000 1

0^yi2K35)

12

0.08
12

$30,688.87

53. (a) 2000 = 1000(1 + 0.07)
2 = 1.07'
ln2 = r In 1.07

In 2

t =

In 1.07

=» 10.24 years

/ 0.07 V^'
(b) 2000 = 1000( 1 + —1

2=1 +

0.007
12

ln2 = 12fln 1 +

0.07
12

In:

121n(l + (0.07/12))

55. (a) 2000 = 1000(1 + 0.085)'
2 = 1.085'
ln2 = rlnl.085
In 2

9.93 years

In 1.085

= 8.50 years

(b) 2000 = 1000( 1 +
2 = (l +

0.085
12

0.085\'2'

12

ln2 = 12rln 1 +

0.085
12

/ =

In 2

12

1 (i ^0-085\

8.18 years

(c) 2000

2
In 2

f =

(d) 2000

2
In 2

/ 0.07 V*"

0.07 Y«'
365/

3651n(ll"(0.07/365))^^-^°y^^

■- lOOOeCO""'
; gO.cni

' O.Olt

In 2
■■ — = 9.90 years

(c) 2000 = 1000 1 +

0.085 V"'

1 +

365 /

0.085 y*"

365 /

, -, -,.. , 1 , 0.085\
ln2 = 365rln( 1 + "ttt" I

t =

365

(d) 2000

0

Inh +
lOOOfOO*"

,0.085;

ln2 ^ ,^

365

e'

hi 2 = 0.085f

_ ln2
' ~ 0.085

= 8.15 years

250 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

57. P = Ce*' = Ce-oow

P(-l) = 8.2 = Ce-o«"<-" ^ C = 8.1265

P = 8.1265e-o™9'

P( 10) = 7.43 or 7,430,000 people in 2010

59. P = Ce'" = Cgoo^^'

P{-\) = 4.6 = Cg0036(-i) => c = 4.7686

P = 4.7686eO'>36'

/'(lO) = 6.83 or 6,830,000 people in 2010

61. If fc < 0, the population decreases.
If ^ > 0, the population increases.

63. P = Ce^ (0, 760), (1000, 672.71)

C = 760

672.71 = 760e'««^

ln(672.7 1/760) . nmmo-7
^ = loOO -0000122

P = 760e-0'»0'22x
When X = 3000, P = 527.06 mm Hg.

65. (a) 19 = 30(1 - e^"*)
30e20t = 11

, = MM„_0.0502
N^ 30(1 - e-00502,)

(b) 25 = 30(1 - e-oo502» )

„- 0.0502/ = _

6

-In 6
^ = ^0050^"'^'^^^^

67. 5 = Ce*/'

(a) 5 = 5 when f = 1

5 = Ce*
lim Ce'/' = C = 30

r — »oo

5 = 30e*

A: = Ing « -1.791S

5 » 30e-'™i8'''

(b) When f = 5, 5 = 20.9646 which is 20,965 units.

(C) 30

69. Ait) = V(r)e-o'0' = 100,000e°^-^e'-''"" = 100,0006"^-^^°""

^ = 100,OOof^ - O.loV"'^''""" = 0 when 16.
dt \Jt I

The timber should be harvested in the year 2014, (1998 + 16). Note: You could also use a graphing utility to graph A{t)
and find the maximum of A(r). Use the viewing rectangle 0 < x < 30 and Q < y < 600,000.

71. ;8(/)= 101og,of/o= 10

/n

10"

(a) /3(10-"') = lOlogjorrr;^ = 20 decibels

(b) )3(10-9) = lOlogioTT^TTi = 70 decibels

10-

10""
(c) /3(10-") = lOlog.o-y^TT^ = 95 decibels

(d) /SdO-") = lOlog.o-— Y^ = 120 decibels

10"
10-

73. R = '"/.„°. / = e«in 10 = io«

(a) 8.3 =

In 10

In/- 0

In 10
/ = 1083 « 199,526,231.5

^ _ g2Rln 10 — gifln 10 _ /^Slu 10)2 = (10" )2

Increases by a factor of e^ "^ '° or 10'*.
^ dR 1

rf/ / In 10

Section 5.7 Differential Equations: Separation of Variables 251

75. False. \iy = Ce^, y ' = Cke^ i- constant.

77. True

Section 5.7 Differential Equations: Separation of Variables

1, Differential equation: y' = ^y
Solution: y = Ce''"^
Check: y' = ^Ce^'' = 4v

3. Differential equation: y" + v = 0
Solution: >" = C, cos x + Cj sin j:
Check: y' = — Cj sin or + Cj cos x

y" = —C^ cos x — C, sin x
y" + y = — Ci cos X — Cj sin JT + Cj cos j: + C2 sin j: = 0

5. y = — cos -t Injsec x + tan j:
1

y' = (—cos a:)

sec jc + tan ;c
cos X

{secx • tanj: + sec'j:) + sin j: Injsec.x + tanjcj

sec j: + tan ;c
= — 1 + sin JT ln|sec -t + tan.ic
1

secx)(tan.t + secj:) + sin x In I seer + tan a: I

y"= (sinx)

(sec.r • tan.t + sec^x) + cosxln|secx + tan.r|

sect + tan.r
= (sinx)(secjt) + cosxln|secj: + tanx|
Substituting,

y" + y = (sin x) (sec x) + cosxln|secx + tanjc| — cos j: ln|secx + tanx|
= tanx.

In Exercises 7-11, the differential equation is j'''*' — 16y = 0.

y = 3 cos X

yW - i6y = -45C0SJC ^ 0,
No.

y = e

v(4)

Yes.

16y = 16e-^ - \6e-^ = 0,

11. y = Cifi^ + C,)?--"^ + C3 sin 2x + C4 cos 2x

y*"' = 16Cie^ + leCjg-^ + I6C3 sin Ix: + 16Q cos 2x

yW - 16y = 0,

Yes.

252 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

In 13-17, the differential equation is xy ' — 2> = x^e'.

13. y = X'^, y' = 2x

xy' -2y = x{2x) - 2(;c2) = 0 ¥= x^e'
No.

15. y = xH2 + e^), y' = xV) + 2x(2 + e^)

xy' - 2y = x[xV + Ixe' + 4x] - 2[xV + 2x^] = xV,
Yes.

17. y = lnx,3'' = -

xy ' — 2y = X

2 In X ^ x^e^. No.

19. V = Ce*^

dy
dx

= Cke'^

Since dy/dx = O.OTy, we have Cke'^' = O.OTCe*".
Thus, /t = 0.07.

21. y- = Cx^ passes through (4, 4)
16 = C(64) => C = i
Particular solution: y^ = j x^ or 4^^ = x^

23. Differential equation: 4yy ' — x = 0

General solution: 4y~ — x- = C

Particular solutions: C = 0, Two intersecting lines
C = ±1, C = ±4, Hyperbolas

\.

.^

^

^^

^"**^*>^-.

p^

"^^--^

"-H^,^

c = ~\ .^

^v

— • ^^*T' ' '

C = -4

25. Differential equation: y ' + 2>' = 0
General Solution: y = Ce~^
y' + 2y = C(-2)e-^ + 2{Ce-^) = 0
Initial condition: y(0) = 3, 3 = Ce° = C
Particular solution: y = 3e~^

27. Differential equation: y" + 9y = 0

General solution: y = C, sin 3a: + Cj cos 3x
y' = 3C| cos 3x — 3C2 sin 3x,
y" = - 9C, sin 3x - QC, cos 3x
y" + 9y = (-9C, sin 3;c - 9C2 cos 3x) +
9(C, sin 3x + C cos 3x) = 0

Initial conditions: y(— -) = 2, y'( — j = 1

2 = C| sin( ^ 1 + Co cosi —

y ' = 3C] cos 3x — 30, sin 3x
1 = 3C, cos( y 1 - 3C2 sin( -|

C. = 2

= -3C,

C, =

1

Particular solution: y = 2 sin 3x — - cos 3x

Section 5.7 Differential Equations: Separation of Variables 253

29. Differential equation: x^y" - 3xy' + By = 0
General solution: y = C,x + Cjx'
y' = C, + 2C2x'^,y" = 6C2X
xy- 3xy' + 3y = x^iec^x) - 3x(C, + BQjc^) +

3{C,x + C^x^) = 0

Initial conditions: y(2) = 0, y '(2) = 4
0 = 2C, + 8C2

C, + 4C2 = 0
C, + I2C2 = 4

Q =2- Ci = -2

Particular solution: y = — 2jr + — jc^

dx

= /

3x^dx = x^ + C

33. $ =

tic 1 +;c2
(m = 1 + x^, du = 2x dx)

J 1 + .t2

d:t = ^ln(l + x^) + C

35.^ = ^^^=1-2
or j: X

/[-!]■

y = I 1 --Idlx

jc - 2 InUI +C = x-\nx~ + C

dy
31. -r- = sin 2j:
(tc

^/^

y = I sin 2x (ic = — - cos 2x + C

{u = 2x,du = 2dx)

dy

39. -;- = xjx - 3 Let u = Vj: - 3, then x = u- + 3 and dx = 2u du.
dx

xjx -3dx= («2

= uy

+ 3)(M)(2M)rfM

2 Km" + 3M2)rf„ = 2^ J + M^j + C = I {x - 3)'/- + 2(x - 3y'- + C

Al.^ = xe^
dx

43.

-'[

xe^dx = -e^ + C

(m = x'^, du = 2xd!.t)

dy x
dx y

lydy = Ixdx

2 2 ^'
y2-;c2 = C

45. ^ = O.OSr
as

j^ = j 0.05 ds

In |r| = 0.05i + Q

47. (2 +:t)y' = 3y

i y J 2 + .r

(ir

In y = 3 ln{2 + .t) + In C = In C(2 + .r)^
y = C(x + 2)3

254 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

49. yy' = sin jc

= sin ;c dx:

v ^ = s

=T- = - cos JC + C,
y^ = -Icosx + C

dy

51. yr^"4? ^ = X
dx

dy =

VT-4?

dx

\-'U

4x^

dx

4/<.-

4x^)-'^H-Sxdx)

y = -i(l - 4;c2)'/2 + C

53. _y In j: — xy' = 0

lf = /

— (fa \ u = In X, du = —

X \ X

liiy = -(ln;c)2 + C,

y = g(l/2)(bu:F + C, = (^g(lnx)V2

55. yy' - e^ = 0

jydy = j,

e^dx

y2 = 2e^ + C
Initial condition: yiO) = 4, 16 = 2 + C, C = 14
Particular solution: y^ = le' + 14

57. >'(x + I) + y' = 0

/?-/'

Iny = + C,

Initial condition: ^(-l) = 1, 1 = Ce-'^^ C = e^^^
Particular solution: v = e[i-(A:+i«/2 = g-(.r^+2r)/2

59. ^(1 +x^)^ = xil +y^)
^ dy = T— — ^ dx

1 +>'-•' l+x^

iln(l +y2) = ^ln(l +;t2) + C,

ln(l + y2) = ln(l + x^) + In C = ln[C(l + x^)]
\ +y^= C{1 + x^)
.v(0) =73:1+3 = C^C = 4
1 + 3,2 = 4(1 + ;r2)

;y2 = 3 + 4;t2

61. — - = Mv sm V-

|3V

— = V sin v^
J " J

dv

Ihm = —-cos v^ + Ci

Initial condition: u(Q) = 1, C =

-1/2

= ^1/2

63. dP - kPdt = 0

/f = f

In P = to + C,

Initial condition: ^(0) = Pg, Pg = Ce° = C
Particular solution; P = P^e'^

Particular solution: u = e(i-':osv^)/2

Section 5.7 Differential Equations: Separation of Variables 255

65.

dy _ -9x
dx " 16y

\\6ydy= - \9xdx

By

-9

2 = -^x2 + C

9 25

Initial condition: y(l)= 1, 8= -- + C, C = —

-9 25

Particular solution: 8y- = ^t-jc^ + ^r.

•^2 2

16y2 + 9x^ = 25

67. m

_ dy 0 -y

/f=H

£tc (x + 2) - X

Iny = -2"^ + C,
y = Ce-^/^

69. /(.r,^) =;c3 - 4x>'2 + y3

f(tx,ty) = i'.c^ - 4rjcf'y^ + t^ y^

= r^U^ - 4x>'2 + y)
Homogeneous of degree 3

71. f{x,y) =
fitx, ty) --

xY

fx'Y

= t=-

xY

Homogeneous of degree 3

73.

f{x. y) = 2 In
/(rx, /y) = 2 In

xy
txty

= 21nr2j:j, =

2(ln f '

+ In

xy)

Not homogeneous

77.

>-' =

x + y
2x '

y = vx

V + X— =
dx

X + vx
Ix

•

dv

1 + V

2

- V

J dv ^

1 - V

'dx

X

-ln(l - v)2 = ln|;c| + In C = ln|Cx|
1

(1 - v2)
1

\Cx\

[1 - (y/x)]

X

2 = \CX\

= \Cx\

(x - y?

\x\ = C(x - y)2

75. f(x, y) = 2 In =

/(a,ry) = 21n— = 21n-
ty y

Homogeneous degree 0

79.

X - y

V = vx

y

X

+

y '

V +

dv
'Tx

=

X
X

+

XV
XV

V

dx +

rdv

^

\_

-

^dx

1 + V

f v+1 , {dx

-ln|v2 + 2v - 1| = -ln|.v| + In Q = in

C,

c

|v^

+ 2v-

^1=:;^

V-

_ V

c

+ 2- -

1

X-

.T

x-

|y- + 2v>'-.T-| = C

256 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

81.

V + X

y

dv

xy

— ^- y = vx

x-^ — y-

.2 _ ^2„2

dx x^ — xf-v

V dx + x dv

1 - V

-dx

/^-/^

-y-j - ln|v| = \n\x\ + InC, = ln|Ci.r|

-1

2v2

= ln|Ci;cv|

-x^
2y'-

= ln|C,);|

y =

= ce-'^'^y'

83. xdy ~ {Ixe'yl^ + y)dx = Q,y = vx

x{vdx + X dv) - {2x6'" + vxjdx = 0

je^'dv = jl

dx

e'' = In C,;c'
«>■/•' = In C, + In x2

e^A = C + lnx^
Initial condition: y(l) = 0, 1 = C
Particular solution: e^^' = 1 + In x^

85. \x sec- + y]dx - xdy = 0,y = vx

[x sec V + xv)dx — xiv dx + x dv) — 0

(sec v + v)dx = vdx + xdv

f , (^

cos V dv = —

sin V = In X + In C,

Initial condition: ^(l) = 0, 1 = Ce° = C
Particular solution: x = e*'"'^/-^*

87.^ = .
dx

/

A:(ic = -j;- + C

rfy
89.| = 4-y

In \4 - y\ = -X + Ci

4-y

y = 4 + Ce"-'

91. ^ = O.Sy.yCO) = 6

93. ^ = O.Olj'llO - j), y(0)

12

I I I I I I { If t I I I I
I I I I I I \ I I I I I I
I I I I I I Ifi I I / i I

/ y / / y ^ y J ^ / / /
y y y y ^ 1( y y y y y y

y y y y

^ y y y y

'■///////
,'/ ///////
■//////////

^y y y y y y y y y y

Section 5.7 Differential Equations: Separation of Variables 257

95. ^ = ky,y = C^
dt

Initial conditions: >'(0) = y^

>'(1620)

2

C

= ^-0

Jo
2

= ^-ofi'"*

t

ln(l/2)

1620

Particular solution: y = ;yQe-'(iii2)/i62o
When f = 25, v =» 0.989yo. y = 98.9% of yo-

99. ^ = ky(y- 4)

97. ^ = kiv - 4)
(fa

The direction field satisfies (dy/dx) = 0 along > = 4; but
not along y = 0. Matches (a).

The direction field satisfies {dy/dx) = 0 along y = Q and y = A. Matches (c).

101.

^ = /t(1200 - w)
dt

J 1200 - w J

ln|1200 - M'l = -fa + C,

1200 - w = e

_ ^-fa+C,

Ce-

w = 1200 - Ce"*'
^(0) = 60 = 1200 - C =* C = 1200 - 60 = 1 140
w = 1200 - 1140e-*'

(a) 1400

103. (a)

f-«--"

J IV - V J
-ln|W- v| = /tr + C,

V = W - Ce^*'
Initial conditions:

W = 20, V = 0 when ; = 0. and

V = 5 when r = 1 .

C = 20,k= - ln(3/4)
Particular solution:

V = 20(1 - ei"'3/4).) = 20( 1

(b) k = 0.8: / = 1.31 years
k = 0.9: t = 1.16 years
k= 1.0: t= 1.05 years

(c) Maximum weight 1200 pounds

V = 20(1 - e-0.2877,)

(b) .? = 120(1 - e-o-28^')(/f
= 20[? + 3.476 l<'-o-S'''

C

Since j(0) = 0. C « - 69.5 and we have
j«20/ + 69.5(e-°-2"'' - 1).

lim w = 1200

/-»oo

258 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

105. Given family (circles): x'^ + y"^ = C
2x + 2yy' = 0

Orthogonal trajectory (lines): y' = '

If'f

dx
x

Iny = \nx + \n K
y = Kx

107. Given family (parabolas): x^ = Cy

2x = Cy'

, _ 2x _ 2x _2y
C x^/y X

Orthogonal trajectory (ellipses):

\

4(

p

®

^y

lly dy = — \xdx

x^ + 2y^ = K

109. Given family: y'^ = Cx^

2yy' = 3Cx^

, _ 3Cx^ _ 3x2 /y2\ _ 3y

2y 2y \x? / 2x

Orthogonal trajectory (ellipses):

y =

2x

3Jydy^-2Jx

-x^ + Ki

3\y dy — —2\xdx

3y2

33,2 + 2x2 = K

111. A general solution of order n has n arbitrary constants
while in a particular solution initial conditions are given
in order to solve for all these constants.

113. M{x, y)dx + N{x, y)dy = 0, where M and A' are
homogeneous functions of the same degree.

115. False. Consider Example 2. v = ^c^ is a solution to
xy ' - 3y = Q,h\Ay = x^ + Ws not a solution.

117. False

f(tx,ty) = tV + t^-xy + 2
*t2f(x,y)

Section 5.8 Inverse Trigonometric Functions: Differentiation 259

Section 5.8 Inverse Trigonometric Functions: Differentiation

1. >> = arcsinjt
(a)

X

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

y

-1.571

-0.927

-0.644

-0.412

-0.201

0

0.201

0.412

0.644

0.927

1.571

(c)

(d) Symmetric about origin:

arcsin(— or) = — arcsinx

Intercept: (0, 0)

3. False.

1 TT

arccos - = T
since the range is [0, tt].

. . 1 77

5. arcsin - = —
2 6

1 -n-
7. arccos t^ = v

9. arctan

73 ^ TT

3 6

11. arccsc(-72)

4

13. arccos(-0.8) « 2.50

15. arcsec(1.269) = arccos 7-77-
\ 1.269

= 0.66

17. (a) sin( arctan 4) = 5

(b) sec( arcsin 5

5

5; ~ 3

19. (a) cot

-^

(b) CSC arctan! - —

11
5

21. y = cos(arcsin 2t)
6 = arcsin 2x

y = cos 0 = J\ - 4a-

23. V = sin(arcsec .r)

6 = arcsec x,Q < 6 < n. 6 i^

y = sinO

The absolute value bars on x are necessary because of
the restriction 0 < 6 < tt. 6 ^ tt/1. and sin 6 for this
domain must always be nonnegative.

260 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

25. y = tanf ;

X

arcsec -

0 = arcsec -

y = tan 6

Jx^'^

27. y = cscl arctan

X

72

6 = arctan

y = CSC I

29. sin(arctan 2r)

2x

yrT4?

/=«

Asymptotes: y = ±\
arctan lie = 0
tan e = Ir

sin e =

2x

vT+4j?

VT+T?

31. arcsin(3x — tt) = 5

3jc — 77 = sin^j)

X = 5[sin(5) + 77] = 1.207

33. arcsinV2x = arccos Jx

v^ = sin(arccos ~Jx)
Jlx = Vl -^ 0 < a: < 1

2x= 1 -;c

3a: = 1

1

^ = 3

35. (a) arccscx = arcsin- |jc| > 1

Let y = arccsc x. Then for

77 77

-— < y < 0 and Q < y < —,

csc>' = ;c => siny = \/x. Thus, y = arcsin(l/j:)
Therefore, arccsc j: = arcsin(l/j:).

1 77
(b) arctan x + arctan - = — , x > 0
X 2

Lety = arctan X + arctan(l/jr). Then,

tan(arctanx) + tan[arctan(l/x)]

tan V

1 - tan(arctanx) tan[arctan(l/x)]

X + (lA)

1 - x(iA)

X + (1/x)
0

(which is undefined).

Thus, >» = 77/2. Therefore, arctan x + arctan(l/x) = 77/2.

37. f(x) = arcsin(x — 1)
X — 1 = sin>'

X = 1 + siny
Domain: [0,2]

Range:

77 77
'2' 2

/(x) is the graph of arcsin x
shifted 1 unit to the right.

n-

-

f-

• /(2-f)

-1

-It'

■("•-f)

39. fix) = arcsec 2x

2x = secy

1
X = — sec y

Domain: (-°°. -|]. [|. '
Range: [0,f),(f,77]

^i-ol

Section 5.8 Inverse Trigonometric Functions: Differentiation 261

41. f(x) = 2 arcsinU - 1)
fix) = ^

Vl - (jt - If V2x-x^

43. g{x) = 3 arccos —

g'ix) =

-3(1/2)

Vl - U-V4) V^^^

45. /(jc) = arctan '

47. g(x)

arcsin 3x

Ma

•^'^""^ 1 + UVa") a

+ x2

g'W

x\hl J\ - ^yr) - arcsin 'hx

3;c - yi - 9j:- arcsin 3.t

49. /!(/) = sin(arccos t) = Vl - t-

//'«=|(l-^2)-'/^(-2f) = -==
2 Vl - r

51. y = X arccos j: - Vl - -t-

>> = arccos X —

VT^

(1 -.r2)-i/2(-2A:)

= arccos x

53. y = - - In + arctan x

2\2 j: - 1

= T[ln(jr + 1) - ln(.t - 1)] + - arctan jc

55. y = X arcsin x + Vl — x-

dx "VVT

+ arcsin A-

VT^

dy _ 1/ 1

1 \ ^ 1/2

dx 4\j; + 1 j: - 1 / 1 + x^ I - x^

1

57. y = 8 arcsin

x ArVl6 — x^

59. J = arctan .t +

1 +x'-

y'=2-

1^ - ^^\ ■'' - f(16 - x^)-/2(-2x)
Vl - U/4)- 2 4

8

Vl6^^

Vl6 - x2

16 - (16 - x^) + x^

2V16 - X-
X-

2V16 - X- Vl6 - x2

3' =

1 (1 + X-) - x(lt)

1 + x^ (1 + x^y-

(1 + X-) + (1 - x^)
(1 + x=)=

(1 + x^r

61. /(x) = arcsin x, a = -

fix)

yr

fix) =

(1 - x2)3/2

.,,,., (l).,.(i)(,_i).

1\ 77 , 2V3/ 1

6^ 3

""2

^2W=/U +/'!

1\ . 1 ,,/l

-^f/lfH-

I\- TT , 2v/3/ 1

^¥('4J

262 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

63. fix) = arcsec x - x

1

fix) =

\x\V^^^l

= Owhen \x\^x'^ - 1 = 1.
jc2(;c2 - 1) = 1

X* — x^ — I = 0 when x'^ = or

/i + Vs

Jf = ± W 2 " ±1-272

Relative maximum: (1.272,-0.606)
Relative minimum: (- 1.272, 3.747)

65. f{x) = arctanx — arctan(j: — 4)

\ + x^ 1 + U - 4)^

1 + .^2 = 1 + (;c - 4)2

0 = -8jc + 16

x = 2

By the First Derivative Test, (2, 2.214) is a relative
maximum.

67. The trigonometric functions are not one-to-one on
(— oo, oo), sot their domains must be restricted to
intervals on which they are one-to-one.

69. y = arccot x,Q < y < tt
X = cot V

1

tany = -

X

So, graph the function

y = arctani - J for ;c > 0 and y = arctan

©-

ioxx < 0.

\

71. (a) cot 0 = -

6 = arccoti -

(b)

de

dt

5 dx

-5 dx

+ (l\ ^' ^' "•" 25 A

If ^ = -400 and a: = 10, ^ = 16 rad/hr.

If ^ = -400 and ;c = 3, ^ = 58.824 rad/hr.
dt dt

73. (a) h(t) = -\(,t' + 256

- 161^ -I- 256 = 0 when t = 4 sec.
h -16f2 4-256

(b) tane

500
0 = arctan

de

500
-8t/125

•lOOOf

dt 1 -h [(4/125)(-/2 + 16)P 15,625 -^ 16(16 - t-Y
When f = 1, de/dt « -0.0520 rad/sec.
When t = 2, rf^M « -0.11 16 rad/sec.

Section 5.9 Inverse Trigonometric Functions: Integration 263

, , tan(arctan x) + tan(arctan y) x + y

75. tan(arctan x + arctan y) = -; -, ; -, r = -; , xy i= I

1 — tan(arctan x) tan(arctan y) 1 — xy

Therefore,

arctan x + arctan y = arctanl ],xy¥' 1.

\ 1 - xyj

Let X = 2 3rid y = 5 .

/1\ (A (1/2) + (1/3) 5/6

arctan^-j + arctan^-j = arctan ^ _ ^^j^^) . (1/3)] = ^'^'^

5/6 , TT

l-(l/6) =arctan3^ = arctanl =-

77. f(x) = be + sin.ic

fix) = k + cosx > QfoTk > I

f'(x) = (t + cos ;c < 0 for A: < - 1

Therefore, /(x) = fcc + sin jc is strictly monotonic and has an inverse for k <

loTk > 1.

79. True

81. True

dr -, 1

—-j arctan jcj = ? > 0 for all x.

dx 1 + X-

d^ , ,T sec^;c sec^x

— H arctan(tan x)\ = ;— = — ;— = 1

dx I + tan-x sec-x

Section 5.9 Inverse Trigonometric Functions: Integration

■•/

79^^

dx = 5 arcsin —] + C

Ki)

3. Let u = 3x, du = 3 dx.

dx = ^\ S3) dx

Jo yr=^? 3jo vi - (3x -'

- arcsin(3x)

0 "Ts

Jt^* = I

4 arctanl -I + C

7. Let u = Zx, du = 2 dx.

■V3/2 , , rV3/2

r^'^=_i , 1 [^"' 2 ,

- arctan(2x)

v^/2

— 2L
0 ~ 6

J xV4x2 - 1 J

2xV(2x)= - 1

tZv = arcsec|2x| + C

r x^ fr t1 r if it ii

11. ;' dr = X - -, , oLv- = X civ - - , civ = -x^ - - ln(x^ + 1) + C (Use long division.)

13. ^=a[v = arcsin(x + 1) + C

J VI - (x + 1)2

15. Let i< = t-.du = 2tdt.

J v'l -t' 2j ,

(2r) dt = - arcsin(f =) + C

'1 - (r

264 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

17. Let u = arcsin x, du

yr^

dx.

0 >

'o Vl -X2

dx

0 32

0.308

19. Let «= 1 - x^, du = -Ixdx.

X

\ -j===dx=M {\-x')-'l\-lx)dx

J-I/2VI - X^ ^J-\/2

-vr^

= -0.134

0 _ 73 -2

-1/2 2

21. LetM = e'^,du = le^ dx.
1 r 2e^

j4 + e^- 2J4 +

1 e^

, ,,,, dx = — arctan -r- + C
{e^f 4 2

25.

f \

J v^ -Jl — X

I J ,{2udu) = l\

J MVl - M- J

dLr. u = ^, X — u^, dx = 2u du
du

= 2 arcsin Vx + C

2 arcsin u + C

-■J7?mw-'i

U-3)

:d:x: +

/:

23. Let M = cosj:, £/m = -s'mxdx.

1 + cos^j:

dx

= - arctan(cos x)\ = —
L Jt/2 4

„ fj: - 3 ^ 1 r 2j; ^ . f 1 _,

27. — ; etc = - -z r dx — 3 \ -z — — dx

J .)c^ + 1 2] X- + \ ] X- + \

1

\n(x^ + 1) - 3 arctan;c + C

V9 - (jc - 3)2 J 79 - U - 3)

. /x- 3

:d[»:

= - 79 - (x - 3)2 - 8 arcsin( -^ ] + C

= - J6x - x^ + 8 arcsini - - 1 ) + C

' }qX- - 2x + 2 Jo 1

1

+ {x- \f

dx =

arctan(ji: - 1

'M

33.

Ix^ + lUn"^ = \x^ + 6x1x3'^ - %^ + L+X3'^ = \x^ + 6x1x3"^ " 4^

= InU^ + 6jj + i3| _ 3 arctan

+ C

{.X + 3)2

-dx

,c r 1 . r 1 _, ■ (x + 2\

35. , = dx = — , , = dx = arcsin — - —

J 7-.r2 _ 4;^ J 74 - (x + 2)2 V 2 ;

37. Let M = -x2 - 4x, d« = (-2x - 4) dx.

j^^^==dx = -||(-x2 - 4x)^'/2(-2x - 4)afe

- 7-x2 - 4x + C

"■ I

2a:- 3

dx =

-274x-x2 + arcsinf^^j = 4 - 273 + -| = L059

74 - (x - 2)

■ dx

Section 5.9 Inverse Trigonometric Functions: Integration 265

41. Let M = ;c2 + l,du = 2xdx.

( X , if 2x

Jx' + lx^ + l 2j(x^+ 1)2 + 1

^ = 2 arctan(;c2 + 1) + C

2m du

43. Let M = Ve' - 3. Then u^ + 3 = e', lu du = e' dt, and "-, "" = dr.

M- + 3

= 2m - 2v^ arctan -^ + C = 2Ve' - 3 - 2^3 arctan
V3

/e' - 3

+ C

45. A f)erfect square trinomial is an expression in x with three terms that factor as a perfect square.
Example: x^ + 6x + 9 = (x + 3)-

47. (a)
(c)

J xj\—x

dx = arcsin x + C, m = ;<:

(b)

yr^^

ate = - Vl -^2 + C, M = 1 - x^

: dt cannot be evaluated using the basic integration rules.

I

49. (a) \Jx-\dx = ^x - 1)3/2 + c, M = X - 1

(b) Let M = V.v- - 1. Then .r = m^ + i and dx = 2m du.
IxV^rn' (& = I (m2 + 1)(m)(2«) dM = 2 I (m-* + u-

)dM = 2(y + j) + C

= :^«^(3«2 + 5) + C = Y^x- l)V2[3(x - 1) + 5] + C = :^(x - l)3/2(3x + 2) + C

(c) Let M = Vx - 1. Then x = m^ + 1 and dx = 2m dM.

|-;r4=^dx = j^^^-i-^(2«) dM = 2 j (m2 + 1) dM = 2[y + M j + C = !«("- + 3) + C = jVx - l{x + 2) + C
Note: In (b) and (c), substitution was necessary before the basic integration rules could be used.

51. (a)

(0,0)

dv

dr \ +T'

y

(0,0)

-/Tf;^^

(0, 0): 0 = 3 arctan(O) + C
y = 3 arctan x

3 arctan x + C

C = 0

3z
2

,.v(3) = 0

55.

= [^ arctanf^^^^ j = ^ arctan(l) = y « 0.3927

266 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

57. Area « (1)(1) = 1
Matches (c)

59. (a) I ^ dtc = 4 arctan x = 4 arctan 1 - 4 arctan 0 = 4( — J -

Let n =

4(0)

(b) Let n = 6.

:dx ~ 4

4 2

+

1 + (1/36) 1 + (1/9) 1 + (1/4) 1 + (4/9) 1 + (25/36) 2

h

3.1415918

(c) 3.1415927

61. (a) —I arcsin - + C ^^, ,

\aj J Vl - {u^/a~)\al Ja^ - u^

1. (a) f [arcsinQ + c] =
[___du__ ^

(b)

Thus^

arcsini - 1 + C.

i " 4- /"I - If u'/a 1 _ 1 r u'

a ^""^a \~ aV\+ {u/a)A ' ai{a^- + u^)/a^i

^, \ du f m' , 1 u ^

Thus, — T = -^ ^ dx = - arctan - + C.

J a~ + u J a'^ + w^ a a

a^ + u^

(c) Assume h > 0.

d_
dx

1 u 1

1

- arcsec — \- C

a a ]

a

u'/a

{u/aU(u/af - 1

™,. r 1^" r «' ,1 i«i „

Thus, — , = — , dx = - arsec -"-J- + C.

J mV«2 - a^ J uju^ - a^ a a

i] =

u^ur — a^)/a^ u^w- — a^

The case « < 0 is handled in a simi-
lar manner.

63. (a) v(t) = -32f + 500

(b) 5(f)

= ^v{t)dt = ^

(-32r+ 500) d;

= - 16f= + 500f + C

5(0) = - 16(0) + 500(0) + C = 0^ C = 0

5(f) = -16f- + 500f

When the object reaches its maximum height,
v(f) = 0.

v(f) = -32f + 500 = 0

-32f = -500

f = 15.625

5(15.625) = - 16(15.625)2 + 500(15.625)

= 3906.25 ft (Maximum height)

—CONTINUED-

Section 5.10 Hyperbolic Functions 267

63. — CONTINUED-

(c)

\l^-'"=-\

kv

dt

Vi2k \V 32 ' '

arctani -. / ^ v'

/32<:f + C

V = tan(c - Vyzici

^ tan(C- V32kt)
k

When r = 0, V = 500, C = arctan(500VA:/32), and we
have

(d) When k = 0.001, v(r) = 732,000 tan[arctan(500V0.00003125) - VO.032 t\

/^ /32

v(f) = ^ / -- tan
k

arctani 500,

v(r) = 0 when t^ = 6.86 sec.

/■6.86 '

(e) /! = 732,000 tan[arctan(500V0.00003125) - 70.032 r] dt

Jo

Simpson's Rule: n = 10; /j « 1088 feet

(f) Air resistance lowers the maximum height.

Section 5.10 Hyperbolic Functions

1. (a) sinh 3 = -
(b) tanh(-2)

10.018

2

sinh(-2) _ e~- - e-
cosh(— 2) e^' + e-

-0.964

5. (a) cosh- '(2) = ln(2 + v^s) - 1.317

04/9)
2/3 /

(b)sech-.(|) = ln(J^4P^U 0.962

3. (a) csch(ln2)

(b) coth(ln5)

2

2 4

^ln2 _ ^-lii2

2 - (1/2) 3

cosh(ln 5)
sinh(ln 5)

gins + g-lnS
^ln5 _ g-]n5

=

5 + (1/5) 13

5 - (1/5) 12

/e^ — e~'\- I ''
7. tanh^ j: + sech- x = I ^ — -r 1 +

e' + g-

£■' + e'

2 ^ir _ 2 + e"^ + 4 e^ + 2 + e'-"

(e' + e-'Y e^ + 2 + e~^

268 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

e' - e^^Xfey + e'A (e^ + e~^\( e^ - e"^
9. sinh X cosh y + cosh x sinh y = \ z II : 1 +

4
= ^{e^y - e-(^+>')] = Y = sinh(A: + .v)

11. 3 sinh a: + 4 sinh' .t = sinh;c(3 + 4sinh^j:) =

2

gX — g-x

3+41

e' — e'

T]

,[3 + e^^ - 2 + e"^] = -(e^ - e-')(e^ + e'^ + 1)

-[e'Jt + e'" + e^ - e' - e'^" - e'"]

sinh(3;c)

13.

sinh a: =

cosh^ ^ ~ ( o ) ~ 1 =^ cosh^ jc = — => cosh x =

3/2 3V13

tanh;ic = —;= — = ,,

713/2 13

CSCh. -3/2 -3

1 2jU
sechjc = — ^ — = —rr—

1 ^

coth X - , , — - ,

3/713 3

15. y =

sinhd - x^)

y' =

-2j:cosh(l - x~)

17. f{x) = ln(sinh;c)

/'W = ^-r(cosh;c) = cothx
sinh

19. y = In tanh

'/2 sech^f^
tanh(V2) \2

1

2 sinh(x/2) cosh(./2)
1

sinh x

= csch;c

21. h{x) = I sinh(2;c) - |

■ ,/ \ 1 , /„ \ 1 cosh(2j:) - 1 . , ,
h'{x) = -cosh(2A:) - - = ^-r^ = sinh^j:

23. fit) = arctan(sinh t)
1

f'(t) =

1 + sinh^ t
cosh t

(cosh t)

cosh' r

= sech /

25. hsty = g{x).

y ^ ycosh X

\ny = cosh a: In j:
\(dy\ _ coshj:

y\dx

+ sinh ;c In ;c

-- = -[cosh X + ;c(sinh x) In x\
ax X

■{cosh X + x(sinh x) In x]

Section 5.10 Hyperbolic Functions 269

27. y = (cosh x - sinh x)^

y ' = 2(cosh X — sinh x)(sinh x — cosh x)
= — 2(cosh X — sinh xY = — 2e"^

29. /(x) = sin X sinh x — cos x cosh x, — 4 < j: < 4

/'(jc) = sin x cosh jc + cos x sinh x — cos x sinh x + sin x cosh jc

= 2 sin x cosh x = 0 when x = 0, ± tt.

Relative maxima: (± tt, cosh tt)
Relative minimum: (0,-1)

f-ff, coshff) i5f;r. coshff)
i — i

31. g(x) = .r sech x =

coshx

(1.20.0.66)

>^

(-1.20.-0.66) -1

Relative maximum: (1.20,0.66)
Relative minimum: (- 1.20, -0.66)

33. y = a sinh x

y' = a cosh x

y" = a sinhx
y '" = a cosh x
Therefore, y'" — y' = 0.

35. /(x) = tanh x
/'(x) = sech-x

/(I) = tanh(l) = 0.7616
1

/'(I) =

cosh-(l)
fU) = -2 sech^x • tanhx /tl) == -0.6397

/',(^) =/(l) +/'(1)(a- - 1) = 0.7616 + 0.42(x - 1)

0.6397

- 0.4200

P.^{x) = 0.7616 + 0.42(x - 1)

-{x - \r-

p./

37. (a) y =10+15 cosh— -15 < x < 15

20 ■•
10

(b) Atx = ±15. y = 10 + 15cosh(l) = 33.146.
At X = 0,y =10+15 cosh(l) = 25.

(c) y' = sinh -j-r. Atx = 15, y' = sinh(l) = 1.175

39. Let M = I - Zx, du = -2dx.

sinh(l - 2x) dx = -^ sinh(l - 2r)(-2) dx

I'

--cosh(l - 2x) + C

41. Let u = cosh(x — 1), du = sinh(x — 1) dx.

i

cosh'(x - 1) sinh(x - Otiv = -cosh^(x - 1) + C

270 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

43. Let u = sinh x, du = cosh x dx.
Tcosh .V

45. Let u = —,du = xdx.

iinhx

■dx = In sinh. V + C

P-f-l

csch^ — \xdx =

-cothy + C

47. Let u — —. du = — rctc.
x x'^

\

csch(l/A-)coth(l/A-)

dx

■[

csch - coth - — x]dx = cscb- + C
X x\ x^ x

49.

25 -A-^ 10 5 - A~" 1015 +,t'

dx + -—\ dx

i"

5 + X

5 - X

o^m'"^ = 5'"^

51. Let u = 2x, du = 2 dx.

r^/2/A
Jo

-dx

rV2/4
Jo

VI - Ax^ Jo VI - [2xf

=(2) dx ■

arcsin(2x)

0 4

53. Let » = x^, du = 2x dx.

J?TT* = i/:

2a , 1 / ■,\ ^

- dx = - arctan(A^j + C

(a^)2 +1 2

55. y = cosh '(3a)

3
^ V9a2 - 1

57. y = sinh~'(tanA)

y' = , (sec^x) = |secA|

Vtan-^ A + 1

59. y = tanh-'(sm2A)
1

1 — sin^ 2a

(2 cos 2a) = 2 sec 2a

61. y = 2Asinh-'(2A) - Vl + 4x^

2

Vl + 4x-

4x

y' = 2a- z + 2 sinh" '(2a) ^=^ = 2 sinh" '(2a)

Vl + 4a2

63. See page 395.

65. y = a sech ' - — Va^ — a-, a > 0
a

-1

^^_

dx (x/a)Vl - (aVq^) ■ Vfl" - a2 aVg^ - A^ ' Va^ - a^ aVq^ - a^

"" + ^

a2 - ^2 - V^5^^

67.

:d^ =

Vl + e^ J e'Vl + (e>p

e^

:a[x = -csch-'(e') + C = -Inl

1 + Vl +

+ C

69. Let iY = Va, du = — tt dx.

2jx

/:

:^A= 2

v^v^l^^ J Vl + (v^)^V2^

— ^) rfA = 2 sinh ' v/J: + C = 2 ln(^ + Vl + a) + C
2Va/

^^•/i^^^ = J(7

-^jr^^'^ = 4ln

(a - 2) - 2

(a - 2) + 2

= 5'"

A- 4

+ C

Section 5.10 Hyperbolic Functions 271

^^' Jl-J-2x^'^'^ - J 3 - 2(1 + l)^'^ - Jj[72U + \)Y - (73)^"^

-1

2^6

In

^{x + 1) - 73

v^U +0 + 73

+ c

276

In

v/2(;c + 1) + 73

72(jt + 1) - 73

+ C

75. Let M = 4jc - 1, ^k = 4 <&.

J 780 + 8^ - ldt= 4J

4 , 1 . /4;c - 1\ ^

, dx = - arcsin — - — + C

781 - {Ax - If 4 \ 9 '

''' y = /sT^'^ = /(- - ^ ^ jrf^)^ = /(- -'^'^^ -«/f^17^

ix

^' . -.20,
— - 4j: + -^ In
2 6

U - 2) + 3

U - 2) - 3

^C=-f-4. + fln

X + 1

X- 5

^ -X . 0,

+ C = — 4.V - — n

2 3

X - 5

.x+ 1

+ C

79. A = 2 sech^atc
Jo •^

81. A

= 2

^/2 + g-xn

dx

Jo ie^^'r +

-dx

8 arctan(e^/2)
8 arctan(e-) - 27r == 5.207

2jo

:dv

7l?FTT

dx

|ln(x-+ ./^^TT)

= - ln(4 + 717) = 5.237

83. l^dt = -^ — dx

J 16 ]x- - lit + 32

3fa^ r

16 J

(^_-^r3^<i^ = ^'"

(.V - 6) - 2

(.V - 6) + 2

+ C = - In
4

.T - 8

X - 4

+ C

When .t = 0: f = 0

C= --ln(2)

When;c = 1:

When t = 20:

t= 10

30A- _ 1
16 ~ 4

In

-3

-4'"(2)=-ln(-

^^ = ^'"6

a(^)-eh4'"^

49 .V - 8

36 2x - 8

eix= 104

104 52 , ^^^ ,
a: = — = -« 1.677 kg

272 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

85. As k increases, the time required for the object to reach the ground increases.

87. y = coshx •
e^ — e~

e^ + e'"

2

= sinhjc

89. y = cosh ' x

cosh y = X
(sinh3')(y') = 1
1

y =:?

smhy Vcosh^ y - 1 ^x^ - 1

91. y = sechx

e" + e^-'
y' = -2(e'^ + e"^)-V - e'")

e^ — e

e' + e ^l\e' + e "

— sech X tanh x

Review Exercises for Chapter 5

1. /(x) = ln;c + 3

Vertical shift 3 units upward
Vertical asymptote: x = Q

12 3 4 5

3- 'n V£^ = 1'"^^" J/+V^' = if'"^^ " ^^ ■" '"^2" + ^^ - '"(^"' + ^^^

5. In 3 + ^ ln(4 - x^) - In jc = In 3 + In 3/4 - x^ - Inx = ln( ^^^
3 V X

7. InVx + 1 = 2

Vjc+ 1 = e2

X + 1 = e"*

X - e" - 1 ^ 53.598

9. g(x) = InVx =2'"^

^'«=t

11. /w = xTinI

1 /; — 1 + 21nx

— j= + V In X = 7=^

2vlnx 2vlnx

13. >- = 75| ln(a + bx) +

^ ^ ±r^ <L

dx if\_a + bx (a +

a + bx
ab

to)" J

(a + bxY

15. y

dy

a \ X / a

1/ fc

a[)c aVa + bx xj x{a + bx)

1

17. u — Ix — 2,du = Idx
1

J 7x - 2 7j

Ix-T ' 7 ' '

Review Exercises for Chapter 5 TTi

, „ f sin .t , r ~ s

19. dx= -\

J 1 + cos j; J 1 +

-dx

1 + cos X J 1 + COS X

-ln|l + cosjtl + C

-r^-K-i)-

X + Inlxl

3 + ln4

23. I secede =

Jo

ln|sec e + tan e|

t/3

= ln(2 + 73)

25. (a) /W = it - 3

1 -,
> = 2-t - 3

2(y + 3)=x .

2(x + 3)=y

r\x) = 2x + 6

(b)

'/

^

>

■^

(c) /-'(/W) =/-'(it - 3) = l{{x - 3) + 6 = ;c
/(/-'W) =/(2;c + 6) = ^(2x + 6) - 3 = j:

27. (a) /U) = v'TTT
y= Jx+ 1
y — 1=0:
x- — \ = y
f-'(x) =x^- \,x > 0

(b)

^

^

■'

(c) /-'(/W) =/-'(7^TT) = VCr - 1)= - 1 = X
■ /(/-W) = /(.t^ - 1) = J{x- - 1) + 1

= v^? = JT for -V > 0.

29. (a) /(x) = 3/.V + 1

V =Vx+ 1
' 3^- 1 =x

x' - 1 =y
/->W=x3- 1

(b)

f-

"^^

y

(c) /-(/W) =/-'(4/^^^) = (^^n^)' - 1 = -r

/(/-'U)) = /(x^ - 1) = ^(.T^ - 1) + 1 = X

31. /U) =y? ^-1

r\x) = U - 2)>/3

(/-')'W = |u-2)-V3

33.

1

35/3

= 0.160

/(.r) = tan .r

/^\ ^ 73

-^Uy 3

f\x) = sec-.r

^U/ 3

3

^ ' \ 3 y /'(7r/6)

4

274 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

35. (a) f{x) = In v^ (b)

y = ln>/t

ey = v^
e'^y = X

e^ = y

/-'W = e^

V

/'V

^

(c) /''(/W) =/"'(ln >A) = e^i"-^ = giBA: = jf
/(/~'W) = Ae^) = In 7^ = In e^ = X

37. y = e-^/2

39. f{x) = ln(e-^) = -jc^
fix) = -lx

41. g(f) = i^e'

g\x) = t-e' + 2te' = te'(t + 2)

43. y = Ve^' + e"^

y' = f(e^ + e-^-)-/2(2e^ - 2e-^) = ' ^ 2.
I ^ e" + e '^

45. 8ix)=~

g'ix)

e^{2x) - JcV x{2 - x)

49. Let M = — 3x^, ^m = —6x dx.

J^-3.^=_i|

£fa:= -7 e-3j:^(-6x)dx= -re"'^ + C

47. y(ln x) + y^ = 0

dy_

(ic x(2y + In x)

51. ^ ^^ Sfa = (e3^-e^ +

e-^)dx

g3jr _ ^ _ g AT ^ ^

3e^ - 3

3e^

+ C

/„-*.-ip.-(-2.,^

53. xe'-^atc

= -/'^ + C

55. Let u = e^ - l,du = e' dx.

e^

1

e^- 1

dx = Ine' - 1 + C

57. y = Wa cos 3x + b sin 3jc)

y ' = e't- 3a sin 3x + 3b cos 3j:) + e*(a cos 3j: + fo sin 3x)

= e^{-3a + b) sin 3x + {a + 3b) cos Sx]
y" = K3(- 3a + b) cos 3jc - 3(a + 3b) sin 3x] + e'[(-3a + b) sin 3x + (a + 3b) cos 3x]
= s^[(-6a - %b) sin 3x + (-8a + 6fe) cos 3x\
y" - 2y ' + lOy = e^{[(-6a - 8fc) - 2(-3a + b) + \Qb] sin 3x + [(-8a + (,b) - 2{a + 3b) + 10a] cos 3a;} = 0

Review Exercises for Chapter 5 275

59. Area

= j xe-'^dx = [-f^'^] = -!(«•"'* - 1) = 0.500

61. y = 33/2

63. y = logjCj: - 1)

65. fix) = 3^-'
/'U) = 3--'ln3

69. g{x) = logjVl - X = - log3(l - x)

g'ix) = ^

1

1

2(1 -;c)ln3 2{x - l)ln3

67. y = x2^-''

Iny = (2x + \)\nx
y' _ Ix + 1

+ l\r\x

y *

(Zx+ 1

3' = >'

+ 2 In .X 1 = ;c^* '( — + 2 In .t

X

71. \{x + l)5(-+»'rfr = 1 As'-*"' + C
J 2 In 5

73. (a) y = x"

y = ax"

75. 10,000 = PefooTXis)
„ 10,000

(b) y = a-
y ' = (In a)<3-'

$3499.38

(c) y = x^ (d) J = a"

In y = JT In ,r y ' = 0

— v' = X • — I- (1) In.v
y- j:

y' = y(l + In.v)

y' = x'(l + In.r)

77. P(/i) = 30e^

P(18,000) = 30e'««»'-- = 15
, _ ln(l/2) _ -ln2

18,000 18,000
Pih) = 30e-<'"° ->/'«•«»
/'(35,000) = 30e-(35.oooin2)/i8,ooo ^ 7 79 inches

79. P = CeOO'S'

2 = eOO'5'
ln2 = 0.015r
In--

t =

0.015

= 46.21 years

81. J = ^i±l

ttc X

h-~l

x^-Ux

y = y + 3 ln|.r| + C

276 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

83. y' -2xy = 0
dv

dx

2xy

\-dy= llxdx
\n\y\ = x^ + Q

y=Ce^

dy v^ ~t~ V*"

85. -^ = ^- (homogeneous differential equation)

dx 2xy

(x^ + y^) dx - 2xy dy = 0

Let y = vx, dy = x dv + v dx.

(x- + vV) dx - 2x{vx){x dv + vdx) = 0

(x^ + vV - 2x^v^) dx-2x>vdv = 0

(x^ — xhP) dx = 2a^v dv

i\-v'^)dx = 2x dv

(dx^ r_2v_

] X J 1 - v2

dv

ln|jc| = -ln|l - v^l + C, = -ln|l - v^l + In C
C C Cx^

1 =

1

— V-

1 -

- {y/xY

X- -

/

Cx

-y2

or

X

x^

^'"x^

-f

87. y = C^x + C^^
y' = Cj + SCjJC'
y" = 6C2X

x^y" - 3xy' + Sy = X'iec^x) - Zx{C^ + SCjX^) + iCyX = C.x^)
= ec^x^ - 3C,x - QCoX^ + 3CiX + SCjX^ = 0

X = 2,y = 0: 0 = 2C, + 8C2 => Cj = -4C2

x = 2,y' = A: 4 = C, + llCj , - '

■ 4 = (-4C2) + I2C2 = 8C2 =» C2 = |, Ci = -2

y = ~2x + -x^

»9.f(x) = 2 arctan(ar + 3)

Review Exercises for Chapter 5 277

91. (a)

Let 0 = arcsin —

sin 6 =

sinl arcsin -I = sm6 = —.

(b)

Let 6 = arcsin

1

sin e = -

cosi arcsin -I = cos 6 = -r~.

93. > = tan(arcsin j:)

vr

95. y = x arcsec x

X

|x|v^^^^

+ arcsec j:

97. y = .t(arcsin x)^ — 2r + 2 Vl — .r- arcsin ;c
2x arcsin x

vr^

+ (arcsin j:)^ — 2 +

2VTZ

2j:

yr

: arcsin x = (arcsin .t)^

99. Let M = e^, (iu = 2e^ dx.

J-TTT^ '^ = \y^ dx = l/yrrU^^^.-) ^ = \ arctan(.-) + C

101. Let « = X-, du = 2xdx.

103. Let M = 16 + X-. du = 2xdx.

(2r) A = — arcsin x^ + C

jjeT^'^^ljleh^^^^'-'-b^^''

+ x^) + C

105. Let u = arctan t- I. <i" = ": ^ <if •

\2/ 4 + x^

. , arctan '-] + C
4\ 2

\Ai -w ■' + <^
\A/ V m

Since y = 0 when f = 0, you have C = 0. Thus,

^\ V

>- = A sin( ^/ - f

109. >- = It - coshv^x

sinhv'x

,^ = 2-^(sinhV^) = 2-^-^

111. Let u = x^. du = 2x dx.

\ ■'' dx = U , ^ ={2x)dx = \\n{x^ + V?~^) + C

278 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

Problem Solving for Chapter 5

1- tan 01 =

tan e, =

10 -X

f{x) = 0, + ^2 ^ arctan(-J + arctanl _

1 +4 ' 1 +

1

36

(10 - x)

= 0

(10 - xf

6

x2 + 9 (10 - xY + 36
(10 - xf + 36 = 2(x2 + 9)
100 - 20;c + x2 + 36 = 2jc2 + 18
;c2 + 20x - 118 = 0

-20 ± V202 -4(-118)
X = = - 10

a = - 10 + v^TS - 4.7648 f(a) =- 1.4153
e = 77 - (0, + e^) « 1.7263 or 98.9°
Endpoints; a = 0: 0 = 1.0304
a = 10: 0 « 1.2793
Maximum is 1.7263 at a = - 10 + 7211 « 4.7648.

/2T8

3. f{x) = sin(lnj:)

(a) Domain: x > Q or (0, oo)

(b) /(x) = 1 = sin(in x) =* In ;c = y + 2^it.
Two values are jt = e'^''^, g(7r/2)+27r_

(c) f(x) = -1 = sin(lnj:)

Two values are x = e""''^, e^''''^.

ln.t = — + 2^77.

(d) Since the range of the sine function is [— 1, 1],
parts (b) and (c) show that the range of/is [— 1, 1].

(e) f'(x) = -cos(lnx)

fix) = 0 ^ cos(!n x) = 0 =^ \nx = — + ktr =i

fie-"'") = 1
/(I) = 0
/(lO) = 0.7440

• Maximum is 1 atx = e^^^ = 4.8105

(f)

[T

lim f(x) seems to be --. (This in incorrect.)

Jr-»0* 2

(g) For the points x = e'"!'^, g-^Vz^ e"'''/^, . . .

we have f(x) = 1.

For the points x = e"''/^, e"^'"!'^, e~^'"'\ . . .

we have f{x) = - 1 .

That is, as j:— ^O"^, there is an infinite number of
points where f{x) = 1, and an infinite number where
f(x) = — l.Thus lim sin(ln a:) does not exist.

You can verifiy this by graphing f{x) on small
intervals close to the origin.

Problem Solving for Chapter 5 279

, , , Area sector t . t , . t

5. (a) 7—r- = T— => Area sector = r— (tt) = -

Area circle Itt Itt 2

, rcosh t

(b) AreaAOP = -{base)(height) - Vjt - 1 dx

rcosh r

1 ^

'^(f) = - cosh t ■ sinh t

\dx

A '(f) = -[cosh- t + sinh- i\ - Vcosh^ t - 1 sinh «

= rCcosh- t + sinh- 1] - sinh^ t
= —[cosh- t - sinh- t] = -

A(f) = -t+C. But, A(Q) = C = O^C = 0

Thus, A(f) = -f or J = 2 A(/).

7. y = In jr

1

y

1,

>> - fe = -(.r - a)
a

y = —X + b — 1 Tangent line
a

lix = Q,c = b - 1. Thus, b - c = b - {b - \) = \.

11. (a)

^
A

,,1-01

jy-""*'= \dt

-0,01

I ^

"-l

1

y0.01

-O.Olr + C

yO.OI

1

C -

- O.Olr

1

y -

(c

- O.Olr)"*

v(0) = l: 1 =^=^ C=l

Hence, y

1

(1 - O.Olr)"

9. Let u = I + Vx, ^ = u - ].x = u^ - 2u + \,
dx = (2m — 2)Jm.

Area = —p dx = — t—

]-,Jx + x h («-!) + ("- - 2u

h u- - u

"I

+ 1)

■du

2 In
2 In 3 - 2 In 2 = 2 Ini
0.8109

(!)

(b)

|y-»*-'rf>-= r

■^— = kt+C,

— E

1

(C - eitr)'/^

.v(0)=.Vo = ^=>C-4

Hence, v

^r

i -)'•■■

For r— > . V— >cx;,

For T = 100, lim v = oo.

r-»r- ■

280 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

13. Since ^ = /t(v - 20),
dt

/t^^-/

dy= \kdt

\n{y - 20) = kt+ C

>> = Ce*' + 20.

When r = 0, y = 72. Therefore, C = 52.

When f = 1, y = 48. Therefore, 48 = Sle" + 20, e* = (28/52) = (7/13), and A: = ln(7/13).
Thus, .V = 52«ti°*''/'3'> + 20.

When f = 5. >- = 52e5M7/i3) + 20 « 22.35°.

JO

15. (a) — = k^S{L - S)

(b) ^ = In m 5(100 - S)

1 -'- Ce-

: is a solution because

rf25 /4

^ = '"19

5|-f) + (100-S)''^

rfr

dr

^= -L(l + Ce-'')-2(-C/te-*')
dt

LCke-

(1 + Ce-'')2

fM

L

Ul

+ Ce-'^-'

(k\

L

CLe-

1 + Ce-*'

U/ 1 + Ce-

L -

1 + Ce-

k^S{L - S), where A:, = — .

L = 100. Also, S = 10 when f = 0 ^. C = 9. And,
S = 20 when t= \ =* fc = -ln(4/9).

Particular Solution. S

100

J + 9gln(4/9)f

100

1 4. 9g-0.8109

(C) 125

^lng)(100-25)f .

= OwhenS = 50or^ = 0.
dt

Choosing S = 50, we have:

50 = 100—

1 + 9ei"M/9)'

2=1+ 9eln(4/9)r

ln(l/9)

ln(4/9)

= r

t = 2.7 months

(d)

(e) Sales will decrease toward the line S = L.

PART II

CHAPTER P
Preparation for Calculus

Section P.l Graphs and Models 282

Section P.2 Linear Models and Rates of Change 287

Section P.3 Functions and Their Graphs 292

Section P.4 Fitting Models to Data 296

Review Exercises 297

Problem Solving 300

CHAPTER P
Preparation for Calculus

Section P.l Graphs and Models

Solutions to Even-Numbered Exercises

2. y= V9 - x^

x-intercepts: (-3,0), (3,0)
y-intercept: (0, 3)
Matches graph (d)

6. y = 6 — 2x

X

-2

-1

0

1

2

3

4

y

10

8

6

4

2

0

-2

4. y = x' — a:

x-intercepts: (0, 0), (- 1, 0), (1, 0)
y-intercept: (0, 0)
Matches graph (c)

8. y = (x - 3)2

X

0

1

2

3

4

5

6

y

9

4

1

0

1

4

9

10. y

x

-3

-2

-1

0

1

2

3

y

2

1

0

-1

0

1

2

12. y = Jx + 2

X

-2

-1

0

2

7

14

y

0

1

V2

2

3

4

5 10 15 20

282

Section P.l Graphs and Models 283

14.

Xmin = -30
Xmax = 30
Xscl = 5
Ymm = -10
Ymax = 50
Yscl = 5

Note that 3; = 10 when x = Ooxx= 10.

16.

j\

J

(a) (-0.5, >-) = (-0.5,2.47)

(b) {x, -4) = (- 1.65, -4) and (x, -4) = (1, -4)

18. y'^ = :? - Ax

^'-intercept: y^ = (fi - 4(0)
y = 0; (0, 0)

jc-intercepts: 0 = x' — 4;c

0 = x{x- 2){x + 2)
x = 0, ±2; (0, 0), (±2, 0)

20. y={x- 1)7^2 + 1

y-intercept: y = (0 - 1)V0^ + 1
y= -1;(0, -1)

x-intercepts: 0 = {x - 1)V?+T
x= 1;(1,0)

22. y

x^ + Zx
[3x + \Y

y-intercept: y =

0^ + 3(0)

[3(0) + 1?
y = 0; (0, 0)

x^ + 7)x

x-intercepts: 0 = 7; — -—

(3x + 1

0

x{x + 3)

(3x + \y

x = 0, -3;(0, 0), (-3,0)

24. >- = 2x - Jx^ + 1

y-intercept: y = 2(0) - VO^ + 1
>.= -1;(0,-1)

x-intercepts: 0 = 2x — Vx^ + 1

2x= Vx^ + 1
4x2 = x2 + 1
3x2 = 1

x^ =-

3

73

' = #(#»

Note: X = — V3/3 is an extraneous solution.

26. >> = x^ - X

No symmetry with respect to either axis or the origin.

28. Symmetric with respect to the origin since
i-y) = (-x)3 + (-x)
—y = — x^ — X
y = }? + X.

30. Symmetric with respect to the x-axis since
x(-yY = xy2 = _ 10.

32. Symmetric with respect to the origin since
(-x)(-^) - V4 - (-x)- = 0

xy - V4--x2 = 0.

284 Chapter P Preparation for Calculus

34. V

a:2 + 1

since y

is symmetric with respect to the v-axis
i-xY- x^

{-xY + 1 .^2 + r

36. I);] — x = 3 is symmetric with respect to the x-axis
since \—y\ — x = ?>

\y\-x = 3.

38.. = -2+2

Intercepts:

(4,0), (0,2)
Symmetry: none

40. 3; = f;c + 1
Intercepts:

(0,1), (-1,0)

Symmetry: none

42. y = x^ + 3
Intercept: (0, 3)
Symmetry: )'-axis

44. y = 2x2 + X = ^(2;f 4. 1)
Intercepts:

(0,0), (-io)
Symmetry: none

50. X = y2 - 4
Intercepts:

(0,2), (0,-2), (-4,0)
Symmetry: x-axis

(-4, Oj

46. y = x^ - 4x
Intercepts:

(0,0), (2,0), (-2,0)

Symmetry: origin

52. y

10

x^+ 1

Intercepts: (0, 10)
Symmetry: y-wdi

-6-4-2 2 4 6

48. y = V9 -x2
Intercepts:

(-3,0), (3,0), (0,3)
Symmetry: y-axis
Domain: [—3, 3]

54. .V = |6 - x|
Intercepts:

(0, 6), (6, 0)

Symmetry: none

Section p. 1 Graphs and Models 285

56. x'^ + Ay- = '^^>y = ±

Va^I?

Intercepts:

(-2,0),(2,0),(0, -1),(0, 1)
Symmetry: origin and both axes
Domain: [—2, 2]

<-.o^

(0,1)

^

(0,-1)

58. 3;c - 4y2 = 8
Af = -ix

y = ±J\x-2

Intercept: Is- 0/
Symmetry: jc-axis

(y^^

'^~^--

60. y - (x + jjU - 2)(j: - j) (other answers possible)

2;c- 13

AX jy —

Lj=f y

3

5x + 3y =

l=>y =

1 -
3

i£

2x- n

1 - 5x

3

3

2x- 13 =

1 - 5x

7.t =

14

X =

2

The corresponding y-

value

isy =

Point of intersection:

(2,-

-3)

-3.

62.

Some possible equations:

x = y^

x=\y\

;c = / + 1

jc^ + >^ = 25

66.

5x - ey = 9 =>

_5x

- 9

6

-7a: + 3y = -18

^y =

7a:-
3

J8

5a: - 9 Ta -

18

6 3

5x - 9 = 14.r -

- 36

27 = 9a:
a: = 3 •
The corresponding v- value is y = 1.
Point of intersection: (3,1)

68. x = 3-f=>f = 3-x

y = X — 1
3 - A = U - 1)2
3 - A = A- - 2i: + 1

0 = .r^ - A - 2 = (x + l)(x - 2)

X = — 1 or X = 2
The corresponding y- values are y = - 2 and y = 1 .
Points of intersection: (- 1, -2), (2, 1)

70. x2 + y2 = 25 ^ >- = 25 - .r=
2x + y = 10=s.y = 10 - 2x
25 - x= = (10 - 2r)=
25 - X- = 100 - 40x + 4x-

0 = 5.v^ - 40x + 75 = 5(x - 3)(x - 5)
X = 3 or X = 5
The corresponding y-values are y = 4 and y = 0.
Points of intersection: (3. 4). (5, 0)

286 Chapter P Preparation for Calculus

72, y = ^ - 4x
y=-{x + 2)

)? - Ax= - (x + 2)
.r^ - 3x + 2 = 0
{x - \Y{x + 2) = 0
x=l or X = —2

The corresponding y-values are >> = — 3 and y = 0.
Points of intersection: (1, -3), (-2, 0)

74.

y = x^ - 2x2 + 1
y = 1 - x2

1 - x2 = x^ - 2x2 + 1

0 = x^ - x2

0 = x2(x + l)(x - 1)

X = - 1, 0, 1
(-1,0), (0,1), (1,0)

"^A

(0.1)/

(-1.0)/"

Yi.o)

76. y = fcc + 5 matches (b).

Use (1, 7): 7 = k{\) + 5 => fc = 2, thus, y = 2x + 5.

y = x^ + A; matches (d).

Use(l, -9): -9 = (1)^ ^k=^k= -10, thus, ); = x- - 10.

y = fcc^''2 matches (a).

Use (1, 3): 3 = kiXf''^ =^ k = 3, thus.y = 3x^/2.

xy = fc matches (c). ' •

Use (1, 36): (1)(36) = it => A: = 36, thus, xy = 36.

78. (a) Using a graphing utility, you obtain

y = -0.1283t2 + 11.0988f + 207.1116

(c) For the year 2004, r = 54 and
y = 432.3 acres per farm.

(b) 500

80. (a) If (x, y) is on the graph, then so is (— x, y) by y-axis symmetry. Since (— x, y) is on the graph, then so is (— x, —y) by

X-axis symmetry. Hence, the graph is symmetric with respect to the origin. The converse is not true. For example, y = x^
has origin symmetry but is not symmetric with respect to either the x-axis or the y-axis.

(b) Assume that the graph has x-axis and origin symmetry. If (x, y) is on the graph, so is (x, — y) by x-axis symmetry. Since
(x, -y) is on the graph, then so is (-x, - (-y)) = (-x, y) by origin symmetry. Therefore, the graph is symmetric with
respect to the y-axis. The argument is similar for y-axis and origin symmetry.

82. True

84. True; the x-intercept is

Section P.2 Linear Models and Rates of Change 287

Section P.2 Linear Models and Rates of Cliange

2. m = 2

4. OT = - 1

£ 40

6. m = -T

10. m

4-2
-2 - 1

_2
3

J.-2. 4)

-2 -(-2)

12. m = ; = 0

4—1

H 1 1 1 l-»-i

(3,-2) (4,-2i

14. m

(3/4) -(-1/4)
(7/8) - (5/4)

1 8

-3/8

16. Since the slope is undefined, the line is vertical and its equation is x = —3. Therefore, three additional points are (—3. 2),
(-3, 3), and (-3, 5).

18. The equation of this hne is
y + 2 = 2(.v + 2)
y = 2j: + 2,
Therefore, three additional points are (-3, -4), (- 1, 0), and (0, 2).

20. (a) Slope =

^ = 1

Ax ~ 3

By the Pythagorean Theorem.
.r= = 30^ + 10^ = 1000
.t = 31.623 feet.

22. (a) m = 400 indicates that the revenues increase by 400 in one day.

(b) m = 100 indicates that the revenues increase by 100 in one day.

(c) m = 0 indicates that the revenues do not change from one day to the next.

288 Chapter P Preparation for Calculus

24. 6;c - Sy = 15

6 -,

Therefore, the slope is m = j and the y-intercept is
(0, -3).

26. y = - 1

The line is horizontal. Therefore, the slope is wz = 0 and
the y-intercept is (0, - 1).

28. jc = - 1

x+ 1 =0

(-1,2)

30.

y = 4
4 = 0

-3 -2 -1

(0,4)

32. y - 4 = -f (x + 2)

5y - 20 = -Ix - 6
3;c + 5y - 14 = 0

4 - (-4) 8 -

y - 4 = 2(;c - 1)
y - 4 = Zx - 2

0 = Ix - y + 2

38. /n =

6-2

-4 -3 -2-1 12

40. wi = 0

y=-2
y + 2 = 0

-I — I — I — I-

12 3 4

(1, -2) (3. -2)

42. m =

(3/4) -(-1/4)
(7/8) - (5/4)

1 8

-3/8

1

^ + 4 3V- 4/
12y + 3 = -32jc + 40
32;c + 12y - 37 = 0

44. m — —-

y = — X + b
a

~x + y = b

a b

Section P.2 Linear Models and Rates of Change 289

= I

— Zx y

~1 2

lx + y= -1
3x + y + 2 = 0

. ^ + ^ =

1

a a

-3+1 =

1

a a

]_ _

1

a

a =

1 => .r + V = 1

X + :y - 1 = 0

50. X = 4

;c-4 = 0

52. y = \x-\

3>' - ;c + 3 = 0

54. y - 1 = 3(;c + 4)
>> = 3;c + 13

56. .t + 2v + 6 = 0

y ~ ~2^ " 3

58.

The lines do not appear perpendicular.

y

/

/\

The lines appear peqjendicular.

The lines are perpendicular because their slopes 2 and -5 are negative reciprocals of each other.
You must use a square setting in order for perpendicular lines to appear perpendicular.

290 Chapter P Preparation for Calculus

60. X + V = 7

y = -x + 7
m= -\

(a) y-2=-lU + 3)
y-l= -x-l

X +y + \ =0

(b) y - 2 = lU + 3)
y - 2 = ;c + 3

x-y + 5 = 0

62.

Ix + Ay

= 7

y

= -!.+

7
4

m

3

~ 4

(a)

y-4 =

-|(x

+ 6)

4y - 16 =

-3;c

- 18

3x +

4^ + 2 =

0

(b)

y-A

= !u

+ 6)

3y- 12

= 4jc + 24

4x-

3y + 36

= 0

64. (a) y = 0

(b);c=-l=>.:c+l=0

66. The slope is 4.50.

Hence, V = 4.5(? - 1) + 156
= 4.5r + 151.5

68. The slope is -5600. Hence, V = -5600(f - 1) + 245,000

= -5600f + 250,600

70.

^

fV'3-0'

l\

1\

You can use the graphing utility to determine that the points of intersection are (0, 3) and (3, 0). Analytically,
X- - Ax + 3 = -x^ + 2x + 3
2jc' — 6jr = 0
2x(x - 3) = 0

X = 0 ^ y = 3 => (0, 3)

• ;c = 3 ^ y = 0 =» (3, 0).

The slope of the line joining (0, 3) and (3, 0) is m = (0 - 3)/(3 - 0) = — 1. Hence, an equation of the line is

>-- 3 = -1(a:- 0)

y = -x + 3.

72.

/Kl

-6

7 -

- 4
0

10

7

mj

11 -

-5

- 4

- 0

7
5

m,

# /Wj

The points

are

not coUinear.

Section P.2 Linear Models and Rates of Change 291

74. Equations of medians:

y = -x

3a + b
c

(x + a)
(x-a)

■' -3a + b
Solving simultaneously, the point of intersection is

m-

(-a.O) , (0.0) (a.O)

(b c\ I a^ - lP-\
76. The slope of the line segment fi-om I -, - 1 to I fc, I is:

[(g^ - b'-)/c\ - (c/3) (3a" - 3fc^ - c^)/(3c) 3a' ~ 3b^ - cr

b - (b/3)

{2b)/3

2bc

The slope of the line segment from I -, - I to ( 0,

b c

2c

[(-g" + fc- + c^)/(2c)] - (c/3) _ (-3a" + 31^ + 3c- - lc')/{bc) 3a' - 3b' - c'

"^ 0-(V3)

Therefore, the points are collinear.

-bl3

2bc

78. C = 0.34x + 150. If .r = 137, C = 0.34(137) + 150 = $196.58

80. (a) Depreciation per year:
^ = $175

>- = 875 - 175a:
where 0 < x < 5.
db) y = 875 - 175(2) = $525

82. (a) V = 18.91 + 3.97.i: (x = quiz score, >> = test score)

(b) "M

(c) 200 = 875 - 175x
175;c = 675

X = 3.86 years

(c) ]fx = 17,y = 18.91 + 3.97(17) = 86.4.

(d) The slope shows the average increase in exam score
for each unit increase in quiz score.

(e) The points would shift vertically upward 4 units. The
new regression line would have a y-intercept 4 greater
than before: v = 22.91 + 3.97.r.

84. 4.1 + 3y - 10 = 0 => c/

|4(2) + 3(3) - 10| _ 7

Va^TY- ~ 5

M. x+ \=Q^>d =

|1(6) + (0)(2) + 1|
^1- + 0"

= 7

88. A point on the line 3x - 4.v = 1 is (- 1, - 1). The distance from the point (- 1, - 1) to 3.v - 4v - 10 = 0 is

1-3 + 4- 10| 9

292 Chapter P Preparation for Calculus

90. y = ntx + 4=>mx + (-\)y + 4 = 0

|Aci + gy, + C\ ^ \m3 + (- 1)(1) + 4|

^ |3m + 3|

Vm^ + 1

The distance is 0 wlien w = — 1. In this case, the line y = —x + 4 contains the point (3, 1).

92. For simplicity, let the vertices of the quadrilateral be
(0, 0), (a, 0), (b, c), and (d, e), as shown in the figure. The
midpoints of the sides are

l»

a + bc\lb + dc + e\ ^Id e

The slope of the opposite sides are equal:

c + e

0

a + b

b + d

£

-1 = £

d b

0

c + e

a + b b + d

111 1

Therefore, the figure is a paralleogram.

(^■^)

94. If ffZi = — l/m,, then mjOTj = ~ 1- Let Lj be a line with
slope mj that is perpendicular to Ly Then miWij = — 1.
Hence, m2 = m-^=^ L^ and Lj are parallel. Therefore, Lj
and Z,j are also perpendicular.

96. False; if m, is positive, then m^= - l/m^ is negative.

Section P.3 Functions and Their Graphs

2. (a) f(-l) = V-2 + 3 = yr = 1
(b) /(6) = V6T3 = 79 = 3

(c) /(c) = v^n

(d) /(jc + Ax) = Jx + Kx + 3

45

4. (a) ^(4) = 4^(4 - 4) = 0

(b) ^(1) = (im - 4) = !(-!)

(c) gic) = c^{c - A) = c^ - Ac^

(d) g(t + 4) = (r + 4)2(f + 4-4)

= (r + A)h = fi + %t^+\6t

6. (a) /(tt) = siniT = 0

(lTr\ . (Itt

(Of

ns

73
2

(b)/l

t) = ^^l'

4

■v^

8.

/(;c) -/(I) 3;c - 1 - (3 - 1) _ 3(;c - 1)

1

1

1

= 3,x^ 1

10. — — = ; = ; = x[x + \),x i= 1

X — \ X — I X — 1

Section P. 3 Functions and Their Graphs 293

12. gix) =x^-5

Domain: (— oo, oo)
Range: [-5, oo)

14. h{t) = cot t

Domain: all t i^ k-rr, k an integer
Range: (-00,00)

16. g{.

X - 1

Domain: (-00, 1), (1, 00)
Range: (-00, 0), (0, 00)

18. f(x) =

x'^ + 2,x < \
2jf2 + 2, ;c> 1

(a) /(-2) = (-2)= + 2 = 6

(b) /(O) = 02 + 2 = 2

(c) /(I) = 1^ + 2 = 3

(d) f{s^ + 2) = 2(52 + 2)2 = 2j^ + 8^2 + 10
(Note: j2 + 2 > 1 for all s)

Domain: (—00, 00)
Range: [2, 00)

20. fix) =

Vx + 4,x < 5
(x - 5)2, X > 5

(a) /(-3) = V-3 + 4 = yr = 1

(b) /(O) = VoT^ = 2

(c) /(5) = ym = 3

(d) /(lO) = (10 - 5)2 = 25
Domain: [-4, 00)
Range: [0, 00)

22. g(x) = -

Domain: (-00, 0). (0, 00)
Range: (- 00, 0), (0, 00)

24. fix) = ;x3 + 2
Domain: (—00, 00)
Range: (—00, 00)

26. fix) =x+ y4-x2
Domain: [—2, 2]
Range:

[-2,272] ==[-2,2.83]
y-intercept: (0, 2)
x-intercept: ( - V2, o)

(-v/2.0>

28. Me) = -5cos^

Domain: (-00, 00)
Range: [-5,5]

30. 7x2-4

0:

V^

y is a function of x. Vertical lines intersect the graph
at most once.

32. x2 + v2 = 4

y = ±74 - X-

y is not a function of x. Some vertical lines intersect
the graph twice.

34. x2 + y = 4 => y = 4 — .r2

y is a function of x since there is one value of y for
each X.

36. xh- - .r2 + 4v = 0 =

.r2 + 4

y is a function of x since there is one value of y for
eachx

294 Chapter P Preparation for Calculus

38. Piix) = x' - X + 1 has one zero. /JjW = x^ - x has
three zeros. Every cubic polynomial has at least one zero.
Given p(x) = Ax^ + Sx^ + Cx + D, we have p -> - oo as
x-^ — oo andp— >oo asx— >oo if i4 > 0. Furthermore,
p— >oo asx— > — oo and/3— > — oo as X— >oo if A < 0.
Since the graph has no breaks, the graph must cross the
X-axis at least one time.

r^

40. The function is/(x) = cx. Since (1, 1/4) satisfies the
equation, c = 1/4. Thus,/(x) = (l/4)x.

42. The function is h(x) = c-y\x\. Since (1, 3) satisfies the
equation, c = 3. Thus, h(x) = 3 V]x[.

2 — 0 1
44. The student travels _ = - mi/min during the first 46. (a)

4 minutes. The student is stationary for the following

6-2
2 minutes. Finally, the student travels

during the final 4 minutes.

10

1 mi/min

,.

500-

^

400-

X

^

V

300-

y

200'

f'

100-

10

20

30

40

50

(b) A(15) == 345 acres/farm

48. (a) g(x)=/(x-4)
g(6)=/(2) = 1
g(0)=/{-4) = -3
Shift /right 4 units

(b) g(x)=/(x + 2)
Shift/left 2 units

(c) g(x)=/(x)+4

Vertical shift upwards
4 units

(d) g(x)=/(x)- 1

Vertical shift down 1 unit

(e) g{x) = 2/(x)

g{2) = 2/(2) = 2
g(-4) = 2/(-4) =

(0 gW = 5/W
g(2) = ^/(2) = i
g(-4) = i/{-4) =

(-4.-6)

50. (a) h(x) = sin(x -I- (n/l)) + 1 is a horizontal shift ir/2 units to the left, followed by a vertical shift 1 unit upwards.
fb) h(x) = - sin(x - 1) is a horizontal shift 1 imit to the right followed by a reflection about the x-axis.

Section P. 3 Functions and Their Graphs 295

52. (a)/(^(l))=/(0) = 0

(b) g(f(\)) = g(l) = 0

(c) g{fm = g{Q) = - 1
(d)/(5(-4))=/(15)= 715

(e)/(^W)=/(x2- \) = J^^X

(f) gifbc)) = g{J~x) = {J~xf - \ = .X - \ {x > 0)

54. fix) = X' - ).,g(x) = cosj:

{f'g){x)=Mx)) =/(cos.v) = cos^v - 1
Domain: (—00,00)
{g'f){x)=g{x^- l) = cos(x2- 1)
Domain: ( — 00, ca)
NoJ'g^ g'f.

56. (/og)(A) =f{V^T2) = -^1=

~Jx + 2

Domain: (—2, 00)

=/)W

.^=7^

2x

You can find the domain of g °/by determining the intervals where (1 + 2x) and x are both positive, or both negative.

-2 -1 _i 0 1 2

Domain: ( — 00, — jj, (0, 00)

58. (a) 25

60. f(-x) = If^ = - l/^ = -fix)
Odd

(b) H'

1.6

0.0021 ^)- + 0.005(^

0.029

= 0.000781 25a- 2 + 0.003125a - 0.029

62. /(-a) = sin^(-.v) = sin(-A) sin(-.t) = (-sin.r)(-sin a) = sin^A
Even > (

64. (a) If/ is even, then (-4, 9) is on the graph.

66. /(-a) = a,„(-Af " + a2„_2(-AP"-^ + • ■ ■ + ^.(-a-)^ + a^

(b) If/ is odd, then (-4, -9) is on the graph.

= /(-v)

Even

68. Let F(a) = /(.v)g(.v) where/is even and g is odd. Then

F(-a) =/(-.v)g(-A) =/(A)[-g(A)] = ~fix)g(x) = -Fix).

Thus, Fix) is odd.

296 Chapter P Preparation for Calculus

70. (a) Let F{x) = fix) ± g{x) where/and g are even. Then, F{-x) = f(~x) ± g{-x) = f{x) ± g(x) = F(x).
Thus, F{x) is even.

(b) LetFW =fix) ± ^W where fandg are odd. Then, F(-x) = f{- x) ± g(- x) = -f(x) + g{x) = -F{x).
Thus, F{x) is odd.

(c) Let FW = fix) ± gix) where f is odd and g is even. Then, F(-x) = fi-x) ± gi-x) = -fix) ± gix).
Thus, Fix) is neither odd nor even.

72. By equating slopes.

y-2 _ 0- 2
0-3 X- 3

6
X- 3

6

74. True

y-2

y =

x-2,

+ 2 =

2x
X- ^'

= Jx^ + y^ = -v/jc^ +

Ix \2

X-2,

76. False; let/W = x^. Then/(3x) = ilxf = 9x^ and 3/W = 3;^^. Thus, 3/U) ^ /(3x).

Section P.4 Fitting Models to Data

2. Trigonometric function

4. No relationship

6. (a) 20

No, the relationship does not appear to be linear.

(b) Quiz scores are dependent on several variables such as
study time, class attendance, etc. These variables may
change from one quiz to the next.

8. (a) s = 9.1t + 0.4

(b) 45

The model fits well,
(c) If / = 2.5, s = 24.65 meters/second.

10. (a) Linear model: H = -0.3323/ + 612.9333
(b) 600

The fit is very good,
(cj When t = 500,

H = -0.3323(500) + 612.9333 « 446.78.

12. (a) S = 180.89jc2 - 205.79;c + 272

(b) 25000

(c) When x = 2, 5 = 583.98 pounds.

Review Exercises for Chapter P 297

14. (a) t = 0.0027 1^2 - 0.05295 + 2.671

(b) 21

(c) The curve levels off for s < 20.

(d) t = 0.002^2 + 0.0346i + 0.183

The model is better for low speeds.

18. (a) H{t) = 84.4 + 4.28 sinf ^ + 3.86
One model is

TTt

C{t) = 58 + 27sin(— + 4.1

(b) 100

20. Answers will vary.

16. (a) T= 2.9856 x lO'V^ _ 0.0641 /j^ + 5.2826p + 143.1

(b) 350

(c) For T = 300°F, p = 68.29 pounds per square inch.

(d) The model is based on data up to 100 pounds per
square inch.

(C) 100

(d) The average in Honolulu is 84.4.
The average in Chicago is 58.

(e) The period is 12 months (1 year).

(f) Chicago has greater variability (27 > 4.28).

Review Exercises for Chapter P

2. y = (.t - l)(.t - 3)

;r = 0 ^ .V = (0 - 1)(0 - 3) = 3 ^ (0, 3) v-intercept

;y = 0 ^ 0 = U - l)(x - 3) => .V = 1, 3 => (1, 0), (3, 0) A-intercepts

4. XV = 4

X = 0 and y = 0 are both impossible. No intercepts.

6. Symmetric with respect to -v'-axis since

y = (-x)-* - (-x)- + 3
v = x" - x~ + 3.

298 Chapter P Preparation for Calculus

8. 4x -2y = 6

y = 2x - 3
Slope: 2
y-intercept: —3

10. 0.02X + 0.15y = 0.25
2x + 15y = 25

12. y = x{6- x)

y = -■i5-« +

Slope: -fj
y-intercept: j

14. y = |;c - 4| - 4

16. y = Si/x

Xmin = -40
Xmax = 40
Xscl = 10
Ymin = -40
Ymax = 40
Yscl = 10

18. • y = ;t -H

(x+ I) -x^ = 7

0 = x'- - a: 4- 6

No real solution
No points of intersection
The graphs of y = Jt -I- 1 and
y = x^ + 7 do not intersect.

20. y = kx^

(a) 4 = kilf =^k = 4andy = 4x^
(c) 0 = M0)3 => any A- will do!

22.

12-
10-

8--

S-

4-
2--

H — I — I — I — I h-

12 3 4 5 6

(7, 12)

—h-»-'
(7,-1)

The line is vertical and has no slope.

24.

(b) 1 = k(-2y =^k= -^ andy = -jx^
(d) -I = k{-iy=^k= l=^y = x^

3-(-l) 3-6

-3 - r -3 -
4 -3

-3-f -11
-44 = 9 -^ 3r
-53 = 3r

Review Exercises for Chapter P 299

26. J, - 6 = OU - (-2))

y = 6 Horizontal line

(-2,6)

I I I I I I > -

28. m is undefined. Line is vertical.

x = 5

6--

4

-4-2 - ■ 2 4
-2--

(5.4)

30. (a)

y-3

r(-t - 1)

Sy - 9 = -2;c + 2
2x -H 3y - 1 1 = 0
(b) Slope of perpendicular line is 1.

>- - 3 = lU - 1)
y = X + 2
Q= X- y + 2
4-3 .

(c)

2 - 1
y - 3 = l(x - 1)
y = x + 2
Q = X- y + 2
(d) y = 3

y- 3 = 0

32. (a) C = 9.25r + 13.50f + 36,500
= 22.75f + 36,500

(b) /? = 30f

(c) 30r = 22.75r + 36,500
7.25r = 36,000

r « 5034.48 hours to break even.

34. X- - V = 0

Function of x since there is one value for v for each x.

36. X = 9 - r

Not a function of x since there are two values of y for
some x.

38. (a) /(-4) = (-4)2 + 2 = 18 (because -4 < 0)
(b) /(O) = |0 - 2| = 2
(c)/(l)= |1 -2| = 1

40. f(x) = 1 - .r= and g{x) = Iv + 1

(a) f[x) - gix) = (1 - .r^) - (2r + \) = -x~ - Ix

(b) f(x)g{x) = (1 - .v-)(2v + 1) = -It^ - .r= + Ir -I- 1

(c) g{f{x)) = ^(1 - .X-) = 2(1 - .v^) + 1 = 3 - li-

300 Chapter P Preparation for Calculus

42. f(x) = x^ -3.xP-

(a) The graph of g is obtained from/by a vertical shift
down 1 unit, followed by a reflection in the jc-axis:

gix) = -[/W - 1]
= -x^ + 3x^+ I

(b) The graph of g is obtained from /by a vertical shift
upwards of 1 and a horizontal shift of 2 to the right.

g(x)=f{x-2) + 1

= (x- 2)3 - 3(x -2)2+1

44, (a) fix) = x^x - 6f

(b) gix) = Jt3(x - 6)2

300

Ai

^ /

(c) hix) = x^U - 6)3

200

4 ■ — ^-r— . ,./ , . 10

46. For company (a) the profit rose rapidly for the first year,
and then leveled off. For the second company (b), the
profit dropped, and then rose again later.

Problem Solving for Chapter P

48. (a) y = - 1.204.t + 64.2667

(b) 70

(c) The data point (27, 44) is probably an error.
Without this point, the new model is

y = -1.4344JC + 66.4387.

2. Let y = mx + 1 be a tangent line to the circle from the point (0, 1). Then
x^ + iy+l)^=l
x^ + imx+ I + l)^= \
irrfi + \)x'^ + 4mx + 3 = 0
Setting the discriminant b^ — 4ac equal to zero,
16m^ - Airrr + 1)(3) =0
\6m- - I2m^ = 12
4m^ = 12
m = ±y3
Tangent lines: y = V^x + 1 and v = - V^x + 1.

Problem Solving for Chapter P 301

4. (a) /(x + 1)

(b) f(x) + 1

(c) lf{x)

(d) f{-x)

(e) -/W

(0 i/wi

4-

:Y^^y:

(g) /(ki)

6. (a) 4y + 3;c = 300 => v

300 - 3a-

w ^ /-, . /'300 - 3.x\ -3.V- + 300x
Aix) = x(2y) = xl^ j =

Domain: 0 < x < 100

(b)

25 50 75 100

(c) A(x) = --{x- - IOOa)

= --(x^ - lOOx + 2500) + 3750

Maximum of 3750 ft" at x = 50 ft, >- = 37.5 ft.

= --(x - 50)= + 3750

A(50) = 3750 square feet is the maximum area,
where x = 50 ft and v = 37.5 ft.

302 Chapter P Preparation for Calculus

8. Let d be the distance from the starting point to the beach,
distance

Average velocity =

time
2d

120 60

120 60
= 80 km/hr

12 3 4 5

3-21 1

(a) Slope = = - Slope of tangent line is greater than —

2-11 1

(b) Slope = _ = - Slope of tangent line is less than -.

, , ^, 2.1 - 2 10 ^, ^ ,. . ^ 10

(c) Slope = — = — . Slope or tangent Ime is greater than — .

(d) Slope = (4 + ;,)_4

vm- 2

, , V4 + /! - 2 V4 + /! - 2 V4 + /! + 2

(e) = • — ==

h h V4 + /! + 2

_ (4 + /;) - 4

//(V4 + /I + 2)
1

JA + h + 2

hi-Q

As /! gets closer to 0, the slope gets closer to — . The slope is - at the point (4, 2).

Problem Solving for Chapter P 303

12. (a)

kl

U-4)2+/ = feV + )^)

(A:2 - ly + {k^ - l)f + &x = 16

If k = 1, then ;c = 2 is a vertical line. So, assume /c' - 1 ¥= 0. Then

( 4 V , 16 16

U + t; — 7 1 +y^ = - r +

k^- 1

x +

A:2- 1

+ / =

fc2 - 1 (F - 1)2

Ak

k^- 1

, Circle

(b) Ift = 3, (;t + ^)" + / = (!)'

(c) For large k, the center of the circle is near (0, 0), and the radius becomes smaller.

14. fix) =y =

1

1 - X

(a) Domain: all -t t^ 1
Range: all v t^^ 0
1

(b) fifix)) =/l

1 - x

1 \ - X x-\

1 — .V — 1 —X X

\ - x) \ - X

Domain: all x i^ 0, 1
(x- 1

(c) f{f{f{x)))=f\

1

1

X j (x- W 1

X I X

Domain: all x t^ 0, 1
(d) The graph is not a line. It has holes at (0, 0) and (1, 1).

{ 1 — »-^

CHAPTER 1
Limits and Their Properties

Section 1.1 A Preview of Calculus 305

Section 1.2 Finding Limits Graphically and Numerically 305

Section 1.3 Evaluating Limits Analytically 309

Section 1.4 Continuity and One-Sided Limits 315

Section 1.5 Infinite Limits 320

Review Exercises 324

Problem Solving 327

CHAPTER 1
Limits and Their Properties

Section 1.1 A Preview of Calculus

Solutions to Even-Numbered Exercises

2. Calculus: velocity is not constant

Distance = (20 ft/sec)(15 seconds) = 300 feet

4. Precalculus: rate of change = slope = 0.08

6. Precalculus; Area = tt[J2)'

8. Precalculus: Volume = 7r(3)-6 = 54tt

10. (a) Area "5+- + - + -« 10.417

(b) You could improve the approximation by using more rectangles.

Section 1.2 Finding Limits Graphically and Numerically

X

1.9

1.99

1.999

2.001

2.01

2.1

fix)

0.2564

0.2506

0.2501

0.2499

0.2494

0.2439

lim ^ 7 = 0.25 (Actual limit is 3.)

x-^2 X 4

4.

X

-3.1

-3.01

-3.001

-2.999

-2.99

-2.9

fix)

-0.2485

-0.2498

-0.2500

-0.2500

-0.2502

-0.2516

lim :; «= -0.25 (Actual limit is -j)

->-3 .T + 3 ^

6.

X

3.9

3.99

3.999

4.001

4.01

4.1

fix)

0.0408

0.0401

0.0400

0.0400

0.0399

0.0392

lim ^^ —, =» 0.04 (Actual limit is ^.)

j:->4 X - 4

X

-0.1

-0.01

-0.001

0.001

0.01

0.1

fix)

0.0500

0.0050

0.0005

-0.0005

-0.0050

-0.0500

cos X — I
lim '■ « 0.0000 (Actual limit is 0.) (Make sure vou use radian mode.)

.C-.0 X

305

306 Chapter 1 Limits and Their Properties

10. lim (;c^ + 2) = 3

jr-»l

12. lim/(x) = lim (;c^ + 2) = 3

14. lim

1

■ does not exist since the

3 a: - 3

function increases and decreases
without bound as x approaches 3.

16. lim sec X = 1

18. lim sin(7rx) = 0

20. C{t) = 0.35 - 0.121- (f- 1)
(a)

(b)

I

3

3.3

3.4

3.5

3.6

3.7

4

C{t)

0.59

0.71

0.71

0.71

0.71

0.71

0.71

lim CW = 0.71

(c)

t

3

2.5

2.9

3

3.1

3.5

4

C{t)

0.47

0.59

0.59

0.59

0.71

0.71

0.71

lim C{t) does not exist. The values of C jump from 0.59 to 0.71 at f = 3.

f— *3.5

22. You need to find S such that 0 < |x — 2| < 5 implies
[fix) - 3| = 1^2 - 1 - 3| = 1^2 - 4| < 0.2. That is,

-0.2 < x2 - 4 < 0.2
4 - 0.2 < x^ < 4 + 0.2
3.8 < x^ < 4.2
JJI < X < JA2

Vl8 - 2 < X - 2 < V4I - 2

So take S = 742 - 2 « 0.0494.
ThenO < |x - 2| < 8 implies

-{742 - 2) < X - 2 < 742 - 2
73^ - 2 < X - 2 < 742 - 2.
Using the first series of equivalent inequalities, you obtain

|/(x) - 3| = |x2 - 4| < 0.2.

2^- IT. '-2

-f

-f

f (. - 4)

< 0.01

< 0.01

< 0.01

0 < |x - 4| < 0.02 = 5
Hence, if 0 < |x - 4| < 8 = 0.02, you have

\U - 4)

-f

< 0.01

< 0.01

< 0.01

|/(x) -L\< 0.01

Section 1.2 Finding Limits Graphically and Numerically 307

26. lim (x^ + 4) = 29

\(x^ + 4) - 29| < 0.01

|a:2 - 25| < 0.01

|(x + 5){x - 5)1 < 0.01

0.01

5 <

U + 5

If we assume 4 < x < 6, then 5 = 0.01/11 = 0.0009.
Hence, if0<|jr-5|<6 = —rr-, you have

1

|.x- 5||a; + 5| < 0.01

|;c2 - 25| < 0.01

(x^ + 4) - 29| < 0.01

|/(x) - L\ < 0.01

(0.01)

28. lim (2;c + 5) = - 1

a:— >-3

Given e > 0:

|(2r + 5)-(-l)| < e
|2x + 6| < e

2\x + 3| < 6

|;t + 3| <|=6
Hence, let 5 = e/2.
Hence, if 0 < |jr + 3| < 5 = -, you have

1^ + 31 < I

i2x + 6| < e

|(2x + 5)-(-l)| < 6

l/W -L\<e

30. lim %x + 9) = 5(1) + 9 = f
Given e > 0:

|V3^ + 9j - y| < 6

i|x-l|<e

k-l|<56
Hence, let 6 = (3/2)e.
Hence, if 0 < |j: — 1 1 < S = je, you have

\x - \\ < ^€

32. lim(-l) = -1

JC— ^2

Given e > 0: |-1 - (-1)| < e
0 < e
Hence, any 5 > 0 will work.
Hence, for any 5 > 0, you have

|{-l)-(-l)| <e
l/W - i| < e

|(|j: + 9)-f| < e
l/W - L| < 6

34. lim v^ = 74 = 2

Jr— >4

Given e > 0: |V^ - 2| < e

|V5 - 2| |v^ -I- 2I < e| V^ + 2I
|x - 4| < e|7c + 2I
Assuming 1 < a- < 9, you can choose 5 = 3e. Then,
0 < \x- A\ < 5 = 3e => |.r - 4| < e| Va + 2I
=> ITx- - 2I < e.

36. lim |.v- 3| =0
Given 6 > 0:

|(.r - 3) - 0| < e

|.T- 3| < e= 5
Hence, let 8 = e.

Hence for 0 < |.r - 3 1 < 6 = e. you have
|.v - 3| < 6
||..-3| -0| <6
|/(.t) -L\<e

308 Chapter 1 Limits and Their Properties

38.

lim {x^
Given e

+ 3x)
> 0:

= 0

\{^

+ 3x) -

- 0|

<

e

\x{x^

3)1

<

6

k-

^-3|

<

w

If we assume - 4 < x < — 2, then 5 = e/4.
Hence for 0 < |j: — (— 3)| < 5 = t. you have

\x + 3\ <-e < i-j-e
' A \x\

\x{x + 3)1 < e

|x2 + 3;t - 0| < 6

l/W - L| < e

40. /W =

x-Z

x^ - Ax + l,

V

h

lim/W = -

The domain is all x i= 1, 3. The graphing utility does not
show the hole at (3, 5).

42. fix) = V4
x'^ — 9

Mm fix) = -

j:— »3 0

3
9 '

44. (a) No. The fact that/(2) = 4 has no bearing on the exis-
tence of the limit of /W as x approaches 2.

(b) No. The fact that lim/(x) = 4 has no bearing on the
value of /at 2.

The domain is all a: # ± 3. The graphing utility does not
show the hole at (3, g).

46. Let p(x) be the atmospheric pressure in a plane at
altitude x (in feet).

lim p(x) = 14.7 lb/in2

48. 0.002

Using the zoom and trace feature, 5 = 0.001. That is, for

0 < Ix - 21 < 0.001,

x-2

- 4

< 0.001.

50. True

52. False; let

f{x) =

Ax, x + A

10,

lim/(x) = lim (x2 - 4a;) = 0 and/(4) = 10 ^t 0

j:-»4 jr->4

54. lim

^

\2

J:->4 X - a

= 1

n

4 + [0.1]"

/(4 + [0.1]")

1

4.1

7.1

2

4.01

7.01

3

4.001

7.001

4

4.0001

7.0001

n

4 - [0.1]"

/(4 - [0.1]")

1

3.9

6.9

2

3.99

6.99

3

3.999

6.999

4

3.9999

6.9999

Section 1.3 Evaluating Limits Analytically 309

56. fix) = mx + b, m i^ Q.l^X e > Qhe given. Take 5 = -j — r.

\m\

IfO <\x- c\ < 8 = 7—7, then

\m.\\x — c\ < e
\mx — mc\ < e
\{mx + b) - (mc + b)\ < e
which shows that lim (mx + b) = mc + b.

58. lim g{x) = L,L > 0. Let e = jL. There exists 5 > 0
such that 0 < |a: - 0| < 8 implies \g{x) - L\ < e = jL.
That is,

-5L < gix) - L < {l

\l < gix) < jL

Hence for x in the interval ic — S, c + S), x ¥= c,
gix) >\l>Q.

Section 1.3 Evaluating Limits Analytically

2. 10

(a) lim gix) = 2.4

(b) lim^U) = 4

x—*0

(a) lim/(f) = 0

/->4

(b) lim /(f) = -5

«W = \._a

.t-9

/(r) = fk - 4|

6. lim x3 = (-2)3 =

a:-»-2

8. lim (3x + 2) = 3(-3) + 2 = -7

x-»-3

10. lim(-r^+ 1) = -(1)^+ 1 = 0

X-*l

2 2

14. lim

3X + 2 -3 + 2

-2

12. lim (3.^3 - 2r= + 4) = 3(1)^ - 2(1)" + 4 = 5

:t-^3 ;c + 5 3 + 5 8

ifi I- V^^^ V3 + 1 ,

18. Imi — = -z :- = -2

a:->3 X - 4 3-4

20. lim4/;c + 4 = ^4 + 4 = 2

x->4

22. lim i2x - ly = [2(0) - 1]' = -1

24. (a) lim fix) = (-3) + 7 = 4

(b) lim gix) = 4^ = 16

(c) lim gifix)) = ^(4) = 16

x—*-3

26. (a) lim/(jc) = 2(4^) - 3(4) + 1 = 21

(b) lim gix) = ^21 +6 = 3

(c) \\mgifix)) = gi2l) = 3

x-*4

28. lim tan X = tan 77 = 0

x-*ir

30. lim sm ^:— = sm — = 1

j:-»1 2 2

32. lim cos 3x = cos Stt = - 1

x—*lr

34. lim cos X = cos ^;r = X

j->57r/3 3 2

« ,. /tT.xA 777 -2v^

36. lim sec —- = sec — - = — :; —

j-»7 V 6 / 6 3

310 Chapter 1 Limits and Their Properties

38. (a) lim [4/W] = 41im/(;c) = 4 - = 6

x^c j:— *c \Z/

(b) lim [fix) + g{x)] = lim/(x) + lim g{x) = ^ + \=2

x—^c j:-+c x-*c L L

(c) lim \S{x)g{x)-\ = riim/Wiriim g{x)\

IB=!

40. (a) XxmlfJQ) = 3/lim/(x) = l/in = 3

Jf-»C

(c) lim [/W]2 = [lim/(x)l2 = (27)2 ^ 729

(d) lim[/(;c)]2/3 ^ riin,/(;c)]2/3 = (27)2/3 = 9

X^ - Zx
42. /W = j: — 3 and h{x) = agree except at j: = 0

1 ;c

44. g{x) = 7 and/U) = -; agree except at j; = 0.

X — I x'- — X

(a) lim h{x) = lim f{x) = -5

jr->-2 jc-*-2

(b) lim/iW = lim/W = -3

(a) lim/(x) does not exist.

jr-»l

(b) lim/(;c)= -1

JC-^O

2;t2 - X - 3

46. /(x) = — and g{x) = 2x — 3 agree except at

JC^ + 1

48. /(j:) = — — — and g{x) = j? — x -\- 1 agree except at
x= -1.

lim f{x) = lim g{x) = -5

lim f{x) = lim g(x) = 3

J—* - 1 JT— » - 1

/

/[

\

,. ,. 1- X -{x- 2)
50. lim -:; = lim • ^ —

;c-*2;c2-4 x^2 (a: - 2)(a: + 2)

= lim

-1

j:->2 X -V 1

^^ ,. ;c- - 5;c + 4 ,. (x - 4)(;c - 1)

52. hm — — = hm 7 -(7 -(

x->4 x2 - 2x - 8 x^A [x - 4)(;c + 2)

= lim 7 rr = ~ = x

x-»4 U + 2) 6 2

,. ,. JlVx-Ji ,. V2 + a; - V2 V2 + ;t + v^

54. lim = lim • — , ^

^^0 X x-^ X Jl + X + J2

2+X-2 ,. 1

= lim 7 — , ;=A— = lim

1 72

^-»o(V2 + X+ J7)x ^-^0 J2 + X + J2 ijl 4

,, ,. v7TT-2 ,. Vx + 1 - 2 Vx + 1 + 2 ,.
56. hm ; = lim ;: • — , — - = hm ■

A- 3

lim -

^**" -^ ~ I'l" -^- . T^ — 11111 , TTr ; TT — 11111 , T"

x-^i X - 3 x^i X - 3 Vx+ 1 + 2 -.-^3 (x - Z\jx + 1 + 2] :c^3 vTTT + 2

_J j_ 4 - (x + 4)

.„ ,. A + 4 4 ,. 4(x + 4) ,. -1 1

58. hm = hm — ^^ '— = hm -7 — -— r = - —

x^o X x-^0 X x->o 4(x + 4) 16

„ ,. (x + Ax)2 - x2 ,. x^ + 2xAx + (Ax)2 - x2 ,. Zb;(2x + Ax) ,. ,, , . ^
60. hm -^ = hm — ^ — = hm — ^^-r = hm (2x + Ax) = 2x

Aar-^O Ax Ax->0 Ax Ax-»0 Ax Ai^O

Section 1.3 Evaluating Limits Analytically 311

^ ,. U + Ajc)3 - x3 ,. x^ + 3x'\x + 3x{Axy + (Axf - jc'

62. lim -. = lim

Ax->o Ajc '^->0 Ax

,. Ax{3x^ + 3xAx + (Ax)^) ,. ,, , , . ,. ,,, , ,
hm — ^^ ; ^^ — — = hm (3r= + 3xAx + (Ax)^) = 3x^

Aar-»0 A.V Ax->0

64. fix)

A- Jx
X - 16

X

15.9

15.99

15.999

16

16.001

16.01

16.1

/w

-.1252

-.125

-.125

7

-.125

-.125

-.1248

Analytically, lim

(4-v^

.r-16 X - 16 x^\(,[J~y_ + 4)(^ _ 4)

= lim — F = --.

-'-'IS Jx-v A 8

It appears that the limit is -0.125.

66. lim ^ = 80

j:->2 X — L

X

1.9

1.99

1.999

1.9999

2.0

2.0001

2.001

2.01

2.1

fix)

72.39

79.20

79.92

79.99

7

80.01

80.08

80.80

88.41

, , . „ ,. .r5 - 32 ,. ix - 2)(.x^ + 2x3 + 4x2 + 8x + 16)
Analytically, lim —- = lim r

■^ -^ x^2 X - 2 x^2 X - 2

= lim (x" + 2x' + 4x^ + 8x + 16) = 80.

j-»2

(Hint: Use long division to factor x^ - 32.)

3(1 - cos.x)

68. lim =^ ^^^^^ = lim [sf^-^^^)] = (3)(0) = 0

„. ,. cose tan 0 ,. sm 0
70. lim = hm — ;— = 1

8->0 0 B->0 6

_^ ,. tan"x ,. sin-x

72. lim = hm ^— = lim

x->0 X Jr->0 X cos' X .t->0

sinx sin x 1
X cos- X J

74. lim (^ sec (^ = tt(— 1) = — tt

= (1)(0) = 0

-. ,. 1 — tanx ,. cosx — sinx

76. lim -: = hm

Jr-.7r/4 sin X — cos X x-^it/4 Sin X cos X — COS" X

(sinx — cos.v)

= lim

v/4 cos x(sin X — cos x)
1

= lim

x->it/4 cosx

= lim (-secx)

.r— nr/4

= -72

_„ ,. sin 2x
78. hm . .,
x-iO sm 3x

r Vsinlv
= hm 2 -r —
-t-»o L V Iv

'-'"\]m-M>-\

312 Chapter 1 Limits and Their Properties

SO.fih) = (1 +cos2/!)

h

-0.1

-0.01

-0.001

0

0.001

0.01

0.1

m

1.98

1.9998

2

?

2

1.9998

1.98

r\/\r\

Analytically, lim (1 + cos 2h) = \ + cos(O) =1 + 1=2.

The limit appear to equal 2.

82. f(x)

X

-0.1

-0.01

-0.001

0

0.001

0.01

0.1

fix)

0.215

0.0464

0.01

?

0.01

0.0464

0.215

Analytically, lim^ = lim^f^^^) = (0)(1) = 0.

j-jO \ X

The limit appear to equal 0.

„. ,. fix + h)- fix) Jx + h- J~x .. Jx + h
84. lim ; — = lim ; = lim —

/i->o h h->o h /i->o h

lim

X + h — x

lim ■

^x + h + Jx
Jx + h + Jx

1

''-*o h[Jx + h + Jx) '"^0 Jx + h + J~x ijx

„, ,• fix + h) - fix) ,. ix + h)- - 4(x + h) - ix^ -Ax) ,. ' x^ + 2xh + h'^ - 4x - 4h - x^ + Ax

86. hm-^^ r — =^-^ = hm -^^— —, — ^^ = Iim ;

h->o h A->o h A-»o h

,. hilx + /i - 4) ,. ,^ , ,, ^
= hm -^^ ; = hm (2x + /z - 4) = 2jc - 4

88. iim [b - \x- a|] < lim/(jc) < lim [b + \x - a|]
b < Wmfix) < b
Therefore, lim/(;t) = b.

90. fix) = |.rsinji:|

6

lim brsin;c = 0

92. fix) = \x\ cos X

94. hix) = X cos —

X

:

Iim X cos X = 0

jc->0

lim X cos - = 0

x-*0 \ X

Section 1.3 Evaluating Limits Analytically 313

x^ - 1
96. fix) = — — j- and g(x) = x + \ agree at all points

except .X = 1.

98. If a function /is squeezed between two functions h and g,
h{x) < fix) < gix), and fi and g have the same limit L as
X— >c, then Vim fix) exists and equals L.

100. fix) = x, gix) = sin^ X, hix)

sm'- X

o^Tn

A

•y

When you are "close to" 0 the magnitude of g is "smaller"
than the magnitude of/ and the magnitude of g is
approaching zero "faster" than the magnitude of/.
Thus, |g|/|/| ~ Owhen jc is "close to" 0

102. sit) = - 16r2 + 1000 = 0 when r = J^ ^^^

16

seconds

.. ■'i 2 j '^^ ,. 0-(-16r2+ 1000)

lim , -p= = lim , -/==

t^sVToh 5V10 _ J. (-»5yio/2 5V10 _

16 f-

lim

125

16

lim

t + ^Y. - ^)

2 /

■»5yio/2 SyiO _ r->5yio/2

5V10

= lim

r^5yio/2

■ leff + ^^^j = -SOyiO ft/sec = -253 ft/sec

104. -4.9f- + 150 = 0 when t =
The velocity at time t = a\s

/l50 ^ /l500
4.9 ~ V 49

= 5.53 seconds.

,. sia) - sit) ,. (-4.9a^ + 150) - (-4.9r + 150) ,. -4.9(a - f)(a + r)
lim = lim = lim-

I-»n a — t r-»n a — t r->n

= lim -4.9(a + t) = -2a(4.9) = -9.8a m/sec.

a - t

Hence, if a = V 1500/49, the velocity is -9.8V1500/49 = -54.2 m/sec.

106. Suppose, on the contrary, that lim gix) exists. Then, since Vim fix) exists, so would lim [fix) + gix)], which is a

.r— *c .x—*c x—*c

contradiction. Hence, lim ^(jc) does not exist.

x-*c

108. Given /(.x) = x", n is a positive integer, then
limx" = limCxx""') = [limJflimx""']

= c[lim ixx"-^ = c[lim.x][limx"--]
= c(c)lim (xa:""') = • ■ ■ = c".

110. Given lim/(x) = 0:

x—*c

For every e > 0, there exists S > 0 such that |/(x) - 0| < ewheneverO < |x - c| < 6.
Now |/(x) - 0| = |/U)| = ||/(.r)| - 0| < 6 for |x - c| < 5. Therefore, lim |/(.t)| = 0.

314 Chapter 1 Limits and Their Properties

112. (a) If lim \f{x)\ = 0, then lim [- |/(x)|] = 0.
-\f{x)\ <f{x) < l/WI
lim[-|/U)|] < lim/W < lim|/W|
0 < Urn fix) < 0

x—>c

Therefore, lim/(jc) = 0.
(b) Given \imf{x) = L:

x—*c

For every e > 0, there exists 8 > 0 such that \f{x) — L\ <e whenever 0 < \x — c\ < S.
Since \\f(x)\ ~ \L\\ < \f{x) - L\< e for \x - c\ < 8, then lim |/(;c)| = \L\.

114. True, lim jc' = (P = 0

x->0

116. False. Let f{x) =

X x i' I
3 x= 1

Then \imf(x) = 1 but /(I) # 1.

X—>1

118. False. Let/{x) = {x'^ and g{x) = x^. Then/W < g{x)
for all ;c ^ 0. But lim/(jc) = lim gix) = 0.

J— *0 x-^O

.-- ,. 1 — COS a: ,. 1 — cosx 1 + cosx
120. lim = hm

j:-*0 X j:-»0 X 1 + COS JT

= lim

1

= lim

i->o;c(l + cosj:) .t"-5bj;(l + cosjc)
sin X sin x

= lim

j->0 ;c

1 + cosj:

lim

x-»0 X

(1)(0) = 0

lim——

jr-»0 1 + cos

l]

122. fix)

secx — 1

(a) The domain of/ is all x t^ 0, ir/l + mr.

(b) 2

^J

The domain is not obvious. The hole at .jc = 0 is not
apparent.

(c) lini/W = -

(d)

sec x — I sec x — I sec j: + 1

sec^j: — 1

x^ sec a; + 1 j:^(secA: + 1)

tan^ X 1 fsin-x\ 1

Hence, lim

;c^(secAr+ 1) cos^j:\ x^ /secj:+ 1

sec ;c - 1 _ 1 /sin^A:\ 1

'o x^ x-^0 cos^ x\ 7? /sec X + 1

124. The calculator was set in degree mode, instead of radian mode.

Section 1.4 Continuity and One-Sided Limits 315

Section 1.4 Continuity and One-Sided Limits

2. (a) lim /W = -2

(b) lim f(x) = -2

(c) lim /W = -2

j:-»-2

The function is continuous at

x= -2.

ii,u 2-^ n,„ 1

1

.'ilV - 4 - iii?* x + 2-

4

4, (a) lim^/U) = 2

(b) lini fix) = 2

l->-2"

(c) lim fix) = 2

j:^-2

The function is NOT continuous at
x= -2.

6. (a) lim fix) = 0

(b) lim /U) = 2

Jc— »-l

(c) lim /U) does not exist.

jr— »— 1

The function is NOT continuous at

x = -1.

,n 1- v^- 2 ,. Ji-2 Ji+2
10. hm — = lim

j:->4- X — 4

= lim

■"- Jc-4 ^ + 2

X- A

4 U-4)(y; + 2)

lim

1

1

'-**' J~x + 1 4

,- ,. k - 2 ,. X - 2
12. lim ■' -^ = lim = 1

jr->2* X — 2 x-»2* X — 2

, ^ ,. (x + Ax)2 + (a: + Ax) - (x^ + x) ,. x~ + 2a;(Ax) + i^xf- -^x + l^-x^-x
14. lim T = lim

lim

Aj->0*

Ax

2x(Ax) + (Ax)- + Ax
Ax

= lim (2x + Ax + 1)
= 2x + 0+l=2x+l

16. lim fix) = lim (-x^ + 4x - 2) = 2

x->2* j->2*

lim fix) = lim (x^ - 4x + 6) = 2

a:->2" x-»2"

18. lim fix) = lim (1 - x) = 0

lim/(x) = 2

j->2

20. lim sec x does not exist since

j:— >Tr/2

lim sec x and lim sec x do not exist.

x->(,r/2)* Ar-.(Tr/2)"

22. lim^ ilx - W) = 2(2) -2 = 2

24. lim 1

= !-(-!) = 2

26. fix)

X + 1

has a discontinuity at x = — 1 since /(- 1) is not defined.

28./(x)= 2,

X < 1

X = 1 has discontinuity atx = 1 since /(I) = 2+ lim/(x) = 1.
2x - 1, X > 1

30. /(f) = 3 - V9 - f^ is continuous on [- 3, 3].

32. ^2) is not defined, g is continuous on [— 1, 2).

316 Chapter 1 Limits and Their Properties

34. fix) = — is continuous for all real x.

x^ + I

36. fix) = cos — is continuous for all real x.

38. fix)

— has nonremovable discontinuities atx = I and x = —I since Vim fix) and lim fix) do not exist.

j:*- — 1 jr-»l x->-l

X - 3
40. fix) = ^ _ has a nonremovable discontinuity at x ■

atx = 3 since

lim/U) = lim = -.

— 3 since lim fix) does not exist, and has a removable discontinuity

42. fix) =

x - 1

U + 2)U - 1)

has a nonremovable discontinuity atx = —2 since
lim fix) does not exist, and has a removable discontinu-
ity at j: = 1 since

lim/(;c) = lim = —.

44. fix)

\x- 3|
jc- 3

has a nonremovable discontinuity at j: = 3 since Vim fix)
does not exist. ^^

46. fix)

-2x + 3, ;c < 1
x\ x > I

has a possible discontinuity at j: = 1 .
l./(l) = 12= 1

2. A?--^^-*^ = AT- (-2x + 3) = 1

lim fix) = lim x^ = 1

3. /(I) = lim/W

lim/U) = 1

/is continuous at a: = 1, therefore, / is continuous for all real x.

\-2x, X < 2

48. fix) = 1 2 _'/! _(_ 1 o ^^ ^ possible discontinuity at j: = 2.

I ^ T'*\ I X y .-V ."^ ^

1. /(2) = -2(2) = -4

lim /W = lirn i-2x) = -4 1 iini/(;c) does not exist.

lim fix) = lim (jc^ - 4;c + 1) = -3j
x^2*-' ^ ' x^y^ ' -^

Therefore, /has a nonremovable discontinuity atx = 2.

50. /W = \ 6

|;:-3| <2
U- 31 > 2

1 < X < 5
X < 1 or j: > 5

has possible discontinuities at x = \,x = 5.

l./(l) = csc|=2 /(5) = csc^ = 2

lim/W = 2

2. lim/W = 2

3. /(I) = lim/U) /(5) = lim/(x)

x-*\ x-*5

/is continuous dXx = \ and jc = 5, therefore,/is continuous for all real x.

Section 1.4 Continuity and One-Sided Limits 317

52. fix) = tan -r- has nonremovable discontinuities at each
2k + 1 , /: is an integer.

54. fix) = 7> -\x\ has nonremovable discontinuities at each
integer L

56. lim/W = 0
lim j{x) = 0

l-»0"

/ is not continuous at x = — 4

\

/

/

,. , , ,.4 sin AT

58. hm g(x) = hm = 4

jr-»0" x-»0" X

lim gix) = lim (a — 2x) = a

x-tO* x->0'

Let a = 4.

60. lim gix) = lim

x'-a^

J:->n X — a

lim (a: + a) = 2a

Fbd a such that 2a = 8 => a = 4.

62. figix))

Jx - 1

Nonremovable discontinuity at j: = 1. Continuous for all .t > 1.

Because/ - g is not defined for x < 1, it is better to say that/ ° g is discontinuous from the right at x = 1.

64. figix)) = sin.r2

Continuous for all real x

66. hix) =

1

U + l)(x - 2)
Nonremovable discontinuity at x

— 1 and x = 2.

m

fcos .V - 1 X < 0
68. fix) = \ X

[Sx. x> 0

/(O) = 5(0) = 0

1- ft \ 1- (cosx - 1)

hm /(x) = lim = 0 -^

j:->0- .t->0- X

lim /(x) = lim (5x) = 0

a:->0* jr->0*

Therefore, lim/(x) = 0 = /(O) and/ is continuous on the entire real line, (x = 0 was the only possible discontinuity.)

x—*0

. . ^r^ '^-^,

/

70. fix) = xVx + 3

Continuous on [-3, oo]

72./U) = ^ -

VX

Continuous on (0, oc)

318 Chapter 1 Limits and Their Properties

74. f(x) =

x-1

The graph appears to be continuous on the interval
[—4, 4]. Since /(2) is not defined, we know that/has
a discontinuity at x = 2. This discontinuity is removable
so it does not show up on the graph.

76. fix) = x^ + 3a: — 2 is continuous on [0, 1].

/(0)= -2and/(l) = 2

By the Intermediate Value Theorem, fix) = 0 for at least
one value of c between 0 and 1.

— 4 TTX

78. f(x) = 1- tan — is continuous on [1, 3].

X o

/(I) = -4 + tan f < 0 and /(3) = -^ + tan ^ > 0.
o 3 o

By the Intermediate Value Theorem, /(I) = 0 for at least
one value of c between 1 and 3.

80. /(x) = x3 + 3x - 2

fix) is continuous on [0, 1].

/(0)= -2and/(l) = 2

By the Intermediate Value Theorem, /(x) = 0 for at least
one value of c between 0 and 1. Using a graphing utility,
we find that x = 0.5961.

%2. hie)= \ + e-zxaae
h is continuous on [0, 1].
MO) = 1 > 0 and /id) «

-2.67 < 0.

By the Intermediate Value Theorem, hid) = Q for at least
one value 6 between 0 and 1. Using a graphing utility, we
fmd that d = 0.4503.

84. fix) = x2 - 6x + 8
/is continuous on [0, 3].
/(O) = 8 and/(3) = - 1

-1 < 0 < 8
The Intermediate Value Theorem applies.
^2 - 6x + 8 = 0
ix - 2)(x - 4) = 0
X = 2 or X = 4
c = 2 (x = 4 is not in the interval.)
Thus,/(2) = 0.

86. fix) =

x'- + x

/is continuous on [j. 4|. The nonremovable discontinuity,
X = 1, lies outside the interval.

/

35 . ,,.. 20
-and/(4)=y

35 ^ 20

The Intermediate Value Theorem applies.

x^ + X .

x2 + X = 6x - 6
x2 - 5x + 6 = 0
(x - 2)(x - 3) = 0
X = 2 or X = 3
c = 3 (x = 2 is not in the interval.)
Thus,/(3) = 6.

Section 1.4 Continuity and One-Sided Limits 319

88. A discontinuity at j: = c is removable if you can define
(or redefine) the function at j: = c in such a way that the
new function is continuous at ;c = c. Answers will vary.

(a) fix)

(b) f(x)

|x-2|
X- 2
sin(x + 2)

X + 2

(c) fix)

1, ifx > 2

0, if-2 < X < 2

1, ifx = -2
0, ifx < -2

90. If/ and g are continuous for all real x, then so is/ + g (Theorem 1.11, part 2). However, //g might not be continuous if g(x) = 0.
For example, let/(x) = xandg(x) = x^ — 1. Then/and g are continuous for all real x, but//g is not continuous atx = ±1.

1.04, 0 < r < 2

92. C = 1 1.04 + 0.36[[r - ll, r > 2, r is not an integer
[l.04 + 0.36(f - 2), r > 2, f is an integer

Nonremovable discontinuity at each integer greater than 2.

You can also write C as
C

c

.1

4-

3--

2-

i4>

1.04, 0 < r < 2

1.04 - 0.36[2 - r], t > 2

94. Let sit) be the position function for the run up to the campsite. 5(0) = 0 (f = 0 corresponds to 8:00 a.m., 5(20) = k (distance
to campsite)). Let rit) be the position function for the run back down the mountain: Hff) = K r(\Q) = 0. Let/(r) = sit) — r(t).

When t = 0 (8:00 a.m.), /(O) = 5(0) - KO) = 0 - /t < 0.

Whenr = 10(8:10 a.m.), /(lO) = 5(10) - r(IO) > 0.

Since /(O) < 0 and /( 10) > 0, then there must be a value r in the interval [0, 10] such that /(f) = 0. If/(r) = 0, then
sit) — rit) = 0, which gives us sit) = rit). Therefore, at some time f, where 0 < t < 10, the position functions for the
run up and the run down are equal.

96. Suppose there exists x, m [a, b] such that/(xi) > 0 and there exists x, in [a, b] such that/Cx,) < 0. Then by the Intermediate
Value Theorem, /(x) must equal zero for some value of x in [x,, .r,] (or [x,. xj if x, < x,). Thus, /would have a zero in [a. b],
which is a contradiction. Therefore, /(.r) > 0 for all x in [a, b] or fix) < 0 for all x m [a, b].

98. Ifx = 0, then/(0) = 0 and Vim fix) = 0. Hence,/is

„ x->0

contmuous at x = 0.

If X T^ 0, then lim/(f) = 0 for x rational, whereas

t—*x

lim/(f) = lim kt = kx i= 0 for x irrational. Hence, /is not
continuous for all x ^ 0.

100. True

1. /(c) = L is defined.

2. lim /(x) = L exists.

Jr-*c

3. fie) = lim/(.v)

All of the conditions for continuits- are met.

320

Chapter 1 Limits and Their Properties

102. False; a rational function can be written as P(x)/Q{x}
where P and Q are polynomials of degree m and n,
respectively. It can have, at most, n discontinuities.

104. (a) s

10 IS 20 25 30

(b) There appears to be a limiting speed and a possible
cause is air resistance.

106. Let V be a real number. If > = 0, then x = O.lfy > 0, then let 0 < Xg < 7r/2 such that M = tan Xq > y (this is possible
since the tangent function increases without bound on [0, 77-/2)). By the Intermediate Value Theorem, /(x) = tan x is
continuous on [0, .^o] and 0 < >• < M, which implies that there exists x between 0 and Xq such that tan ;c = y. The argument
is similar if y < 0.

108. 1. /(c) is defined.

2. hm/{A:) = lim /(c + Ax) = /(c) exists.
[Let a: = c + Ax As;ic— >c, A;c-^0]

3. lim/(x)=/(c).

Therefore, /is continuous at JT = c.

110. Define /(jc) = /^(.t) — f^ix). Since /i and/, are continuous on [a, b], so is/
f(a) =Ua) -f,{a) > 0 and f{b) = fp) - Mb) < 0.
By the Intermediate Value Theorem, there exists c in [a, b] such that /(c) = 0.
/(c) = Mc) - /,(c) = 0 => /,(c) = /2(c)

Section 1.5 Infinite Limits

2. lim = oo

x-»-2* X + 2

lim

2-;c + 2

4. lim sec -— = oo

x->-2+ 4

lim sec -— = — oo
x-*-2- 4

6. fix

X

-3.5

-3.1

-3.01

-3.001

-2.999

-2.99

-2.9

-2.5

fix)

-1.077

-5.082

-50.08

-500.1

499.9

49.92

4.915

0.9091

\im fix) = —oo

lim fix) = oo

Section 1.5 Infinite Limits 321

8. /W = sec-

X

-3.5

-3.1

-3.01

-3.001

-2.999

-2.99

-2.9

-2.5

fix)

-3.864

-19.11

-191.0

-1910

1910

191.0

19.11

3.864

\\m f(x) = -oo
lim fix) = oo

jr-*— 3'*'

10. lim

.'-^r (x - 2)5 ^

lim 7 rrr = — oo

x-^2- ix - 2)3

Therefore, .r = 2 is a vertical asymptote.

14. No vertical asymptote since the denominator is never zero.

,^ ,■ 2+j: ,. 2+x

12. hm -r- = lim r = oo

x->o- ;t-(l - x) x-^o* x-i\ - x)

Therefore, x = 0 is a vertical asymptote.

,. 2+x

hm -:n- : = oo

/-^i-;c2(l - x)

V 2+.X
iim -57; r

— —00

Therefore, x = 1 is a vertical asymptote.
16. lim his) = —00 and lim his) = 00.

J— *-5 5— >-5*

Therefore, s = - 5 is a vertical asymptote,
lim his) = — 00 and lim his) = 00.

s—^5 s—*5 *

Therefore, .r = 5 is a vertical asymptote.

18. fix) = sec TTt

cos TTX

has vertical asymptotes at

2« + 1

-, n any integer.

22. fix]

4(.r- + .t - 6) 4U + 3)(.t - 2)

20. gix) =

(l/2).t3 - .t- - 4.t _ 1 a:(.v- - 2x - 8)
3x- - 6.t - 24 ~ 6 j:- - 2jc - 8

-JC,

;c?t -2,4

No vertical asymptotes. The graph has holes at x
and x = 4.

T,xi= -3.2

.r(x3 - Ix' - 9x + 18) xix - 2){x~ - 9) xix - 3)'
Vertical asymptotes at .r = 0 and .v = 3. The graph has holes at jt = - 3 and x = 2.

24. hix) =

(x + 2)(.t - 2)

f(f-2)

;c3 + 2x- + .r + 2 U + 2)(.t2 + 1)
has no vertical asymptote since

lim hix) = lim ^ = — -r.

x-»-2 x-,-2X- + 1 5

f =t 2

^ ' (f - 2)(r + 2)it- + 4) (r + 2)(f- + 4)'
Vertical asymptote at r = — 2. The graph has a hole at

322 Chapter 1 Limits and Their Properties

,„. tan 0 sin S , . ,

28. g(0) = — — = has vertical asymptotes at

V 6 cos 6

30. lim - — ^V^ = lim U - 7) =

jt->-l X + \ Ar-»-l

-8

(2m + 1)77 77

6 = = — + nv, n any integer.

There is no vertical asymptote at 0 = 0 since

tan e
hm— — = 1.

e->o Q

Removable discontinuity at x = — 1

,, ,. sm(:ic + 1)
32. lim — —^ = 1

jr->-l X + I

Removable discontinuity at ,3
x= -1

34. hm = -00

x->l* I — X

36. lim

x-*4- x~ + \6 2

38. lim

6jc2 + ;c - 1

lim

3jc- 1 5

x->-(i/2)* Ax^ - 4x - i j->-(i/2)+ 2x - 3

40. lim ^-T^ = \

42. lim x^ - - = 00
x^o- \ X

44. lim

x-tMl)-^ cos j:

46. lim i^i^ = lim [(x + 2)tan ;c] = 0

;t->0 cot X x-»0

48. lim x^ tan ttj: = 00 and lim x^ tan ttj:
Therefore, lim x^ tan ttx does not exist.

a:->(l/2)

50. /U) =

X^- 1

X^ + X + I

lim /U) = lim U - 1) = 0

x-*l x—*l

52. fix) = sec -

lim fix) = — 00

I

J.

Pi

in

54. The line x = c is a vertical asymptote if the graph of/
approaches ±co asx approaches c.

56. No. For example, fix) =
vertical asymptote.

•has no

58. P

lim — = fe(oo) = 00 (Tn this case we know that k > 0.)
V-.0* V

Section 1.5 Infinite Limits 323

,77 IOOtt „ ,
60. (a) r = SOtt sec- - = -^— ft/sec
6 3

(b) r = 50-n- sec^ ^ = 200-77 ft/sec

(c) lim [5077 sec- 6] = co

62. m =

Vl - (vVc^)

lim m = lim , ° ^^

64.

(a) Average speed =

Total distance
Total time

50 =

2d

id/x) + (d/y)

50 =

2xy

y + x

50}-

f 50^ =
50,v =
50.r =

2xy

2xy - SOy

2yU - 25)

X

25;t

- 25

y

Domain:

X > 25

66.

(a)

A=^bh

-¥'

= ^(10)(10tar

50 tan e - 50 0

Domain: 0

X

30

40

50

60

y

150

66.667

50

42.857

(b)

(c) lim — = oo

x^25* .X — 25

As X gets close to 25 mph, y becomes larger and larger.

(b)

0

0.3

0.6

0.9

1.2

1.5

fie)

0.47

4.21

18.0

68.6

630.1

(c)

(d) lim A = oo

e->-ir/2"

68. False; for instance, let
x^- 1

f(x)

X - \

70. True

The graph of/has a hole at (1, 2), not a vertical
asymptote.

72. Let/(.r) = "J and gix) = -j, and c = 0.

lim -^ = oo and lim —7 = 00, but

x-)0 AT x->0 X

1 1

jt->o v.-r j:^/ j-»o

i(^) =

-00 ^t 0.

?{v)
74. Given lim f{x) - 00. let ? (.v) = 1 . then lim ^^-r- = 0

X -tr .1 ->c J{X)

by Theorem 1.15.

324 Chapter 1 Limits and Their Properties

Review Exercises for Chapter 1

2. Precalculus. L = J{9 - If + (3 - 1)^ == 8.25

4.

X

-0.1

-0.01

-0.001

0.001

0.01

0.1

fix)

1.432

1.416

1.414

1.414

1.413

1.397

Urn fix) = 1.414

6. six

2x
x-2

(a) lim gix) does not exist.

jr-»2

(b) lim^W =0

x—*0

8. lim 7^ = 79 = 3.

jr->9

Let 6 > 0 be given. We need

I V5 — 3| < 6 => \~/x + 3|| VJ "■ 3| < e| V5 + 3 1

\x - 9\ < e\Vx + 3\

Assuming 4 < x < 16, you can choose 6 = 5e.
Hence, for 0 < |j^ - 9| < S = 5e, you have

|;c - 9| < 56 < \Vx + 3|e
I V5 - 3| < 6
l/W -L\<e

10. Um 9 = 9. Let e > 0 be given. 5 can be any positive
number. Hence, for 0 < |j: — 5 1 < 5, you have
|9 - 9| < e
\f(x) - L| < e

12. Iim3|>' - ll = 3|4- ll = 9

14. lim

l->3 f — 3 r-»3

= lim (r + 3) = 6

,^ ,. V4 + ;c - 2 ,. ^4 + X - 2 ^4 + ;f + 2

16. lim = lim • , — r

j:-»o j: j^o X ^4 + X + 2

1

lim - , = -

^^0 V4 + X + 2 4

18. lim-

j->0

(i/Vi + s)- ] ^ ,.^ [(i/Vi +5) - 1 _ (i/Vi + j) + 1

^ (l/Vl + s)+ I.

Um

^^^^ [1/(1 + .)] - 1 ^ ^.^^^ -1

^-^0 5[(i/v'T+7)+ 1] ^™(i +4(i/yrT7)+ 1]

20. lim

x^-2 a:-' +

= lim

(;t + 2)ix - 2)
2 (x + 2)(a:'- - 2;c + 4)

x-2

4x 4(7r/4)

22. um = — ; — = TT

j>-^(jr/4) tanx 1

= lim , --, — y

x-.-2X^ - 2x + 4

4_

12

Review Exercises for Chapter ] 325

24. lim

A:t->0

cos(7r + Ax) + 1
Ax

lim

cos TT COS Ajf — sin TT sin Ajc + 1

Ax

(cos Ax -
Ax

= -0 - (0)(1) = 0

= lim

Ax-fO

D] ,. r . sinAx]
— — lim sm 17 — : —

J Ax-.0|_ Ax J

26. lim [/(x) + 2g{x)] = -I + 2(f) = ^_

28. f(x)

X- 1

(a)

X

1.1

1.01

1.001

1.0001

fix)

-0.3228

-0.3322

-0.3332

-0.3333

lim

1* X - 1

= -0.333 (Actual limit is -|.)

, 1-3/J ,. 1-3/^ l+V^+ilGf

lim — = lim — • -p — } Jy

x-,r X - 1 x-,V X - 1 1 + 3/^ + (i/x}^

-™* (x - l)[l + 4/1 + (^']

= lim

1

'* 1 + ^/5 + (^'

(b)

30. j(f) = 0 => -4.9r- + 200 = 0 => f= » 40.816 ^ t = 6.39 sec
When t = 6.39, the velocity is approximately

lim^^^^-=^ = lim-4.9(^ + r)

t-*a a — t t—*a

= lim -4.9(6.39 + 6.39) = -62.6m/sec.

32. lim |.ic - IJ does not exist. The graph jumps from 2 to 3

X— *4

at X = 4.

34. lim g{x) =1 + 1=2.

36. lim f{s) = 2

38. fix)

"ix- — x — 1

lo,

X - 1

X ^ 1
x= 1

3r^ - t - 2
lim/(.x) = lim — '—, — -

.t-»l .t-»l X — 1

= lim (3x + 2) = 5 ^ 0

Removable discontinuity at x = 1
Continuous on (-co, 1) u (1, cx>)

326 Chapter 1 Limits and Their Properties

40. fix) =

5 - X, X <2
2jc - 3, X > 2

lim (5 — jc) = 3

lim (2a: - 3) = 1

Nonremovable discontinuity at x = 2
Continuous on (- oo, 2) U (2, oo)

n ^/l

lim

1 + - = oo

Domain: (-oo, - 1], (0, oo)

Nonremovable discontinuity at x = 0
Continuous on (— oo, — 1] U (0, oo)

44. f{x) =

lim

x+ 1
2x + 2

X + 1 1

:r-.-l 2{X +1) 2

Removable discontinuity atx = —I
Continuous on (— oo, — 1) U (- 1, oo)

46. fix) = tan 2x

Nonremovable discontinuities when
(2n + l)7r

Continuous on

/(2w - l)7r (2n + Dtt'
V 4 ' 4

for all integers n.

48. lim (x + 1) = 2

lim (x + 1) = 4

jr-»3"

Find ii and c so that lim (x^ + to + c) = 2 and lim (x^ + to + c) = 4.

Consequently we get \ + b + c = 1 and 9 + 3i + c = 4.
Solving simultaneously, Z? = - 3 and c = 4.

50. C = 9.80 + 2.50[-[-xI - 1], x > 0
= 9.80 - 2.50DI-xl + 1]
C has a nonremovable discontinuity at each integer.

52. fix) = V(x - l)x

(a) Domain: (-00, 0]u [1, oo)

(b) lim_/(x) = 0

(c) lim^/(x) = 0

54. hix) =

4x

4-x2
Vertical asymptotes at x = 2 and x = — 2

56. fix) = CSC TTX

Vertical asymptote at every integer k

58. lim

(1/2)* 2x - 1

60. lim -J = lim

-= -i
-"1- x" - 1 " x^-\- i^ + l)(x - 1) ~ 4

^, ,. x^ - 2x + 1

62. hm ; = 00

j:-»-1* X + 1

64. lim

^l/^F^^A

,, ,. secx

66. lim — ■ — = 00

JT-'O* X

68. lim

j:-»0" X

Problem Solving for Chapter 1 327

70. f(x) =

tanlr

X

-0.1

-0.01

-0.001

0.001

0.01

0.1

fix)

2.0271

2.0003

2.0000

2.0000

2.0003

2.0271

(a)

, . tan 2;c
hm = 2

Ji-»0 X

(b) Yes, define

/(.

rtai

tan 2x , X i' 0

x = 0
Now/(x) is continuous atjc = 0.

Problem Solving for Chapter 1

2. (a) Area APAO = ^bh = |(1)W = |

Area APBO = ^bh = |(1)().) = f = f

^, , , Area APBO a:72

(b) a(x) = , „. ^ = — 7— = X

' Area APAO .x:/2

X

4

2

1

0.1

0.01

Area APAO

2

1

1/2

1/20

1/200

Area APBO

8

T

"

1/2

1/200

1/20,000

a{x)

4

2

1

1/10

1/100

(c) lim a(x) = lim a: = 0

l->0* jr->0*

4. (a) Slope =

4-0 ^ 4
3 -0 ~ 3

3 3

(b) Slope = - - Tangent line: y — 4 = — ^x — 3)

3 25

-4^ + T

(c) Let e = (x, y) = (.v, V25 - x^)

725 - .t^ - 4
X- 3

,. V25 -X--4 V25 - .r- + 4

(d) lim m^ = lim r • — ,

.t_>3 -^ _j^3 .V - 3 V25 - .r= + 4

25 - .v= - 16

= lim

3 (x - 3)(V25 - x^ + 4)

,. (3 - x)(3 + x)

-'^3 (.V - 3)(V25 - .r= + 4)

,,„, -(3+.T) -6 _ 3

— Mm — , — 7 — — —

^^3 V25 - x^ + 4 4 + 4 4

This is the slope of the tangent line at P.

J a + bx - J3 Va + to - V3 Va + to + V3

■v Va + to + V3

(a + to) - 3

A:(Va + to + v^
Letting a = 3 simplifies the numerator.
Thus,

to

,. vT+to- V3 ,.

lim = lim / . = T^r

•"^^o .x: -<^o.v(V3 + to- + 73)

■*o V3 + to- + v/3'

Setting ■

73 + V3
Thus, 0 = 3 and ii = 6

= >/3, you obtain b = 6.

8. lim fix) = lim (a= - 2) = a= - 2

x-tO- .r->0-

o-t / tan .V

lim fix) = lim = a because lim = 1

j:->o' .t-i-o*tanj: V .>->o x

Thus,

a- — 2 = a

a~- a-2 = 0

{a - 2)ia +0 = 0

a= -1.2

328 Chapter I Limits and Their Properties

10.

-• -2--

o« -

0>

o«

(a) /(i) = W = 4

/{3) = i = 0

/(I) = lU = 1

(b) lim fix) = 1

lim /W = 0
lim/(jc) = -oo

x—*0~

lim fix) = oo

(c) /is continuous for all
real numbers except

x = 0,±l, ±i±i. .

12. (a) V- =

192,000

2 192.000 , 2 .-

= v2 - Vq^ + 48

192,000
'" V - Vq- + 48

192,000

lim r = — r

v->o 48 - Vf,^

(b)

Let Vq = V48 = 4 V3 feet/ sec.
, 1920

1920

+ Vq^ - 2.17

Vo^ + 2.17

14. Let a T^ 0 and let e > 0 be given. There exists 8, > 0
such that if 0 < l;c - 0| < S, then \fix) - i| < e.
Let 8 = 5J\a\. Then for 0 < |x - 0| < S = 8J\a\,
you have

lax < 8,

l/M - L| < 6.
As a counterexample, let /U) =

Then Wmfix) = 1 = L,

but \m\fiax) = liin/(0) = 2.

X7t 0
x = 0'

1920

v2 - vo^ + 2.17

1920

hm r = -—-z ^

v^o 2.17 - v^

(c)

Let Vq = V2.17 mi/sec (= 1.47 mi/sec).
10,600

Vo2 + 6.99

lim r =

10,600
6.99 - v„

Let Vq = V6.99 = 2.64 mi/sec.

Since this is smaller than the escape velocity for earth,
the mass is less.

CHAPTER 2
Differentiation

Section 2.1 The Derivative and the Tangent Line Problem . . 330

Section 2.2 Basic Differentiation Rules and Rates of Change 338

Section 2.3 The Product and Quotient Rules and

Higher-Order Derivatives 344

Section 2.4 The Chain Rule 350

Section 2.5 Implicit Differentiation 356

Section 2.6 Related Rates 361

Review Exercises 367

Problem Solving 373

CHAPTER 2
Differentiation

Section 2.1 The Derivative and the Tangent Line Problem

Solutions to Even-Numbered Exercises

2. (a) m

1

4
(b) m = 1

/(4)-/(3) 5 - 4.75

4-3

1

0.25

Th..e /(4)-/(l), /(4)-/(3)
^"«- 4-1 > 4-3

(b) The slope of the tangent line at (1, 2) equals /'(I).
This slope is steeper than the slope of the line
through (1, 2) and (4, 5). Thus,

«

-»

^<;' -('«</•(.).

6. g{x) = -x+ I IS SL line. Slope = -

8.

Slop. . ,0.1). lim * + '^'-'<»

Aa^o Aj:

,. 5 - (2 + Axy - 1
= hm ,

Ajr-»0 Ax

,. 5 - 4 - 4(Ax) - (Ax)' - 1
= lim 1

Ai->0 Ax

= lim (-4 - Ajc) = -4

Ax^O

10. Slope at ( 2,7)-lim'^^-2 + ^;)-'^(-

-2)

12.

8(x} = -5

,. (-2 + At)2 + 3 -
= lim 1

Ar—O Af

_7

^'« = il5o^^^^^^

B ,. 4 - 4(Ar) + (Ar)2
= lim T

Ar->0 .Af

- 4

= lim-^-/-^)

Ar->0 Ax

= lim (-4 + Ar) = -4

= lim -^ = 0

Ai:->0 Ax

14. fix) = 3a: + 2

16

. fix) = 9 - |;c

/'W=lim^(^^t^)-^«

Ar-»0 Ax

/'(x)=lim^(^ + ^)-/«

■^ Ax->0 Ax

- lim '■^^'^ + ^) + 2] - [3x + 2]

Ar^O AjC

,.^^^ [9 - (1/2)U + Ax)] - [9 - (l/2);c]

At-»0 Ax

,. 3Ax
= lim -; —

Ai->0 Ax

= lim -- = --

Ac^o V 2/ 2

= lim 3 = 3

AX-.0

330

Section 2. 1 The Derivative and the Tangent Line Problem 331

18. fix) =l-x^

nx + ^x)-f{x)

^(^^ = i,'?o K^

,. [1 - {x + Axn - [1 - x"]

= lim :

Aj-»o Ax

,. I - x^ - IxAx - [Axf - \ + }?

= lim ;

Aa:->o Hoc

-2xAx - iiyx)' ,. , „ .X
= hm : = lim ( — 2x — Ajc) = —2x

20. f{x) =x^ +x^

■' Ai-»0 Ax

,. [U + A.t)3 + {x + Ax)2] - [x^ + .t2]

= lim :

Ai-»0 Ax

= lim

■T^ + 3.T^A.x + 3x(Ax)- + (A-x)^ + .^ + ltA.r + (A.x)- - x-' - x^

Ax

,. S.x^A-x + 3x(Ax)2 + (A.x)' + IxAx + (A.x)-
= lim -.

A^-»0 Ax

= lim (3x^ + 3xAx + (Ax)^ + Ix + (Ax)) = 3x^ + Zx

22. fix) = ^

fix) = lim

= lim

A;[->0

/(x + Ax)-/(x)

A.X

1

1

(x + Axf x"
Ax

,. X- - (x + Ax)^

~ lim -: — ; . ., -,

^x^o Ax{x + Ax) XT

— lim

-2xA.x - (Ax)^
A^^o A.x(x + Ax)V

,. -2x - Ax

lim 7 — , K \-> ■>
tix-^o (x + Ax)-x"

-Ix

X*

_2_
x3

24. /(x) = -^

vx

/'(x) = hm

Ax— »0

/(x + A.x) - /(x)
Ax-

= lim

Ax-»0

Jx + A.X J~x
Ax

= lim

4^ - 4Vx + A.X ( Jx + Jx + Ax

^-^0 AxVxVx + A.V Vv'x + Jx + Ax
Ax - 4(x + A-x)

= lim

A^^o Axv^Vx + A.x(v''x + Vx + A.x)

-^^o VxVx + A.x(^/x + Vx + Axr)

-4 _ -2

v/xVx(Vx + v'x) xvx

332 Chapter 2 Differentiation

26. (a) /(*) = ^2 + 2;c + 1

f{x + ^x)-fix)
f{x) = lim

,. [{x + Ax)^ + 2{x + Ax) + l]-W + 2x+ I]
— jjjj,

Ar^O Ax

= lim

Ax-»0

2xAx + (Axf + 2Ax
Ax

= lim (2;c + A;c + 2) = 2;c + 2

Aji->0

At (-3, 4), the slope of the tangent line is w = 2(- 3) + 2 = —4.

The equation of the tangent line is

y - 4= -4(x + 3)
y = -Ax - 8.

28. (a) fix) = x3 + 1

f'(x) = lim

fix + Ax) -fix)
Ax

= lim

Ai-^O

lim

[U + AxY + 1] - (;c3 + 1)
Ax

jc^ + 3j:^(A;c) + 3.t(Aj:)^ + (Ax)^ + 1 - x^ - 1
Ax

= lim [3x2 + 3^(^) + (^)2] = 3^2

Ar-»0

At (1, 2), the slope of the tangent line is m = 3(1)^ = 3.
The equation of the tangent line is

:y - 2 = 3(x - 1)
y = 3x - 1.

(b)

(-3,4rt

/

\

(b)

30. (a) fix) = Vx - 1

■' Ax->0 Ax

,. Vx + Ax - 1 - Vx - 1 / Vx + Ax - 1 + Vx - 1

= iim : • — . = ,

t^^o Ax \Jx + Ax - 1 + Vx - 1

(x + Ax - 1) - (x - 1)
'^^o Ax(Vx + Ax -T + Vx - l)

1 1

= lim — , , - ,

Aat^ Vx + Ax - 1 + Vx - 1 2Vx - 1

At (5, 2), the slope of the tangent line is

1

1

til — , — .

2V5 - 1 4
The equation of the tangent line is

(b)

y - 2 = -(x - 5)

Section 2. 1 The Derivative and the Tangent Line Problem 333

32. (a) /W = ^^

(b)

f'{x) = lim

/(;c + M-/U)
Ax

lim

Ax-^O

lim

X + Ax + 1 -t + 1
Ax

(x + 1) - (x + Ax + 1)

Ai^o Ax(x + zlx + l)(x + 1)
Ax-^O (x + Ax + l)(x + 1)

^ 1

At (0, 1), the slope of the tangent line is

The equation of the tangent line is y = — x + 1.

34. Using the limit definition of derivative, /'(x) = 3x'. Since
the slope of the given line is 3, we have

3x- = 3

x2 = l=>x = ±1.

Therefore, at the points (1, 3) and (- 1, 1) the tangent
lines are parallel to 3x - y - 4 = 0. These lines have
equations

.V - 3 = 3(x - 1) and y - 1 = 3(x + 1)

y — 3x y = 3x + 4

(0. rp

"^

\

36. Using the limit defmition of derivative, fix) =

-1

2(x - D''^-

Since the slope of the given line is --, we have

1

1

2(x - l)'-^ 2

1 = (x - D^^^'

1 = X - 1 => X = 2

At the point (2, 1), the tangent line is parallel to
X + 2y + 7 = 0. The equation of the tangent line is

y-l = --ix-2)

y = -^x + 2

38. h(- 1) = 4 because the tangent line passes through (- 1, 4) 40. /(x) = x- => /'(x) = 2x (d)
6-4 2 1

hX-\) =

3-(-l) 4 2

42. /' does not exist at x = 0. Matches (c)

44.

Answers will var>'.
Sample answer: y = x

334 Chapter 2 Differentiation

4^ . ^ V r fix + 2Ax)-f(x) .. f(x + ^x)-f{x)

46. (a) Yes. lim r-- = lim -. = / (x)

Ax^O 2Dx Ax->0 Ax

(b) No. The numerator does not approach zero.

f(x + Ax)-f{x-Ax) fix + Ax) - fix) - fix - M + fix)

(c) Yes. lim — = lim —

aj:-»o 2Aa; Ai^o 2Aj:

= lim

Al->0

fix + Ax) -fix) fix -Ax) -fix)
lAoi 2i-Ax)

= \rix)+\f'ix)=f'ix)
(d)Yes. lim^^^-^y-^^^^^/'M

AX-.0 AjC

48. Let (jcq, >(,) be a point of tangency on the graph of/. By the limit definition for the derivative, fix) = 2x. The slope of the line
through (1, - 3) and (xg, yg) equals the derivative of /at x^:

%

2Xn

-3 -Jo = (1 - JCo)2xo

3 ;cq"- 2Xg
- 2xo - 3 = 0

^r 2
■^^0

(Xg - 3)(A;g + 1) = 0 => Xg = 3, - 1

Therefore, the points of tangency are (3, 9) and (- 1, 1), and the corresponding slopes are 6 and - 2. The equations of
the tangent lines are

y + 3 = 6U - 1) y + 3 = -2(x - 1)

y = 6x — 9 y = —2x — I

50. (a) fix) = x2

fix)

lim

Ax-fO

= lim

= lim

Ajr->0

= lim

Aj:-»0

fix + Ax) -fix) •

Ax

ix + Ax)^ - x^
Ax

x^ + 2xiAx) + iAx)^ - x^
Ax

Ar(2jc + Ax)
Ax

= lim (2x + Ax) = 2x

Aj->0

At X = - !,/'(- 1) = — 2 and the tangent line is
y - 1 = -2(x +1) or y = -2.x - 1.
At X = 0,/'(0) = 0 and the tangent line is y = 0.
Atx = 1,/'{1) = 2 and the tangent line is y = 2x

\

/

\

\

For this function, the slopes of the tangent lines are
always distinct for different values of x.

(b) g'ix) = lim

= lim

Ax^O

= lim

Ai->0

= lim

Ax->0

g(x + Ax) - g(x)
Ax

(x + Ax)^ - x'
Ax

x^ + 3x^(Ax) + 3x(Ax)^ + (Ax)^ - x^
Ax

Ax(3x^ + 3x(Ax) + (Ax)^)
Ax

= lim (3x2 + 2xiAx) + (Ax)^) = 3x^

Ax->0

At X = - 1, g'i— 1) = 3 and the tangent line is

y + 1 = 3(x + 1) or y = 3x + 2.
At X = 0, g'iO) = 0 and the tangent line is y = 0.

At X = 1, g'il) = 3 and the tangent line is

y - 1 = 3(x - 1) or y = 3x - 2.

For this function, the slopes of the tangent lines are
sometimes the same.

Section 2. J The Derivative and the Tangent Line Problem 335

52. fix) = kx^

By the limit definition of the derivative we havsf'ix) = x.

X

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

m

2

1.125

0.5

0.125

0

0.125

0.5

1.125

2

fix)

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

54. g(x)

fix + Om)-f(x)
0.01
= {3Vx + 0.01 - 3v/3)lOO

56. /(2) = ^23) = 2,/(2.1) = 2.31525

/'(2) » "'f^ ~ ^ = 3.1525 [Exact:/'(2) = 3]

The graph of g(x) is approximately the graph of/'(x)-

58. fix) = --l>x and/'U) = -x~ - l

60. fix) = ;c

/(2 + Ax)-/(2) (2 + A^)+^^^
5a^ W ^^ U - 2) + /(2) = -^ ix - 2) + -

2(2 + M- + 2-5(2 + M. 5 (2A.X + 3) _ ,. , 5

2(2 + Ax) A.X ^ ''2 2(2 + Av) '2

(a)Ajt= 1: 5^, = -(jf- 2) +- = -J: +

5 5 5
2 = 6" -"6

Ar = 0.5: 5^ = 7(x - 2) +

5 4

:.v + —

16, ~~ 5 ^ 2^ 41^

2 ~ 21-^ "^42

(b) As Aat— >0, the line approaches the tangent line to/at (2. ?).

/Vx = 0.1:5^ = -(.v-2) + - = -. + 42

r X

Y

)^/r\

62. ^(.r) = .rCt - 1) = x^ - .t, c = 1

»,x ,. six) - g(l) X- - .t - 0 ,. .v(x - 1)

g (1) = lim ; — = lim ; — = lim —

J:->1 X - 1 .t-.l X - 1 v-»: X - 1

= lim X = 1

336 Chapter 2 Differentiation

64. f(x) = j^ + Zx,c= I

r,„. ,. /(-t)-/(l) ,. x^ + 2x-Z ,. (x - 1)U^ + ^ + 3) ,. .2^ ■ ,. -
/ (1) = lim ~ = lim ; = lim ; = lim (x^ + x + ?>) = 5

x-,\ X - \ x->\ X - \ J->1 X - \ jr-»l

66. f(x) = -, c = 3

X

x->3 X — 3 jr->3 jc - 3 j:->3 3a: X — 1> -r-»3 V 3;c/ 9

68. gW = (x + 3)'''3,c= -3

g'(-3) = lim ^W^/ll^= lim^-^^n'°= l™ ^^
Does not exist.

70. f(x) = |j; - 4|, c = 4

^,(4) = ,im^W^= lim ^^1^ = lim ^^^

^4 X - A x^i X — 4 x^4 X - 4

Does not exist.
72. /(.t) is differentiable everywhere except at x = ±3. (Sharp turns in the graph.)
74. f{x) is differentiable everywhere except atx = 1. (Discontinuity)
76. fix) is differentiable everywhere except at x = 0. (Sharp turn in the graph)
78. fix) is differentiable everywhere except at jc = ±2. (Discontinuities)
80. fix) is differentiable everywhere except at x = 1. (Discontinuity)

82. fix) = yp^

The derivative from the left does not exist because

,. /W-/(l) ,. Vl - x2 - 0 ,. Vl -x2 Vl -;c2 ,. 1 + r .,, ■ ,

lim ; — = lim ; = lim ; — • . = lim , = — oo. (Vertical tangent)

x-^i' x — I x->i- X - \ x->i- j: — 1 VI - Jc^ -t^'^ J\ — x?-

The limit from the right does not exist since/is undefined for jt > 1. Therefore,/is not differentiable at x = 1.

X, X < \
x^, X > I

84. /U) ,,

The derivative from the left is

,. /U)-/(l) ,. x-1 ,. , ,

lim ^ = lim = lim 1 = 1.

x->i' ;c — 1 x^\~ X — I x^i-

The derivative from the right is

fix) -/(I) ,. jc^ - 1 ,. , ,, ^
hm = -^-^-^^ — f!^ = lim = lim ix + \) = 2.

jr-»r X — 1 Jr-»1* X — I Ar-»r

These one-sided limits are not equal. Therefore, / is not differentiable at j: = 1.

Section!.] The Derivative and the Tangent Line Problem 337

86. Note that/ is continuous at x = 2.f(x) =
The derivative from the left is
lim

3* + 1, .r < 2
V^, X > 2

r-»2- X - 2 x->2- X — 2 x-y2- X — 2 2

The derivative from the right is

,. fix) - m ,. v/5-2 72^ + 2

lim — = lim — • — 7=

x-^2* x-2 x->2^ X -2 J2x + 2

1- 2.V-4 2ix - 2) 2 1

- iiJ?* (.-2)(72^ + 2) = .'i'?M.-2)(72^ + 2)= ii-^^T^T^ = i

The one-sided limits are equal. Therefore, / is differentiable at j: = 2. (/'(2) = 5)

88. (a) fix) = x^ and/'U) = 2x

(b) gix) = ;c' and g'U) = 3.r^

(c) The derivative is a polynomial of degree 1 less than the original function. If hix) = x", then /('(.t) = nx" '.

(d) If/W = A then/'U) = lim ^^ + ^ - /^

,. U + A.r)-* - ;c^

= Imi :

Aj->o Ax

lim

Ax->0

= lim

Ax-*0

.r* + 4.r3(Aj:) + 6xHAx)- + 4.T(A.r)^ + (A.r)^ - x*
Xx

Ax(4.t3 + ex'jAx) + 4x(A.r)^ + (Ax)^)
A.V

= lim (4x3 + 6.i2(A.x) + 4.r(Ax)= + (Ax)^) = 4x^

Ajt-»0

Hence, if/(x) = x^, then/'(x) = 4x3 which is consistent with the conjecture. However, this is not a proof, since you must
verify the conjecture for all integer values of n,n> 2.

90. False, y = |x - 2| is continuous at x = 2, but is not differentiable at x = 2. (Sharp turn in the graph)

92. True — see Theorem 2. 1

94.

As you zoom in, the graph of v, = x- + 1 appears to be locally the graph of a horizontal line, whereas the graph of
J": ~ kl + 1 always has a sharp comer at (0, 1). v, is not differentiable at (0, 1).

338 Chapter 2 Differentiation

Section 2.2 Basic Differentiation Rules and Rates of Change

2. (a) y = jc-'/2

(b) y = X-'

y'--hr

3/2

y'= -AT

y'W = -5

y'W=-i

4. fix) = -2

6. >' = ;c«

fix) = 0

>-'= 8;c''

(c) y = x-'/2

(d) y = x-2

y'=-!^-

5/2

y'= -2x-3

^-'d) = -1

y'W = -2

8. , = ! = .-

10. :y = 4^ = x^'

y' = Sx"

V' = -X-3/'* = ^—

-*^ 4 4xV4

12. gW = 3;c - 1

14. y =

r^ + 2? - 3

16. y

= 8-x3

18. fix) = 2x3 -

g'(x) = 3

y' =

2f + 2

y'

= -3x2

fix) = 6x2 _

20. g(r) = IT cos f

22. .v =

5 + sinx

24. V

— /„ M + 2 cos X — „x 3 + 2 COS X
(2x)3 8

g'{t) = -TTsinf

>'' =

COSJC

y'

= |(-3)x-

"^ - 2 sin X = — -J — 2 sin X

Function

Rewrite

Derivative

Simplifv

''■ y = 3^ .

2 -2
>- = JX 2

r--\.-'

4
^ - 3x3

^'■y^iZy

y--'-r-'

27r

y - 9x3

30. . = ^3

y = 4;c3

y'=\2£-

y' = 12«2

32. /W = 3-|.(f,2)

34. y =

= 3x3-6, (2, 18)

36. /(x) = 3(5 - x)\ (5, 0)

m-h

• y'-

-- 9x-

= 3x2 _ 30;t + 75

3''(2) =

= 36

fix) = 6x - 30

. /'(M

/'(5) = 0

38. g{i) = 2 + 3 cos f, (rr.

-1)

40. /U) =

x^-3x- 3x-2

42. fix) = X + x-2

g'W = -3sinf

fix) =

2x - 3 + 6x-3

/'(x) = 1 - 2x-3

gV) = 0

=

2X-3+4

X?

--?

, , , 2x2 _ 3^ + J
44. hix) = = 2x - 3 + X-'

, . , - 1 2x2- 1

ft (x) = 2 r =

X^ x2

46. y = 3x(6x - 5x2) = ig^2 _ 15^^
y' = 36x - 45x2

48. fix) =Vx+Vx = xi/3 + xi/5

50. fit) = r2/3 - ri/3 + 4

/'W4x-2/3+ix-/3._l_^_^

2 1 2 1

/W 3^ 3^ 3,1/3 3^/3

Section 2.2 Basic Differentiation Rules and Rates of Change 339

2
52. fix) = ^7= + 3 cos X = 2jt-'/5 + 3 cos Jt

— 2 -2

/'W = — Jc""*/^ - 3 sm;c = ^^ - 3 sinjc

54. (a) y = x> + X
y' = 3jc2 + 1

At (-1,-2): y'= 3{-\Y +1=4.
Tangent line: y + 2 = 4U + 1)

4jc - y + 2 = 0

(b)

/

'J

/i

56. (a) y = (x2 + 2x)(;c + 1)
= x? + 2,x- + 2x
y' = Zx- + 6x + 2

At (1,6): y' = 3{\)- + 6(1) + 2 = 11.
Tangent line: y - 6 = ll(.x - 1)

0= \\x- y - 5

(b)

T^^

58. y = ;<^ + .X

;y'= 3;c2 + 1 > 0 for all x.

Therefore, there are no horizontal tangents.

60. v = ;c- + 1

>>' = Iv = 0 => .I = 0
At;c = O.y = 1.
Horizontal tangent: (0. 1)

62. y = JTix + 2 cos at, Q < x < Itt
y' = ^ - ls\r\x = Q

J3

TT 2tT

x = yor-

TT 73Tr+ 3

At :ic = — , >> =

277 2v^Tr - 3
At j: = — , y = ^ .

Horizontal tangents:

TT JItt + 3\ /27T 2x/3Tr - 3

3"

3 '

64. k ~ X- = — 4jc + 7 Equate functions
— 2r= —4 Equate derivatives

Hence, a: = 2 anclA:-4=-8 + 7=>it = 3

66. kjl(. = j: + 4 Equate functions

Equate derivatives

2j~x

Hence, k = 2>/x and

(27x)v^ = x + 4^>lt = .r + 4=>.v = 4=i.fc = 4

68. The graph of a function/
such that/' > 0 for all x and
the rate of change the function
is decreasing (i.e./" < 0)
would, in general, look like the
graph at the right.

/

(P?^

yi

•7

-X^

340 Chapter 2 Differentiation

70. gix) = -5fix) => g'ix) = -Sf'ix)

72.

If/ is quadratic, then its derivative is a linear function.
fix) = ay?- + bx + c
fix) = 2ax + b

74. m, is the slope of the line tangent to >> = x. mj is the slope of the line tangent toy = l/x. Since

y = x

y' = 1 => m^ = \ andy =

^ =1^

The points of intersection ofy = x and y = l/x are

1
jc = —

X

±\.

Atjc = ±l,m2=— 1. Since wij = — \/m^, these tangent lines are perpendicular at the points intersection.

76. fix) = -, (5, 0)

fix)

_1_
x^

x^ 5 — X

- 10 + 2x = -x^y
.V2

■ 10 + 2x = -X-

•10 + 2x = -2x
4x = 10

x = -^,y

The point (j, j) is on the graph of/. The slope of the
tangent line is/'^j) = -js-

Tangent line: y ~ 1

'25 ^ 2

25y - 20 = -8x + 20
8x + 25>' - 40 = 0

78. /'(4) = 1

dy/dx=l

^

Section 2.2 Basic Differentiation Rules and Rates of Change 341

80. (a) Nearby point: (1.0073138, 1.0221024)

1.0221024 - 1

Secant line: y - I

1.0073138 - 1
y = 3.022(x - 1) + 1
(Answers will vary.)

<;c-l)

/(l.l)

f

(b) fix) = 3x^

Tix) = 3{x- I) + \ =2x -2

(c) The accuracy worsens at you move away from (1, 1).

2

/■"

'f

/

(d)

Ax

-3

-2

-1

-0.5

-0.1

0

0.1

0.5

1

2

3

fix)

-8

-1

0

0.125

0.729

1

1.331

3.375

8

27

64

T{x)

-8

-5

-2

-0.5

0.7

1

1.3

2.5

4

7

10

The accuracy decreases more rapidly than in Exercise 59 because y = x^ is less "linear" than y = x''^.

82. True. If/W = g{x) + c, then/'W = gXx) + 0 = g'{x).

86. False. If f{x) = — = x'", then f'{x) = -njc""-' = -^

88. f(t) = f - 3, [2, 2.1]
fit) = 2t
Instantaneous rate of change:

(2, 1) => /'(2) = 2(2) = 4
(2.1, 1.41) => /'(2.1) = 4.2
Average rate of change:

/(2.1)-/(2) 1.41-1

84. True. If y = x/n = (I/tt) • .x, then rfy/otc = (l/7r)(l) = I/tt.

2.1 - 2

0.1

4.1

92.

s(t) = - 16/2 - 22r + 220
v(f) = -32r- 22
v(3) = -118 ft/sec
s(t) = - 16f2 - 22f + 220

= 112 (height after fallmg 108 ft)
- 16/2 - 22/ + 108 = 0
-2(/ - 2)(8/ + 27) = 0
/ = 2
v(2) = -32(2) - 22
= - 86 ft/ sec

■[»■!]

90. fix) = sin j:,

fix) = cos;c

Instantaneous rate of change:
(0, 0) ^ /'(O) = 1

73

77 2.

6' 2

/'

0.866

Average rate of change;

/(7r/6)-/(0)_ (1/2) - 0 ^ 3 ^

(77/6) - 0 (7r/6) - 0 TT

94. s(t) = -4.9/2 + v'(,/ + 5o

= -4.9/2 + 5o = 0 when / = 6.8.
5o = 4.9/2 = 4.9(6.8)2 = 226.6 m

342 Chapter 2 Differentiation

96.

2 4 6 8 10

Time (in minutes)

(The velocity has been converted to miles per hour)

98. This graph corresponds with Exercise 75.

(0,0)

2 4 6 8 10

Time (in minutes)

100. s{t) = --at^ + c and s'{t) = -at.

Average velocity:

5(fo + M) - 5(^0 - ^t) _ [-(l/2)a(fo + M)^ + c\- [-(l/2)a(fo - Af)^ + c)]
% + Af) - (ro - Af) 2Af

-{\/2)a%^ + 2foAf + (Af)^) + (l/2)a(fo^ - IfpAf + (Af)^)

dV
102. V = j^ — = 3j2
as

dV
When 5 = 4 cm, —- = A% cm^.
ds

2Ar

- 2afoAr
2Af

-ato

s '(tg) Instantaneous velocity at f = r,,

104. C = (gallons of fuel used)(cost per gallon)

15,000\, ^ 18,750

dC^ 18,750
dx x^

X

10

15

20

25

30

35

40

C

1875

1250

537.5

750

625

535.71

468.75

dC
dx

-187.5

-83.333

-46.875

-30

-20.833

-15.306

-11.719

The driver who gets 15 miles per gallon would benefit more from a 1 mile per gallon increase in fuel efficiency.
The rate of change is larger when x = 15.

106. ^=K(T- TJ

Section 2.2 Basic Differentiation Rules and Rates of Change 343

108. >> = -, ;c > 0

X

At (a, b), the equation of the tangent line is

y = — t(x -a) OT y

a a

The j:-intercept is (2a, 0).

2

a^ a

The y-intercept is I 0

1 1 (1
The area of the triangle is A = -bh = -(2a) -

2 I \a

110. y = x^

y' = 2x

(a) Tangent lines through (0, a):

y - a = 2x(x - 0)
X- — a = 2x~

±V— a = X

The pwints of tangency are (± ^--a. - a). At ( V-a, - a) the slope is y '( -J— a) = 2 v— a. At i

>''(-y^^ = -2^-0.

Tangent lines: jy + a = 2V— a(A- — -J— a) and _y + a = — 2V— a(.v + v — a)
' V=-2, '

-a. - a) the slope is

-ax + a

y = 2^ —ax + a y

Restriction: a must be negative,
(b) Tangent lines through (a, 0):
y - 0 = 2x(a: - a)
7? = 2x^ — 2mx
0 = jr — 2ajc = x(x — 2a)
The points of tangency are (0, 0) and (2a, 4a-). At (0, 0) the slope is .v '(0) = 0. At (2a, ^<^) the slope is >■ '(2a) = 4<3.
Tangent lines: v - 0 = 0(.r - 0) and y - 4a- = 4a(.r - 2a)
>> = 0 V = 4a.r — 4fl-

Restriction: None, a can be any real number.

112. /i(j:) = Isinjcl is differentiable for all x i= mr, n an integer.
/2(jc) = sin|x| is differentiable for all .r i= 0.
You can verify this by graphing/, and/j and observing the locations of the sharp turns.

344 Chapter 2 Differentiation

Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives

2. f{x) = (6x + 5)(x5 - 2) 4. g{s) = JKa - 5^) = s''\A - P)

fix) = (6. + 5)(3.^) + (^ - 2)(6) ^,^^^ ^ ^^^^^_^^ ^ ^^ _ ^,^1^_,,, ^ _^^3,, ^ l-_^^

= 18x3 + 15a:2 + 6x3-12
= 24x3 + 15^2 _ 12

6. g(x) = Vx sin X

g '(x) = Vx cos X + sin x( — -7= J = Vx cos x +

10. h{s)

4- 552

/s- 1

VW

25'/2

-»<"=g^

1

^ sinx

2Vx

,, . (2f - 7)(2f) - (r2 + 2)(2)
. '^'^- {2t-ir-

12.f(t) = y

2f _ 14; _ 4
(2f - If

,. t\- sm t) - cos t{Zt^)

r sin f + 3 cos r

(V^ - 1)(1) - .(|.-'/2J .w (^3)2

V^ - 1 — — v«

2^" V;-2

[Js-\f 2(7^ - if

14. /(x) = (x2 - 2x + l)(x3 - 1) 16. fix) = J^

^-(,) = (,. _ 2, + i)(3x^) + (x3 - l)(2x - 2) ^_^ - 1)(1) - (. + 1)(1)

= 3x2(x - 1)2 + 2(x - l)2(x2 + X + 1) ^^'''

= (x- 1)2(5x2 + 2X + 2)
/'(I) = 0

(x-

1)^

X — 1 — X -

- 1

(X - 1)2

2

{X - 1)2

2

= -2

^'(2)= (2-1)

18. /(x)=^

„, , (x)(cosx) - (sinx)(l)

/w-

X cos X - sin X

r2

/7r\ (7r/6)(V3/2) - (1/2)
■^U/ TJ-2/36

37377- 18

772

3(7377-6)

772

Function Rewrite

DenVarive

Simplifv

5x2-3 5 ,
20. y = ^ y- ^x^-

3
4

, 10

, 5x

Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 345

Function

Rewrite

22- y = L^

, = |.-

24..= ^^^-^

3 2 5

26. f{x) - ^ _ J

(.x= - mx^ +

3) - (x3 + 3x + 2)(2x)

J w -

U^ - 1)^

.t* - 6x2 _ 4^

-3

{x- - D-

30. fix) = ^(v^ + 3) = x>/3(x'/2 + 3)

* _1/-|\ . / !/'> . n\l _„ — 2/3

/'W = x^'H-^x'^^^j + U>/2 + 3)1 jx

= 7X-'/6 + x-2/3
6

' +^

6x1/6 ^2/3

Alternate solution:

/(X)= 3/^(v/^ + 3)
= x5/6 + 3x1/3

/'(x) = fx-'/6 + X-2/3

5 +^

Derivative

= --r-3

Simplify

y =

5x3

6x

y =7^

28. /W = -^

fix) = x^
= 2x3

x+ ij U+ ij

(x + 1) - (x - 1)1 [x - 1

ix + \r

:" + X - 2]

H

X + 1

(4.r3)

2x^
(.

32. /i(x) = (x^ - 1)= = x^ - 2x2 + 1
/i'{x) = 4x3 - 4x = 4x(x2 - 1)

(^U6 ^2/3

34. g{x)=xi- ^

= 2x -

g'W

X X + 1 / X + 1

(x + l)2x - x^d) _ 2(x2 + 1-c + 1) - x2 - 2x _ x2 + It + 2

U + 1)2

U + 1)^

(X + 1)2

36. /(x) = (x2 - x)(x2 + l)(x2 + X + 1)

fix) = [Ix - 1)(X- + l)(x2 + X + 1) + (x2 - X)(2x)(x2 + X + l) + (x2 - x)(x2 + l)(2x + l)

= (2x - l)(x'* + .v3 + 2x2 + X + 1) + (x2 - x)(2r3 + 2x2 + 2x) + (x2 - x)(2t3 + x2 + 2v + 1)

= 2x5 + x'' + 3x3 + X - 1 + 2x-5 - 2x2 + 2x^ - x'* + .t3 - x2 - X

= 6x5 + 4x3 - 3_^2 _ 1

38. /(x) = £1 ^■

fix) =

(c2+.r2)(-2v)-(c2-x2)(2v)

ic~ + X2)2

40. fie) = (e+ i)cos0

fie) = ie+ i)(-sin e) + (cos e)(i)
= COS e - (e + 1) sin e

— 4xc2

(C^ + x2)2

346 Chapter 2 Differentiation

42. fix) =

f'(x)

X COS X - sin ;i:

44. y = X + cotx

y' = 1 — csc^jc = -cot^;c

46. /z(j') = 10 CSC j'

h '(s) = — J + 10 CSC 5 cot ;

48. y

sec a:

, X sec X tan j: — sec x

y = 72

sec.t(xtanx — 1)

50. y = x sin x + cos x

y' = x cos ;c + sin ;c — sin X = jt cos x

52. /(j:) = sin x cos x

/'W = sin;c(-sinj:) + cos;c(cosx)
= cos 2x

54. h{e) = Sdstcd + dime

h'{e) = SOsec etan 0 + 5 sec e + Ssec^ 9 + tan e

58. /(e)

/'(e)

sin 6

1 - cos I
1

cos d — \

cos 0 - 1 ~ (1 - cos 9)2

(form of answer may vary)

56. fix) = (^VfT^)^^' "^ ^ + ^^

;f5 _|_ 2;i^ + 2;c2 - 2
f'(x) = 2 1 2 j^ \\2 (form of answer may vary)

60. f{x) = tan ;c cot x = 1

fix) = 0
/'(I) = 0

62. fix) = sin x(sin x + cos x)

fix) = sin j:(cos jt — sin ;c) + (sin x + cos x)cos x
= sin j: cos x — sin^ ;c + sin x cos x + cos^ x
= sin 2x + cos 2x

lir\ . 77 . •"■ 1

64. (a) fix) = ix- \)ix-^ - 2), (0, 2)

fix) = ix- 1)(2a:) + ix^ - 2)(1) = Zx^ - 2x - 2

/'(O) = -2 = slope at (0,2).

Tangent line: y - 2 = -2x => y = -2x + 2

(b)

66.(a)/(.) = ^,(2,|

. ^ ix + 1)(1) - (x - 1)(1) ^ 2
■^ ^ ^ (;c + 1)2 ix + 1)2

/'(2) = I = slope at (2, | ,.

12 2 1

Tangent line: y - - = -ix - 2) =!> y = -x - -

(b)

y

___

/

Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 347

68. (a) fix) = sec;c, \^-,2
fix) = sec jt tan jf

/||j = 2y3 = slopeat^|,2J.
Tangent line:

673x -3y + 6- 1^-n = 0

70. fix)

x^+ 1

.,, , _ U^ + l)(2x) - (x^)(2x) Ix

fix) = 0 when x = 0.
Horizontal tangent is at (0, 0).

(.V- + 1)2

(b)

V.

/

^^

/ ^

-.-. ,-,/ X xizosx - 3) - (sinj: - 3a:)(1) xcosx — sinx
72. / (x) = ^3 = —^

,, , x(cosx + 2) — (sinx + 2x)(l) xcosx — sinx

g U) = ; = 1

, , sin X + 2x sin X — 3x 4- 5x ,, .
gix) = = = fix) + 5

X X

/ and g differ by a constant.

_. ., . cosx

74. fix) = = X " cos X

•^ X"

/'(x) = — x~" sinx — MX"""' cosx

= — x"''"H^sinx + ncosx)

X sin X + « cos X

x" +

76. V = 7rr2/i= -rrit + 2)(^Jt

= |(r3/2 + 2ri/2)^

V"(f) = ^tI:^''^' + t"'''- W = -7-775-77 cubic inches/sec

2\2

4ri

When/! = 1: fix) = -

X sin X -(- cos X

When n = 2: /'(x) = -

When /I = 3: /'(x) =

When « = 4: fix) =

X sin X + 2 cos x
x3

X sin X -^ 3 cos x

X sin X + 4 cos x

78. P = -

^ , ^,/ > X sin X + n cos x
For general njix) = 1^7:^ .

80. fix) = sec X

^(x) = CSC X, [0, 27r)
fix) = g'ix)

1 sinx

sec X tan X , cosx cosx

sec X tan X = — esc x cot x => \ — = — 1

sin^x

cos^x
317 Itt

CSC X cot X

= - 1 =* tan^ X = - 1 =* tan X = - 1

1 cos X

sin X sm X

-1

348

Chapter 2 Differentiation

82. (a)n(f) = -9.6643r2 + 90.7414f + 77.5029

v(f) = -276.4643r2 + 2987.6929/ + 1809.9714

vW ^ -276.46f- + 2987.69f + 1809.97
^ -* «(f) -9.66/2 + 90.74f + 77.50

A represents the average retail value (in millions of
dollars) per 1000 motor homes.

40A6{x^ - 2.09;c + 17.83)
(c) A (t) ,^2 _ ^ 3^^ _ g Q2)2

ix'

,/ s x-^ + 2j: — 1
86. fix) = = X

fix) = 1+-,

/'« = -^

2-i

84. fix) =x\^
x'-

fix) = 1

64

fix)

192

88. /W = secx

fix) = secjctanx

/"W = sec jc(sec^ jc) + tan A;(sec j: tan ;c)

= sec ;c(sec^ x + tan^ x)

90. fix) = 2 - 2x-'
/'"(x) = 2x-2 = 4

92. /<'»'W = 2;c + 1
/'='W = 2

/fe)W = 0

94. The graph of a differentiable func-
tion/such that/ > 0 and/' < 0
for all real numbers x would in
general look like the graph below.

96. fix) = 4 - hix)
fix) = -h'ix)
/'(2) = -;!'(2) = -4

98. fix) = g(;c)/r(jc)

/'(;t) = gix)h'ix) + hix)g'ix)
/'(2) = gi2)h'i2) + hiDg'H)
= (3)(4) + (-l)(-2)
= 14

100.

It appears that/ is quadratic; so/'
would be linear and /"would be
constant.

102. sit) = -8.25/2 + 66/
v(/) = - 16.50/ + 66
ait) = - 16.50

/(sec)

0

1

2

3

4

sit) (ft)

0

57.75

99

123.75

132

v(/) = 5 '(/) (ft/sec)

66

49.5

33

16.5

0

a(/) = v'it) (ft/sec2)

-16.5

-16.5

-16.5

-16.5

-16.5

Average velocity on:

[0, 1] is ^^j^= 57.75.

[1,2] is 2^-15^.4,25.
[2,3]isi^Ml_99^24^3

M32- 123.75 ^g^3_
4-3

Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 349

104. (a) f(x) = x"

f (x) = n(n - l)(n - 2) ■ ■ • (2)(1) = n\

Note: n\ = n(n - I) ■ ■ -3 • 2 • \ (read "n factorial.")

106. [xf{x)]' = xf'ix) +f(x)

[xfix)]" = xf"{x) +f'{x) +f'{x) = xf"{x) + 2f'{x)
[xfix)]'" = xf"'ix) +f"{x) + 2f"(x) = xf"'ix) + 3/"U)
In general, [xf{x)}"^ = xf^"Kx) + «/'"-"U)-

108. f(x) = sin x

4f) =

/'(;c) = cosx fU =^

f"ix) = -smx /"(jj^"^

(a) FlU) =r{a)ix - a) + f{a) = 0[j: - -1 + 1 = 1

(b) f(x) = -

("jfr^ =

/'"'W

(-l)"(n)(n- !)(» - 2) ■ ■ ■ (2)(1)

/'2W = ^/"(fl)U - ar +f'(a)(x - a) + f(a) = -(- 1)U - - + 1

1 l/ ^
2\ 2

(c) Po is a better approximation than P,.

(d) The accuracy worsens as you move farther away from x = a

(-!)"«!

(b)

(?■')

IT

/"'N

V

V

110. True, y is a fourth-degree
polynomial.

112. True

d"y
dx"

0 when n>A.

116. (a) (fg ' - f'g) '=fg" + f'g'- f'g '-f'g
= fg"-f"g True

(b) (fg)"={fg'+f'g)'

= fg" + f'g' + f'g' + f'g
= fg"+2f'g'+f"g
=^fg" + f"g False

114. True. If v(f) = c then
a(t) = v'(r) = 0.

350 Chapter 2 Differentiation

Section 2.4 The Chain Rule

y=figix))

4. y = 3 tan(T7x^)

gix)

U = X + I

y=fiu)

y = M-'/2
y = Jitanu

6. y = cos

3x

3x

2 2

8. y = {2^+ \Y

y' = 2(2^' + l)(6;c2) = nx\2>? + 1)

12. /(f) = (9f + 2)2/3

/^(.) . |(9. . 2)-./3(9) ^ ^^

y = cos M

10. y = 3(4 - ^2)5

y' = 15(4 - x'^)\-2x) = -3044 - x'^Y

1/2

14. g(x) = V5 -3x = (5 - 3;c)

.w4(5-3.)-(-3) = ^^^

16. g(x) = V;c2 -lx+ \ = J{x - 1)2 = |x - 1|

1, X > 1

^'(^) =

•1, JC < 1

18. f{x) = -3V2 -9x
fix) = -3(2 - 9x)V4

/'(.) = -|(2-9.)-3/^(-9)=^(^^

20. s{t)

1

;2 + 3f - 1

40 = (;2 + 3f - 1)-'

j'(0 = -l(f2 + 3f- \)-\it + i)

-(2f+3)
(f2 + 3f - 1)2

22. y =

(f + 3)3
y = -5{t + 3)-3

y' = \5{t + 3)-'' =

15

(f + 3)^

24. .w=vS ;' ,,

g(f) = (f2 - 2)-'/2 ,

g'it) = -^(/2 - 2)-3/2(2f)

(,2 _ 2)3/2

26. /(x) = x{-ix - 9)3

/'(x) = x[3(3x - 9)2(3)] + Ox - 9)3(1)
= (3x - 9)2[9x + 3a: - 9]
= H{x - 3)2(4x - 3)

28. V = 42V16 - z2

y' = |^'(|(16 - x2)-'/2(-2x) ) + x(16 - x2)i/2

-^

2yi6"

+ Wl6 - x^

-xQx^ - 32)
2Vl6 - x2

30. y =

Tzn

1,

(x4+ 4)1/2(1) - x^(x^+ 4)-'/2(4x3)

x*+ 4
x* + 4 - 2x'' 4 - jc"

(jc" + 4)3/2 (^4 + 4)3/2

32. hit) =

,2 \2

?3 + 2

h'it) = 2( 3

f2 \/(f3 + 2)(2f) - f2(3f2)\ ^ 2g2(4f - f^) ^ 2r3(4 - ;3)

f3 + 2/1 (f3 + 2)2 / (f3 + 2)3 (r3 + 2)3

Section 2.4 The Chain Rule 351

«-« = (f^)'

Y ^ J3x^ - 2\V(2x + 3)(6x) - {3x^ - 2)i2)\
8 W = ^I^TTTT) I ?^7TTu j

2;c + 3

(2;t + 3)2

3(3;c2 - 2)2(6.t' + 18a: + 4) ^ 6(3jc2 - 2)-(3x2 + 9a: + 2)
(2a: + 3)'' (2a: + 3)''

36. y

^/

2x

x+ 1
1

^ 72^(X + 1)3/2

y ' has no zeros.

38. f(x) = 7^(2 - a:)2

(;c - 2)(5;c - 2)

/V) =

2v^

The zeros of/' correspond to the points on the graph of
/ where the tangent lines are horizontal.

y

W

40, y= it-- 9)Vr + 2
, 5t^ + 8r - 9

y =

2jt + 2

The zero of y ' corresponds to the point on the graph of y
where the tangent line is horizontal.

//

K

/ j

42. g{x) = V.v - 1 + V.r + 1
g'(x)

1 + »

2Vx - 1 2jx + 1
g ' has no zeros.

44. y = a:^ tan -

X

dy ^ \
-r = 2a: tan -

,1
sec--

X X

The zeros of y ' correspond to the points on the graph of
y where the tangent lines are horizontal.

W

46. (a) y = sin 3a:
y ' = 3 cos 3a:
y'(0) = 3
3 cycles in [0, 2Tr]

(b) y = sin!

;)cos(

y'(0) = ^

HaIfcyclein[0. 27r]

The slope of sin ax at the origin is a.

352 Chapter 2 Differentiation

48. V = sin TTx

dy

-f- = TT COS TT

dx

50. h{x) = sccix'^)

h '(x) = 2x secU^) tan{x^)

52. y = cos (1 - 2a;)2 = cos ((1 - 2x)^)

y' = -sin (1 - 2xY(2(\ - 2x)(-2)) = 4(1 - 2x) sin(l - 2xY

54. g{e) = secf ^ej tm[^9

g'ie) = sec(|ej sec2(|ej| + tan(|ej sec(|0J tan(^0j^

4secge[sec^|eUtan^ie

-, / X cosy

56. g(v) = = cos V • sm V

CSC V

g '(v) = COS v(cos v) + sin v(— sin v) = cos^ v — sin^ v = cos 2v

58. y = 2 tan^ x

y ' = 6 tan^ x • sec^ x

60. g(t) = 5 cos^ irr = 5 (cos Trrp
g'(f) = 10 cos 7rf(— sin Trt)i'7r)

= — 10iT(sin -n-t)(cos TTf) = — Sirsinlirf

62. h(t) = 2cot2(7r/ + 2)

/;'(«) = 4cot(7rr + 2)(-csc2(T7r + 2)(7r))
= -ATTCOtiTTt + 2) csc2(Trf + 2)

64. y = 3x - 5 cos(7rj:)^

= 3;i: - 5 cos(7r^;t2)

£ = 3 + 5sin(ir2x2)(2i72x)

= 3 + 10tt2;c sin(i7:t)2

66. .V = sin;c'/3 + (sinx)'/^

y' = cos ;c'/'( -x~^/^ + —(sin a:)~^/' cos x

cosx'/^ COSJf 1

;c2/3 "^ (sin;c)2/3j

68. >- = (3x3 + 4^)i/5_ (2, 2)

>'' = |(3x^ + 4x)-''/5(9x2 + 4)

9x- + 4
5(3x3 + 4jj)4/5

y(2) = 2

70. /W = (^^ = (.- 3x)- (4,^

fix) = -2(x2 - 3x)-3(2x - 3)

^'(^) = -32

-2(2x - 3)
(x2 - 3x)3

72. /(x)=^^, (2,3)

fix)

2x- 3'

(2x - 3)(1) - (x + 1)(2)

(2x - 3)2

(2x - 3)2

/'(2) = -5

TA 1 ^ / h- 2

74. V = — I- Vcos X, — , —

X \2 IT.

1

-f2 2 Vcos X

y'i'jT/2) is undefined.

Section 2.4 The Chain Rule 353

76. (a) fix) = |Wx2 + 5, (2, 2)

fix) = ^;c

(x2 + 5)-'/2(2jc)

+ ^U^ + 5)'/2

+ ^Jx' + 5

3V^?T5 3
/'(2)^^4(3) = ^
Tangent line:

>■ - 2 = y (x - 2) => 13x - Q)' - 8 = 0

(b)

^

IZ

78. (a) f(x) = tan^x, i^, 1
/'(x) = 2 tan X sec^ x
/'(l) = 2(1)(2) = 4
Tangent line:

(b)

y-i=4U--

A

4x - y + (1 - tt) = 0

80. /W = (x-2)-'
/'W = -(x-2)-2 =

fix) = 2U - 2)-3

1

U - 2)^

2
(x - 2)3

82. /(x) = sec^ TTX

fix) — 2 sec 7rx(ir sec ttx tan ttx)

= 2ir sec^ ttx tan i7x
/"(x) = 277 sec^ 7rx(sec^ 7rx)(7T) + 27rtan 7rx(2Trsec' ttx tan ttx)
= 2ir^ sec'' ttx + 47r^ sec^ ttx tan^ ttx
= 2-77^ sec^ TTx(sec- ttx + 2 tan^ vx)
= Itr- sec'^ '77x(3 sec' t7x — 2)

84.

86.

/is decreasing on (— oo, — 1) so/' must be negative
there. /is increasing on (1, oo) so/' must be postive
there.

The zeros of/' correspond to the points where the graph
of/has horizontal tangents.

88. g(x)=/(x2)

g'(x) =/'(x-)(2x) => g'(.v) = 2X-/V-)

354 Chapter 2 Differentiation

90. (a) g{x) = sin2;c + cos^;^ = 1 ^> g'W = 0
g '(x) = 2 sin j: cos x + 2 cos x(— sin x) = 0
(b) tan^x + 1 = sec^.n:
g{x) + l^f(x)
Taking derivatives of both sides,

Equivalently,/'(x) = 2 sec x • sec x • tan x and
g '(x) = 2 tan X • sec* x, which are the same.

92. y = \ cos 12f - 5 sin 12r

V = >-' = |[- 12 sin 12f] - \[U cos 12r]

= -4 sin lit - 3 cos 12f
When t = 77/8, y = 0.25 feet and v = 4 feet per second.

94. >• = A cos wt

3.5

(a) Amplitude: A = -r- = 1.75
y = 1 .75 cos wr

r. ■ J in 277 TT

Penod: 10 => w = — — = —

V = 1.75 cos — r

(b) V = >>'= 1.75

77 . irf

= -0.35 77 sin

77f

96. (a) Using a graphing utility, or by trial and error, you
obtain a model of the form

(TTt

T{t) = 64.18 - 22.15 sini — + 1

(b)

(c) rW = -22.15 cos(y+l)(|

TTt

= -11.60 cost— + 1

(d) The temperature changes most rapidly when r = 4.1
(April) and t = 10.1 (October). The temf>erature
changes most slowly (T'(t) = 0) when r = 1.1
(January) and r = 7.1 (July).

98. (a) g{x) =f{x)-2^g '{x) = f'(x)

(b) h{x) = lf{x)=^h\x) = 2f\x)

(c) Ax) =f{-3x)=>r'{x) =f'i-3x){-3) = -3/'(-3x)
Hence, you need to know/'(-3j:).

r'(0) = -3/'(0) = (-3)(-|) = 1
r'(-l) = -3/'(3) = (-3)(-4)= 12

(d) six) = f(x + 2) ^ s'ix) = fix + 2)
Hence, you need to know/'(A: + 2).

s'{-2)=fX0)

X

-2

-1

0

1

2

3

/'W

4

2
3

1
3

-1

-2

-4

g'ix)

4

2
f

1
3

-1

-2

-4

h'{x)

8

4
3

~3

-2

-4

-8

r\x)

12

1

s'ix)

1

3

-1

-2

-4

-3. etc.

Section 2.4 The Chain Rule 355

100. fix +p)= fix) for all x.

(a) Yes,/'(j: + p) = fix), which shows that/' is
periodic as well.

(b) Yes, let gix) = filx), so g'ix) = 2/'(2x).
Since/' is periodic, so is g'.

102. \ffi-x) = -/(x), then

£[/(-^)] = f [-/W]
/'(-x)(-l) = -/'(x)
/'(-x)=/'(x).
Thus,/'(x) is even.

104.

|[|«|] = j^[J^] = |m-'/^(2««0 = ^ = „'^.„ ^ 0

106. /{x) = |x2 - 4|

108. fix) = |sinx|

/'(x) = cos x( 7-; — r ],x¥'klT
' sinx '

110. (a) fix) = sec(2x)

/'(x) = 2(sec2x)(tan2x)

fix) = 2[2(sec 2x)(tan 2x)] tan 2x + 2(sec 2x)(sec2 2x)(2)
= 4[(sec 2x)(tan- 2x) + sec-* 2x]

(b)

4f)=K?

il

2sec|f tanf =4V3

/"( Ij = 4[2(3) + 2^] = 56

P,(x) = 4V3(x - -f j + 2

P^Cx) = |(56)(x - 1)' + 4^3 (x - f) + 2

= 281 X - -^ r + 4V^( X - ^ I + 2

TT\2

6y

(c) P2 is a better approximation than P^
112. False. If/(x) = sin^ It, then/'(x) = 2(sin lx)(2 cos Ix).

(d) The accuracy worsens as you move away
from X = ir/6.

114. False. First apply the Product Rule.

356 Chapter 2 Dijferentiation

Section 2.5 Implicit Differentiations

2. x^-f=\e

2x - lyy' = 0

, X

6. x'y + y^x = - 2

x^' + 2xy + y^ + lyxy' = 0

{x^ + 2xy)>''= -{y^ + Iry)

y

-y(y + 2t)
;c(;c + 2y)

jc' + / = 8
3;c2 + 3y2y' = 0

(xy)'/2 - X + 2^ = 0
\{xy)-'l\xy' + y) - 1 +2>''=0

-^y' + -^

1 + 2)-' = 0

2v^ ijxy

xy' + y - ijxy + 4v^>'' = 0

2V^-y
4vxy + X

10. 2 sin X cos y = 1

2[sinx{-siny)y' + cosy(cosx)] = 0

cos X cos y

y = ^ :

sin X sm y

= cot X cot y

12. (sin TTx + cos Try)^ = 2

2(sin TTX + C0S7ry)[Tr cos ttx — 7r(siniTy)y'] = 0

77 cos TTX — TT^sin '77y)y ' = 0

COS TTX

y =

Sin Try

14. cot y = X — y

(— csc^y)y' = 1 — y'
1

y' =

1 — csc^y
1

-cot-y

= -tan-^y

16. X = sec-

y' 1 1

1 = —=^ sec- tan -

r y y

sec(l/y) tan(l/y)

-y2cos(^)cot(^)

18. (a) (x2 - 4x + 4) + (y2 + 6y + 9) = -9 + 4 + 9

(x - 2)2 + (y + 3)2 = 4 (Circle)

(y + 3)2 = 4 - (x - 2)2

(b)

= -3±V4 - (x - 2)2

(c) Explicitly:

^ = ±^[4-(x-2)2]-i/2(-2)(x-2)
ax 2

^(x

-2)

(74-

(;c - 2)2

-(^

-2)

±V4-

(;c - 2)2

-(;c-2)

-3 ± 74 - (x - 2)2 + 3

-(;c-2)
y + 3

(d) Implicitly:

2x + 2yy'- 4 + 6y'= 0

(2y + 6)y' = -2(x - 2)

, -{x-D

Section 2.5 Implicit Differentiation 357

20. (a) 9y'^

x^ + 9

(b)

y^ = ^+l =

x^ + 9

±Jx^ + 9

.^p ,., ^y 4^'^'r^''^'^^ ±x ±x X

(0 ExpLctly: - = = j^^=== = ^(^ = ^

(d) Implicitly: Qy^ - x^ = 9
ISyy' - 2x = 0
18>7'=2x

2x a:

ISy 9y

22.

/ = 0

2i - 3y^' = 0

3/

y =

At (1,1): y' = -.

24. (x + yy = }^ + f

x^ + 3;c^ + 3xf + f = x^ + y^
3x^ + 3ry2 = 0
x^y + xy^ = 0

x^' + 2xy + 2xyy' + y^ = 0

{x- + 2xy)y'= -(y^ + 2xy)

y(y + It)
"jc(.t + 2y)

At(-1, 1): y' = -1.

26.

X3 +

4ry + 1

3j;^ + 3y^y ' = ^xy ' + 4v
(3>'^ - 4x)y ' = 4>' — 3x2
, 4y - 3x2

At (2, !),>.' =

(3y2 -- 4x)
4-12 8

28. X cos y = 1

x[— y'siny] + cosy = 0

y -

cos y
X sin V

1 cot V
= — cot V = -

X ' X

*"^'f>>' = 5^-

30. (4 - x)y2 = x3

(4-x)(2)yO+r(-l) = 3x2

3X2 + y2

y =

2y(4 - x)

At (2, 2): y' = 2.

32. x' + y3 - drv = 0

3x2 + 3^^,' - 6,rv.' - 6y = 0

y'(3>-2 - dv) = 6y

, 6v

3.x2
3.t2

3^•2 - dr

M 8\ , ^ (16/3) - (16/9) ^ 32
'^^ \y ir ■ (64/9) - (8/3) 40

358 Chapter 2 Differentiation

34. cos y = X

— sin y • y ' = 1

•1

y ' = -: — , 0 < V < TT
sin 3;

sin^y + cos^y = 1

sin-y = 1 — cos^y

siny = Vl — cos-^y = Vl - x^

-1

^ yn^'

36.

-1 < ;c < 1

xY - 2x = 3

2x^' + 2xr - 2 = 0

jc^yy ' + xy^ —1=0

1 -xy^

->
xy

2xyy' + x2(y'P + x^" + 2xyy' + / = 0

4xKy' + x^iy^ + x^" + / = 0

4-4xy^- , (1 - xy^)^ , 2, - , 2 n

=^ H 7-7 1- x'Vy + y^ = 0

X xy^

4xy^ - 4xY + 1 - 2xy2 + ^.ly^ + ;t4y3y" + ^2^4 ^ q

x'^y^y" = 2xy — 2xy~ — 1
„ 2xY -2xf-\

38. \ — xy = X — y

y — xy = x — 1

X- 1

y' = 0
/'=0 ■

A'

40. / = 4^

2yy' = 4

, 2

y"=-2y-Y =

r-2]
LrJ

2 _ -4

y y

, X — 1

42. >^ = JTT

2>y'

(x^ + 1)(1) - U - l)(2x)

{x^ + 1)2

X2+ 1

- 2x2 + 2x

(;c2

+ 1)^

1 + 2x

-x2

2y{x2 + 1)2

1+4-4

1

"^''^•5 }"' [(275)/5](4 + 1)2 1075-

V5

1

Tangent line: > ~ ~^ = /^U - 2)

lOVSy - 10 = X - 2
X - lOySy + 8 = 0

Section 2.5 Implicit Differentiation 359

44. a:^ + y^ = 9

At (0, 3):

Tangent line: ^^ = 3
Normal line: x = 0.

At (2, VS):

_2
Tangent line: v - Vs = —i={x - 2) => 2a: + VSy -9 = 0

Normal line: y - V5 = -^{x - 2) ==> VSJc - 2y = 0.

(0.31

1

f A

V

y

/--

>t'

>i

J^

46. >»* = 4x

2xy'=4 , _

y' = -= 1 at (1,2)

Equation of normal at (1, 2) is y - 2 = -\{x — 1), v = 3 - x. The centers of the circles must be on the normal and at
a distance of 4 units from (1,2). Therefore,

[x - 1)2 + [(3 -x)- If = 16

2(a: - \)- = 16

x= \ ± ijl.

Centers of the circles: (l + 2^2, 2 - 2V2) and (l - 2^2, 2 + 275)

Equations: [x - \ - ijl)- + (v - 2 + ijl]- = 16

(x - 1 + ijlf + (.V - 2 - 272)- =16 -

48. 4x2 + y - 8x + 4.^, + 4 = 0
8x + lyy' - 8 + 4y'= 0

8x 4 - 4x

" 2>' + 4 .V + 2
Horizontal tangents occur when x = 1 :

4(1)2 + y2 - 8(1) + 4^ + 4 = 0

y2 + 4y = y(>' + 4) = 0 => y = 0, -4
Horizontal tangents: (1, 0), (1, -4).

Vertical tangents occur when y = —2:

4a- + (-2)2 - 8x + 4(-2) + 4 = 0

4x2 - 8.T = 4x(x - 2) = 0 => X = 0, 2

Vertical tangents: (0, -2), (2, -2).

360 Chapter 2 Differentiation

50. Find the points of intersection by letting y^ = x^ in the equation 'bp- + 3>'-^ = 5.
Ix'^ + lx^ = 5 and 3x5 + Ix^ - 5 = 0
Intersect whenx = 1.

Points of intersection: (1, ±

1)

/ = ^:

2x2 + 3^2 -.

= 5:

2yy' = 3a:2

Ax + 6yy ' =

= 0

2x

'^-yy

At (1, 1), the slopes are:

r-\

2

y =-3.

At (1, - 1), the slopes are:

3

y =-2

, 2

^ =3-

Tangents are perpendicular.

[21^+3/ =5]

/"

~>((1. 1)

v^

[>=;■

52. Rewriting each equation and differentiating,
jc^ = 3(y - 1) xOy - 29) = 3

^ ,

. = y + l

>» = jj"'

-K!-'

3' =

1

|jr(3>-29) = 3|,5 [x_3=3>.-

A

A

J

^

f

For each value of jt, the derivatives are negative reciprocals of each other. Thus, the tangent lines are orthogonal at both points
of intersection.

54.

X2 +y2= C^

y = Kx

2x + 2yy' = 0

y' = K

• X

'^~y

c

7K'

k

X

At the point of intersection {x, y) the product of the

slopes is {—xly)(fC) = (—x/Kx){K) = — 1. The curves are orthogonal.

K = -] \^-^ c =

56. ^2 - 3xy2 + y = 10

(a) 2;c - 3/ - 6xyy' + 3y^' = 0

i-6xy + 3/)y'= 3y2 - 2;c

y=3/-2.
^ 3y^ - 6xy

*>-f-*'f-'^vf = °

(2x - 3/)^ = {6xy - 3y^^

58. (a) 4 sin x cos >> = 1

4 sin x(— sin y)y ' + 4 cos x cos y = 0

cos X cos y

>" = —■ ■■

sin X sin y

I \dy dx

(b) 4smjc(— siny)— + 4 cos x— cosy = 0
at at

dx

.-.^dy

cos JT cos y-r = sinjcsiny-,
dt dt

Section 2.6 Related Rates

361

60. Given an implicit equation, first differentiate both sides
with respect to x. Collect all terms involving y' on the left,
and all other terms to the right. Factor out v ' on the left
side. Finally, divide both sides by the left-hand factor that
does not contain y '.

62.

64.

>/x + Vv = v/c

' +-^^=0

2Vx iVy'^
di
dx

V-y

Tangent line at (xq, yg):

yo

^U - Xg)

77jn

Use starting point B.

.x-intercept: [xg + v^Vyo> O)
y-intercept: (o, Vq + V^v5o)
Sum of intercepts:

Uo + V^VS^) + (.Vo + v^v^) = -^o + 2V^V>\J + Jo = (^ + n/v^)" = (Vc)" = c.

Section 2.6 Related Rates

2. y = 2(x2 - 3x)

Tt^^^'-^^t

dx

1 dy

dt 4x - 6 dt

(a) When A; = 3and^ = 2,^ = [4(3) - 6](2) = 12

(b)When.= ,andf=5,f = ^(5) = -f

4.

.r^ + / = 25

2x

dx
dt

■ly

di
dt

di
dt

dx

dt

x^x
yldt

y\dy_
' xldt

(a) When x = 3, y = 4, and dx/dt = 8.

dt 4

(b) When .t = 4, y = 3. and dy/dt = -2.

dx
dt

r(-2)

362 Chapter 2 Dijferentiation

6. y

1 +x2

dx
dt

= 2

-2x

dx

dy_

dt L(l+x2)2jdr

(a) Whenx = -2,

dy -2(-2)(2) 8 ,

■ = :^ = TT^ cm/sec,

dt 25 25 '

(b) When x = 0,

— = 0 cm/ sec.
ar

(c) When x = 2,

dy _ -2(2)(2) _ -8

dr

25

25

cm/sec.

8. y = sinj:

dx

, = cos X—r
dt dt

(a) When x = 7r/6,

= ( cos — 1(2) = >/3 cm/sec.
(b) When x = 7r/4,

— = (cos— 1(2) = V^ cm/sec.

(c) When x = 7r/3,

-j- = (cos-r-j(2) = 1 cm/ sec.

dx . dy .

10. (a) — negative ==> — negative
dt dt

„ dy . . dx

(b) — positive => — positive
dt dt

12. Answers will vary. See page 145.

14. D = Jx^ + /
dt

j£^

— — = - (x^ + sin^ x) '/^{2x + 2 sin X cos x)^
flf 2 dr

16. A = trr^

dx jc + sin j: cos x dx 2 + 2 sin a; cos x

dA , dr

— - = lirr—
dt dt

If dr/dt is constant, dA/dt is not constant.

— - depends on r and -r.
dr dr

v^

+

sin-

;c dr
18.

V

dr
dt

dV
dt

Vx2 +

= 2
= 4Trr2

sin"^x

dr
dr

dK

(a) When r = 6, — = 477(6)^(2) = 28817 in Vmin.
dt

dV
When r = 24, — = 4ir(24)2(2) = 460877 inVmin.

(b) If dr/dt is constant, dV/dt is proportional to r^.

20. K = ;c3

dr
dV

- = 3;c2 -
dr dr

(a) When ;c = 1
dV

dt

= 3(1)2(3) = 9 cmVsec.

(b) When x = 10,
dV

dr

= 3(10)2(3) = 900cmVsec.

Section 2.6 Related Rates 363

22. V = ^TTr% = ^TrrK3,r) = Trr^

dt

dV - ^dr
dt dt

24. v = -.r^;, = -,.— ;,3=__,3

/ r h 5 , \

I By similar triangles, - = — => r = —h.j

dt

(a) When r = 6,

dt

= 3-7r(6)2(2) = 21677 inVmin.

(b) When r = 24,
dV

dt

= 377-(24)2(2) = 345617- inVmin.

fify ^ ^^2 ^ ^ = / 144 W
</? ~ 144 dt^ dt~ \25TTh^jdt

dh \44 9

26. V = -foA(12) = 6bh = 6^2 (since b = h)
( ) ^= nh— — = — —

When /j = 1 and — - = 2, -- = -r77r(2) = - ft/min
fi!/ dt 12(1) 6

(b) If -r = o in/min = — - ft/min and h = 1 feet, then
af o 32

f = (l^^^) = !^'Vmin.

28. a:^ + y^ = 25

„ dx „ dy
2x— + 2y-Y = 0
dt dt

dx y dy O.lSy . dy

— - = —^•-f= ^ smce -r- = 0.15.

dt X dt X dt

When X = 2.5,

y = yi8J5,^ = -^41^ 0-15 = -0.26m/sec
dt 2.5

30. Let L be the length of the rope.

(a)

L2 = 144 + x^

^.dLdx

2L-— = 2x-r

dt dt

dx L dL AL . dL

— = — = smce -3- = -4ft/sec.

dt X dt X dt

WhenL= 13,

>;4fl/sec

X = JL^ - 144 = Vl69 - 144 = 5 ^

i!|iy

dx 4(13) 52 ^Q_
dt 5 5 - 10.4 ft/sec.

" i^ft''^

(b) If^= -4, andL = 13,
dt

Speed of the boat increases as it approaches the dock.

dL xdx

dt Ldt

= ^>-4)

^^--

AsL ^ 12 + ,

dL
dt

decreases, so

the speed decreases.

364 Chapter 2 Differentiation

32. x^ +y^ = 5^

^ dx „ ^ ds

2x— + 0 = 2s —

dt dt

dy

since -r = 0
dt

dx _ s ds
dt X dt

When s = \0,x= VlOO - 25 = V75 = 5 V3

^ = ^(240) = ^ = 16073 = 277.13 mph.
dt 5 V 3 V 3

. s^-

= 90^ + x2

x =

= 60

dx _
dt "

= 28

ds _
dt '

X dx
''~s"dt

Home

When X = 60,

s = V902 + 602 = 30VT3

d^
dt

60

30yi3

(28)

56

13

15.53 ft/sec.

36. (a)

20 _ y
6 y ~ X

20^-

- 20x = 6^

14y = 20a:

10

dx_

-5

rfr

£f> 10 (ic 10, ^, -50^,

^(y - x) _dy dx _-50 , ._ -50 35 _

15

ft/ sec

38. x{t) = - sin 77?, x^ + y2 = 1

277

(a) Period: — = 2 seconds
77

(b) When x = -y

VHI

2 4

Lowest point: 0

enx

3
= T0'^ =

v-

= 4 ^

3
10

3 .
= — sm 77/

1

^ sm 77/ = -

1
=^' = 6

dx _
dt

3

— 77 cos 77/

;c2 + / = 1

2x— + 2>'— = 0^— = 3-.

rf/ rf/ dt y dt

dy _ -3/10 3
15/4' 5

-977 -97577

'^"^■^ = 7Tp-I"'^°^'^

Speed =

2575 125

-9V577

125

= 0.5058 m/sec

Section 2.6 Related Rates 365

40.

1 = -1 J-
R A J R-i

dR^
dt

dR.
dt

1.5

R^' dt ~ /?,2 ■ dr "^ T?,^ ■ Jf
When /f , = 50 and /?2 = 75,
R = 30

42. rgtanO = v^

32r tan 0 = v^, r is a constant.

32rsec20— = 2v—
A dt

dv 16r

A

sec- 0 —
dt

Likewise, — = -rr- cos- 0

dt 16r

rff'

dt

1(50)2^'^"^ (75)2

= (30)-| 777^(1) + 7:;iT2(l-5)

0.6 ohms/sec.

44.

sin fl = —

X

^ = (-l)ft/sec

cos d\

de

dt

dd
dt

-10 dx

X- ' dt

10 dx

X- dt

{see e)

= ~'0f 25 ^ 10 1 ^ 2 ^ 2V2T

25- V25- - 10= 25572T 25^21 525

= 0.017 rad/sec

46. tane = -

— - = 30(2 tt) = 607Trad/min = 17 rad/sec
dt

sec^e

^\ ^ Udx

dtj 50\ dt

dx ,„ , Jdd
— = 50 sec- e\ —
dt \dt

(a) When 6 = 30°, ^ = ^^ ft/sec.

(b) When e = 60°, — = 200Tr ft/sec.

(c) When 6 = 70°,^ = 427.43Tr ft/sec.

48. sin 22°

0 = -— . ^ + i . ^

y- dt y dt

^ = - • ^ = (sin 22°)(240) = 89.9056 mi/hr

50. (a) dy/dt = 3(dx/dt) means that y changes three times as
fast as X changes.

(b) y changes slowly when v = 0 or .v = L. y changes
more rapidly when .v is near the middle of the
interval.

366 Chapter 2 Differentiation

52. L? = 144 + x^; acceleration of the boat

£x
dt^-

^- . ■ ■ ^r dL ^ dx

First denvative: 2L—- = 2x—-

dt dt

^dL dx

L-r- = X —
dt dt

, , . . d-L dL dL d^x dx dx

Second denvative: L -pr H — r ' ~r = ^<^tt '^ ~r ' ~T

dt^ dt dt dt^ dt dt

d^x
dt""

I'f-(f)"-(f

dx

dL

dL

d^L

When L = 13, j: = 5, -r = - 10.4, and —- = — 4 (see Exercise 30). Since —- is constant, -pr = 0.

dt

dt

dt

dt^

g = |[13(0) + (-4)2 -(-10.4)2]

= |[16 - 108.16] = |[-92.16] = - 18.432 ft/sec2

54. y(i) = -4.9r2 + 20

dy
dt

y(\) = -4.9 + 20 = 15.1

y\\) = -9.8

T, • M • , 20 y

By similar triangles, — = — r

X X — \2

20x - 240 = xy.

Whenjy = 15.1, 20;c - 240 = .x(15.l)

(20 - 15.1);c = 240

240

4.9'

20;t - 240 = ry

,„(& dy dx

^"^t^^t^^t

(0.0)1

Atr = 1,

dx

x

dy

dt

20 - y

dt

dx

240/4.9

I

dt 20 - 15.1

(-9.8) = -97.96 m/sec.

Review Exercises for Chapter 2 367

Review Exercises for Chapter 2

2. fix) =

x+ 1
X- 1

x + ^x + \ _ X + 1

,. fix + Ax) -fix) ,. ;c + Ax - 1 j: - 1
/'W = lim -^ — ^-^ = hm T

= lim

Ax->0

ix + Ax+ l)ix-\) - ix + Ax- Dix + 1)
AtU + Aj: - l)(jc - 1)

~ i^^o M^c + Ajc - l)(;c - 1)

lim

-2Ax

= lim

^"o Axix + Ax- 1)U - 1) ^"o (;c + Ajc - l)(x - 1) (jc - 1)-

4./W = -

■' Ai->o Ajc

6. /is differentiable for all jc t* —3.

lim

X + Ax X

Aj:-»0 Ax

2x: - (2x + 2Ax)

Aj:-»0 AjcU + Ajc)jC

-2Ax

Ax-^0 Ajc(x + Ajc)jc

r -2 -2

= lim 7 : — ^ — — TT

iijr->0 (X + Ax)x X~

, , _ Jx^ + 4x + 2, ifx < -2 '
»--^W-jj_4^_^2 if., >-2

(a) Nonremovable discontinuity at .V = — 2.

(b) Not differentiable at .r = - 2 because the function is
discontinuous there.

10. Using the limit defmtion, you obtain h '(«) = ^ - 4.t.

AtJC=-2,;z'(-2) = |-4(-2) = y.

-2
12. (a) Using the limit definition, fix) = -, -tt.

(jc + \y

At jc = 0./'(0) = -2. The tangent line is

>- - 2 = -2U - 0)

y = -Ix + 2

(b)

^

t^'

^

\

368 Chapter 2 Differentiation

14. /'(2) = lim

/U)-/(2)

2 X - 2

1

= lim

1

x+ I 3

= lim

2 JC - 2

3 -a:- 1

x->2 (x - 2)U + 1)3

-1

1

lim / , ,N-

x->2 (x + 1)3 9

18. y = - 12

y' = 0

20. gW=xi2
g'W = 12x"

16.

22. /(r)= -8^5
fit) = -40^

24. g(j) = 4s^ - 5s^
g'is) = I6s^ - \0s

30. g{a) = 4 cos a + 6
g '(a) = - 4 sin a

26. /(jc) = ;c'/2 - ;t-

1/2

28. h{x) = |x-2

/'W = |.-'/^ + 1.-/^ = ^

-4 _ -4

, , 5 sing
32. g(a) = — 2a

,. . 5cosa
g (a) = — 2

34. 5 = - 16f2 + So
First ball:

-16/2+ 100 = 0

Aoo 10 ^^ J u-
f = - / — -^ = — = 2.5 seconds to hit ground
V lo 4

Second ball: . ■ '

-16r- + 75 =0

f2 =

ns 573

2.165 seconds to hit ground

Since the second ball was released one second after the first ball, the first ball will hit the ground first. The second ball will hit
the ground 3.165 - 2.5 = 0.665 second later.

36. 5(f) = - 16;^ + 14,400 = 0
' 16/2 = 14400

r = 30 sec ■

Since 600 mph = g mi/sec, in 30 seconds the bomb will move horizontally (gjCSO) = 5 miles.

Review Exercises for Chapter 2 369

38.

(a) y

32 . ,,

32
— ^x

64

(b) v'= 1 -^x

0 if r = 0 or X

32'

Projectile strikes the ground when x = v^jlsl.

Projectile reaches its maximum height vXx = v^/M.
(one-half the distance)

32 / 32 \
(c) y = x ^.r' = x\\ ;.v = 0

when ,v = 0 and x = Xq-/32. Therefore, the range is
-v = Vq-/32. When the initial velocity is doubled the
range is

(2vo)^
32

32

or four times the initial range. From part (a), the
maximum height occurs when x = Vo^/64. The
maximum height is

y\

64

32 /vo'

64

Vo-\64

64 128 128

If the initial velocity is doubled, the maximum
height is

2 >

V

(2vo)
64

J 128 \128

(d) Vq = 70 ft/sec

iW-

Range: x = - ^^

153.125 ft

Maximum height: y = 7^ = ^:^ " 38.28 ft

or four times the original maximimi height.

40. (a) y = 0.14;c= - 4.43a + 58.4

(b) 320

(C) 2

(d) If.v = 65, V = 362 feet.

(e) As the speed increases, the stopping distance increases at an increasing rate.

370 Chapter 2 Differentiation

42. g{x) = (x3 - Zx){x + 2)

g\x) = (x3 - 3x)(l) + {x + 2)(3;c2 - 3)
= x^ - Ix + ix^ + 6x^ - 3x- 6
= 4.^ + ex'^ - 6x- 6

46./W = ^^l
f\x) =

^ {x- 1)2
50. fix) = 9(3^2 - 2x)-'

Z'U) = -9(3x2 _ 2x)-2(6x - 2)

54. y = 2jc — ;c2 fan ^^

y' = 2 — x^ sec^ x — 2xVwix

x -

- 1

{x-

- 1)(1) -

-(;c +

1)(1)

(x-

-IF

-2

18(1 - 3x)
(3a:2 - 2xY

44. /(f) = t^cost

fit) = t\-smt) + cosf(3f2)
= — f' sin f + 3?2 cos t

(x2 + 1)(6) - (6a: - 5)(2x)
(;c2 + 1)2

2(3 + 5x- ?,x^)
ix" + 1)2

fix)

sin x

52. y = ^

, _ (j:2)cosx — (sinx)(2x) _ x cos jc — 2 sin jr

xf>

X?

56. y

1 + sinx
1 — sin X

, _ (1 — sinx) cosx — (1 + sinx)(— cosx)
^ " (1 - sinx)2

2cosx

(1 — sinx)2

58. v(/) = 36 - t\0< t < 6

ait) = v'it) = -2r

v(4) = 36 - 16 = 20 m/sec

a(4) = - 8 m/sec

62. hit) = 4 sin f — 5 cos t

h 'it) = 4 cos t + 5 sin t
h"it) = -4 sin f + 5 cos t

60. fix) = Ux^'"'
fix) = 3X-3/4

-9 —9

^ W 4 ^ 4^7/4

64.

y

(10 — cosx)

xy + cos X = 10
xy ' + y — sin X = 0

xy ' = sin X — >>
xy' + y = (sin x — >>) + y = sin x

66. fix) = (x2 - l)'/3

fix) = |(x2 - 1)-V3(2x)

2x

3(x2 - 1)2/3

68. /(x) = (x2 + ^)'

Review Exercises for Chapter 2 371

70. h{e}

h'{9)

(1 - er

{i-ey-e[3(i- en-i)]
(1 - er

(1 - 8)^(1 - 6 + 36) ^ 26+ 1

(1 - ef (1 - 6Y

74. y = CSC 3x + cot 3x

y' — —3 CSC 3.t cot 3j: — 3 esc- 3x
= - 3 CSC 3x(cot 3x + CSC 3x)

72. y = I - COS 2j: + 2 cos^ x
y' = Is'mlx — 4 COS jc sin j:

= 2[2 sin j: cos x] - 4 sin x cos x

= 0

76. >■

sec^;

7 5

)> ' = sec^ a:(sec x tan x) — sec'' ar(sec j: tan x)
= sec^ X tan x(sec^ .t - 1)
= sec^ X tan^ x

78. f(x)

fix)

3x

Vx=+T

1,

3(x2 + l)'/2 - 3x|(x2 + l)-'/2(2x)

x2+ 1
3(x^ + 1) - 3x'

(X^ + 1)3/2

3

(x2 + 1)3/2

80. y

cos(x — 1)
X - 1

-(x - 1) sin(x - 1) - cos(x - 1)(1)

(x - ly-

1

U - 1)-'

[(x - 1) sin(x - 1) + cos(x - 1)]

82. fix) = [(x - 2)(x + 4)Y = (x2 + 2t - 8)2
fix) = 4(x3 + 3x- - 6x - 8)

= 4(x - 2)(x + l)(x + 4)
The zeros of/' correspond to the points on the graph of/ where the tangent line is horizontal.

w

84. ^(x) = x(x- + 1)1/2
2x2 + 1

g'ix)

VxH^

g' does not equal zero for any value of x. The graph of g
has no horizontal tangent lines. .

1

86. y = V3x(x + ly

3(x + 2)2(7x + 2)

y =

273x

y ' does not equal zero for any x in the domain.
The graph has no horizontal tangent lines.

f

372 Chapter 2 Differentiation

88. y = 2 csc^ (v^)

y' = ;pcsc^ Vxcot Vx

Jx

The zero of y ' corresponds to the
point on the graph of 3; where the
tangent line is horizontal.

T

90. y = j;-' + tanjc

y' = —x~'^ + sec^jr

y" = 2x~^ + 2 sec jc(sec x tan x)

= —; + 2 sec^ j: tan ;<:

96. h{x) = xjx^ - 1
2x^-1

x{2jP- - 3)
(;c2 - 1)3/2

92. y = sin^ x

y' = 2 sin X cos x
y" = 2 cos 2jc

sin 21

94. ,W = ^

;c2+ 1

2(-3a:^ + 5;c + 3)
(x?- + 1)2

(x2 + 1)3

98. V = J2^ = 72(32);; = 8v^

dh Jh

(a) When /i = 9, ^ = ^ ft/sec.
an 3

(b) When h = A,— = 2 ft/sec.
an

100. x2 + 9y2 - 4x + 3y = 0
2x + ISyy'- 4 + Sy' = 0

3(6y + l)y' = 4 - 2x

4 - 2x

y =

3(6y + 1)

102. ■f = x'-x^ + xy-y^

0 = x3 — x?y + xy — 2y^

0 = 3x2 _ j,2y ' _ 2xy + xy' + y — 4yy '

(x2 — X + 4y)y' = 3x2 _ 2xy + y
, _ 3x2 _ 2xy + y

y =

x^ — X + Ay

1©4. cos(x + y) = X

-(1 +y')sin(x + y) = \

— y'sin(x + y) = 1 + sin(x + y)

, _ 1 + sin(x + y)
sin(x + y)

= -csc(x + 1) - 1

106. x2 - y2 = 16

2x - 2yy' = 0
/ X

'^y

At (5, 3): y' = \

Tangent line: y - 3 = -(x - 5)
5x - 3y - 16 = 0

Normal line:y — 3 = - •r(x - 5)

3x + 5y - 30 = 0

108. Surface area = A = 6x2, j, length of edge.

dt

^ = 12x^ = 12(4.5)(5) = 270cmVsec
dt dt

Problem Solving for Chapter 2 373

110. tan e = ;c

—r = 3(27r) rad/min
2 Jd6\ dx

dx
dt

(tan2 e + l)(67r) = Gt^jc^ + 1)

When;c

1 <ic , /I ,\ IStt, , . ,^„ , ,,
= -, — = 0771 T + 1 = —z- km/ mil) = 450irlcm/hr.

2 at \4 / 2

Problem Solving for Chapter 2

Let {a, a^) and {b, - fc- + 2i> - 5) be the points of tangency.

For y = x^,y' = 2x and for y = —x^ + 2x-5,y'= - 2r + 2.

Thus, 2a= -2i) + 2=*a + i'= l,ora= 1 - b. Furthermore, the slope of the common
tangent line is

a"- - {-b^ + 2b - 5) ^i\ - by + b^ -2b + 5
a- b {\ - b)- b

\ -2b + Ir + b- -2b + 5

-2b + 2

2b

= -2b + 2

=>2b- - 4b + 6 = 4b^ - 6b + 2
=i' 2b- -2b- 4 = 0
=>/r-fo-2 = 0
=^(b- 2)(b + 1) = 0
* = 2, - 1
For b = 2, a=l— b=— I and the points of tangency are (— I, 1), (2, -5). The tangent line has slof)e —2:

y - 1 = -2(x = 1) ^y = -2r - 1
For b=—l,a=\—b = 2 and the points of tangency are (2, 4) and {- 1, — 8). The tangent line has slope 4:
y - 4 = 4(x - 2) ^ y = 4jt - 4.

4. (a) y = ^-, y ' = 2jc. Slope = 4 at (2, 4).
Tangent line: y - 4 = 4(x — 2)
y = 4x — 4

(b) Slope of normal line: — |.
Normal line: y - 4 = -jU — 2)

1 1 9

y = -4X + 2
y=-\x + l = x^^4x-+x-\S = 0=^ (4.x + 9)^ - 2) = 0
j: = 2, — 4. Second intersection point: (— 3, ig)

(c) Tangent line: y = 0
Normal line: .r = 0

—CONTINUED—

374 Chapter 2 Differentiation

4. —CONTINUED—

(d) Let {a, a'), a ^ 0, be a point on the parabola y = x^. Tangent line at {a, a^) isy = 2a{x — a) + a^. Normal line at [a, a^) is

y = - -— (x — a) + cP-. To find points of intersection, solve
2a

x^ = -^Jc - a) -^ a^

x'- + ^x = a^ + l
2a 2

11,11

4a ~\ 4a

X + -— = a + —^=>x = a (Point of tangency)
4a 4a

x + -—=—\a + --\ =>x = —a

4a \ 4a) 2a 2a

!£■+ 1

2a

The normal line intersects a second time at x =

6. f(x) = a + b cos ex

f'{x) = —be sin ex

At (0, 1): a + ^ = 1 Equation 1

. /77 3\ , /c-n-\ 3 ^ . ^

At I —, - I: a + b cosi ~r] ~ '^ Equation 2

— iicsLnl — 1=1 Equations

From Equation \, a = \ — b. Equation 2 becomes {\ — b) + b cos \~7']~'^^=^~i' + b cos "T = T

From Equation 3, fc = ;^ — r. Thus -, — r- H ; — r cos — r = —

. eiA . etrX . eiA \4 ) 2

l-cos(f) = icsin(^

Graphing the equation g{c) = - c sin ( — - J + cos ( -— ) — 1, you see that many values of c will work.

13 3 1

One answer: e = 2,b = —-,a = — =^f(x) = - - - cos 2x
2 2 2 2

Problem Solving for Chapter 2 2nS

8. (a) by- = x3(a - x);a,b > Q

2 _ -^^(0 ~ x)

b^

Graph >>! =

V^(a — x)

and;y2

V;c3(a - x)
b

(c) Differentiating implicitly.

lb'^yy'= -ix-ia - x) - x^ = lax- - Ax^
, (3ax^ - Ay?) __ „

' ~ 2b^

— \j

iax^ = 4x3

3a = 4x

3a

fc^2

-mi'

-^)

21a^\
64 U

«)

/

270"
256b- ~^

y = ±

373a2
16ft

Two

. (3a
points: 1 — ,

3 73 a
16fc

W--

373 a
16ft

-)

10. (a)

3- =

1 =

dx

dt '

x>/3 => ^ =
dt

1(8)-V3|
= 12 cm/sec

. 1,-=.

,dx
dt

D =

dD

-{,-.

y^){2x

dx
dt

+ 2v

dy\
dt)

(b)

Vx^ + .V' =

dx^
dt

dt

7t= +

1

.V"

8(12) + 2(1)

^_

98

49

(b) a determines the x-intercept on the right: (a, 0).
ft affects the height.

764 + 4 768 717

cm/sec.

dy dx

^ y ., ^ de ^dt~^dt

(c) tan 0 = ^ => sec- 8 ■ —- = :;

x dt X-

From the trrangle, sec 6

68 „ dd 8(1) -2(12) -16 -4 ,.

^- "^"'^^ ^ = ,/68\ = -68" = "F '"'^'''
64

376 Chapter 2 Differentiation

12. £'W = lim ^^' + ^^ ~ ^^'

A.T->0 Ax

£(x)E(Ax) - Ejx)

= lim

A.t-»0 Ax

,. ^, JE{Ax) - 1
= lim E(x){-^

Ax^O V Ax

^, , ,. £(Ax) - 1
= £(x) hm -^

A.r->0 Ax

But, £'(0) = lim EM_im . ,i^ MM^ = 1.

A.v->0 Ax A.t->0 Ax

Thus, E'{x) = E(x)E'(Q) = E(x) exists for all x.
For example: E{x) = e*.

14. (a) v(t) = - Y' + 27 ft/sec

27
a(t) = — T- ft/sec^

27 27

(b) v(f) = — -t + 27 = 0 => — r = 27 ^> f = 5 seconds

S(5) = -^^(5)- + 27(5) + 6 = 73.5 feet

(c) The acceleration due to gravity on Earth is greater in
magnitude than that on the moon.

CHAPTER 3
Applications of Differentiation

Section 3.1 Extrema on an Interval 378

Section 3.2 Rolle's Theorem and the Mean Value Theorem . 381

Section 33 Increasing and Decreasing Functions and

the First Derivative Test 387

Section 3.4 Concavity and the Second Derivative Test .... 394

Section 3.5 Limits at Infinity 402

Section 3.6 A Summary of Curve Sketching 410

Section 3.7 Optimization Problems 419

Section 3.8 Newton's Method 429

Section 3.9 Differentials 434

Review Exercises 437

Problem Solving 445

CHAPTER 3
Applications of Differentiation

Section 3.1 Extrema on an Interval

Solutions to Even-Numbered Exercises

2. fix) = cos

fix) = ~2^"*Y

/'(O) = 0

/'(2) = 0

fix) = - 3xjx + 1

fix) = -3x
3

^ix + 1)-'/^

+ Jx + l(-3)
ix + l)-i/2[;t + 2ix + 1)]

-fix + l)-'/2(3x + 2)

/t-3 =0

6. Using the limit definition of the derivative,

\x\) - 4

lim ^«^ = lim (i-
X - 0 x->o-

j-»0

X

\x\)

= 1

lim ^«^ = lim (^

x->0+ X — 0 x-*0* X — 0

4

1

/'(O) does not exist, since the one-sided derivatives are
not equal.

8. Critical number: x = 0.
X = 0: neither

10. Critical numbers: ;c = 2, 5
X = 2: neither
X = 5: absolute maximum

12. gix) = x'^ix^ - 4) = ;r* - 4je
g'ix) = 4x3 - 8;c = 4^(^2 _ 2)

Critical numbers: x = Q,x = ± Jl

14. fix)

4x

x^+ 1

^ (;c^ + 1)(4) - (4;c)(2x) ^ 4(1 - ;c^)
^ ^'" ix'- + 1)2 ix^ + 1)2

Critical numbers: x = ±1

16. /(0) = 2 sec 0 + tan e, 0 < e < 277

/'(e) = 2 sec 0 tan 0 -I- sec^ 6
= sec 0(2 tan 0 -1- sec 0)

sec 6

2f^U

1

\C0S 0/ COS 0.

sec2 0(2 sin 0 -H 1)

Ttt IItt

On (0, 2Tr), critical numbers: 0 = —r, 0 = —r-

o 0

378

Section 3.1 Extrema on an Interval 379

18. fix) = ^^, [0, 5]

f'{x) = - => No critical numbers

Left endpoint: (0,-1 Minimum
Right endpoint: (5, 5) Maximum

20. f{x) = x^ + 2;c-4, [-1, 1]
fix) = 2x + 2 = 2U + 1)
Left endpoint: ( - 1 , - 5) Minimum
Right endpoint: (1, - 1) Maximum

22. fix) =x^ - \2x, [0, 4]

fix) = 3^2 - 12 = -iix^ - 4)
Left endpoint: (0,0)
Critical number: (2, - 16) Minimum
Right endpoint: (4, 16) Maximum
Note: X - - 2 is not in the interval.

26. y = 3 - |t- 3|,[-1,5]

From the graph, you see that f = 3 is a critical number.

Left endpoint: (-1,-1) Minimum

Right endpoint: (5, 1)

Critical number: (3, 3) Maximum

24.

gix) = VTx, [-

1,1]

Left endpoint:

— 1, — 1) Minimum

Critical number

(0,0)

Right endpoint:

(1,1) Maximum

28

hit) = ^, [3, 5]

h'(t)=7:^.

Left endpoint: (3, 3) Maximum

Right endpoint: (5,-1 Minimum

30. gix) = secx
g'ix) = secttanjc
Left endpoint:

6' I

E 2.
6' V3

r, 1.1547

Right endpoint: ( t> 2 I Maximum
Critical number: (0,1) Minimum

32. y^x--l- cos;c, [-1.3]
>>' = 2a: — sinjT
Left endpoint: (- 1, - 1.5403)
Right endpoint: (3, 7.99) Maximum
Critical number: (0, - 3) Minimum

34. (a) Minimum: (4, 1)
Maximum: (1,4)

(b) Maximum: (1, 4)

(c) Minimum: (4, 1)

(d) No extrema

36. (a) Minima: (-2. 0) and (2. 0)
Maximum: (0, 2)

(b) Minimum: (—2.0)

(c) Maximum: (0, 2)

(d) Maximum: (l. v^)

380 Chapters Applications of Differentiation

38. fix

2- x\ 1 < jc < 3
2 - 3;c, 3 < x < 5

Left endpoint: (1,1) Maximum
Right endpoint: (5, - 13) Minimum

."•"

\(3. -■')

\(5.-13)

40. fix) = , [0, 2)

2 — X

Left endpoint: (0,1) Minimum

'

(0.1)

42. (a) 3

(b) fix) = 3;cV3 - X, [0, 3]

fix)

Maximum: I 2, -

Minimum:
(0, 0), (3, 0)

x(^)(3-x)-i/2(-l) + (3-;c)>/2(l)

3:

= |(3-;c)-'/2(|)[-x + 2(3-x)]

^ 2(6 - 3x) ^ 6(2 - x) ^ 2(2 - j:)
~ 373 - X ~ 3V3 -X ~ V3 -X

Critical number: x = 2

/(O) = 0 Minimum

/(3) = 0 Minimum

/(2) = !

Maximum: 2

44. /(x) =
/W =
fix) =

/"W =

1

[1-

x^+ 1'

-2x

2

(;c2 + 1)

-2(1 - 3x2)

(x2 + 1)3

24jc - 24x3

(x2 + !)-»
Setting/'" = 0, we have x = 0, ± 1.

if"(l)| ~ 7 is the maximum value.

46. fix)

x^+ 1

[-1,1]

24x - 24x3
/"'W = / 2 -L 114 (^^^ Exercise 44.)

/<%) =

/'5)(x) =

(x^ + D"

24(5x^ - lOx^ + 1)
(x2 + 1)5

-240x(3x^ - lOx^ + 3)

(x2 + 1)6
[/'"•'(O)! = 24 is the maximum value.

48. Let/(x) = 1/x. /is continuous on (0, 1) but does not
have a maximum. /is also continuous on (— 1, 0) but does
not have a minimum. This can occur if one of the
endpoints is an infinite discontinuity.

\

V

Section 3.2 Rolle's Theorem and the Mean Value Theorem 381

50.

52. (a) No
(b) Yes

54. (a) No
(b) Yes

v-sin2e 77 377

dt

IS constant.

(by the Chain Rule)

^ V- cos 20 tie
\6 dt

In the interval [77/4, 3 77/4], d = 77/4, 377/4 indicate
minimums for dx/dt and 0 = 77/2 indicates a maximum
for (ic/iir. This implies that the sprinkler waters longest
when 6 = 77/4 and 377/4. Thus, the lawn farthest from
the spinkler gets the most water.

58. C=2x + ^0^^,l.x<300

C(l) = 300,002
C(300) = 1600

X-
2x2 = 300,000
x2 = 150,000

X = lOOyn == 387 > 300 (outside of interval)
C is minimized when x = 300 units.
Yes, if 1 < jr < 400, then x = 387 would minimize C.

60. fix) = W

The derivative of/ is undefined at every integer and is
zero at any noninteger real number. All real numbers are
critical numbers.

62. True. This is stated in the Extreme Value Theorem.

64. False. Let/(x) = a;^. x = 0 is a critical number of/.

g(x)=f(x-k)

= (x- kV- ■
X = i- is a critical number of g.

Section 3.2 Rolle's Theorem and the Mean Value Theorem

2. Rolle's Theorem does not apply to/(x) = cot(x/2) over
[77, 377] since/is not continuous at x = 277.

4. fix) = x(x - 3)

x-intercepts: (0, 0), (3, 0)

f'(x) -2x-3 = 0atx =

6. /(x) = -3xVx+ 1
x-intercepts: (-1.0), (0, 0)

fix) = -3xi(x + l)-'/2 - 3(x + 1)'/^ = -3(x + l)->/2(| + (x + i:

fix) = -3(x + l)-'/2(|x + l) = Oatx = -|.

382 Chapters Applications of Differentiation

8. fix) = x^ - 5x + 4, [1, 4]
/(l)=/(4) = 0

/is continuous on [1, 4]. /is differentiable on (1, 4).
RoUe's Theorem applies.

fix) = 2x-5

Ix- 5 = Q ^> x = -
c value: —

10. fix) = {x- i){x + 1)2, [- 1, 3] •

/(-l)=/(3) = 0

/is continuous on [— 1, 3]. /is differentiable on (- 1, 3).
Rolle's Theorem applies.

fix) = (x- 3)(2)(;c +\) + {x+ 1)2

= (x+ l)[2x - 6 + x+ \\

= (x+ \){ix - 5)

c value: —

11. fix) = 3 - |;c- 3|,[0, 6]

/(0)=/(6) = 0

/is continuous on [0, 6]. /is not differentiable on (0, 6)
since /'(3) does not exist. Rolle's Theorem does not apply.

16. fix) = cos X, [0, I-it]

/(0)=/(27r) = 1

/is continuous on [0, 2Tr]./is differentiable on (0, 2Tr).
Rolle's Theorem applies.

fix) = — sinj:

c value: tt

r2 -

■,[-1,1]

14. fix) =

/(-i)=/(i) = o

/is not continuous on [— 1, 1] since /(O) does not exist.
Rolle's Theorem does not apply.

18. fix) = cos 2x.,

2

'K'f]

/-

12

^i-S*A?

Rolle's Theorem does not apply.

20.

fix) = secx.

JL 2L
"4'4

/'

-f\

V2

/is continuous on [- Tr/4, Tr/4]./is differentiable on
(— 7r/4, tt/4). Rolle's Theorem applies.

fix) = secjctanjc

sec jc tan X = 0

x = 0

c value: 0

22. fix) =x- x^l^, [0, 1]

/(0)=/(1) = 0

/is continuous on [0, l]./is differentiable on (0, 1).
(Note: /is not differentiable ■Ax = 0.) Rolle's Theorem
applies.

/'W = 1 -

1

7>l/^

0

1 =

33/?
1

27

1 _ 73
27 9

/3
c value: ^ = 0.1925

Section 3.2 Rolle's Theorem and the Mean Value Theorem 383

24. X(x) = --sin-
/(-l)=/(0) = 0

[-1,0]

/is continuous on [- 1, 0]./is differentiable on (- 1, 0).
Rolle's Theorem applies.

1 TT

/(x) = ---cos- = 0

TTX 3

-— arccos— [Value needed in (— 1, 0).]

« -0.5756 radian
c value: —0.5756

28.

J

t'o .;

26. CW= 10(- + ^^
(a) C(3) = C(6) = y

(b)

^'« = iof-^^o^

1

= 0

x^ + 6.x + 9 X-
2x:2 - 6x - 9 = 0
6d

a 08

6 ±673 3 ±373

In the interval (3, 6): c = ^ "^.^ = 4.098.

30./(;c)= |.x-3|, [0,6]

/is not differentiable at x = 3.

32. f{x) = x{x^ — j: — 2) is continuous on [— 1, l] and
differentiable on (— 1, 1).

/(I) -/(-I)

1

l-(-l)

f\x) = 3.t2 - 2.t - 2 = - 1
){x - 1) = 0
1

34. f{x) = (x + l)/.v is continuous on [1/2, 2] and
differentiable on (1/2, 2).

/(2) -/(1/2) (3/2) - 3

2 - (1/2)

3/2

= -1

/V) = -^

XT = 1

c = 1

-1

384 Chapters Applications of Differentiation

36. fix) = j^ is continuous on [0, 1] and differentiable on
(0, 1).

/(l)-/(0) 1 -0

1-0 1

fix) = 3;c2 = 1

V3

= 1

X = ±-

In the interval (0, 1): c

V3

40. fix) = x - 2 sin j: on [- TT, ir]
(a)

tangent^

^y^^'^canl

^

'tangent

(b) Secant line:

,i fi2ILlflzA^IJ:J-A=■^

TT — (— V) 2tT

y — TT = l(x — tt)
y = X

38. fix) = 2 sin X + sin 2x is continuous on [0, it] and dif-
ferentiable on (0, tt).

= 0

/M-/(0)_0-0

TT — 0 TT

fix) - 2 cos X + 2 cos 2j: = 0

2[cosj: + 2cos2j: - 1] = 0

2(2 cos a: - l)(cosx + 1) = 0

1

cos X = —

cos X = —I

IT Sir

In the interval (0, tt): c = — .

(c) fix) = 1 - 2 cos jc = 1
cos X = 0

^ = -2' Airi"'

+ 2

Tangent lines: v -(— -2] = i(j:- —

y = X — 2

y = X + 2

42. /(x) = -jc^ + 4x3 + 8^2 + 5^ (0, 5), (5, 80)
80-5

(b) Secant line: y - 5 = 15(x - 0)

0 = 15x - >> + 5

fix) = -4x3 + 12x2 + 16x

/(5)-/(l)_,,
5-1

-4c3 + 12c2 + 16c = 15

0 = 4c3 - 12c2 - 16c + 15

c = 0.67 or c = 3.79

(c) First tangent line: y - fie) = m(x - c)

y - 9.59 = 15(x - 0.67)

0 = 15x - y - 0.46
Second tangent line: y - fie) = mix — c)

y - 131.35 = 15(x - 3.79)

0 = 15x - >' + 74.5

Section 3.2 Rolle's Theorem and the Mean Value Theorem 385

44. 5(?) = 200( 5
(a)

\ 2 + r/
5(12) - 5(0) _ 200[5 - (9/14)] - 200[5 - (9/2)] _ 450

12-0

(b) S'(t) = 200

12

(2 + tf

450

7

1

1

(2 + tY 28
2 + ? = 277

t = 2V7 - 2 = 3.2915 months
S'{t) is equal to the average value in April.

46. f{a) = f(b) and /'(c) = 0 where c is in the interval (a, ii).

(a) g{x) = fix) + k (b) gix) = fix - k)

gia) = gib) = fia) + k gia + k) = gib + k) = fia)

g 'ix) = fix) =* g '(c) = 0 g Xx) = fix - k)

Interval: [a, b] g 'ic + k) = /'(c) = 0

Critical number of g: c . Interval: [a + A, i? + ^]

Critical number of g: c + k

■fia)

(c) gix) = fikx,

4f ) = ^(f
g'ix) = kf'ikx)

g'{j\ = kf'ic) = 0

Interval:

a b
k' k

Critical number of g:

48. Let Tit) be the temperature of the object. Then 7(0) = 1500° and r(5) = 390°. The average temperature over the
interval [0, 5] is

390 - 1500
5-0

-222°F/hr.

By the Mean Value Theorem, there exists a time to, 0 < t^ < 5, such that Ty^) = —222.

50. fix) = 3 cos^

(a)

-> '"■-^

fix) = 6 cos

TTXW TT

-37rcos^) sin

2 JJ\2

TTX

U>M

\M\ft

nf

jm

(b) /and/' are both continuous on the entire real line.

(c) Since /(- 1) =/(l) = 0, Rolle's Theorem applies on
[- 1, 1]. Since /(I) = 0 and/(2) = 3, Rolle's
Theorem does not apply on [1, 2].

(d) lim fix) = 0

-t— >3

lim fix) = 0

386 Chapters Applications of Differentiation

52. /is not continuous on [—5, 5].
'l/x, x¥^0

Example: f(x) =

0, jc = 0

54. False. /must also be continuous and differentiable on each
interval. Let

56. True

58. Suppose/(.r) is not constant on (a, b). Then there exists Xi and ^Cj in {a, b) such that/(j:|) i= f[x^. Then by the Mean Value
Theorem, there exists c in {a, b) such that

. ^ f{Xo) - fix,)

f'ic) = "^^^ — ^-^-^ ^0. .

This contradicts the fact that/'(jc) = 0 for all x in (a, b).

60. Suppose/U) has two fixed points c, and Cj. Then, by the Mean Value Theorem, there exists c such that
f (c) = = = 1.

This contradicts the fact that/'(j:) < 1 for all x.

62. Let/W = cos x.f'is continuous and differentiable for all real numbers. By the Mean Value Theorem, for any interval [a, b\
there exists c in (a, b) such that

-fie)

fib)-f{a)
b- a

cos b - cos a . '

; = — sm c

b - a

cos b - cos a = (-sin c)ib — a)

\cosb — cos a I = |-sinc||ii - a\

Icos b - cos a\ < \b - a\ since | — sin c\ < 1.

Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 387

Section 3.3 Increasing and Decreasing Functions and the First Derivative Test

2.

y=-{x+l)^

Increasing on: (-

oo,

-1

Decreasing on: (-

-1,

oo)

6

x^

^ x+1

4. f(x) =^-lx-^

Increasing on: {- 1, 0), (1, oo)
Decreasing on: (-oo, - 1), (0, 1)

Critical numbers: x = Q, —2 Discontinuity: jc = — 1

Test intervals:

-oo < X < -2

-2 <x < -\

-1 < X < 0

0 < a: < oo

Sign of /'W:

y' > Q

y' < 0

y' < 0

y' > Q

Conclusion:

Increasing

Decreasing

Decreasing

Increasing

Increasing on (-oo, -2), (0, oo)
Decreasing on (-2, - 1), (- 1, 0)

8. h{x) = 21x- X?

h'{x) = 21 - l>x- = 3(3 - x)(3 + x)

h\x) = 0

Critical numbers: x = ±3

Test intervals:

— oo < j: < —3

-3 <.T < 3

3 < v < CO

Signof;z'(^):

/i' < 0

h' >Q

/i' < 0

Conclusion:

Decreasing

Increasing

Decreasing

Increasing on (—3, 3)
Decreasing on (— oo, —3), (3, oo)

4
10. y = ;c + -

X

, {x - 2){x + 2)

Critical numbers: jc = ±2 Discontinuity: 0

Test intervals:

-oo < X < -2

-2 < .t < 0

0 < .t < 2

2 < .r < oo

Sign of y ':

y' > 0

y' < 0

y' < 0

y' > 0

Conclusion:

Increasing

Decreasing

Decreasing

Increasing

Increasing: (— oo, —2), (2, oo)
Decreasing: (-2,0), (0,2)

388 Chapter 3 Applications of Differentiation

12, f(x) = x^ + &C+ 10
fix) = 2x + 8 = 0
Critical number: x = -4

Test intervals:

-oo < X < -4

-4 < X < oo

Sign of/'U):

/' < 0

/'>0

Conclusion:

Decreasing

Increasing

Increasing on; (— 4, oo)
Decreasing on: (— oo, — 4)
Relative minimum: (—4,-6)

14. fix) = -{x^ + Sx+ 12)
fix) = -2x - 8 = 0
Critical number: x = —4

Test intervals:

-oo < X < -4

-4 < ;c < oo

Sign of /'(x):

/'>0

/'<0

Conclusion:

Increasing

Decreasing

Increasing on: (— oo, — 4)
Decreasing on: (-4, oo)
Relative maximum: (-4,4)

16. f{x) = jc^ - 6*2 + 15

fix) = 3x^ - \2x = 3x(x - 4)
Critical numbers: x = Q,4

Test intervals:

-oo < X < 0

Q < X < 4

4 < a: < oo

Sign of /'(jc):

/'>0

/'<o

/'>0

Conclusion:

Increasing

Decreasing

Increasing

Increasing on (— oo, 0), (4, oo)
Decreasing on (0, 4)
Relative maximum: (0, 15)
Relative minimum: (4, — 17)

18. fix) = ix + mx - 1)
fix) = Ixix + 2)
Critical numbers: x = —2,0

Test intervals:

-oo < x < -2

-2 <;c < 0

0 < ;t < oo

Sign of /'(x):

f'>0

/'<0

/'>0

Conclusion:

Increasing

Decreasing

Increasing

Increasing on: (— oo, —2), (0, oo)
Decreasing on: (—2,0)
Relative maximum: (-2,0)
Relative minimum: (0, - 4)

Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 389

20. f(x) = y - 32r + 4

fix) = Ax^-7>2 = 4(jr3 - 8)

Critical number: x = 2

Test intervals:

-oo < X < 2

2 < X < CO

Sign of/'(x):

/'<o

/'> 0

Conclusion:

Decreasing

Increasing

Increasing on: (2, oo)
Decreasing on: (-oo, 2)
Relative minimum: (2, - 44)

22. f(x) = x^'^ - 4

/'W

±r-'/3

3" 3jc'''3

Critical number: j; = 0

Test intervals:

— oo < JC < 0

0 < JT < oo

Signof/'U):

/'<o

/'>o

Conclusion:

Decreasing

Increasing

Increasing on: (0, oo)
Decreasing on: (- oo, 0)
Relative minimum: (0, - 4)

24. fix) = {x- l)'/3
1

fix)

3{x - 1)V3
Critical number: x = 1

Test intervals:

-oo < A < 1

1 < .v < oo

Sign of/'U):

/'> 0

/'>o

Conclusion:

Increasing

Increasing

Increasing on: (—00,00)
No relative extrema

26. /(x) = |x + 3| - 1

F + 3| [-1, X <
Critical number: j: = — 3

-3
-3

Test intervals:

-00 < X < -3

-3 < .r < 00

Sign of/'(.r):

/'<0

/'>0

Conclusion:

Decreasing

Increasing

Increasing on: (—3, 00)
Decreasing on: (—00, —3)
Relative minimum: (—3, — 1)

•^W%.1

fix) ^^ ^ ^^^'^

- (;c)(l)

^^^^ (x +

1)^

Discontinuity: x =

-1

1

{x + \y

Test intervals:

-00 < x < -\

- 1 < .t < 00

Sign of/'(.r):

/'>0

/'>0

Conclusion:

Increasing

Increasing

Increasing on: (-00, - 1), (— 1, 00)
No relative extrema

390 Chapters Applications of Differentiation

30

/w =

x + 3
x^

x

+

3
x^

fix) =

1
x"

6

-{x +
x'

3.

Critical number: x

=

-6

Discontinuity:

X =

0

Test intervals:

-oo < j: < -6

-6 < X < 0

0 < JC < oo

Sign off'ix):

/'<o

/'>0

/'<0

Conclusion:

Decreasing

Increasing

Decreasing

Increasing on: ( - 6, 0)
Decreasing on: (— oo, —6), (0, oo)

Relative minimum: ( —6, — —

32. /W

/'W =

3;c -4

X- 2
(x - 2)(2x - 3) - jx^ -3x- 4)(1) x^ - 4a: + 10

(^ - 2)2

Discontinuity: x = 2

Increasing on: (— oo, 2), (2, oo)
No relative extrema

(x - 2Y

Test intervals:

-oo < X < 2

2 < X < oo

Sign of/'W:

/'>o

/'>0

Conclusion:

Increasing

Increasing

34. fix) = sin X cos x = - sin 2x, 0 < x < 27r

fix)

cos 2x = 0

„ . . , , 77 377 5t7 777

Critical numbers: x = — , — -, — -, — —

4 4 4 4

Test intervals:

0<x<^
4

77 377

4 4

377 577

4 4

577 777

4 4

-— < X < 277

4

Sign of/'(x):

/'>0

/'<0

/'>0

/'<0

/'>o

Conclusion:

Increasing

Decreasing

Increasing

Decreasing

Increasing

Increasing on: (o, Jj, (^, ^j, (^, 277)

„ . /77 377\ /577 777

Decreasing on: [-^,^)\-J,-^

Relative maxima:

77 1 \ /577 1

4'2/'V4'2

Relative minima:

377 1\ /777 1

2/'\4' 2

Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 391

36. f{x) = ;-, 0 < X < Itt

1 + cos-'j:

.,, , cos x{2 + sinlt)
(1 + cos-x)^

Critical numbers: x = — , — —

2 2

Test intervals:

0<x<^

TT 377

y<x<Y

377

— < j: < 277

Sign of /'W:

/'>o

/'<o

/'>0

Conclusion:

Increasing

Decreasing

Increasing

Increasing on: (0, — ),(—, 277

Decreasing on:

77 377
2' 2

Relative maximum: | -r-. 1

Relative minimum:

377

38. f(x) = 10(5 - Jx'- - 3x + 16), [0, 5]
5(2x - 3)

(a) /'W =

v^

3.V + 16

(b)

(c)

5(2;c - 3)

vv

3,t + 16

Critical number: -v

(d) Intervals:

(»■!)

(I-)

fix) > 0

fix) < 0

Increasing

Decreasing

/is increasing when/' is positive and decreasing
when/' is negative.

40./(a-)=^ + cos ^,[0,477]
(a)/'W=|-|sm^

2ji 3« 4ji

(c)

1

r

sm

—

0

2

2
r

sm

2

1

.V

77

2

0

Critical number: .v = 77

(d) Intervals:

(0, 77) (77. 477)

fix) > 0 f\x) > 0

Increasing Increasing

/is increasing when/' is positive.

392 Chapters Applications of Differentiation

42. /(f) = cos^t - sin^r = -2 sin^f = g{t), -2 < t < 2
fit) = -4 sin f cos t = -2 sin 2f
/symmetric with respect to y-axis

zeros off. ±—

Relative maximum: (0, 1)
Relative minimum: ( — — ,

.i.if.-.

44. f(x) is a line of slope ~ 2 =>/'W = 2.

46. /is a 4" degree polynomial =>/' is a cubic polynomial.

48. /has positive slope

In Exercises 50-54,/'(ar) > 0 on (-oo, -4),/'(x) < 0 on (-4, 6) and/'(x) > 0 on (6, oo).

50. g{x) = 3f(x)-3 .. 52. g(x) = -f{x) 54. g{x) ^ f{x - \0)

. g'(x) = 3/'W .. g'ix) = -fix) g'ix) =f'{x - 10)

g'(-5) = 3/'(-5) >0 g'(0) = -f'(0) > 0 ■ g'(8)=/'(-2) < 0

56. Critical number: x = 5

/'(4) = -2.5 =>/is decreasing atx = 4.

/'(6) = 3 =>/is increasing at j: = 6.
(5,/(5)) is a relative minimum.

58. s{t) = 4.9(sin e)!^

(a) v(r) = 9.8(sin e)r speed = |9.8 (sin e)t\

(b) If 6 = Tr/2, the speed is maximum,

v(f) = 9.8 1.

60. C =

(a)

3f

27 + f3'

? > 0

f

0

0.5

1

1.5

2

2.5

3

C{t)

0

0.055

0.107

0.148

0.171

0.176

0.167

The concentration seems greater near t = 2.5 hours.

(b) 025

The concentration is greatest when t = 2.38 hours.

, , „, _ (27 + f3)(3) - (3f)(3r^)
^^ (27 + f3)2

_ 3(27 - 2f^)
(27 + ;3)2

C = 0 when ; = 3/ 4/2 = 2.38 hours.

By the First Derivative Test, this is a maximum.

Section 3.3 Increasing and Decreasing Functions and the First Derivative Test 393

62. P = IMx

P' = 2.44

X'

20,000

X

5000, 0 < ;c < 35,000

0

10,000
X = 24,400

Increasing when 0 < jc < 24,400 hamburgers.
Decreasing when 24,400 < j: < 35,000 hamburgers.

Test intervals:

0 < JC < 24,400

24,400 < X < 35,000

Sign of/":

P' > 0

P' < Q

M. R= VO.OOIT'' -47+100
0.0047-3 - 4

(a) R' =

IJO.OOIT* - 47-+ 100
7=10°,/? = 8.3666n

= 0

(b)

The minimum resistance is approximately
/?« 8.37a at r= 10°.

66. f(x) = 2 sin(3.i:) + 4 cos(3a:)

The maximum value is approximately 4 All. You could use calculus by finding /'(x) and then observing that the maximum
value of/occurs at a point where /'(jf) = 0. For instance, /'(0. 154) » 0, and/(0.154) = 4.472.

68. (a) Use a cubic polynomial

fix) = a-jX^ + a^x' + a^x + a^.

(b) fix) = 3ayX^ + 2a2X + a^

(0, 0): 0 = ao (/(O) = 0)

0 = a, (/'(0) = 0)

(4, 1000): 1000 = 64^3 + \6a, (/(4) = 1000)

0 = 48a3 + Sa. (/'(4) = 0)

(c) The solution is Qq = a, = 0, ^2

375

125

f/\ -125,, 375,
fix) = —f-x^ + —X'-

(d)

\

-(4, laxi)

A

lO.O) \

394 Chapters Applications of Differentiation

70. (a) Use a fourth degree polynomial /W = a^x^ + a^x^ + a^x'^ + a^x + Aq.
(b) f\x) = 4a^ + 3a^x^ + 2a.^ + fl;

(1, 2): 2 = a^ + a^ + a2 + a^ + Oq

0 = 4a^ + 3a3 + laj + Qj

(— 1,4): 4 = (34 — 03 + Oj — fli + Oq
0 = — 4a4 + Sflj — 2a2 + Oi

(3, 4): 4 = 81^4 + 27^3 + Qqj + 3a, + flo

0 = 108^4 + 27a3 + 6^2 + a,

(/(I) = 2)
(/'(I) = 0)
(/(-I) = 4)

(/'(-i) = o)

(/(3) = 4)
(/'(3) = 0)

(c) The solution is a^ = T, a, = —5, Oj

2' "4

/W = -ix^ + y+ \x^

3 , 23

(d)

(-1.4)

(3.4)
(1.2) \

/

\

72. False

Let h{x) = /UJgU) where/(x) = g(x) = ;c. Then
h{x) = X- is decreasing on (—00, 0).

74. True

If /(x) is an «th-degree polynomial, then the degree of
/'(x)isn - 1.

76. False. . •

The function might not be continuous.

78. Suppose /'(jc) changes from positive to negative at c. Then there exists a and b'ml such that/'W > 0 for all j: in {a, c) and
f'{x) < 0 for all X in (c, b). By Theorem 3.5, /is increasing on (a, c) and decreasing on (c, fc). Therefore, /(c) is a maximum of
/on (a, i>) and thus, a relative maximum of/.

Section 3.4 Concavity and the Second Derivative Test

2. y= -x^ + 3x^ - 2, y" = -6;c + 6
Concave upward: (— oo, 1)
Concave downward: (l,oo)

4. /U) =

2x+ 1'-^ {2a: + 1)3
Concave upward: (-ido, — j)
Concave downward: (— 5, oo)

6.y

^ {-3x^ + 40.j;3 + 135x), y" = -^ x(x - 2)(x + 2)

270

Concave upward: (-00, -2), (0, 2)
Concave downward: (-2, 0), (2, 00)

8. h(x) = ^ - 5x + 2
h'(x) = 5x^-5
h'\x) = 20^3
Concave upward: (0, 00)
Concave downward: (-00, 0)

Section 3.4 Concavity and the Second Derivative Test

395

10. y = X + 2 CSC X, (- 77, ir)
y' = 1 — 2 CSC ;c cot a:
y" = — 2cscx(-csc^j:) - 2 cot a:(-csc jrcot^)

= 2(csc-'jc + cscj:cot^x)
Concave upward: (0, it)
Concave downward: (-tt, 0)

12. f{x) = 2x^ -Zx- - \lx + 5

fix) = (>x^-6x- n
f"(x) =\lx-(i

f"{x) = 12j: - 6 = Owhen.r =

Test interval

-oo < x < \

5 < .r < oo

Sign of /"W

fix) < 0

fix) > 0

Conclusion

Concave downward

Concave upward

Point of inflection: (5. ~t)

14. f(x} = 2x^ - 8x + 3

f'(x) = 8x^-8

fix) = 24x- = Owhenx = 0.

However, (0, 3) is not a point of inflection since /"(jc) > 0 for all x.
Concave upward on (-00, 00)

16. fix) = .r^U - 4)

fix) = x> + 3.r2(x - 4) ■ ■

= x'[x + 3(x - 4)] = 4xHx - 3)
fix) = 4x^ + Mx - 3) = 4x[x + 2ix - 3)] = 12a:{x - 2) = 0

fix) = 12xU - 2) = 0 when jr = 0, 2.

Test interval

-00 < X < 0

0 < X < 2

2 < .V < 00

Sign of /"W

fix) > 0

fix) < 0

fix) > 0

Conclusion

Concave upward

Concave downward

Concave upward

18.

Points of inflection: (0, 0), (2, - 16)

fix) = x^x + 1, Domain: [— 1, 00)

fix) = W^U + 1)-'/^ + V7TT = :^j=

I 2vx + 1

fix)

6VxTT - (3x + l)ix + 1)-'/^ 3x + 4

4(x +1) ~ 4(x + 1)3/2

/"(x) > 0 on the entire domain of/ (except forx = - 1, for which /"(x) is undefined).
There are no points of inflection.

Concave upward on (— 1, 00)

X + 1
20. fix) = — 7=- Domain: jc > 0

fix) =

X- 1

2x^/2

nx) = ^

Point of inflection: 3

Test intervals

0 < -v < 3

3 < v < oc

Sign of /"(.r)

/" > 0

f < 0

Conclusion

Concave upward

Concave dow nw ard

V3

4./3

396 Chapter 3 Applications of Differentiation

3x
22. f(x) = 2 CSC — , 0 < X < 2tt

,,/ s , 3;c 3;c

fix) = -3cscycoty

9/ 3x 3x 3x\

f"(.x) = -| csc^ — + CSC — cot'^ — I T^ 0 for any x in the domain of/.

Concave upward: ( 0. ~^ ). I ~^. 2'7t

Concave downward: I — , — 1
No points of inflection

24. f(x) = sin jc + cos j:, 0 < jc < 27r

fix) = cos-t - sinx
fix) = — sinjc — cosj:

f"{x) = Owhen;c

377 Ttt
4 ' 4 ■

Test interval:

3-n-

3-77 777

77r

-— < X < 277

4

Sign of/"U):

fix) < 0

/"W > 0

/"W < 0

Conclusion:

Concave downward

Concave upward

Concave downward

Points of inflection: ( "Z- ^ )' ( "4^' ^

26. /W = ;c + 2 cos X, [0, 27r]
f'{x) = 1 — 2 sin X
/"(x) = — 2 cos X

/"(x) = Owhenx = ^,^
•' ^ ' 2 2

Test intervals:

0<x<f

77 377

2<^<T

3l7 377

2 <^< 2

Sign of/"(x):

/"< 0

/">o

/"< 0

Conclusion:

Concave downward

Concave upward

Concave downward

Points of inflection:

77 77\ (377 377

2' 2

2 ' 2

28. /(x) = x2 + 3x - 8
/'(x) = 2x + 3
/"(x) = 2
Critical number: x = -5

/"(-!) > 0

Therefore, (-|, -7-) is a relative minimum.

30. /(x)= -(x-5)2
fix) = -2(x - 5)
/"W = -2
Critical number: x = 5

/"(5) < 0
Therefore, (5, 0) is a relative maximum.

Section 3.4 Concavity and the Second Derivative Test 397

32. f(x) =x^ -9x- + llx

fix) = 3;c2 - 18x + 27 = 3U " 3)-

f"{x) = 6(x - 3)

Critical number: x = 3

However, /"(3) = 0, so we must use the First Derivative
Test./'(x) > 0 for all x and, therefore, there are no
relative extrema

34. g{x) = -$x + 2)Hx - 4)2

,, , -(x-4)(x- $ix + 2)

g U) = ^

g"{x) = 3 + 3x - 1^2

Critical numbers: x = —2, 1, 4

g"(-2) = -9 < 0
(- 2, 0) is a relative maximum.

^"(1) = 9/2 > 0
(1, — 10.125) is a relative minimum.

^"(4) = -9 < 0
(4, 0) is a relative maximum.

36. f(x) = Jx- + 1

Vjc- + 1
Critical number: j: = 0

/"W = (^2 + 1)3/2

/"(O) = 1 > 0

Therefore, (0, 1) is a relative minimum.

38. f(x)
fix)

X

X - 1

-1

U - 1)^

There are no critical numbers and x = 1 is not in the
domain. There are no relative extrema.

40. f(x) = 2 sin.r + cos 2v, 0 < .v < 27r

f'(x) = 2 cos ;t - 2 sin It = 2 cos x - 4 sin jt cos x = 2 cos .t:(l - 2 sin x) = 0 when x = —,—, —-, -r-.

o i 0 2

f"(x) = -2 sinx - 4 cos 2x

r(f)>o

/<t) < »
r(f ) > 0

Relalive maxima: (f.5).(^.|)

Relative minima: ( T- 1 ). ( ^. ~ 3

398 Chapters Applications of Differentiation

42. f(x) = x^Ve^^-, [- V6, V6]

3x(4 - x^)
(a) /U) = ,

V6 - AT-

f'(x) = 0 when ;c = 0, a: = ±2.

6(x* - 9x2 + 12)

■^"^""^ (6 - x2)3/2

/"(x) = 0 when x = ±

h - 733

(b) /"(O) > 0 => (0, 0) is a relative minimum.
/"(±2) < 0 =» (±2, 4V2) are relative maxima.
Points of inflection: (±1.2758, 3.4035)

44. fix) = VS sin X, [0, 27r]

(a) /'(x) = x/2x cos X +

Critical numbers: x = 1.84, 4.82

/"(x) = -VSsinx +

cosx cosx smx

V2x v2x 2xV2x
2cosx (4x^ + l)sinx

72x

2x72^

_ 4xcosx — (4x2 + l)sinx
2x>/2x

(b) Relative maximum: (1.84, 1.85)

Relative minimum: (4.82,-3.09)

Points of inflection: (0.75, 0.83), (3.42, -0.72)

(c)

The graph of/ is increasing when/' > 0 and
decreasing when/' < O./is concave upward when
/" > 0 and concave downward when/" < 0.

(c)

/is increasing when/' > 0 and decreasing when
/' < O./is concave upward when/" > 0 and
concave downward when/" < 0.

46. (a)

/' < 0 means/
decreasing

/' decreasing means
concave downward

(b)

/' > 0 means/
increasing

/' decreasing means
concave downward

48. (a) The rate of change of sales is increasing.
S" > 0

(b) The rate of change of sales is decreasing.

S' > 0, S" < 0

(c) The rate of change of sales is constant.

S'= C,S"= 0

(d) Sales are steady.

5= C,S' = 0,5"=0

(e) Sales are declining, but at a lower rate.

S' < 0. 5"> 0

(f) Sales have bottomed out and have started to rise.

S' > 0

50.

Section 3.4 Concavity and the Second Derivative Test 399

52.

54.

56.

(b) Since the depth d is always increasing, there are no

relative extrema./'U) > 0

(c) The rate of change of d is decreasing until you reach
the widest point of the jug, then the rate increases
until you reach the narrowest part of the jug's neck,
then the rate decreases until you reach the top of the
jug-

60. (a) f(x) = Vx

fix) = |x-V3
fix) = -ix-V3
Inflection point: (0, 0)
(b) /"W does not exist at ^ = 0.

62. f(x) = ax^ + bx- + ex + d
Relative maximum: (2, 4)
Relative minimum: (4, 2)
Point of inflection: (3, 3)
f'(x) = 2ax' + 2bx + c,/"(.v) = 6ax + 2b

/(2) = ia + Ab + 2c + d = A
/(4) = 64a + 16Z? + 4c + rf = 2

56a + 12fc + 2c = -2 => 28a + 6fc + c = - 1

/'(2) = \2a + Ab + c = 0. /'(4) = 48a + 8t + c = 0, /"(3) = 18a + 2fo = 0

28a + 6fc + c = - 1 18a + 2fc = 0

12a + 4fc + c = 0 \6a + 2b= -\

\6a + 2b =-\ 2a =1

a = ^, b = — T, c= 12, d = —6

fix)

tX

ix~ + IZt

400 Chapters Applications of Differentiation

(-1000. 60)

(1000.90)
B

(0. 50)

64. (a) lineOA: 3' = -0.06x slope: -0.06

line CB: y = 0.04;t + 50 slope: 0.04

f(x) = ax^ + bx- + CX + d
fix) = 3ax2 + 2bx + c
(- 1000, 60): 60 = (- 1000)3a + (1000)2fc - 1000c + d

-0.06 = (1000)2 3a _ 2000fc + c
(1000, 90): 90 = (1000)^0 + (1000)^^ + 1000c + d

0.04 = (1000)2 3a + 2OOO6 + c
The solution to this system of 4 equations is a = - 1.25 x 10"^ b = 0.000025, c = 0.0275, and rf = 50.
(b) J = - 1.25 X 10-V + 0.000025^2 + o.0275.)t + 50 (c)

66. 5 =

(d) The steepest part of the road is 6% at the point A.
5.75573 8.52ir2 , 0.654r

+ 0.99987, 0 < r < 25

10^ ID'S 10^

(a) The maximum occurs when T == 4° and S « 0.999999.
(b)

(c) 5(20°) - 0.9982

5 10 15 20 25 30

68. C = 2x +

300,000

C = 2 - ^^^ = 0 when x = 100715 « 387

By the First Derivative Test, C is minimized when
X = 387 units.

70.S^^^,t>0

(b) S'it)

13,000f

(65 + f2)2

_ 13,000(65 - 3r2) _
^ ^'' ~ (65 + t^y "

t = 4.65

5 is concave upwards on (0, 4.65), concave
downwards on (4.65, 30).

(c) 5 '(f) > Oforf > 0.

As t increases, the speed increases,
but at a slower rate.

Section 3.4 Concavity and the Second Derivative Test 401

y\

72. /W = 2(sinx + cosx), /(O) = 2

/'W = 2(cosx- sinx), /'(O) = 2

f%x) = 2(-sinA: - cosx), /"(O) = -2

/'iW = 2 + 2(jc - 0) = 2(1 + x)
P/W = 2

P2W = 2 + 2U - 0) + 5(-2)U - 0)2 = 2 + 2x - x2
Pj'U) = 2 - 2x
P^'W = -2

The values of/, P,, Pj- and their first derivatives are equal at x = 0. The values of the second derivatives of/ and P2 are equal
at X = 0. The approximations worsen as you move away from x = 0.

74. /(x) =
Ax) =

X - 1'

-(x+l)
2v^(x - 1)='

/(2) = 72

/'(2) =

372

272

•^ ^""^ 4.r3/2(x - 1)3 • -^ ^-^ 872 16

„,, /-^( 372 V 372 572
P,(x) = 72 + — (x - 2) = — X + — -

Pi Xx) =

372

P,(x) = 72 + ( -^ |(x - 2) +

(-¥>-)- K¥)<—»= = ^ ^ ^u -» ^ ^u - 2,

p ,, . 3^^2372

P2 (x) = ;— + ,^ (x - 2)

P2't^)

4

2372
16

16

The values of/, P,, P; and their first derivatives are equal at x = 2. The values of the second derivatives of/ and P; are equal
at X = 2. The approximations worsen as you move away from x = 2.

76. /(x) = x(x - 6)- =x^ ' 12x- + 36x

/'(x) = 3x- - 24x + 36 = 3(x - 2)(x - 6) = 0

f"(x) = 6x - 24 = 6(x - 4) = 0

Relative extrema: {2, 32) and (6, 0)

Point of inflection (4, 16) is midway between the relative extrema of/.

402 Chapters Applications of Differentiation

78. p(x) = ax^ + bx^ + ex + d

p'{x) = 3ax^ + 2bx + c

p'U) = 6ax + 2b

6ax + 2b = 0

b
■'^'Ta

The sign ofpXx) changes at x = —b/3a. Therefore, {-b/3a, p{—b/3a)) is a point of inflection.

M_/_^UJ^Ucf-AU. '" '^

3a) "V 27aV \9aV \ 3aj 27a^ 3a

Whenp(x) =x^ - 3x^ + 2,a= \,b= -3, c = 0, andd = 2
-(-3)

--T^ + d

Xn =

= 1

■°" 3(1)

2(-3P (-3)(0)

+ 2= -2-0 + 2 = 0

^0 27(1P 3(1)

The point of inflection of p(jc) = x^ — 3jr^ + 2 is (xq, y^) = (1, 0).

80. False. /(x) = 1/x has a discontinuity at x = 0.

82. True

y = sin(to)
Slope: y' = b cos{bx)

-b < y' < b (Assumed > 0)

84. False. For example, let f{x) = {x - 2Y.

Section 3.5 Limits at Infinity

2. fix) =

2x

v^?T2
No vertical asymptotes
Horizontal asymptotes: y = ±2
Matches (c)

4. f(x) = 2 + •

x^+ 1
No vertical asymptotes
Horizontal asymptote: y = 2
Matches (a)

6. fix) =

2x^ - 3x + 5
x'^+ 1

No vertical asymptotes
Horizontal asymptote: y = 2
Matches (e)

8. fix) =

2x^

X + 1

X

10°

10'

102

103

10*

105

106

fix)

1

18.18

198.02

1998.02

19,998

199,998

1,999,998

lim fix) = CO (Limit does not exist.)

Section 3.5 Limits at Infinity 403

10. f[x) =

8x

JjF-1,

X

10'

102

103

10^

W

10«

10^

fix)

8.12

8.001

8

8

8

8

8

lim f{x) = 8

12. f{x) = 4 + -

x^ + 1

X

lO"

10'

102

103

10*

10^

10^

fix)

5

4.03

4.0003

4.0

4.0

4

4

lim f{x) = 4

14. (a) /iW = ■'-^-^ = = 5.r - 3 + -

X X X

lim h(x) = CO (Limit does not exist)

x—*oo

c 3 7

5 - - + —

X X-

(b) h(x)

fix) _ 5x- - 3.r + 7

lim h(x) = 5

(c) /jU)

fix) ^ 5x- - 3x + l ^ 5 _ ^ 7

X^ X^ X x^ .t^

16. (a) lim .? , ^ = 0

(b) lira

3.r3

- 1

3 -

Ix

ix -

- 1

3 -

2;c-

Zx - 1

lim ftW = 0

18. (a) lim , , , . = 0

(b) lim

x^o= 4^2 + 1

5x3/2

(c) lim

x^oo 4.v3/2 + 1 4

5.r3/2

=c4yx + 1

oo (Limit does not exist)

in r 3.r3 + 2 3 1

20. lim -—-, T-^; = - = -

x-^ao 9x3-2x2 + 7 9 3

22. lim 4 + - = 4 + 0 = 4

jr-»cxi V Xj

/l 4\
24. lim -X ^ = - oo (Limit does not exist)

x-.a= \2 X-J

y 1

26. lim , = lim , , .- (for x < 0, x = - v/?)

•t->-°= Vx + (l/x)

-1

404 Chapters Applications of Differentiation

28. lim — ^|^= lim 3 + (1 A) (f„, ^ ^ Owphavp-.

x^-oc Vl + (1/x)

,„ ,. X ~ COS j: ,. / , cos a:
30. hm = lim 1

Jl->co X x-tao V X

=1-0=1
Note:

32. lim cos - = cos 0 = 1

x->oo \X/

cos X
lim = 0 by the Squeeze Theorem since

_1 <- '^0^^ ^ i
a: ~ X ~ X

34. /u:

3x

lim /(jc) = 3
lim /U) = -3

_, ,. 1 ,. tanr

36. lim X tan - = lim = lim

X ->oo X I ->o* t t ->o*

s'mt 1
t cos /

(1)(1) = 1

(Letjf = 1/r.)

Therefore, y = 3 and y
horizontal asymptotes.

= - 3 are both

38. lim {2x - V4x^ + 1 )

lim

X — »oo

{2x - J\x^ + Ij

2x + jAx' + 1
2x+ V4a;2 + 1

= lim

1

JT ^oo 2x + V4;c2 + 1

= 0

40. lim (3;c + J^x^ - ;c) = lim

X— > — oo

(-^^^'•1^^^

/ T

-°o 3;c - V9x^ — X

= lim

= lim

79?

(for JC < 0 we have x = — V? )

- ^x~
1

x^-^l + V9 - (l/jc) 6

42.

j:

IQO

10'

W

103

lo-*

10^

10«

/W

1.0

5.1

50.1

500.1

5000.1

50,000.1

500,000.1

lim

- xjx^ — X x^

+ xVx

1 x'

Limit does not exist.

+ xV^^

lim

x^

X ' -♦oo X'

+ x^fx

Section 3.5 Limits at Infinity 405

44.

X

10"

10'

102

103

10*

10^

106

fix)

2.000

0.348

0.101

0.032

0.010

0.003

0.001

lim — = 0

^-"^ x~Jx

46. jc = 2 is a critical number.
f\x) < Oforx < 2.
fix) > 0 for ;c > 2.
lim f{x) = lim /W = 6

For example, let/U)

0.lU-2)2+ 1

+ 6.

48. (a) The function is even: lim f(x) = 5

j:— >-oo

(b) The function is odd: lim f{x) — —5

50. y

Intercepts: (3, 0), ( 0, -

Symmetry: none

Horizontal asymptote: y = 1 since

lim

X - 2

1

lim

J:-»oo X ~ 2

Discontinuity: x = 2 (Vertical asymptote)

52. y^

2x

9-x^
Intercept: (0, 0)
Symmetry: origin
Horizontal asymptote: v = 0
Vertical asymptote: ;c = ±3

54. y

x^ -9
Intercept: (0,0)
Symmetry: y-axis
Horizontal asymptote: v = 1 since
x"

lim

1 = lim

,x^-9 ' x-Xiox^-9'
Discontinuities: x = ±3 (Vertical asymptotes)
Relative maximum: (0, 0)

J

L

406 Chapters Applications of Differentiation

56. v =

2x2

x^ + 4
Intercept: (0,0)
Symmetry: y-axis
Horizontal asymptote: y = 2
Relative minimum: (0, 0)

58. x^ = 4

Intercepts: none

Symmetry: y-axis

Horizontal asymptote: y = 0 since

4 4

lim —^ = 0 = lim

:->-<x>X x->ooX'

.2-

Discontinuity: x = 0 (Vertical asymptote)

60, y =

2x

1

Intercept: (0, 0)

Symmetry: origin

Horizontal asymptote: y = 0 since

lim

2x

0 = lim

2x

1 -X2

Discontinuities: x = ±1 (Vertical asymptotes)

62, y = 1 +

1

Intercept: (-1,0)

Symmetry: none

Horizontal asymptote: y = 1 since

lim (l +-) = 1 = lim (l +-

JT— >-oo\ x/ j: ->oc \ X

Discontinuity: x = 0 (Vertical asymptote)

64. y = 4n - ^

Intercepts: (±1,0)
Symmetry: y-axis
Horizontal asymptote: y = 4
Vertical asymptote: x = 0

66. y =

Domain: (-oo, -2), (2, oo)

Intercepts: none

Symmetry: origin

Horizontal asymptotes: y = ± 1 since

lim , .

X ->tx. ^x- - 4

1, lim

Vertical asymptotes: x = ±2 (discontinuities)

-1.

L

I I I I I > »

J-L J-4_t

Section 3.5 Limits at Infinity 407

68. fix)

EH^^[^

.:l„

E^r

)

.,, . U^ - \){2x) - ;c^(2r) -2x ^ , ^

/ W = (^,2 _ 1)2 = (^2 _ 1)2 = 0 when.t = 0.

.„. ^ ^ (;c^ - l)-(-2) + lr(2)U^ - l)(2;c) ^ 2(3.t-- + 1)
^ ^ ' (x^ - D" (x^ - 1)3

Since /"(O) < 0, then (0, 0) is a relative maximum. Since /"(jt) ^ 0, nor is it undefined in the domain of/, there are no points
of inflection.

Vertical asymptotes: ,i: = ± 1

Horizontal asymptote: y = 1

70. f(x)

1

1

x- - x - 2 {x + l)(.r - 2)

.,, , -{2x- 1) ^ . 1

/W = (,2 _,_ 2)2 = 0 when. = -.

/"W =

(x^ -x- 2P(-2) + (2t - l)(2)(;c^ -x- 2)(2x - 1)

(a-2 - X-2Y

6(;c^ - ;c + 1)

^

^

{x--x- ly

Since /"(t) < 0, then (5, —5) is a relative maximum. Since /"(x) =?^ 0, nor is it undefined in the domain of/, there are no
points of inflection.

Vertical asymptotes: x = —\,x = 2

Horizontal asymptote: y = Q

2(x5 + 3jc' - \)
fix) = ^-. ,J = 0 when. « 0.5321, -0.6527, -2.8794.

ix- + X + \y
/"(o) < 0

Therefore, (0, 1) is a relative maximum.

f"i-2) > 0
Therefore,

-2,-r

is a relative minimum.

Points of inflection: (0.5321, 0.8440), (-0.6527, 0.4491) and (-2.8794, -0.2931)

Horizontal asymptote: y = 0

(-0.6527,0.4491)

2 (0J321. 0.8440)

(-2.8794.-0.2931)

408 Chapters Applications of Differentiation

74. g(x) =
g'U) =
g'U) =

2x

2
(3;c2 + 1)3/2

-18;c

(3x2 + 1)5/2

No relative extrema. Point of inflection: (0, 0).

T, ■ , 2

Honzontal asymptotes: v = +—i=

No vertical asymptotes

s

^^

76. fix) = liE^ Hole at (0, 4)

fix)

4x cos 2j: — 2 sin 2x

There are an infinite number of relative extrema. In the interval
(-277, 2Tr), you obtain the following.

Relative minima: (±2.25, -0.869), (±5.45, -0.365)

Relative maxima: (±3.87,0.513)

Horizontal asymptote: y = 0 -

No vertical asymptotes

6

(b) fix)

x^ -2^2 + 2
2jc2

J^_2x^ _2_
2^2 2x2 + 2xi

1

X + 1

gix)

(c)

The graph appears as the slant asymptote y = — j-^ + 1-

80. lim 100

V,/V2 ->00

^-Kfe] = >o*-o]-

100%

82. >- =

3.351f2 + 42.46 U - 543.730

f

(a) 5

(b) Yes. lim y = 3.351

Section 3.5 Limits at Infinity 409

S4.S = J^,t>0

(b) Yes. lim 5 = ^ = 100

86. lim 4^ = lim ^Sf "17^1"
Divide p{x) and q{x) by .t".

Case 1: Hn < m: lim^-rr- = lim —

xa= q{x) X-

Case 2: If m = n: lim '—r-r = lim

b„ +

a. +

b„ +

0 +

bo

fffl — 1 yffl

+ 0 + 0 0

+ ■ • • + 0 + 0 b„

= 0

+ 0 + 0 a„

'1 ''o b„ + ■ ■ ■ + 0 + 0 b„

Case 3: If n > m: lim '-rr = lim

fc„ +

fc„ + • • • + 0

A_ + ^

= ±oo.

xT-

88. False. Let y^ = v^TT, then Vi(0) = 1. Thus, v, ' = 1/(2 V.t + l) and v, '(0) = 1/2. Finally, ■

3'/'=-^(7TTF5and.v,'t0)=4.

ht\. p = or- + bx + 1, thenp(O) = l.Thus,p'= lax + bandp'(O) = 5 ^> fc = 5. Finally, p"= laandpT^O) =
Therefore,

/'W =

(-l/Sy + (l/2)x+ 1, .r < 0
.VxTT, ;c > 0

and/(0) = 1,

f(l/2) - (I/4).r, x <<d ^ \

1 / / \ and/'(0) = -, and

I l/(2v^^rT), .V > 0 2

- 1/4 , ;c < 0 _ , 1

and/'tO) = — .

L-i/(4(x+ \y/~), x> 0 4

/"W =
/"(a:) < 0 for all real x, but/(jr) increases without bound.

410 Chapters Applications of Differentiation

Section 3.6 A Summary of Curve Sketching

2. The slope of/ approaches oo as j:— >0 , and approaches
— coasx—^0*. Matches (C)

4. The slope is positive up to approximately x = 1.5.
Matches (B)

6. (a) Xq,X2,x^
(c) Xi

(e) X2,Xj

(b) X2,X3

(d) j:,

8. y

x^ + 1

1 - JC2 (1 - x){x + 1)

(x2 + 1)2 {x^ + 1)2

0 when.x = ±1.

>'''=-^^ = 0when;. = 0,±73.
Horizontal asymptote: y = 0

y

y'

y"

Conclusion

- oo < .r < - 73

-

-

Decreasing, concave down

x=-73

73
4

-

0

Point of inflection

-73 < x < -\

-

+

Decreasing, concave up

x= -\

1
2

0

+

Relative minimum

-1 < a: < 0

+

+

Increasing, concave up

;c = 0

0

+

0

Point of inflection

0 < j: < 1

+

-

Increasing, concave down

;c= 1

1
2

0

-

Relative maximum

1 < a; < 73

-

-

Decreasing, concave down

x=73

73
4

-

0

Point of inflection

73 < X < oo

-

+

Decreasing, concave up

Section 3.6 A Summary of Curve Sketching 411

10. y =

y =

x^

+ 1

x^

- 9

-

IQx

{x^

_9)2

60(

x2 + 3)

U^ - 9)3

0 when x = 0

< 0 when x = 0

Therefore, | 0, —^1 is a relative maximum.

Intercept: (0, --

Vertical asymptotes: j: = ±3
Horizontal asymptote: y = \
Symmetric about y-axis

12. /(,)=^±2^,^2

X X

-2
f'(x) = ^r < 0 when x i- Q.

■' x~

fix) = J ^ 0

Intercept: (-2,0)
Vertical asymptote: x = Q
Horizontal asymptote: y = 1

5
4
3--

H — I — I — !-»-«

14. fix) =x + ^

^. . , 64 (x - 4)(x^ + 4j: + 16) ^ .

/ (;c) = 1 " ~J = ~3 = 0 when x = 4.

192
fix) = ^ > 0 if X ?^ 0.

Therefore, (4. 6) is a relative minimum.
Intercept; (-2 4/4, o)

Vertical asymptote: x = 0
Slant asymptote: y = x

,, ,, , x3 4x

16. / x) = -T— - = X + -^ -

X — 4 X — 4

/'(x) = •''"^•r ~,!?^ = 0 when X = 0. ±2V3

(x- - 4)-

/ (•'^) = "7"^ 7u~ = 0 when x = 0

(x- - 4)3

Intercept: (0. 0)

Relative maximum: (-2v^, -3v^)

Relative minimum: (2^3,3^3)

Inflection point: (0, 0)

Vertical asymptotes: x = ±2

Slant asymptote: y = x

412 Chapters Applications of Differentiation

18. y = = 2x - 1 +

y' = 2-

x-2
3

x-2
2x2-8^ + 5

(x - 2)3

{x - 2Y (x - 2)2

+ 0

= 0 when x =

4+ V6

U-Jl
Relative maximum: I , - 1.8990

Relative minimum:
Intercept: (0, -5/2)

2
4+76

, 7.8990 j

Vertical asymptote: x = 2
Slant asymptote: y = 2jc — 1

fi+v?

.-1.899)
H 1-^

-8 -4

■kzu

-4 /•' ^ 6

20. gU) = xV9 - X Domain: ;t < 9

3(6 - x)

&\x)

2V9 -;c

0 whenj: = 6

^"W = 4(9^ _ ^)3/2 < 0 whenx = 6

Relative maximum: (6, 6V3)
Intercepts: (0, 0), (9, 0)
Concave downward on (- oo, 9)

1 (6. 6V3 )

22. y = xJXfs - x^ Domain: -4 < x < 4

2(8 - x2)

y =

716"

= 0 whenx = ±272

„ 2x{x-^ - 24) ^ . ^

>" = (16 _ ^2)3/2 = 0 when;: = 0

Relative maximum: (272, 8)
Relative minimum: (-272,-8)
Intercepts: (0, 0), (±4, 0)
Symmetric with respect to the origin
Point of inflection: (0, 0)

24. y = 3(x - 1)2/3 _ (^ _ 1)2

2 2 - 2(;c - l)''/3

y' ^ (x - l)'/3 " ^^"^ " ^^ = (^ 1 1)1/3 = 0 whenx = 0,2

(y 'undefined for X = 1)

^"=3(7^^^"^^°^°'^"''^^

Concave downward on (-oo, 1) and (1, oo)

Relative maximum: (0, 2), (2, 2)

Relative minimum: (1,0)

Intercepts: (0, 2), (1, 0), (- 1.280, 0), (3.280, 0)

Section 3.6 A Summary of Curve Sketching 413

26. y = -jU^ -3^ + 2)

y' = —x^-\- 1 =0 when x = ±\
y"= -2x = Owhenx = 0

y

y'

y"

Conclusion

-oo < x < -\

-

+

Decreasing, concave up

-t = -1

4
3

0

+

Relative minimum

-1 < * < 0

+

+

Increasing, concave up

x = Q

"3

+

0

Point of inflection

0 < X < \

+

-

Increasing, concave down

x= 1

0

0

-

Relative maximum

1 < j: < oo

-

-

Decreasing, concave down

28. fix) = iU - D' + 2

fix) = ix- 1)2 = Owhenjc = 1.
fix) = 2(;c - 1) = 0 when a; = 1.

fix)

fix)

fix)

Conclusion

-oo < X < \

+

-

Increasing, concave down

X = 1

2

0

0

Point of inflection

1 < j: < oo

+

+

Increasing, concave up

30. fix) = ix+ \)ix- 2)(;c - 5)

fix) = ix+ l)(.r - 2) + ix + 1)U - 5) + ix - 2)ix - 5)

= 3(x2 - 4;t + 1) = 0 whenjc = 2± VJ.
fix) = 6(.r - 2) = 0 when x = 2.

fix)

fix)

fix)

Conclusion

-oo < .X < 2 - yi

+

-

Increasing, concave down

jc = 2 - 73

673

0

-

Relative maximum

2 - V3 < x: < 2

-

-

Decreasing, concave down

X = 2

0

-

0

Point of inflection

2 < j: < 2 + VI

-

+

Decreasing, concave up

a; = 2 + 73

-673

0

+

Relative minimum

2 + 73 < .T < oo

+

+

Increasing, concave up

(2-^.6v^)

C+v^.-fr/J)

Intercepts: (0, 10), (- 1, 0), (2, 0), (5, 0)

414 Chapters Applications of Differentiation

32. y = 3.t^ - 6x^ + |

>>'= 12x3 - 12x = 12x(x2 - 1) = Owhenx = 0,jc = +1.

/3
y"= 36*2 - 12 = 12(3x2 - 1) = 0 whenx = ±^.

y

y'

y"

Conclusion

-oo < X < -1

-

+

Decreasing, concave up

x= -1

-4/3

0

+

Relative minimum

-l<x<-f

+

+

Increasing, concave up

V3
'- 3

0

+

0

Point of inflection

-f <x<0

+

-

Increasing, concave down

x = 0

5/3

0

-

Relative maximum

0<x<f

-

-

Decreasing, concave down

V3
^- 3

0

-

0

Point of inflection

f <X<1

-

+

Decreasing, concave up

x= 1

-4/3

0

+

Relative minimum

1 < X < oo

+

+

Increasing, concave up

34. /(x) = X* - 8x3 + ig^ _ i6x + 5

fix) = 4x3 - 24x2 + 36x - 16 = 4(x - 4)(x - 1)^ = 0 when x = 1, x = 4.
/"(x) = 12x2 _ 4g^ + 36 = I2(x - 3)(x - 1) = 0 whenx = 3, x = 1.

fix)

fix)

fix)

Conclusion

-oo < X < 1

-

+

Decreasing, concave up

x= 1

0

0

0

Point of inflection

1 < X < 3

-

-

Decreasing, concave down

x= 3

-16

-

0

Point of inflection

3 < X < 4

-

+

Decreasing, concave up

x = 4

-27

0

+

Relative minimum

4 < X < oo

+

+

Increasing, concave up

J 1(0. 5)

^ (I.O) 1(5.0)

(4. -27)

Section 3.6 A Summary of Curve Sketching 415

36. y=(x- ly

y'= 5{x- !)■' = Owhen;c= 1.
y"= 20U- 1)3 = Owhenx= 1.

>-

y'

y"

Conclusion

-oo < .I < 1

+

-

Increasing, concave down

X = 1

0

0

0

Point of inflection

1 < j: < oo

+

+

Increasing, concave up

I

(1.0)

1 2 3

38. y=\x^-6x + 5|

,^2{x- 3)(x^ - 6a: + 5) _ 2(x - 3)(x - 5){x - 1)
^ \x^ - 6x + 5| \(x - 5){x - 1)1

= 0 when jc = 3 and undefined when x = 1, .r = 5.

„ _ 2(x- - 6x + 5) _ Ijx - 5)(.r - 1

\x^ - 6x + 5| |(.r - 5)(;c - 1)

undefined when.x = 1, .r = 5.

12 3 4 5 6

v

y'

y"

Conclusion

-OO < .T < 1

-

+

Decreasing, concave up

x= 1

0

Undefined

Undefined

Relative minimum, point of inflection

1 < .V < 3

+

-

Increasing, concave down

.v = 3

4

0

-

Relative maximum

3 < X < 5

-

-

Decreasing, concave down

x = 5

0

Undefined

Undefined

Relative minimum, point of inflection

5 < j: < oo

+

+

Increasing, concave up

40. y = cos X - — cos 2x, 0 < x < Itt

y' = — sin ;c + sin 2j: = — sin .r(l — 2 cos x) = 0 when x = 0. tt,
y" = — cos j: + 2 cos 2x = -cos x + 2(2 cos-;r - 1)

77 Stt
y 3 "

= 4 cos^ X — cos j: — 2 = 0 when cos .r

1 ± V33

= 0.8431, -0.5931.

Therefore, x = 0.5678 or 5.7154, x = 2.2057 or 4.0775.

T, , • ■ /•^ 3\ /57r 3\

Relative maxima: I T' 7 )> 1 ~T'' T I

Relative minimum: tt.

Inflection points: (0.5678, 0.6323), (2:2057, -0.4449), (5.7154, 0.6323), (4.0775, -0.4449)

416 Chapter 3 Applications of Differentiation

42. y = 2(x - 2) + cot X, 0 < ;c < 77

y = 2 — csc^ X = Q when x = —, — -

4 4

y " = 2 csc^ X cot .T = 0 when x = —

„ , . . /377 377

Relative maximum: ( — , — 5

Relative minimum: I T' ":7 ~ 3

Point of inflection: ( — , 77 — 4 1
Vertical asymptotes: x = Q, tt

44. y = sec^l — 1 - 2 tan

(f)

1, -3 < * < 3

'-^-<f)'"(f)(f)-^-'(f)(f)^»--^

Relative minimum: (2, — 1)

-5-4-3-2-1

(2.-1)

46. g(;c) = ATCOtX, — 277 < JC < 277

sin x cos JC — j:

g'W

sin^j:

p'(0) does not exist. But lim x cot j: = lim = 1.

jr->o ^-»o tan X

Vertical asymptotes: x = ±277, ±77

-"«pK-T-4(-f-»MI''MT-°

Symmetric with respect to y-axis.
Decreasing on (0, 77) and (77, 277)

48. fix) = 5

1

X - 4 X +

h)

/h

X = - 2, 4 vertical asymptote
y = 0 horizontal asymptote

50. fix)

Ax

Jx^ + 15

y = ±4 horizontal asymptotes
(0, 0) point of inflection

52. /"is constant.
/' is linear.
/ is quadratic.

Section 3.6 A Summary of Curve Sketching 417

54.

(any vertical translate of/ will do)

56.

X

A

(any vertical translate of the 3 segments of/ will do)

58. If/'(jc) = 2 in [-5, 5], then/(.r) = 2x + 3 and/(2) = 7 is the least possible value of/(2). If/'(.r) = 4 in [-5, 5], then
f{x) = 4.t + 3 and/(2) = 1 1 is the greatest possible value of /(2).

60. g{.x)

3^ - 5x + 2,

x^ + 1
Vertical asymptote: none
Horizontal asymptote: v = 3

The graph crosses the horizontal asymptote v = 3. If a
function has a vertical asymptote at x = c, the graph
would not cross it since /(c) is undefined.

62. g{x) =

x^ + x-2
x - 1

{x + 2)(jc - 1) _ \x + 2.

if .r ^ 1
[Undefined, if.r = 1

X - 1
The rational function is not reduced to lowest terms.

hole at (1,3)

64. g(x) = ; = 2.V + 2 -

X - 5

X- 5

t

]{

T

The graph appears to approach the slant asymptote

y = 2x + l.

418 Chapters Applications of Differentiation

66. f(x) = tan(sin ttx)
(a) 3_

(c) Periodic with period 2

(e) On (0, 1), the graph of/ is concave downward.

(t>) fi—x) = tan{sm(- ttx)) = tan(-sin irx)

= — tan(sin ttx) = —f{x)

Symmetry with repect to the origin

(d) On (— 1, 1), there is a relative maximum at [\, tan l)
and a relative mmimum at (—5, -tan l).

68. Vertical asymptote: x = —3

Horizontal asymptote: none

x^
^ x + 3

70. Vertical asymptote: x = 0
Slant asymptote: y = —x

y = — X +

1 \ - x^

72. fix) = |M2 - (ax) = ^{ax)(ax - 2), a # 0

f'(x) = a^x — a = a(ax — 1) = 0 when x = —.

a

■ f"{x) = a^ > Oforall;c.

(a) Intercepts: (0,0), (-0

Relative minimimi: (~ ~t
Points of inflection: none

(b)

74. Tangent line at P: y - yQ= f'(xQ){x - x^
(a) Let>' = 0: -Jq = f'{xQ){x - x^)
f'(Xf)x = xJXxq) - .Vo

x = x -^^ = x -^^
° f'ixo) ""^ fix,)

x-mtercept: \Xg

fM

_ jy-^n.

fW

, 0

(c) Normal line: y - yo= ~j;7~\ix " ^0)

Let y = 0: -^q

■M

Z'Uo)

-^o/'W = -x + Xq

x = Xq + yof'ixo) = Xo+ fixo)f'{xo)
;t-intercept: (xq + /(xo)/'^. 0)

(e) \BC\ =

/fa)

fix.)
fix,)

(g) \AB\ = |a:o - fa +/fa)/'fa))| = |/fa)/'fa)|

(b) Letx = 0: >> - yo =/'fa)(~-*o)
3' = >'o - ^o/'fa)
>" = /fa) - -To/'fa)
J- intercept: (0,/fa) - xj'ix,))

(d) Let j: = 0: y - j-q = 7^("-*o)

>' = >'o +

/'fa)

y-intercept: ^0, y, + t^J
m |pc|2-,2 + /'/falU/fa)irfa)i+/(xa)!

iPCp =

[ffa)Vl + \fix,)f

/'fa)
(h) |APp=/fa)V'fa)- + V

\AP\ = |/fa)|Vl + [/'fa)]2

Section 3. 7 Optimization Problems 419

Section 3.7 Optimization Problems

2. Let X and y be two positive numbers such that x + y = S.
P = xy = x(S - x) = Sx- x^

-r- = S - Ix = 0 when x = -.
dx 2

-r^ = - 2 < 0 when Jc = -.
dx^ 2

P is a maximum when x — y = S/1.

4. Let X and y be two positive numbers such that xy = 192.

192
S = x + 2,y = ^- + 7,y

dS

0 when y = 8.

192
dy y-

d^S 384
S is minimum when y = 8 and x = 24.

-rrr = — 5- > 0 when y = 8.

6. Let X and y be two positive numbers such that
j: + 2v = 100.

P = xy = y(100 - 2y) = lOOy - If-
dP
dy

d^P

dy^

P is a maximum when x = 50 and y = 25.

100 - 4y = Owheny = 25.
= -4 < 0 when V = 25.

8. Let -t be the length and y the width of the rectangle.
2;c + 2y = P

P-2x P

A = xy = x\- - xj = —X - x-

dA P P

-— = — — 2r = 0 when x = — .
dx 2 4

d-A ^ ^ , P

—rr = — 2 < 0 when x = —.
dx- 4

A is maximum when x = y = P/4 units. (A square!)

10. Let X be the length and y the width of the rectangle.
xy = A
A

P = lx + 2y = 2x + 2[- j = 2x + —

dP ^ 2A ^ ^ r-

-— = 2 ;- = 0 when x = VA.

d\ AT

d^P AA

> 0 whenx = Va.

dx^ x^

P is minimum when x = y = ^/a centimeters.
(A square!)

420 Chapters Applications of Differentiation

12. fix) = Jx- 8, (2, 0)

From the graph, it is clear that (8, 0) is the closest point on
the graph of/ to (2, 0).

14. fix) = ix+ 1)2, (5, 3)

16. F

dF

dv ''

22 + 0.02v2
22 - 0.02v2

(22 + 0.02v2)2
= 0 when v = yiToO - 33.166.

By the First Derivative Test, the flow rate on the road is
maximized when v = 33 mph.

d= VU-

- 5)2 + [ix + 1)2 - 3]2

= ^jr-

- lOx + 25 + (;c2 + 2x - 2)2

= v^-

- lOx + 25 + y* + 4;c3 - 8x + 4

Vx* + 4^3 + ^2 - 18x + 29

Since d is smallest when the expression inside the radical
is smallest, you need to find the critical numbers of

gix) = ;c^ + 4a^ + x2 - 18;c + 29

g'ix) = Ax^ + 12x2 + 2x- 18

= 2(;c - l)(2x2 + 8;c + 9)

By the First Derivative Test, x = \ yields a minimum.
Hence, (1, 4) is closest to (5, 3).

18. 4jc + 3^ = 200 is the perimeter, (see figure)

: , , /200 - 4;c\ 8 ,^„ „

A = lxy = 2x( ^ 1 = -(50x - x^)

4^ = I (50 - 2a:) = 0 when.t = 25.
dx 3

d^A
dx^

16

< 0 when x = 25.

100

A is a maximum when jc = 25 feet and >" = "3" feet.

20. (a)

(c)

Height, X

Length & Width

Volume

1

24 - 2(1)

1[24 - 2(1 )]2 = 484

2

24 - 2(2)

2[24 - 2(2)]2 = 800

3

24 - 2(3)

3[24 - 2(3)]2 = 972

4

24 - 2(4)

4[24 - 2(4)]2 = 1024

5

24 - 2(5)

5[24 - 2(5)]2 = 980

6

24 - 2(6)

6[24 - 2(6)]2 = 864

dV
dx

(b) V = ;c(24 - 2;c)2, 0 < ;c < 12
(d) 1200

The maximum volume seems to be 1024.

d^V
dx^

d^V
dx^

= 2xi24 - 2jc)(-2) + (24 - 2jc)2 = (24 - 2x)(24 - 6x)

= 12(12 - x)i4 - x) = 0 when a; = 12, 4 (12 is not in the domain).

= 12(2jc - 16)

< 0 when x = 4.

When X = 4,V = 1024 is maximum.

Section 3.7 Optimization Problems 421

22, (a) P = 2x + 2irr

= 2x + 2^1)
= 2x + Try = 200

■f

» r

200 -2x 2
• >> = = —(100 - X)

(b)

Length, x

Width, y

Area, ry

10

-(100 - 10)

77

(10)- (100 - 10) == 573

TT

20

-(100 - 20)

TT

(20)- (100 - 20)= 1019

TT

30

-(100 - 30)

TT

(30)- (100- 30)- 1337

TT

40

-(100-40)

TT

(40)-(100 - 40) - 1528

TT

50

-(100 - 50)

IT

(50)-(100 - 50) - 1592

TT

60

-(100 - 60)

TT

(60)-(100 - 60) = 1528

TT

The maximum area of the rectangle is approximately 1 592 m^
(c) A= xy = x-(100 - ;c) = -(100;c - jc^)

TT TT

(e) 2000

Maximum area is approximately
1591.55 m2 (jc = 50 m).

24. You can see from the figure that A = xy and y =

A = x(^] = |(6;c - x').

6-x

dA 1

dx 2
dx"

(6 - 2j:) = 0 when x = 3.
= - 1 < 0 when x = 3.

1 ; 1 4 5 6

A is a maximum when x = 3 and y = 3/2.

422 Chapters Applications of Differentiation

26. (a) A = - base x height

= ^{2Vir=i5)(4 + h)

= Vl6 - h^{4 + h)

=^ = 4(16 - h'')-"\-2h){A + h) + (16 - h^y^
ah 2

= (16 - h'^Y^'\-h(A + h) + (16 - h^)]

-2(h^ + 2/! - 8) -2{h + 4)ih - 2)

Vl6 - h^

Vl6 - h^

dA

— - = 0 when h = 2, which is a maximum by the First Derivative Test.

ah

Hence, the sides are 2Vl6 — /i^ = 4V3, an equilateral triangle. Area = 12V3 sq. units.
4 + h j4~Th

(b) cos a

V8V4 + h

V8

tan a =

Vie

4 + /i

Area = 2(|j(v'l6 - ^^{4 + h)

- {4 + hYtana
= 64 cos" a tan a
A '{a) = 64[cos'' a sec^ a + 4 cos' (— sin a)tan a] = 0
^ cos" a sec- a = 4 cos' a sin a tan a
1=4 cos a sin a tan a
1 . ,

sin a = — => a = 30° and A = 12 V3.

(c) Equilateral triangle

28. A = 2ry = 2x^r'^ — x^ (see figure)

dA
dx

2{r^ - 2x^)

0 when x

72r
2 ■

By the First Derivative Test, A is maximum when the rectangle has dimensions

V2rby(72r)/2.

U 7:317
\

(-r. 0)

.^/P^]

(r.O)

Section 3. 7 Optimization Problems 423

30. xy = 36 => y = —

36

A = {x + 3)(v + 3) = U + 3)1 — + 3

= 36 + M + 3^ + 9
x

dA -108

Dimensions: 9x9

+ 3 = 0 => 3j:2 = 108 => X = 6, .V = 6

32. V = Trr-^/i = Vq cubic units or /i =

V„

5 = 2wr- + lirrh = 2( Ttr^ + -^

^ = lilirr - -^) = 0 when r= ^^ units.
dr \ r~ I \ 2ir

h =

Vn

Voi^nT-'' 2Vo

1/3

2r

77{ yV2^)2 77^/3 (2^)1/3

By the First Derivative Test, this will yield the minimum surface area.

34. V = Trr^x

x + 2TTr= 108 => X = 108 - 27rr (see figure)

V = 7rr-(108 - 27rr) = 77<108r- - 2m^)

dV

dr

7r(216r — Gvr-) = 6Trr(36 — irr)

= 0 when r = — and x = 36.

i2t/ "Si*

— -r = 77(216 - 127rr) < 0 when r = — .
dr- TT

Volume is maximum when x = 36 inches and r = 36/ tt = 1 1.459 inches.
36. V = TTX^h = 'TTx-\2^/r — jr) = 2itx-Jr~ — x~ (see figure)

f =4"'

i)(r- - x^y-(-2x) + 2a- Vr- - x^

2ttx

K2r2 - 3;c2)

= 0 when x - 0 and x- = -—- => x = — r— ■

3 3

By the First Derivative Test, the volume is a maximum when

X = -^ — and n = — ;=.

3 73

Thus, the maximum volume is
4Trr^

'--{Hih

73/ 373'

UT^^l

U-V^.

424 Chapters Applications of Differentiation

38. No. The volume will change because the shape of the container changes when squeezed.

40. V = 3000 = -Txr' + irr-h

, 3000 4

h = T - T'"

7rr' 3

Let k = cost per square foot of the surface area of the sides, then 2k = cost per square foot of the hemispherical ends.

C = 2k(4Trr^) + kilirrh) = k\ Strr^ + l-rrr

V

3000 4

Trr'

]-[

16 , , 6000
3 r

dr

^TTr - ^^1 = 0 when r = ^/-^ - 5.636 feet and /i = 22.545 feet.
.3 r^ ] V 2iT

By the Second Derivative Test, we have

dr^

32 12,000
^r-T + -, —

> 0 when r = ^i

'1125

277 •

Therefore, these dimensions will produce a minimum cost.

42. (a) Let x be the side of the triangle and y the side of the
square.

(b) Let X be the side of the square and y the side of the pen-
tagon.

A = -(cot^W2 + -|cot-jj>'- where 3x + 4y =20

^,.W5-U;o...^.

A' =

f'-(-J')H)=»

60

4V3 + 9

When x = 0,A = 25, when x = 60/(473 + 9),

A « 10.847, and when x = 20/3, A == 19.245. Area is

maximum when all 20 feet are used on the square.

(c) Let X be the side of the pentagon and y the side of the
hexagon.

A = jfcot ^|;i:2 + (cot ^\y^ where 5x + 6y = 20

= 5U^U,2(^)

20 - 5xV

'4' = |(cot|)j: + 3

^(-1

6

20 - 5x

0< x<A.

= 0

A = tIcoI^W^ + -lcot^Mwhere4;c + 5>' = 20

4 \2

= x^ + 1.72047741 4 - -^j , 0 < x < 5.

A'=2x- 2.75276384(4 - ^.x] = 0

X = 2.62

When.1 = 0,A = 27.528, when.x = 2.62, A = 13.102,
and when x = 5.A ~ 25. Area is maximum when all 20
feet are used on the pentagon.

(d) Let X be the side of the hexagon and r the radius of the
circle.

'?)^

A = -I cot — 1x2 + nr~ where 6x + 2'nr = 20
3v^ , ^ AO 3x\2 10

A'=3v^-6fi«-^) = 0

\ 77 77/

x = 1.748

;c = 2.0475

When ;c = 0, A = 28.868, when x - 2.0475,

/4 « 14.091, and when ;c = 4, -4 = 27.528. Area

is maximum when all 20 feet are used on the hexagon

When j; = 0,-4 = 31.831, when;c = 1.748, A = 15.138,
and when x = 10/3, A = 28.868. Area is maximum when
all 20 feet are used on the circle.

In general, using all of the wire for the figure with more
sides will enclose the most area.

Section 3.7 Optimization Problems 425

44. Let A be the amount of the power line.

A = h-y + Ijx- + /

dy

j^r+

= 0 when y

V3-

d^A 2;c2 ^^ X

T^ = T^; ;tt7; > 0 tor V = — ?=.

dy- {x- + ffl^ ■' 73

The amount of power line is minimum when y = j:/ 73 .

(-;t, 0) (.t. 0)

46. fix) = |;c2 ^W = i^;c^ - \x^ on [0, 4]

(a) 9

-^

(b) rfU) = f(x) - g(x) = 5X^ - [j^x* - \x^) = x^- YiX'^
d'{x) = 2x-\x^ = Q=^?,x = T'

=>;c = 0,272 (in [0,4])

The maximum distance is d = 4 when x = 2 V2.

(c) fix) = X, Tangent line at [ijl, 4) is

y - 4 = 272(;c - 2V2)
y = 2V2X - 4.
g'{x) = \ic' - X, Tangent line at (2V2, o) is
y-Q = (i(2v^)3 - 2ji){x - 272)

y = ijlx - 8.
The tangent lines are parallel and 4 vertical units apart.

(d) The tangent lines will be parallel. \i dix) = f(x) — g{x),
then d'{x) = 0 = f'(x) — g'(x) implies that/'(x) = g'(x)
at the point x where the distance is maximum.

48. Let F be the illumination at point P which is x imits from source 1.

0 when

F =

X-

U-,

id

-xT-

dF
dx

-2W,
x'

+

2U2
id - xy

/^..

X

r' id- xy

i/K~ d-x
id - x) Vfi = X l/T,

dVT,=x{i/T,+ VQ
di/T,

l/T,+ k/L

d^F 6W,

2
6W0

dx^ x^ id - xY
This is the minimum point.

> 0 when x

dl/T,

l/T,+ l/T.

5o.(a)r=-^^ + i^

dT

(b) :7r =

—CONTINUED—

dx 2jx~ + 4 4
.t \_

v^?Tl~ 2

Z\~ = .r= + 4
x^ = 4
x = 2

7T2) = V2 + ^ hours

- - = 0

426 Chapters Applications of Differentiation

50.

— CONTDWE

(c) r =

D—

^2

Jx^ + A

A

dT_
dx

X

1

= 0

V, ^X^ + 4 V2

X

V2

Vx2 + 4

sine =

V2

T

d depends on — only.

52.

Jx^ + d
^1

ll+V^r

V2

^)^

dT

X

+

X -

a

dx v,Vx2 + rfi2 VjVrf,^ + (a - x)-
Since

= 0

Jx^ + rfi^
we have

sin S] and , . ^ , , = - sin e.

Jdi + {a- xY

sin 01 sin St „ sin 6, sin 0-,
= 0 => = -.

V, V, V, V,

Since

d^T _ d{^

d^

dx^ V,(;c2 + rf,2)3/2 vld.2 + (^ _ ^)2]3/;

this condition yields a minimum time.

-> 0

(d) Cost = Jx^ + 4 C, + (3 - x)C2

7FT4 (3 - x)

(1/C,)
From above, sin 9 =

(I/C2)
1/Ci _ Q

i/Q c,

54. CW = Ikjx^ + 4 + A:(4 - x)

C'W =

Vx2 + 4

A: = 0

2x = V^^ + 4

4;c2 = jc2 + 4

3x2 = 4

2
73

Or, use Exercise 50(d): sin 0 = — = -

e = 30°

Thus, X

73-

56. V = ];TTr^h = ^nr^JXAA - r^
3 3

= |Jr2[|](144 - r2)-'/2(-2r) + 2rVl44 - rA

By the First Derivative Test, V is maximum when r = 4^6 and h = 4V3.

Area of circle: A = 7r(12)2 = 14477

Lateral surface area of cone: S = 'n{A^)J{A^Y + (473)^ = 48>/6i7

Area of sector: 14477 - 48^677 = - Or^ = 128

14477 - 48n/677 277/ f-:\ , , ^,

e = -—^ — = -—(3 - V6) " 1.153 radians or 66°

72 3 ^ '^

Section 3.7 Optimization Problems 427

58. Let d be the amount deposited in the bank, i be the interest rate paid by the bank, and P be the profit.
P = (0.12)rf - id
d = ki^ (since d is proportional to i^)

p = (o.i2)(w2) - i(ki^) = kio.m^ - (3)

^ = A:(0.24i - 2i^) = 0 when / = ^ = 0.08.
di 3

d^P
di

T^ = k(0.24 - di) < 0 when / = 0.08 (Note: k > 0).

The profit is a maximum when i = 8%.

60. P = -—s^ + 6s' + 400

dP 2 3

(a) — = -— J= + I2s = -— j(5 - 40) = 0 when a: = 0, i = 40.
ds 10 10

d-P ^ 3
di^ ~ 5

d^-P

s + n

_, , (0) > 0 =^ J = 0 yields a minimum.

d-P

-p;-(40) < 0 => i = 40 yields a maximum.

ds-

The maximum profit occurs when s = 40, which corresponds to $40,000 {P = $3,600,000).

(b)~= -Is +12 = 0 when s = 20.

ds^ 5 .

The point of diminishing returns occurs when 5 = 20, which corresonds to $20,000 being spent on advertising.

62. 5, = |4m - 1| + \5m - 6| + |10m - 3|

Using a graphing utility, you can see that the minimum occurs when m = 0.3.
Line ^ = 0.3^:

^2 = |4(0.3) - 1| + |5(0.3) - 6| + 1 10(0.3) - 3| = 4.7 mi.

428 Chapter 3 Applications of Differentiation

64. (a) Label the figure so that r^ = x^ + h^.

Then, the area A is 8 times the area of the region
given by OPQR:

h

y^^N

V

■wryiWv": : V

x :;

if

V-

%

V ii

:ii^ ^

^ ii

m y

= 8 he- + (jc - h)h

= ^\(r^ - x") + [x- >/;^^^)vr2rrpj

= &xVr^ - x^ + 4x2 - 4r2

A '(x) = 8vV— 12 -

8^2

8^2

7^

+ 8x = 0

Vr^

= 8x + 8Vr2 - x^

x'^ = xVr^ - ^2 + (r2 - x2)
2^2 - /^ = xVr^ - x~
4jc^ - 4xV + A^ = x^(r^ - x^)
5x^ - 5x^r^ + r^ = 0 Quadratic in x^.

Sf- ± 725/ - 20/^ _ r^t r^-i

10 - inP ± ^5J-

lO"-

Take positive value.

/TTT!

= 0.85065r Critical number

-2 v2

(c) Note that x^ = — (s + Vs) and r^ - x^ = — (s - Vs).

a u fi X

Co) Note that sin - = - and cos - = -. The area A of the

2 r 2 r

cross equals the sum of two large rectangles minus
the common square in the middle.

A = 2(2x){2^) - 4^12 = %xh - 4h^

= 8/^ sin - cos - — 4r^ sin^ -
2 2 2

A(x) = 8x7^2 - x2 + 4x2 _ 4^

= 8[^(5 + 75)^(5 - 75)]"' + 4^(5 + 75) - 4;-

= iro^^'^]

'''2 2 r-

+ 2r^ + -75r2 _ 4^

= 1^75 - 2.2 + ^^

= 2r2

5^--f

2.2(75 - 1)

= 4r2 sin 0 - sin2

A'(d) = 4r2(cos e - sin -cos ~ ) = 0

a a

COS 0 = sin - cos - = 2 sin 0

2 2

tane = 2

e = arctan(2) » 1.10715 or 63.4°

Using the angle approach, note that tan 0 = 2, sin 0 = — ^ and sin2| - 1 = — (1 - cos 6) = -\\

J.

75

Thus, A(e) = 4.21 sin 0 - sin2:

V75 A* V5

^ 4^^ - 1) , ,^(^ _ ^)

Section 3.8 Newton's Method 429

Section 3.8 Newton's Method

2. /U) = 2x2-3
fix) = 4x

x, = l

n

^n

/uj

/K)

X ^^'"^
" f'ix„)

1

1

-1

4

I

4

5
4

2

0.125

5.0

0.025

1.225

4. /(x) = tanjc
fix) = sec-j:
jci = 0.1

n

x„

/u„)

/'UJ

" /'UJ

I

0.1000

0.1003

1.0101

0.0993

0.0007

2

0.0007

0.0007

1.0000

0.0007

0.0000

6. /W = x5 + ;c - 1
fix) = 5.^+\

Approximation of the zero of/ is 0.755.

n

■l^n

/UJ

/'UJ

" fiXn)

1

0.5000

-0.4688

1.3125

-0.3571

0.8571

2

0.8571

0.3196

3.6983

0.0864

0.7707

3

0.7707

0.0426

2.7641

0.0154

0.7553

4

0.7553

0.0011

2.6272

0,0004

0.7549

8. fix) = x- 2jx + 1
/'W = 1 ^

Vx+ 1
Approximation of the zero of/ is 4.8284.

n

^.

/(^J

/'(xj

/'UJ

1

5

0.1010

0.5918

0.1707

4.8293

2

4.8293

0.0005

0.5858

.00085

4.8284

10. fix) =1-2x3
/'W = -&r2
Approximation of the zero of/ is 0.7937.

n

x„

/U„)

/'UJ

n-\)

fK)

" /'UJ

1

1

-1

-6

0.1667

0.8333

2

0.8333

-0.1573

-4.1663

0.0378

0.7955

3

0.7955

-0.0068

-3.7969

0.0018

0.7937

4

0.7937

0.0000

-3.7798

0.0000

0.7937

430 Chapters Applications of Differentiation

12. fix) = ^x'

3x

fix) = 2jc3 - 3

Approximation of the zero of/ is -0.8937.

Approximation of the zero of/ is 2.0720.

n

x„

/U„)

nx„)

fix,)
f'(x„)

X -^^^"^
" f'ixj

1

-1

0.5

-5

-0.1

-0.9

2

-0.9

0.0281

-4.458

-0.0063

-0.8937

3

-0.8937

0.0001

-4.4276

0.0000

-0.8937

n

J^.

fixj

fK)

fix,)
f'ixJ

X ^^'"^
" fix,)

1

2

-1

13

-0.0769

2.0769

2

2.0769

0.0725

14.9175

0.0049

2.0720

3

2.0720

-0.0003

14.7910

0.0000

2.0720

14. f{x) = x^ — cos X
fix) = 3x^ + slnjc
Approximation of the zero of/ is 0.866.

n

Xn

fix,)

fix,)

fix,)
fix,)

X ^^^"^
" fix,)

1

0.9000

0.1074

3.2133

0.0334

0.8666

2

0.8666

0.0034

3.0151

0.001 1

0.8655

3

0.8655

0.0001

3.0087

0.0000

0.8655

16. h{x)=f{x)-g{x) = 3-x-

x^+ 1

h'ix)

■1 +

Ix

(x^ + 1)2

Point of intersection of the graphs of /and g occurs
whenx« 2.893.

n

x.

hix,)

h'ix,)

hix,)
h'ix J

X ^^^"^
" h'ix,)

1

2.9000

-0.0063

-0.9345

0.0067

2.8933

2

2.8933

0.0000

-0.9341

0.0000

2.8933

18. hix) = fix) - gix) = x^
h'ix) = 2x + sin;ic

cosx

One point of intersection of the graphs of/ and g occurs
when X ~ 0.824. Since /U) = x^ and gix) = cos x are
both symmetric with respect to the y-axis, the other point
of intersection occurs when x ~ — 0.824.

n

X,

hix,)

h'ix,)

hix,)
h'ix,)

X '"^"""'^
""" h'ix,)

1

0.8000

-0.0567

2311 A

-0.0245

0.8245

2

0.8245

0.0009

2.3832

0.0004

0.8241

20. fix)=x"-a = 0
fix) = nx"-^

X,." - xn + a _ in - l)x," + a

Section 3.8 Newton's Method 431

_xl±J_

i

1

2

3

4

Xi

2.0000

2.2500

2.2361

2.2361

75 = 2.236

24. j:,,,=-^

2x:,3 + 15

3.r,'

I

1

2

3

4

.r,

2.5000

2.4667

2.4662

2.4662

Vl5 = 2.466

26. /(jc) = tan ;c
/'W = sec^jc
Approximation of the zero: 3. 142

n

x„

fiXn)

n\)

f'ixj

1

3.0000

-0.1425

1.0203

-0.1397

3.1397

2

3.1397

-0.0019

1.0000

-0.0019

3.1416

3

3.1416

0.0000

1.0000

0.0000

3.1416

28. y = 4x^-\2x^+l2x-3= f{x)
y'= 12x- - 24x + 12 = /'(x)

3

/'Uj) = 0; therefore, the method fails.

n

■^^

/UJ

/'UJ

/'UJ

" /'UJ

1

3

2

3

2

3

1

2

1

2

1

1

0

—

—

30. f{x) = 2 sin ;c + cos 2x
fix) = 2 cos X - 2 sin 2x

377

Fails because/'(.x,) = 0.

377

2

/UJ

/'UJ

32. Newton's Method could fail if f'{c) = 0, or if the initial value x^ is far from c.

34. Let g{x) = /U) - X = cot .r - jc
g'ix) = -csc^x - 1.
The fixed point is approximately 0.86.

n

x„

8{x„)

g'ix„)

H.vJ
^'U„)

, ^UJ

" ^'UJ

1

1.0000

-0.3579

-2.4123

0.1484

0.8516

2

0.8516

0.0240

-2.7668

-0.0087

0.8603

3

0.8603

0.0001

-2.7403

0.0000

0.8603

36. f(x) = sin X, f'(x) = cos x
(a)

2

(b) .t, = 1.8

x. = X, - ^r4 = 6.086

/Ui)

(c) A-, = 3

X, = X, - ^7^ - 3.143
/U,)

(d)

2-

; \ (1.8. 0.974)

1 ■

>OL (3,0.141)

. /'^Vy...^,_^l6.086. 0)

/

■'f'/V^

-1-

(3.143. 01 V-^

-2-

\

The .r-intercepts correspond to the values resulting
from the first iteration of Newton's Method.

(e) If the Initial guess .v, is not "close to" the desired zero
of the function, the .v-intercept of the tangent line may
approximate another zero of the function.

432 Chapter 3 Applications of Differentiation

38. (s0x„,,=x„{2-2x„)

(h) x„^i =xll- llxj

(■

1

2

3

4

^,

0.3000

0.3300

0.3333

0.3333

(

1

2

3

4

^i

0.1000

0.0900

0.0909

0.0909

\ = 0.333

40. f(x) = ;c sin jt, (0, it)

f'(x) = x cos X + sin jc = 0

Letting F{x) = /'(x), we can use Newton's Method as follows.
\F'{x) = 2 cos JC — a: sin x\

n = 0.091

n

^n

F{x^)

nxr)

X ^^'"^
" F'{x„)

1

2.0000

0.0770

-2.6509

-0.0290

2.0290

2

2.0290

-0.0007

-2.7044

0.0002

2.0288

Approximation to the critical number: 2.029

42. y=f{x)=x\ (4,-3)

d= J{x -AY + {y + 3)2 = V(x - 4)2 + (jc^ + 3)2 = Jx!^ + 7x^ - 8x + 25

rf is minimum when D = x^ + Ix^ — 8x + 25 is minimum.

g{x) = D'= 4x3 + 14a: - 8
g'ix) = 12x2 + 14

n

J^.

gixj

^'UJ

H^„)

X ^(^^^

1

0.5000

-0.5000

17.0000

-0.0294

0.5294

2

0.5294

0.0051

17.3632

0.0003

0.5291

3

0.5291

-0.0001

17.3594

0.0000

0.5291

X = 0.529
Point closest to (4, - 3) is approximately (0.529, 0.280).

(4, -3).

44. Maximize: C

C' =

3f2 + t
50 + t^

-2t* - 2f3 + 300r+ 50

= 0

(50 + r3)2
Let/(x) = 3/^ + 2«3 - 300; - 50
fix) = 12r3 + 6f2 - 300.

Since /(4) = -354 and/(5) = 575, the solution is in the mterval (4, 5).
Approximation: t = 4.486 hours

n

x„

fix J

f'ixj

f'ixj

X ^^'"^

" f'W)

1

4.5000

12.4375

915.0000

0.0136

4.4864

2

4.4864

0.0658

904.3822

0.0001

4.4863

Section 3.8 Newton's Method 433

46. 170 = O.SO&c^ - n.974x^ + 71.248;c + 110.843, 1 < jc < 5
Let/W = 0.808;c3 - 17.974jc2 + 71.248;c - 59.157
fix) = 2.424x2 _ 35,948;( + 71.248.

From the graph, choose x^ = I and Jt, = 3.5. Apply Newton's Method.

n

x„

/UJ

/'UJ

no

" /'UJ

1

1.0000

-5.0750

37.7240

-0.1345

1.1345

2

1.1345

-0.2805

33.5849

-0.0084

1.1429

3

1.1429

0.0006

33.3293

0.0000

1.1429

n

Xn

/UJ

f'U„)

fUn)

X -^^'"^

1

3.5000

4.6725

-24.8760

-0.1878

3.6878

2

3.6878

-0.3286

-28.3550

0.01 16

3.6762

3

3.6762

-0.0009

-28.1450

0.0000

3.6762

The zeros occur when x = 1. 1429 and x = 3.6762. These approximately correspond to engine speeds of 1 143 rev/min and
3676 rev/min.

48. True

50. True

52. f{x) = J A - x^ sin(jc - 2)
' Domain: [-2, 2]
x = -1 and x = 2 are both zeros.
fix) = J A - x- cosU - 2) -
Let ;c, = - 1.

V4 - X-

= sin(jr — 2)

n

Xn

/(.vj

/'UJ

" /'UJ

1

-1.0000

-0.2444

-1.7962

0.1361

-1.1361

2

-1.1361

-0.0090

-1.6498

0.0055

-1.1416

3

-1.1416

0.0000

- 1.6422

0,0000

-1.1416

Zeros: x = ±2,x= -1.142

434 Chapters Applications of Differentiation

Section 3.9 Differentials

2. fix) = - = 6x-^

fix) = - \lx-^

■12

Tangent line at 2

3 -12

-f (. - 2)

-" 2 2

X

1.9

1.99

2

2.01

2.1

f{x) =

6

1.6620

1.5151

1.5

1.4851

1.3605

T{x) =

3

-f

1.65

1.515

1

1.5

1.485

1.35

4. fix) = ^x
1

fix)

lj~x

Tangent line at (2, Jl):
y - fi2) = f'i2)ix - 2)

1

72

2^

U-2)

^ + 1

272 v^

x

1.9

1.99

2

2.01

2.1

f{x) =

-Ji

1.3784

1.4107

1.4142

1.4177

1.4491

T{x)-

->.^

1

72

1.3789

1.4107

1.4142

1.4177

1.4496

6. fix) = CSC ;c

fix) = -cscjTCOtj:

Tangent line at (2, esc 2): y - /(2) = /'(2)(;c - 2)

y — CSC 2 = (—esc 2 cot 2)(x - 2)

>; = (-CSC 2 cot 2)(.x; - 2) + CSC 2

x

1.9

1.99

2

2.01

2.1

fix) = cscx

1.0567

1.0948

1,0998

1.1049

1.1585

r(;c) = (-csc2cot2)(x-

- 2) + esc 2

1.0494

1.0947

1.0998

1.1048

1.1501

8. y = fix) = 1 - 2x\f'ix) = -Ax,x = Q,Lx = dx =

Ay = fix + Ax) - fix)

= /(-0.1)-/(0)

= [1 - 2(-0.1)2] - [1 - 2(0)2] = _o.02

-0.1

dy=f'ix)dx
= /'(0)(-0.1)
= (0)(-0.1) = 0

10. y=fix) = 2x+ \J'ix) = 2,x = 2,Ax = dx = 0.01

Ay = fix + Ax)- fix) dy = fix) dx
= /(2.01)-/(2) =/'(2)(0.01)

= [2(2.01) 4- 1] - [2(2) + 1] = 0.02 = 2(0.01) = 0.02

Section 3.9 Differentials 435

12. y = 3jc2/'

dy = Ix'^'^dx = -jj^dx

14. y = V9^^

tfy = -{9 - jc2)-'/2(-2x)<ic = Z^^t
2 V9 -;c2

16. y = v^ +

v^

-^ 'iv^ IxJ'x)''^ IxJ'x'^

18. y = j: sin a:

dy = U cos ;c + sin j:) (it

20. y

^^+ 1

rU^ + 1)2 sec^jftanx - sec^j:(2j:)1 ,

^^ = [ x^f-^. J^

[2 sec-j:(jr^tan j: + tan x — x)1 ,
U^TTp J^

22. (a) /(1. 9) =/(2 - 0.1) -/(2) +/'(2)(-0.1)

» 1 +(-l)(-0.1)= 1.1
(b) /(2.04) =/(2 + 0.04) =/(2) +/'(2)(0.04)

= 1 + (-1)(0.04) = 0.96

24. (a) /(1.9) =/(2 - 0.1) »/(2) +/'(2)(-0.1)
« 1 + O(-O.l) = 1
(b) /(2.04) = /(2 + 0.04) « /(2) + /'(2)(0.04) ,
= 1 + 0(0.04) = 1

28. (a) g(2.93) = ^(3 - 0.07) - g(3) + g'(3)(-0.07)

« 8 + 5(-0.07) = 7.65
(b) g(3.1) = g(3 + 0.1) = g(3) + g'(3)(0.1) ^
= 8 + 5(0.1) = 8.5

26. (a) 5(2.93) = ^(3 - 0.07) « g(3) + 5'(3)(-0.07)

- 8 + (3)(-0.07) = 7.79
(b) g(3.1) = g(3 + 0.1) = g(3) + g'(3)(0.1)
- 8 + (3)(0.1) = 8.3

30. A = \bh, i) = 36, /! = 50

db = dh = ±0.25
oW = ^fc d/z + U rfi

^A^dA = 3(36)(±0.25) + 3(50)(±0.25)
= ±10.75 square centimeters

32. j: = 1 2 inches

l^x = dx = ±0.03 inch

(a) V = x^

dV= 3x'dx = 3(12)2(±0.03)
= ± 12.96 cubic inches

(b) 5 = 6x2

dS= \2xdx= 12(12)(±0.03)
= ±4.32 square inches

34. (a) C = 56 centimeters

AC = rfC = ± 1 .2 centimeters
C

C = 2n-r :

A = Trr* = M::

77/ 47r

dA = -^CrfC = -^(56)(±1.2) = —

277 277 77

dA

33.6/77

A [1/(477)1(56)-

=^ 0.042857 = 4.2857%

dA (\/27T)CdC 2dC
^^T- (1/477)C= --^^0.03

^ , 0^ = 0.015 = 1.5%

436 Chapter 3 Applications of Differentiation

36. P = (500.x - x^) - i^-x^ - llx + 3000], x changes from 1 15 to 120

dP = (500 -2x- X + ll)dx = (577 - 'ix) dx = [577 - 3(115)](120 - 115) = 1160
Approximate percentage change: —(100) = (100) = 2.7%

38. V = f 777^, r = 100 cm, dr = 0.2 cm

AV = dV = Airr^dr = 4-77(100)2(0.2) = SOOOTrcm^

40. E = IR

R

^ _

E
I

dR = -~dl

R

-{E/ndi

dl

E/I

I

dR
R

=

dl
J

=

dl
I

42. See Exercise 41.

A = |(base)(height) = |(9.5cot e)(9.5) = 45.125 cot 6
dA = -45.125 csc^erffii

csc^ e de de

cot 6 sin 6 cos 6

^ 025^

(sin 26.75°)(cos 26.75°)

^ 0.0044

(sin 0.4669)(cos 0.4669)

= 0.0109 = 1.09% (in radians)

44. /z = 50 tan 9

9 = 71.5° = 1.2479 radians
dh = 50 sec^ d • dd

50 sec2(1.2479)

50tan(1.2479)
9.9316

■de

2.9886

■de

< 0.06

< 0.06

\de\ < 0.018

46. Let/(;c) = ^,x = 21,dx= - 1
fix + A;c) - fix) + /'(x)dc = ^ +

3^

etc

^^ =- ^7 + 3-i^(- 1) = 3 - ^ - 2.9630
Using a calculator, ,^26 = 2.9625

48. Let fix) =x^,x = 3,dx= -0.01.

fix + Ax) '-fix) ■+f'ix) dx = x' + 3x^dx
fix + Ax) = (2.99)3 = 33 + 3(3)2(-0.01) = 27 - 0.27 = 26.73
Using a calculator: (2.99)^ = 26.7309

Review Exercises for Chapter 3 437

50. Let/(jt) = tanjc,;c = 0, ^ = 0.05,/'(x) = sec^x
Then

/(0.05)»/(0)+/'(0)<it

tan 0.05 «= tan 0 + sec^ 0(0.05) = 0 + 1(0.05).

52. Propagated error = f{x + Aa:) - f(x).

relative error =

, and the percent error

X 100.

54.True,^ = ^ = a
Ax dx

56. False

Let/(j;) = -/x, X = \, and Ax = aLc = 3. Then
^y =f(x + Ax) -f(x) =/(4) -/(I) = 1
and

..=/W^ = ^(3) = f.
Thus, dy > Ay in this example.

Review Exercises for Chapter 3

2. (a) /(4) = -/(-4) =
(c)

-3

At least six critical numbers on (— 6, 6).

^■fi^)-^^,^^^^-\

fix) = X

1

+ (x2 + l)-'/2

(b)/(-3)= -/(3)= -(-4) = 4

(d) Yes. Since/ (-2) = -/(2) = -(-1) = 1 and

/(I) = — /(— 1) = —2, the Mean Value says that there
exists at least one value c in (—2. 1) such that

f\c)

/(l)-/(-2) -2-1

1 - (-2)

1 +2

= -1.

(e) No, lim/(x) exists because /is continuous at (0, 0).

j:-»0

(0 Yes, /is differentiable at x = 2.

6. No. /is not differentiable at .t = 2.

(x2 + 1)3/2

No critical numbers ■ -

Left endpoint: (0, 0) Minimum
Right endpoint: (2,2/^5) Maximum

8. No; the function is discontinuous at .r = 0 which is in the interval [—2, 1].

10. /(x) = - 1 < X < 4

fix) = --,

f(b)-f(a) _{\/A)- 1 _ -3/4
b - a 4-1 3

12.

f(x) = Vx - 2r. 0 < X < 4

"4

fix) =

2J-.-'

m

b -

-/(«)
- a

-6-0
4-0

/'(c) = -i
c

1 1

no = ^

c = 2

c = 1

438 Chapter 3 Applications of Differentiation

14. fix) = lx^-2,x+ \

fix) = Ax-3

f(b)-f(a) _2\ - 1 _
b - a 4-0

f\c) = 4c - 3 = 5

c = 2 = Midpoint of [0,4]

16. g(x) = (x + 1)3
g'U) = 3U + 1)2
Critical number: x = —\

Interval

-oo < x < -\

-\ < X < CO

Signofg'U)

g'(x) > 0

g'{x) > 0

Conclusion

Increasing

Increasing

18. f(x) = sinj: + cosjc, 0 < ;c < 2ir
/'U) = cos j: — sin j:

Critical numbers: x = —,x

4

5lT

4

Interval

0<x<^

4

TT Sir
4 4

— < X < 27r

Sign of fix)

fix) > 0

/'W < 0

fix) > 0

Conclusion

Increasing

Decreasing

Increasing

20. gix) = ^sm(f-l], [0,4]

^'U) = |(f cos(f -1

2 2

= 0 when x = 1 + -, 3 + —

TT IT

Relative maximum: (1 H — , -

2 3
Relative minimum: I 3 H — . — -

TT 2

Test Interval

0 < X < 1 +-

77

l+-<x<3+-

3 +- < X < 4

TT

Signofg'W

g'W > 0

g'ix) < 0

g'ix) > 0

Conclusion

Increasing

Decreasing

Increasing

22. (a) y = /I sin(Vfc7^f) + Bcos{Vk/mt)

y' = A ^/kjm cosy Jkjmt) - B^/kJmsiny^/kJmt)

s'm^k/mt A i r—r- \ A

••tan(VfeAwr) = -.

= 0 when

Therefore,

zosjkjmt B

sin( Vfc/m t) =
cosy Vk/mt) =

B

JA^ + B^'

When V = y ' = 0,
>1

y = A

vM^+S2

VA^ + 52

VA2 + 52

(b) Period:

27r

Frequency:

1

1

iTT/Vk/m 277

/k/m

Review Exercises for Chapter 3 439

24. fix) = (x + 2)Hx - 4) = x3 - lit - 16
fix) = 3;c2 - 12

fix) = 6x = Owhen^ = 0.
Point of inflection: (0, - 16)

Test Interval

-OO < AT < 0

0 < a: < OO

Sign of /"(or)

f"(x) < 0

fix) > 0

Conclusion

Concave downward

Concave upward

26. h(t) = t - 4jt+l Domain: [-l,oo)

9

h'{t)=l

h"(t)

Vr + 1

0 => f = 3

1

(/ + 1)3/2
1

/i"(3) = o > 0 (3, —5) is a relative minimum.

28.

12 3 4 5 6 7

30. C=(£)..(f),

x^ ^ 2

32. (a) S = -0.1222f3 + 1.3655f- - 0.9052r + 4.8429

fb) 35

(c) S'{t) = 0 when t = 3.7. This is a maximum by the
First Derivative Test.

(d) No, because the t^ coefficient term is negative.

Iv 2/v

34. hm - ., , ^ = lim . _^ ^ , , = 0

36. lim , .," = lim , , =

ar->oo Vx^ + 4 Jr-.oo Vl + 4/.t-

= 3

38. g{x)

x= + 2

lim

5.t=

= lim

.t ^oc X^ + 2 ;t -.OO 1 + (2/x2)

Horizontal asymptote: y = 5

= 5

40. f(x)

3x

lim . ., = hm , ., — =

^-»°o VAT + 2 ■'-►OO Vx- + 2/vj:

= lim

^-^0° Vl + (2/.v=)

= 3

lim

3j:

j:-»-oo ^x^ + 2 x-«-oc

3x/x

3

= lim — ^

X-----V1 +(2/.r^)

= -3

Horizontal asymptotes: v = ±3

440 Chapters Applications of Differentiation

42. fix) = \x^ - 3.r2 + 2x| = \x{x - l)ix - 2)|
Relative minima: (0, 0), (1, 0), (2, 0)
Relative maxima: (1.577, 0.38), (0.423, 0.38)

44. g{x) = — — 4 cos X + cos 2x

Relative minima: {Itrk, 0.29) where k is any integer.
Relative maxima: {(2k - 1)tt, 8.29) where k is any integer.

46. fix) = 4x^ - xf = x^i'i - x)

Domain: (—oo, oo); Range: (— oo, 27)
fix) = 12.x:2 - 4x^ = 4;c2(3 - x) = 0 when x = 0, 3.
f'U) = 2Ax - \2x- = 12t(2 - ;c) = 0 when.x: = 0, 2.
/"(3) < 0

Therefore, (3, 27) is a relative maximum.
Points of inflection: (0, 0), (2, 16)
Intercepts: (0, 0), (4, 0)

48. fix) = (x2 - 4)2 ■

Domain: (-oo, oo); Range: [0, oo)
fix) = Axix- - 4) = 0 when a- = 0, ±2.
fix) = 4(3.t- - 4) = 0 when;c = ±^.

/"(O) < 0
Therefore, (0, 16) is a relative maximum.

/"(±2) > 0
Therefore, (±2, 0) are relative minima.

Points of inflection: (±273/3, 64/9)
Intercepts: (-2, 0), (0, 16), (2, 0)
Symmetry with respect to y-axis

50. fix) = ix- 3)(;c + 2)3

Domain: (-cx), oo); Range: [-^^6^,oo)
fix) = ix- 3)(3)(;c + 2)2 + ix + If

= (4a; - l)ix + 2)2 = 0 when;c = -2,\.
fix) = iAx - l)i2)ix + 2) + (x + 2)2(4)

= 6(2;c - \)ix + 2) = 0when;c = -2,\.

Therefore,

16,875

■2se~) is a relative minimum.

Points of inflection: (-2, 0), (5,
Intercepts: (-2, 0), (0, -24), (3, 0)

m5\

16)

Review Exercises for Chapter 3 441

52. f(x) = (x- ly^^x + 1)2/3

Graph of Exercise 39 translated 2 units to the right (x replaces by j: - 2).
(— 1, 0) is a relative maximum,
(l, - l/i) is a relative minimum.
(2, 0) is a point of inflection.
Intercepts: (-1,0), (2,0)

54. fix) =

2x

1 +x'-

Domain: (—oo, oo); Range: [—1,1]

.,, , 2(1 - x)il + x) ^ . _^,

/ (x) = ,. . -^.^ = 0 when ;c = ± 1.

(1 + x-y

f"(x) = ~,^^^ ~.f •* = 0 when x = 0, + ^.

(1 + x-y

/"(I) < 0
Therefore. (1, 1) is a relative maximum.

/"(-I) > 0
Therefore, (— 1, — 1) is a relative minimum.

Points of inflection: (-^3, -^3/2), (0, 0), (73, 73/2)

Intercept: (0, 0)

Symmetric with respect to the origin

Horizontal asymptote: y = 0

56. f(x) =

Domain: (— oo, oo); Range:

"■{

(1 + X*){2x) - xMx>) ^x{l - x)(l + x){\ + X-)
/ W = — (TTTP = (TTl^p = 0 when.v = 0. ±1.

,,. ^ (1 + x^yq - 10:c^) -{2x- 2r^)(2)(l + .x^^Ax^) 2(1 - 12v^ + 3.t^) ^ .

/"(±1) < 0

Therefore, I ± 1, - 1 are relative maxima

/"(O) > 0
Therefore. (0, 0) is a relative minimum.

J 6 ± 733

Points of inflection

Intercept: (0,0)
Symmetric to the >'-axis
Horizontal asymptote: v = 0

.(.7^,o.4(.^

733

0.40

-I 1 f— NtH 1 (->

442 Chapters Applications of Differentiation

58. fix) = .r2 + - = ^^^^

X X

Domain: (-00, 0), (0, oo); Range: (-00,00)
/ U) = 2x - ^ = ^; = 0 when a; = ^.
/"W = 2 + ^ = ^^^^^^ = 0 when;c = - 1.

Airi^ > »

/ 1 3 \
Therefore, — ;=, —p^ is a relative minimum.

Point of inflection: (-1,0)
Intercept: (-1,0)
Vertical asymptote: x = 0

60. /(jc) = |;c- 1| + |;c- 3|

Domain: (-00,00)
Range: [2, 00)
Intercept: (0,4)

y

4 1[ (0, 4)

/

3-
2-
1-

v

J

/

—

h-

1

— H —

2

— 1 —
3

1 ^^

4

62. f(x) = —(2 sin ttj: - sin l-nx)

Domain: [- 1, 1]; Range:

-373 373

277 ' 277

f\x) = 2(cos TTx - cos Itm) = -2(2 cos im + l)(cos Trt - 1) = 0

Critical Numbers: x = +-, 0

f"(x) = 2it(— sin TTx + 2 sin 277x) = 277 sin 77x(— 1 + 4 cos •nx) = 0 whenx = 0, ±1, ±0.420.

2 -373^

By the First Derivative Test:

3' 277

2 3V3

is a relative minimum.

is a relative maximum.

v3' 277

Points of inflection: (-0.420, -0.462), (0.420, 0.462), (±1, 0), (0, 0)
Intercepts: (- 1, 0), (0, 0), (1, 0)
Symmetric with respect to the origin

64. f{x) = x", n is a positive integer.
(a)/'(x) = /j^-'

The function has a relative minimum at (0, 0) when n is even,
(b) f"{x) = nin- \)x"-^

The function has a point of inflection at (0, 0) when n is odd and n > 3.

Review Exercises for Chapter 3 443

f

66. Ellipse: frr + fz = 1' >" = tV144 - jc^
144 16 3

A = (It)! |Vl44 - X- 1 = ^;cVl44 - x'^

dA ^41" ^

dx 3LVl44 -a:-

_ 41" 144 - 2jc^
~ 3Lvi44-x-

= + yi44

^]

= Owhenx= 772 = 672.

The dimensions of the rectangle are Ix = 12^2 by y = -Vl44 — 72 = 4^2.

68. We have points (0, y), [x, 0), and (4, 5). Thus,

y- 5 5-0 5x

m = 7 = or V

0-4 A- x ■ X - 4
5x \2

Let/U) = l2 = ;c2 +

X - 4

fix) = 2;c + 50|

X- 4

X - 4 — X

. (x - 4Y

] = „

100^ n
X — -, -;t = 0

U - 4)3
■:i(x - 4)3 - 100] = Owhen.r = Oor^: = 4 + 3/lOO.

L =

V^

25.r=

(x - 4)2 X- 4

J{x - 4Y- + 25

yiOO + 4

yioo

7100-/3 + 25 = 12.7 feet

70. Label triangle with vertices (0, 0), (a, 0), and {b, c). The equations of the sides of the triangle are y = (c/b)x and
y = [c/(b — a)]ix — a). Let (x, 0) be a vertex of the inscribed rectangle. The coordinates of the upper left vertex are
(x, {c/b)x). The y-coordinate of the upper right vertex of the rectangle is {c/b)x. Solving for the .t-coordinate x of the
rectangle's upper right vertex, you get

c

7

— a
(b — a)x = bijc — a

b- a

-(x-a)

X =

a-b
X + a = a ; — X.

Finally, the lower right vertex is
Width of rectangle: a — x — x

(b.c)

^l.x-a)

(0.0) (.1,0) / (aO)

(o-ilf*..0)

Height of rectangle: -rx (see figure)
b

A = (Width)(Height) = [a- ^-j^x - ;cj(^.v] = (a - ^xj^x

dA I a \c I c

-Ix^V-'bTb^^^'

a\ ac lac „ , b

- r = T^x = 0 when .r = —

bl b kr 2

Al) = {" - \%%% = (f)(f) = r^ = %'''] = |(Area of triangle)

444 Chapters Applications of Differentiation

72. You can form a right triangle with vertices (0, y), (0, 0), and (x, 0). Choosing a point {a, b) on the hypotenuse
(assuming the triangle is in the first quadrant), the slope is

b b-0

■y

-bx

0 — a a — X

Let fix) = L^ = x^ + y^ = x^ +

fix) = 2x + 21

a — x

-bx \2

a - X

-bx
a — X

-ab 1

a - xy\

2x[{a - xy + ab^] . , . ^ .r-r^

— ^=^ -^-: = 0 when x = 0,a + </ab^.

(a - xy

Choosing the nonzero value, we have y = b + l/a^b.

L = J{a + i/^Y + {b + i/^y

= (a^ + 3a''''3&2/3 + 3^2/3^4/3 + ijiyr-
= (a2/3 + ^2/3)3/2 meters

74. Using Exercise 73 as a guide we have L, = a esc 9 and Lj = ^ sec d. Then dL/dd — — a esc 0 col 0 + b sec Oiaa 0 = Q when

. ^r-rr „ 7^2/3 + ^2/3 7^2/3 + ^2/3
tan 6 = </a/b, sec 6 = 1-775 , esc 6 = —^ and

^1/3

L — L^ + L2 = a CSC d + b sec 6 = a
This matches the result of Exercise 72.

,1/3

(a2/3 + j,2/3)l/2 (a2/3 + ^2/3)1/2

a'/3 + ^ /,l/3

(^2/3 + ^2/3)3/2^

76. Total cost = (Cost per hour)(Number of hours)

'2 ^/110\ llv 825

50 V

^^^•^«/Vv

dT^ 11^_ 825 ^ 11 v^ - 41,250
dv ~ 50 v2 ~ 50v2

= 0 when v = V3750 = 25 V^ « 61.2 mph.
d'^T 1650

dv^

0 when v = 25 V6 so this value yields a minimum.

78. fix) =x^ + lx+ \

From the graph, you can see that/(x) has one real zero.

fix) = 3^2 + 2
/changes sign in [— 1, 0].

n

Xn

/UJ

/'UJ

fi^n)

1

-0.5000

-0.1250

2.7500

-0.0455

-0.4545

2

-0.4545

-0.0029

2.6197

-0.0011

-0.4534

On the interval [-1,0]: ;c == -0.453.

Problem Solving for Chapter 3

445

80. Find the zeros of/(jc) = smirx + x — \.

f'(x) = TTCOS ITX + 1

From the graph you can see that/U) has three real zeros.

n

K

/U„)

/'UJ

/'UJ

X ^^'"^

1

0.2000

-0.2122

3.5416

-0.0599

0.2599

2

0.2599

-0.0113

3.1513

-0.0036

0.2635

3

0.2635

0.0000

3.1253

0.0000

0.2635

n

■\

/UJ

/'U„)

f'UJ

" /V„)

1

1.0000

0.0000

-2.1416

0.0000

1.0000

n

-"^n

fix J

/'UJ

nx„)

" fix J

1

1.8000

0.2122

3.5416

0.0599

1.7401

2

1.7401

0.0113

3.1513

0.0036

1.7365

3

1.7365

0.0000

3.1253

0.0000

1.7365

The three real zeros of/(.r) are a: = 0.264, x = 1, and x ^ 1.737.

82, y = V36 - x-

dx 2^ 736 -:<^

afy =

v'^

=5 (it

84. p = 75 - jx

4

Ap = ;7(8) - p(7)

^5-«Ul75

[Ap = o'p because p is linear]

Problem Solving for Chapter 3

2. (a) dV = 3x2a[x: = 3x2Ax

AV = (x + Ax)3 - ;c^ = 3.t2AA: + 3.r(A.t)= + (Ax)'

AV- dV = 3x{Axy + (AxY = [3xAx + (Ax)2]Ax

^ ^ ^

e

= eAjt, where e — > 0 as Ax — ) 0.

Av
(b) Let fi = -^ - /'(x). Then e^O as Ax->0.

Furthermore, Ay — dy = Ay - f'ix)dx = eAx.

446

Chapter 3 Applications of Differentiation

4. Let h{x) — g{x) — fix), which is continuous on [a, b] and
differentiable on [a, b). h(a) = 0 and h{b) = gib) - fib).

By the Mean Value Theorem, there exists c in ia, b)
such that

h'ic)

hib) - hja) gib) - fib)

b — a b — a '

Since h '(c) = g '(c) — /'(c) > 0 and b - a > 0,
gib) -fib) >0 ^ gib) >fib).

3a

One point of inflection.

6. (a) /' = lax + b,f" = 2a i= 0. No points of inflection.

(b) /' = 3ax^ - 2bx + c,f" =6ax + 2i = 0=> x =

(c) y' = kyiL — y) = kLy — IqP-

y" = kLy' - 2kyy' = ky'iL - 2y)

If y = —, then y" = 0 and this is a point of inflection because of the analysis below.

++++++

.V": 1-

d = JWn?, sin d = -.
a

Let A be the amount of illumination at one

of the comers, as indicated in the figure. Then

kl . „ klx

(13^ + x")

sin 0 ■

(132 + ;c2)3/2

(x2 + 169)3/2(1) - .r I I(x2 + 169)'/2(2x) "^
A'W = kl ^^^^, = 0

=^ (x^ + 169)3/2 = 3xHx^ + 169)'/2
x^ + 169 = 3x^
2x2 = 169

13

72

9.19 feet

By the First Derivative Test, this is a maximum.

Problem Solving for Chapter 3 447

10. Let T be the intersection of PQ and RS. Let MN be the perf>endicular to SQ and PR passing through T.
Let TM = X and TN = b - X.

SN MR
b - X X

SN = - — -MR

X

b - X X X

SQ = ^^^{MR + PM) = ^^^d

AW = Area = i^. + {(' " 'd)ib - .) = ^^d[x + ^' ' ""^'^

A'U)=^d

'x(4x - 2b) - [Ix- - 2bx + tr)'

[ x^ J

2x2 _ 2bx + b-

A \x) = 0 ^ 4^2 - Ixb = 2x- -2bx + Ir

2x^ = IP-
b

X = — 7=

v/2

Hence, we have SQ = d = 7—= — '-d = I v 2 -

X b/j2

\)d.

Using the Second Derivative Test, this is a minimum. There is no maximum.

S N Q

12. (a) Let M > 0 be given. Take N = ^/m. Then whenever x > N = Jm,
you have

f(x) = x~ > M.

(b) Let 8 > 0 be given. Let M
you have

Then whenever x > M

e .r'

1

(c) Let E > 0 be given. There exists N > 0 such that |/(.r) - L\ < e whenever .v > N.

Let 6 = — Let x = -.

N. y

If 0 < V < S = — , then - < — => .v > Af and

N X N

[fix) - L\

/(;)-^

448 Chapters Applications of Differentiation

14. Distance = J^^Tl? + 7(4 - xY + 4^ = f(x)

X 4 - X

f'(x)

0

74^ + x^ V(4 - x)2 + 42
xV(4 - x)2 + 42 = U - 4)742 + ;c2
x2[16 - 8a: + ^ + 16] = (x2 - fo + 16)(16 + ;c2)

32x2 _ 8^3 + j;" = x'' - 8;c3 + 32;c2 - 128x + 256
128x = 256
x = 2
The bug should head towards the midpoint of the opposite side.
Without Calculus; Imagine opening up the cube:

The shortest distance is the line PQ, passing through the midpoint.

16. (a) s =

v^^lOOO-^
hr \ km

3600

sec
hr

18

V

20

40

60

80

100

s

5.56

11.11

16.67

22.22

27.78

d

5.1

13.7

27.2

44.2

66.4

d{t} = 0.071^2 + 0.389i + 0.727
(c) 1°

T = -(O.O7I52 + 0.3895 + 0.727) + —
s s

The minimum is attained when 5 = 9.365 m/sec.

(b) The distance between the back of the first vehicle
and the front of the second vehicle is d{t), the safe
stopping distance. The first vehicle passes the given
point in 5.5/ s seconds, and the second vehicle takes
d(s)/s more seconds. Hence,

^ d{s) 5.5
s s

(d)

T(s) = 0.0715 + 0.389 +

6.227

T'{s) = 0.071

6.227

, ^ 6.227
■ ■^ 0.071

• 5 ~ 9.365 m/sec

r(9.365) « 1.719 seconds

9.365 m/sec • ^^ = 3.37 km/hr

(e) 49.365) = 10.597 m

18. (a)

X

0

0.5

1

2

1

1.2247

1.4142

1.7321

71 +x

f-

1

1.25

1.5

2

(h) Let f(x} = 71 + X. Using the Mean Value Theorem on the interval [0, x],
there exists c,0 < c < x, satisfying

1 /U)-/(0) _ 7m^- 1

271 + c X - 0 x

Ac) =

Thus 71 + X

— , + 1 < — 1-1 (because 7l + c > 1).

271 + c 2

CHAPTER 4
Integration

Section 4.1 Antiderivatives and Indefinite Integration 450

Section 4.2 Area 456

Section 4.3 Riemann Sums and Definite Integrals 462

Section 4.4 The Fundamental Theorem of Calculus 466

Section 4.5 Integration by Substitution 472

Section 4.6 Numerical Integration 479

Review Exercises 483

Problem Solving 488

CHAPTER 4

Integration

Section 4.1 Antiderivatives and Indefinite Integration

Solutions to Even-Numbered Exercises

dx\ X XT

dx\ 37^ J dx\3

x^ - 1
= tA/2 _ ^-3/2 = ± i

4. '^'

r= -nG^ C

Check: — [ttO + C] = 77

8.? = 2;c-3
ax

;>' = — — + C = — r + c

-2

Check:

+ C

= 2x-

Given

10.

Rewrite

J^^ : ]■

X ^cbc

Integrate

-1

+ C

Simplify

12. L(a:2 + 3) ^ I (jc3 + 3x) die j + sfyj + C jx'' + ^x^ + C

"•J(3^ i if

-^dx

I/JC"

9V-1

+ C

4 2

9x

+ C

/'

16. (5 - x)^ = 5x - y + C

Check: ^5^-Y + C'=5-a:

/

18. (4x3 + 6^2 _ 1)^ = ^4 + 2x3 - X + C

Check: -f [x" + 2x3 - x + C] = 4x3 +6x2-1
ax

/<

20. I (x3 - 4x + 2)rfx = — - 2x2 + 2x + C

Check: 4"

dx

x^

— - 2x2 + 2x + C

4

= x3 - 4x + 2

22./

^^ " ^) ^ = /(^"^ " r'') "^-Iti^ iQ ^ ^ - ¥'' - ^"' ^ ^

Check: -^f |x3/2 + x'/2 + c) = x'/^ + ix-'/2 = v^ + -^
dx\Z I 2 2Vx

450

Section 4. 1 Antiderivatives and Indefinite Integration 451

i. Uv^ + i)dx= I (.

24. I (i/x^ + l)dx= I (jc^/" + 1) dx = V/t +X+ C
Check: ^f^^^'''' + ;c + c) = .x^/" + 1 = V? + 1

26.

1^- = /

;c-''dx = — + C

Check: "rl "A + c) = ^

2x'

+ C

28. r'"^y — -dx = 1^-2 + 2x-3 - 3x-'')d:x 30. j (2/2 - 1)^ dt = j (4r' - 4?^ + 1) rfr

-1

X~' 2x~2 "Ir-J

-1 -2 -3

J + - + C

Check:

dx

X x'^ jc"

^ x-2 + 2x'^ - 3x-'*
x^ + 2x-2

X*

4. 4

f3 + ; + C

Check:

^(1,5 _ 1,3

dt\5 3

r^ + ; + c = 4r* - 4/2 + 1

(2/2 - 1)2

32. 1(1 + 3f)r2d; = j(;2

+ 3/3) rfr = |/3 + I'' + C

Check: ^(j/' + I/-* + C) = /2 + 3/' = (1 + 3/)/2

34. 3 A = 3/ + C

Check: —(3/ + C) = 3
dt

36. I (f2 - sin /) dt = -/^ + cos / + C

Check: — -/^ + cos / + C = /2 - sin /
dt\3 J

38.

(e- + sec2 e) (ie = -e^ + tan e + c

Check: — (-e^ + tan 6 + c] = 6- + sec2 e

40. sec ^(tan y — sec y) dy = (sec y tan y — sec2 y) dy

= sec y - tan y + C

Check: -;-(sec y — tan y + C) = sec y tan v — sec2y
dy

= sec y(tan y - sec y)

42. f-^^?iViv= f^iT= ff-^y^ix

J 1 — cos-x J sin-.x J \sin.x/Vsin.x/

= esc X cot X d.x = - CSC X + C

_, , li r ^T 1 cos X

Check: -H -esc x + CI = esc x cot x + — — • — —
dx sin X Sin x

cosx

1 — cos2x

44. f(x) = Jx

46. /'(x) = X

48. fix) =

fU)

+ C

452 Chapter 4 Integration

50. ^ = 2U - 1) = 2x - 2, (3, 2)
ax

= \2{x -

l)dx = x^ -2x+ C

2 = (3)2 - 2(3) + C => C = - 1
y = x^-2x-l

52.

dy
dx

y

= -x-2- (1,3)

/-

;c-2cic = - + C

3 = Y+C=> C = 2

v = - + 2, x>0

54. (a)

x"

1. (-1,3)

y=j-x+C
3=^-(-l) + C

3 = --+ 1 + C

C =

56. g'(^) = 6x2,g(0)= -1

g(0) = - 1 = 2(0)3 + C =* C ^
^(;c) = 2^3-1

58. f'is) = 6s - Ss\ /(2) = 3

fis) = \{6s - 8s^)ds = 3^2 - 2j'' + C

/(2) = 3 = 3(2)2 _ 2(2)t + C= 12- 32 + C^C = 23
/(i) = 3*2- 2*^+23

60. /"(x) = x^
/'(O) = 6
/(O) = 3

/'W = |.

x^dx = h^ + Ci

/'(O) = 0 + Ci = 6^ Ci = 6

/'(;c) = 3^ + 6

=10'

/(;c) = \{-x^ + 6]dx = —x" + 6x + C2

/(O) = 0 + 0 + Q = 3
fix) = — x" + 6j: + 3

C2 = 3

62. fix) = sinx
/'(O) = 1
/(O) = 6

/'W = I

sin X £& = — cos X + C

/'(O) =-l+Ci = l=>Ci = 2
/'(x) = — cos X + 2

•(^) = J

fix) = I (- COS X + 2)dx = — sinx + 2x+ C2

/(O) = 0 + 0 + C2 = 6 => Q = 6
fix) = -sinx + 2x + 6

Section 4.1 Antiderivatives and Indefinite Integration 453

64. ^ = k^t, 0 < f < 10
dt

Pit)

= ffa'/2A =

rfa3/2 + C

P(0) = 0 + C = 500 => C = 500

PW = t/c + 500 = 600 => A: = 150

Pit) = T{150)f3/2 + 500 = i00r3/2 + 500
P(7) = 100(7)3/2 + 500 = 2352 bacteria

66. Since/" is negative on (-co, 0),/' is decreasing on
(-00, 0). Since/' is positive on (0, co),/' is increasing
on (0, oo)./' has a relative minimum at (0, 0). Since/' is
positive on (-oo, oo),/is increasing on (-oo, oo).

68. /'to = ait) = -32ft/sec2
/'(O) = Wo
/(0) = ^o
fit) = v(r) = I
/'(O) = 0 + C, = vo =» C,

32d?= -32r + C,

/'(f) = -32f + vo

fit) = j(0 = J (-32r + vo) dt= - 16f2 + VqI + Q

/(O) = 0 + 0 + Cj = io =* Q = •5o
/(r) = - 16r- + vof + Jo

70. Vo = 16 ft/sec
ig = 64 ft

(a) j(f) = -16f'+ 16f + 64 = 0

-I6(t^- f-4) = 0

1 ± yi7

Choosing the positive value,

1 + yr?

(b)

~ 2.562 seconds.

v{t) = s'it) = -32r+ 16

<^)

1 + yn

2 / -32^^1 + 16

-16yi7 = -65.970 ft/ sec

72. From Exercise 7 l,/(r) = -4.9?^ + 1600. (Using the
canyon floor as position 0.)

fit) = 0 = -4.9f2 + 1600

4.9^2 = 1600

, 1600

4.9

f = 7326.53 = 18.1:

74. From Exercise 71, f(t) = -4.9F + Vor + 2. If
/(f) = 200 = -4.9f2 + vof + 2,
then

Kr)

-9.8f + Vo = 0

for this f value. Hence, f = Vo/9.8 and we solve

-'■'[ts) ^ "its) ^ ^ = 2o«

(9.8)2 +98 198

-4.9vo= + 9.8vo2 = (9.8)2 198
4.9 vo^ = (9.8)2 198

Vo2 = 3880.8 => Vo = 62.3 m/sec.

454 Chapter 4 Integration

76.

V dv =

-GM

-dy

r

¥^

y

Wheny ■■

= R,v--

= Vo-

y

GM
R

+ C

c

-k

GM
R

2^

GM 1 , GM

IGM

v2 = Vq^ + 2GM|

e-a

78. x(t) = {t- l)(t - 3)2 0 < r < 5

= t^ - 7f2 + 15f - 9

(a) v(r) = x'it) = 3r2 - Ut + 15 = (3f - 5)(r - 3)
ait) = v'(r) = 6f- 14

(b) v(r) > 0 when 0<r<-and3<?<5.

(c) a{t) = 6t - 14 = 0 when / = -.

^«l)-)(l-)-H)-

80. (a) a{t) = cos t

v{t) = J a(t) dt = I cos t dt = sin t + Ci = sin t (since Vq = 0)

/W = I v(r) rff = I sin f rff = -cos t + Cj

/(O) = 3 = -cos(O) + C, = - 1 + C2 => C2 = 4
fit) = - cos r + 4
(b) v(r) = 0 = sin f for f = kir, k = 0,\,2,. . .

82. v(0) = 45 mph = 66 ft/sec

30 mph = 44 ft/ sec
15 mph = 22 ft/sec
ait) = -a
v(l) = -at + 66

■y(f) = -|f2 + 66t (Leti(O) = 0.)

v(?) = 0 after car moves 132 ft.

—at + 66 = 0 whenr

66

a '

■66\ ^ a (66

2\a

+ 661

?)

= 132 when a

33

= 16.5.

ait) = - 16.5

v{r) = - I6.5t + 66

sit) = -8.25t2 + 66f

(a)

- 16.5r + 66 = 44

22

/ 22 \
<,6.5)-^3.33ft

(b)

- 16.5r + 66 = 22

44
t = ^^^^ ^ 2.667

.(^) = 117.33 ft

(c)

#

^

73.33 117.33
feet feet

It takes 1.333 seconds to reduce the speed from 45 mph to
30 mph, 1.333 seconds to reduce the speed from 30 mph
to 15 mph, and 1.333 seconds to reduce the speed from
15 mph to 0 mph. Each time, less distance is needed to
reach the next speed reduction.

Section 4. 1 Antiderivatives and Indefinite Integration 455

/■30 r30

84. No, car 2 will be ahead of car 1. If v,(f) and V2(f) are the respective velocities, then \v2{t)\dt > \v^(t)\dt.

Jo Jo

86. (a) V = 0.6139f3 - 5.525r" + 0.0492f + 65.9881

,^^ ,, , ,,^ 0.6139?* 5.525/^ , 0.0492^2 ^^„„„,
(b) i(f) = v{t)dt = : : — + z + 65.988U

= jvO

4 3 2

(Note: Assume ^(0) = 0 is initial position)

s(6) = 196.1 feet

88. Let the aircrafts be located 10 and 17 miles away from the aiiport, as indicated in the figure.
v^(t) = Z:^ r - 150 Vg = kgt - 250 Airpon

■Sa(') = h^ '- - 150: +10 Sb = ^kg e- - 250t + 17

A -•— B

H h-

(a) When aircraft A lands at time r^ you have

^aUa) = ^a ?a - 150

100

50

^AiO = ^I^A tl - 150r. + 10 = 0

"A^'-Al t'^A 'a

1/50^,

^ 150r^=-10

125f^ = 10

= i°-
'^ " 125-

50

125

625 ,

/t^ = — = 50(— -) = 625 =^ S^it) = 5,W = ^^t-- 150f + 10

Similarly, when aircraft B lands at time tg you have

135

Vgitg) = kg tg " 250 = " 1 1 5 => Ag =

SB{tg)=-kgtl-250tg+ 17 = 0

U—]ti - 250tg = - 17

365

2 *

2Vr '''^

34
'« " 365-

135 ,.,,/365\ 49,275 ^,, ^,, 49,275, .,,„

(b) 20

(c) d = Sgit) - s^{t)

Yes, tf < 3 for f > 0.0505.

20

^

3

/

— ^1

90. True

92. True

456 Chapter 4 Integration

94. False. / has an infinite number of antiderivatives, each
differing by a constant.

96. j^{s{x)Y + [c{x)i\ = 2s{x)s'(x) + 2c{x)c'{x)

= ls{x)c{x) - 2c{x)s{x)

= 0

Thus, \s{x)Y + \c{xJY = k for some constant k. Since,

5(0) = 0 and c(0) = 1, A: = 1.

Therefore,

[six)f + [cixW = 1.

[Note that s{x) = sin x and c{x) = cos x satisfy these
properties.]

Section 4.2 Area

2, ^k(k-2) = 3(1) + 4(2) + 5(3) + 6(4) = 50

*=3

4.^1=1+1 + 1 = 47
^•,4y 3 4 5 60

4

I

1 = 1

6. ^[(/ - 1)2 + (i + 1)3] = (0 + 8) + (1 + 27) + (4 + 64) + (9 + 125) = 238

15 5

10.

l['-(i

16. ;£(2(-3) = 22'-3(15)

i=l i=l

= 2

15(16)

45 = 195

9 "

12. -y

[■-(!-')!

10 10 10

18. 2(r-l) = X''-Il

1=1 /=1 (=1

rio(ii)(2i)"

L 6

"•;sv^

10 = 375

10

1

i=l

10 10

20. 2*^ + 1) = 2^'' + E'

1=1 1=1

102(11)2 , r 10(11)

L-

3080

22. sum seq(x 03 - 2x, x, 1, 15, 1) = 14,160 (71-82)
■g(;3 _ 2i) = (15)'('5 + ^y _ ; 15(15 + 1)

4
(15)2(16)2

15(16) = 14,160

24. 5 = [5 + 5+4 + 2](1) = 16
5 = [4 + 4 + 2 + 0](1) = 10

26. 5 =

2 1
5+2+1+-+-

2 + 1+l + Ul"
3 2 3

55
6

28. 5(8)

1 , „\1

+ 2^. L/l + ^U. /!.zU.(yr.z)l

lV2,V3,,,V5,V6,y7

--/^-^

= 7! 16 + ^ + ^ + ^+ 1 +-V + -^ + ^+ VIH 6.038
4V222 222'

5(8) = (0 + 2)- +

'1 _\1

-a*'Ia*

v^h

7 + 2)7 -5.685

4 14

Section 4.2 Area 457

3...,5,=.(i)./Tf(i).y
-i

-'Di^v-gmi^v-erd

Ifj ^ V24 ^ v^ _^ yi6 _^ 79"

= 0.859

„ ,- r/64\n(« + 1)(2« + 1)1 64,. \2n^ + ?,n~ + n
32. hm — H^ ^ ^ = -r'™ 3

34. hm -r „ =xlim ^ = -(i) = -

6^ -^ 3

36. 2^^V^ = -3E(4/- + 3) =

An(n + 1)

+ 3/j

2/1 + 5

= 5W

5(10) = § = 2.5

5(100) = 2.05

5(1000) = 2.005

5(10,000) = 2.0005

,„ ^APJi- 1) 4^3 .,

n4

+ 1)2 n{n + l)(2n + 1)

]

^ 4r»^ + 2«- + n _ 2n- + 3« + 1]
" /j4 4 6 1

= T-?[3«3 + 6^2 + 3« - 4/j- - 6« - 2]
3«

= Jir[3n' + 2n^ - 3n - 2] = S{n)

5(10) = 1.056
5(100) = 1.006566
5(1000) = 1.00066567
5(10,000) = 1.000066657

40. lim y - - = lim — "V (' = lim

4 /n(w +
nA 2

limfl+iU2

n— *oo Z\ /I/

'>iV/2

42. lim y 1 +- - = lim ^Y (/j + 2/)-

n->oo/e^,\ nj \n/ n->oo /I-",^,

= lim 4rn3 + (4„)(^?(^^ + 4(.)(. + 1)(2„ +0

r 2 4 2 2 1

2 lim 1+2+- + - + - + -^
n->cc L /I 3 /I 3n-J

=21+2+

4\ 26

458 Chapter 4 Integration

2iY 2

44. Urn 2[i +-)[-] = 2\im -^^{n + 2i)

nj \nl n-»oo n^

1 ^

2 lim -7 y (n^ + 6nH + I2ni^ + Si^)

n->oo n^ i^,

2 1im (l+3+- + 4 + - + ^ + 2+- + 4
n->oo \ n n n'- n n''

= 2 lim I 10 +

/I— »co

20

46. (a) y

(b) Ax =

3-1 2

n n

Endpoints:

n n n

■(!)

i<i.iiei<i + 2g)<..-<,+(,-i)g)<i.„g)

(c) Since y = j: is increasing, /(m,) = /(j;,_ ,) on [x,^ ,, jc,].

1/

M:[

iM.--i)(^

i.(.-i)^^

(d)/(M,.)=/(Ac,)on[;c,_„A:,]
(e)

j:

5

10

50

100

^W

3.6

3.8

3.96

3.98

5(«)

4.4

4.2

4.04

4.02

,![-"-«

(f) lim y\\ +0- 1)

limp

n->oo \n

n +

2/«(« + 1)

")]

lim 2

1 + (•

,. [„ 2n + 2 4] ,. r, 21 ,
= hm 2 + = hm 4 =4

n^oo L n n\ n-^oa L n]

^] = lim ^n + P)^^^l
nJ n^oo n\_ \n) 2 J

= lim [2 + 21:^^1=,, [4^21 ^

Section 4.2 Area 459

48. y = 3a; - 4 on [2, 5]. [Note: Ax = = - j

5W = 2/2 +

3A/3

i=lL

3(2 + ^1-4

3A .18 ^3(1^2' -12

, 27/(/j + \)n\ ^ 111, , 1

Area = lim S{n) = 6 +

27 39

SO. y = x^ + 1 on [0, 3]. Note: Ax

3-0 3

*)=i/(?)e)=i,[(!r-B

97 n •J n

^ 27 n(w + l)(2n + 1) , 3 , , ^ 9 2/i^ + 3« + 1
w^ 6 n 2 n'^

Area = lim S{n) = -(2) + 3 = 12
« — *oo 2

1 »-x

SI. y = \ - x~ on [- 1, 1]. Find area of region over the interval [0, 1]. I Note: A.r = -

*.=i/e)(M[.-(i)l

, 1 ^.. , n(>z + l)(2n + 1) 1/9,3,1

= 1 - ^ X(- = 1 —. = 1 - 7 2 + - + —

rv' /e', 6n^ 6\ n n-

1 12

- Area = lim s(n) = 1 - t = -

Area

4

54. >' = Ix: - x3 on [0, 1]. [Note: Ax = = -j

Since y both increases and decreases on [0, 1], T(n) is neither an upper nor lower sum.

= lV-lv-3 = "(" + 1) _ 1 \"Hn + 1)']

1 +

1 1

1

4 4n 4n-

Area = lim T(n) =1-7 = 7

n->oo 4 4

460 Chapter 4 Integration

56. y = )? - ^ oni- 1, 0]. [Note: Aa: = ^^ = -j

.»)=i/(-.4)(^i[(-V-(-i)i

iW^'i^f-'m-^-nrn:'-74;

2 In 3 n "*" 3«3 4 2n 4n2

,-■ / ^ . 5 4 1 7
Area = hm s{n) = 2-- + --j = T:r

58. gCy) =^y,2<y<4. (Note: Ay = ^^^ — - = -

Sin) = 2 ^(2 +

n +

1 /;(« + 1)

2 +

« + 1

Area = lim S{n) = 2+1=3

12 3 4 5

60. f(y) = 4y-yM <y <2. f Note: Ay = = -

^W = E/(i+f)(i

1 "

4„.i- ,.i

= ^3n+^^
n\_ n

n n

+ 1) 1 n{n + l)(2n + 1)

3 +

« + 1 (n + l)(2n + 1)

Area = lim S{n) = 3 + 1

i = ii
3 3

12 3 4 5

62. /;(>■) = y + 1, 1 < >- < 2 I Note: Ay = -

«")=IX'+^)(i)

n /^[ \ n' n'^ n

In +

1 nKn +\Y , 3 «(« + 1)(2» + 1) ^ 3 n{n + 1)

+ ■

. , (« + IP , 1 (n + l)(2n + 1) , 3(n + 1)
n24 2 n^ 2/i

Area = lim Sin) = 2 + 7+l+| = ^

n -trjj 4 2 4

Section 4.2 Area 461

64. fix) =x^ + 4x,0 < X < 4,n = 4

Let c, =

Xi + x.

66. f{x) = sin X, 0 < X < -, n = 4

Let C; =

Aj: = L c, = -, c.

Area=|;/(c,)Ax= ^^r + 4c,](l)

i")^(!-)-(f-»)^(f-*

53

.77 77 Stt

_ Stt _ Vtt

Area « ^-^^'''^ ^ = E (^'" ^^^

= 1

77-/' . 77 377 -577 . 777\

68. /(x)

^2+ 1

on [2, 6].

«

4

8

12

16

20

Approximate area

2.3397

2.3755

2.3824

2.3848

2.3860

70. fix) = cosv/x on [0, 2].

n

4

8

12

16

20

Approximate area

1.1041

1.1053

1.1055

1.1056

1.1056

72. See the Definition of Area. Page 259.

74. fix) =l/x,Q < X <

n

10

20

50

100

200

sin)

10.998

11.519

11.816

11.910

11.956

Sin)

12.598

12.319

12.136

12.070

12.036

Min)

12.040

12.016

12.005

12.002

12.001

(Note: exact answer is 12.)

76.

78. True. (Theorem 4.3)

a. A = 3 square units

462 Chapter 4 Integration

lir

80. (a) d =

(b) sin 0 = -

r

h = rsin 6

A = -bh = -r{r sin 9) = -i^ sin 0

(c) A„ = n\ -r~ sm — = — sin — = -nr-^f

1 , . 27r\ r-n . 2it
-r~ sm — = — - sin —

2 n J 2 n

lir/n

Let.r = 2Tr/n. As « — > oo, x — > 0.

lim A„ = lim Trr- 1 = irr^l)

n-»oo jr->0

82. (a) J]2i = n(n+ 1)

The formula is true for « = 1: 2 = 1(1 + 1) = 2
Assume tliat the formula is true for n = k:

k

^2i = k{k+ 1).

k+l
I

Then we have ^ 2i = ^2/ + 2(/t + 1)

= k{k + 1) + 2(A: + 1)
= {k+ l)(/t + 2)
Which shows that the formula is true for n = A: + 1.

The formula is true for n = 1 because
P(1 + 1P_4_
4 4

Assume that the formula is true for n = k:

e{k + 1)2

k+\ k

Then we have ^ i^ = "^P + {k + 1)

i=l ;=1

k\k + 1)2

4

ik + 1)2

4 ^^

ik 4- 1)2„

+ ()t + 1)3

{k^ + 4(k + 1)]
ik + 2)2

which shows that the formula is true foTn = k+ 1 .

Section 4.3 Riemann Sums and Definite Integrals

2. f(x) =l^,y = 0,x = 0,x=l,c^ = ^

Ax,

i^ a - 1)3 3(2 - 3t + 1

n-" w

n n /;3 r^,-2 _ ^; 4. 1 "1

lim 2 /(c,) Ax, = lim X V T ^^ T^

n-.cc ,^j n->os ,^j V n^ L 1 J

lim ^2(3'-'- 3'' + 0

= lim —

n->oo n

= lim

lim —

1— »oo /Z

n2(„ + 1)2

Jnjn + l)(2n + 1)\ ^ n(n + 1)1

3n* + 6n3 + 3«2 _ 2n^ + 3n^ + n n- + n]
4 2 "^ 2 J

3/2" , n^ „2-| 1-3 1 1 ] 3

Section 4.3 Riemann Sums and Definite Integrals 463

4. >> = JT on [-2, 3]. Note: Ax =

3 - (-2) _ 5

n n

► 0 as n-^oo

2/W^,= 2/(-2.^B = 2U.|)(5) = -,o.iy,

r

-,„.(|)*fil = -,o.f(,.i).^i

j:a:x= lira - + — =-

, n^oo \2 2n/ 2

6. y = Ix'^ on [l, 3]. (Note: Ajc =

3-1 2

2A/2

X3 1 +

n i\n
liVIl

n \n

>0 as «^oo

4 n(n +1) 4 «(n + l)(2n + 1)

« "I :: 1 — ^ ::

= 6+12^^ + 4(^^±il%l^

r

3x'dx = lim

. _^ 12(n + 1) ^ 4{n + 1)(2« + 1)

= 6 + 12 + 8 = 26

8. y = 3x= + 2 on [- 1, 2]. (Note: Ax =

|/Wi., = |/(-i.f)(2

2-(-l)_3

is

3(-l +-1 +2
«/

3|l-^ + ^K2
n n

>0 as «— >oo

3n

18 n(n + 1) ^ 27 n(n + 1)(2« + 1)

3„]

£.'"=

= 15 _ 27(n + 1) _^ 27 (n + Djln + 1)
n 2 «'

+ 2).. = lim [15 - 27^^^^ + ^k±_feM)l

n ->oo L /I 2 n' J

= 15 - 27 + 27 = 15

10. lim V6c,(4- qpAt, = 6x(4-x)2dr

IAI-.0 ,/e, Jo

on the interval [0, 4].

12. lim vfAKv, = \ ^dx
on the interval [1. 3].

Jo

(4 - 2x) (ic

16. f'x-

Jo

dx

18. f ^

(it

r»/4
20. t:

Jo

tan X dx

464 Chapter 4 Integration

Jo

22. {y-2Ydy

24. Rectangle

A = bh = 2(4)(a)

J -a

4dx = ^a

5-

": 24- -S^ Rectangle

f m
pmm — -.

-a I a

28. Triangle

A = ^bh = |(8)(8) = 32

30. Triangle

A = |m = ^(2a)a

Jo

x)dx = 32

A =

x\) dx-- a^

J '4 r4 r4

x^dx = 60,\xdx = 6,\dx = 2.
2 Jz J2

26. Triangle

A=\bh = |(4)(2)

32. Semicircle

A = ^Trr^

A= fy

a,.

r^ - x^dx--

y

u

r

Semicircle

jpy^''

*^

^3^ = 0

r4 r4 ("4

38. (a:3 + 4)a:r = Legate + 4 Lie = 60 + 4(2) = 68

42. (a; \ f{x) dx = \f{x)dx+ \f{x)dx = 4 + (-1) = 3
Jo Jo h

(b)

\f{x)dx= -J

f(x)dx= -(-1) = 1

(c) fix) dx=0
(d)

-5fix) dx= -5\ fix) dx= -5i-l) = 5

36. I 15dx=l5\ dx= 15(2) = 30

/•4 /•4 r4 r4

40. (6 + 2;i: - jc^)^ =6 dx + 21 xdx - x^dx

= 6(2) + 2(6) - 60 = -36

44. (a) j fix) dx= \ f(x)dx- ifix)dx = 0-5 =

(b) fix) dx- \ fix) dx = 5 - i-5) = 10

(c) j 3/(;c) dx = 3j fix) dx = 3(0) = 0

(d) 3fix) dx = 3\ fix)dx = 3(5) = 15
Jo Jo

Section 4.3 Riemann Sums and Definite Integrals 465

46. (a) [f(x) + 2]dx = f{x) dx + 2dx = 4+ 10= H (b) f(x + 2)dx= f(x) dx = 4- (Let // = ;c + 2.)

Jo Jo Jo J-2 Jo

1>

(c) fix) A = 2 fix) dx = 2(4) = 8 (/ even)

(d) fix) dx = 0 if odd)

48. The right endpoint approximation will be less than the
actual area: <

50. The average of Exercise 39 and Exercise 40 consists of a
trapezoidal approximation, and is greater than the exact
area: >

52. fix) = \x\/x is integrable on [— 1, l], but is not contin-
uous on [— 1, 1]. There is discontinuity atx = 0. To see
that

r

M

dx

is integrable, sketch a graph of the region bounded by
fix) = \x\/x and the .t-axis for - 1 < .x < 1. You see that
the integral equals 0.

-i — 1^

H 1-

56. >

1: 2' 3: * 5! 6: 7: 8; *

c. Area = 27.

Jo

60. I X sin ,r dx

54.

1 — I — *-'

b. A = 3 square units

Jo -1^- + J

dx

n

4

8

12

16

20

Lin)

2.8186

2.9985

3.0434

3.0631

3.0740

Min)

3.1784

3.1277

3.1185

3.1152

3.1138

Rin)

3.1361

3.1573

3.1493

3.1425

3.1375

n

4

8

12

16

20

Lin)

7.9224

7.0855

6.8062

6.6662

6.5822

Min)

6.2485

6.2470

7.2460

6.2457

6.2455

Rin)

4.5474

5.3980

5.6812

5.8225

5.9072

62. False

f..v^ ...(/;...-)(/;

64. True

'x dx

66. False

i:

V dx = 6

466 Chapter 4 Integration

68. f(x) = sin;c, [0, 2Tr]

Xq = 0, Xy = ~r, X2 = —, Xj = 1T,X^ = LIT
TT TT '2.TT

TT

TT 2t7

377

-'"6' ^2 -3. -3 3-M 2

i /(c,) .y., = /(f) Ax, + /(f) ^, + /(l^) ^X, + /(f) Ax,

=(i)(f)^(f)te)^(f)(f)^(-)(^)-o-

70. To find /o W dx- use a geometric approach.

H M It 1 1 — »

1 2 3

Thus,

Jo

ldjc= 1(2 - 1) = 1.

Section 4.4 The Fundamental Theorem of Calculus

2. f{x) = cos X

TT

cosxdx = 0

Jo

4. fix) = .tV2 - X

r2

r

,rV2 — X dx is negative, -z

\.v = [3v];

3(7) - 3(2) = 15

|V3„

+ 4) (fv =

--v^ + 4v

■fH-20 -(-6 + 8) = -f

-I

10. (3jc2 + 5x- A)dx

, 5x2

x3 + 4x

2

3 / 45

= 27 + 12

I \ 2

1+2-^

12.|V-90.. = [i.-f,2];_

= 38
1 9\ /I 9

4 2/ V4 2

-i:('-z^)-[^^i:

;=(i-,)-(.-iu-.

Section 4.4 The Fundamental Theorem of Calculus 467

£v'/3rf, = gv^/3]'^^ = l[(yr3)4] _ (yz3)4] = o

^f

^dx= J2\ x''/^dx =

72(2)x'/2l* = [272^1* = 8 - 2v^
20. I (2 - t)Jtdt = f (2f>/2 - r3/2) ^, = r|,3/2 _ 1^/2!^ = [7^(20 - 6r)l^

22. I ^^^ Cix = I I U2/3 - ;c5/3) dx

J-8 2^ 2j_8

¥<»--)=¥

3 3

5 8

^n

80

;24 - 15;c)

' = -^(39) + §(144) = ^5^^

80' ' 80'

80

24. {I - \x-l\)dx= [i + {x-?,)]dx+ [3 - (x - 3)] (&

(6 - jc) cit

.2.

3

1

[-f]

/9
I2"

4)-

f

(24-

4+16-18

9 ^ n

2 ~ 2

26. fV^ - 4;c + 3 1 ir = \\x^- - Ax + 3) dx - \\x^ - Ax + I) dx + \\x^ - Ax + 3) dx *^P'« "P *e integral at the zeros
Jo Jo Ji J3 -": = 1- 3)

J - 2x2 + 3;c

.^3

- Zx~ + 3x

x"

2x- + 3x

= [j - 2 + 3 j - (9 - 18 + 9) + f| - 2 + 3 j + (y - 32 + 12 ) - (9 - 18 + 9)

4 4 4

=T-0+T+T-0=4
3 3 3

28.

30.

■7T/4

de= I de =

0 4

Jo COS' e Jo

r-^n ^ r yn

(2 — csc^ x) dx = \2x + cotxl = (ir

32. £\2. + cos t) dt = [r^ + sin rj^;;^^ = (f + ^ " (t " '

+ 0)-(f+l)=f-l=^

34. P = - I s[nddd =

1T

-]7r/2

•1 -

r

36. A = (1 -;c'»)atc= x-

40. A = (.T + sin x) dx = \-i — cos .v

0 yi'^ 1 ■)

-- cos e I = --(0 - 1) = - - 63.7%

38. A = \dx = -- " = -^ + 1 = ^
Ji-t- L A-Ji 2 2

= — + -) = '""""'"'*

468 Chapter 4 Integration

42. Since y > 0 on [0. 8],

Area= (1 + x^'^)dx =
Jo

44. Since y > 0 on [0, 3],

r

(3x - x^dx

X + -x^l^
4

3 , y?
2 3

= 8 + -(16) = 20
0 4

3^9

0 2-

r

46. I -^dx

■2;c2

^=4.? = 4

2 2

/(c)(3 - 1) = 4

I-
-I

r77-/3 r -|7r/3

48. cos j: (ic = sin j: = V3

J-,r/3 L J-^/3

373

52.

cos

'^- 277

c^±

0.5971

1 f'^

dx =

'2 1"'^
— sinx =

.77 Jo

2

/ ,„v „ cos A

(it/2) - Ojo

77

2
Average value = —

77

2
cos j: = —

77

X - 0.881

54. (a) I fix) dx = Sum of the

areas

= |(3 + 1) + |(1 + 2) + |{2 + 1) + (3)(1)

c = ^1 --- 1.6510

50.

3 - 1

^d!x = 2f(l +x-2)dtc = 2L

26

3

(b) Average value =

i'

fix)dx

8 4

7-1 63

(c) A = 8 + (6)(2) = 20

20 10
Average value = "7" = ~;~

\a.

2. --

12 3 4 5 6 7

12 14 5 6 7

Section 4.4 The Fundamental Theorem of Calculus 469

^f

56, I fix) dx = (area or region B) = \ f{x) dx - fix) dx 58. - 2fix) dx = -ll fix) dx

Jo Jo Jo Jo

-2(-1.5) = 3.0

= 3.5 - (-1.5) = 5.0

= U^

60. Average value = - 1 /(x) tic = -(3.5) = 0.5833

64. P = 5(7f + 30'
(a)

62.

^ojy

2\ J '^

«'-!

" _ 2kR-

0 " 3

t

1

2

3

4

5

6

p

155

157.071

158.660

160

161.180

162.247

1 QC4 150

Average profit = 7(155 + 157.071 + 158.660 + 160 + 161.180 + 162.247) = — -. — == 159.026
6 o

(b) I j 5(7f + 30) dt = ^ si^^'- + 30r

6.5

954.061

J5.0 6

(c) The definite integral yields a better approximation.

66. (a) R = 2.33f^ - 14.67f3 + l.61i^ + 70.67r
(b) 100

= 159.010

Jo

(c) Rit) dt =

2.33f5 14.67;* 3.67f^ 70.67f-

5 4

= 181.957

68. (a) histogram

N
,.
18--
16--
14--
12

123456789

(b) [6 + 7 + 9 + 12 + 15 + 14 + 11 + 7 + 2]60 = (83)60 = 4980 customers

(c) Using a graphing utility, you obtain

Nit) = -0.084175f3 + 0.63492r2 + 0.79052 + 4.10317.
(d)

Jo

(e) Nit)dt'^ 85.162
Jo

The estimated number of customers is (85. 162)(60) = 5110.

(f) Between 3 p.m. and 7 p.m., the number of customers is approximately
Hence, 3017/240 = 12.6 per minute.

( I Nit) dtjieO) =

(50.28)(60) = 3017.

470 Chapter 4 Integration

70. F{x)

= i^ + 2t-2)dt =

B-'

" X

- 2r

_2

= ( J + ;c2 - 2;c) - (4 + 4 - 4)

= ^ + ;c2 _ 2;c - 4
4

F(2)

= 4 + 4-4-4 = 0 Note: F(2) = ifi + It - 2) dt = 0

h

F(5)

= ^ + 25-10-4= 167.25
4

F(8)

= ^ + 64-16-4= 1068
4

72. Fix)

P-2 P ll
= -^dt=- 2t-^dt = \

1 1

2 X^ 4

F{1)

44-

F(5)

=^4=-^=-«-

F(8)

_ 1 1 _ 15
~ 64 4 ^ 64

74. Fix)

■x -pr

= sin 9^9= -cose = -cos j: + cosO = 1 - cos;c
Jo Jo

Fil)

= 1 -cos 2- 1.4161

F(5)

= 1 - cos 5 ^ 0.7163

F(8)

= 1 - cos 8 = 1.1455

76. (a) tif- + 1) dr = (r^ + t) rfr =
Jo Jo

d

1^ 1,

^ 1 1 x^

= -.X* + -x^ = — (;c2 + 2)
0 4 2 4^ ■*

(b)

cic

1 4_u 1 2

4 2

x^ + ;c = ;c(jc2 + 1)

78. (a)

P..=

2

_^/2

3

= f.3/2_|.|,3/2_8)

I sec r tan t (it = s

Jv3 L

80. (a) I sec r tan t (it = I sec t

V3

= sec X — 2

ir/3

(b)f 1x3/^

16

= rl/2 =

(b) —[sec X - 2] = sec X tan X

82. F(x)

rff

F'(x) =

84. Fix) = I Vtrft
F'(x) = ifx

Jo

86. F(x) = sec' t (if
Jo

F'(x) = sec'x

x2+ 1

Section 4.4 The Fundamental Theorem of Calculus 471

n,) = £,>.,4g>o

F\x) = 0

Alternate solution

F{x) = \ t^dt

= t^dt+ \ t^dt

= -) fidt+ \ ^
Jo Jo

dt

F'(x) = -(-xn-l) + {x^) = 0

90. Fix

.) = f r3.. = [^l = [-^]; = ^ 4^ F'W = 2.-

Alternate solution: F'(.r) = (x^)~\2x) = 2x'

= I sin d-c
Jo

91. F{x) = I sin e-dd

F'ix) = sin (x2)2 (2x) = Zrsinjc"

94. (a)

X

1

2

3

4

5

6

7

8

9

10

gix)

1

2

0

-2

-4

-6

-3

0

3

e

(c) Minimum of ^ at (6, - 6).

(d) Minimum at (10, 6). Relative maximum at (2, 2).

12

(e) On [6, 10] g increases at a rate of — = 3.

(f) Zeros of g: x = i,x = &.

96. (a) g{t) = 4 - ^

(b)

lim g(t) = 4

r->oo

Horizontal asymptote: v = 4

(b) A{x)

r

4-1).

4>
4r + - =
t]\

4.t + - - 8

4x- - 8.T + 4 4(.t - 1)-

lim A(x) = lim4.r + --8=oo + 0-8 = oo

X— >o<: x->o<: \ .t /

The graph of A(x) does not have a horizontal asymptote.

98. True

100. Let F{t) be an artiderivative of/(r). Then,

d^

dx

J-i-(-t)
/(f) dt
uW

/(f) dt

F{t)

■iti

4-c)

Rv(.r)) - F(,u{x))

u(x)

= ^f(v(.t)) - F{u{x))

= F'(v(.t))v'(x) - F'(M(:t))«'(^)
= f(v(,x))v\x) - f{u{x))u\x).

472 Chapter 4 Integration

102. G{x)

Jo I Jo

f(t)dt

(a) G(0)

Jo L Jo

ds

fit) dt

ds = 0

(c) G"(x) = X ■ fix) + \fit)dt

Jo

(d) G"(0) = 0-/(0)+ \ fit)dt = Q

Jo

104. xit) = it- Dit - 3)2 = r' - 7^2 + 15r - 9
x'(r) = 3f- - 14f + 15
Using a graphing utility,

(b) Let Fis) = J /(f) dt.
Jo

G(x) = F(5) ds
Jo

Total distance

Jo

\dt ~ 27.37 units

Section 4.5 Integration by Substitution

\figix))g'ix)dx u = gix) du = g'ix)dx

2. |xV;c3 + \dx

4. sec 2;c tan 2;c ^

^ + 1

2x

6. I^^dx

2d:x

COS j: dx

8. [u^

9)3(2;c) dx = ^ ^ ' + C

Check: ^

ax

(x^ - 9)"
4

+ C

4(^2 - 9)3

(2x) = ^2 - mix)

../,

10. I (1 - 2x2)'/3(-4x) <fe = ^(1 - 2x2)''/3 + c

Check: -^

^(1 - 2x2)4/3 + C

G'(x) = Fix) =xlfit)dt
Jo

G'(0) = o( /Wdr = 0
Jo

= I • |(1 - 2x2)i/3(-4x) = (1 - 2x2)i/3(-4x)

../.

x^U' + 5)''d:t = ||"(x3 + 5)4(3x2) dx = l^^^y^ + C = ^^^-Js^ "^ ^

Check:

^[(xL+ili , J ^ 5(x3 + 5[
dxl 15 J 15

3 5
5)^(3x2) _ ^

x3 + 5)V

14. |x(4x2 + 3)3 ^ = M (4x2 + 3)3(8j,) ^ = i

(4x2 + 3)4-

.,c = (^^i±^ + c

32

Check:

otc

(4x2 + 3)4

32

+ C

4(4x2 + 3)3(8;t)
32

= x(4x2 + 3)3

Section 4.5 Integration by Substitution 473

ft'VFT5dt = ^f{

16. I t^VFTldt = tI (r» + 5)'/2(4r3) dt = \^^^jp^ '^^^h''^ ^^'''^ "^ ^

Check: 4

7(r* + 5)3/2 + c
6

1 1
6 ' 2

(f* + 5)i/2(4f3) = (f + 5)'''2(t3)

S. mV«^ + 2 ^" = T ("^

18. I «V«3 + 2du = ~\{u^ + 2)'/2(3«2) rftt = ^^ ^r^' +C= ^ ^ ' + C

+ c] = I • |(«3 + 2)'/2(3«=) = («3 + 2)'/2(«=)

20.

'^7)5^ = ^J(l+^")-W)^=4(l+^)"' + C = ^^j

+ c

Check:

dx

x^

(1 + x^P

2^- /(T6^'^ = 4/('^ - -^)-=(-3-)^v = -|[^^^^]

+ C

1

3(16 -x3)

+ C

Check:

dx

^ +c]=|(-l)(16-x3)-2(3x2)= •»

3(16 - x^)

(16 - x^r-

24.

/^-i/'

x^ 1 r 1(1+ x^)'/- /I + v^

Check:

dx

ym?

+ c

4 1/2

-•-(1 +.v4)-'/2(4.v3) = --=^==
2 2 VTT^

26.

/['-i5V]-=/('-H-=f-l(9-=?-^^--

r'- 1
9x

+ C

Check:

<&

^-r' + C

, 1 _. , 1

Check: ^[V^ + C] = ^
30. y-^-T^'it = I (i'/2 + 2f3/2) A = |f3/2 + |rV2 + c = ^^^''-(5 + 6f) + C

Check: 4

2 4

£f3/2 + 1^/2 + c

;l/2 + 2r3/2 =

f + 2f2

Check:

¥*i?

474 Chapter 4 Integration

34. j 2773^(8 - f/^) dy = iTrU^y - f'^) dy = ItUy^ - ^fA + C = ^(14 - y^/^) + C

Check:

dy

477 -r.

14 - y3/2) + c

d_
dy

277(4/ -^7/2 I + C

= \6TTy - l-nfl'^ = (277)')(8 - y^/z)

1. V = , dx

= y|(l + y?)-^l\Zx^) dx

_ ior(i+^3)i^i

1 1/2 J

20

= ^yrT^ + c

40. (a)

M^'^SM

-I--

-3-

M^?W

38. j;

X- 4

--dx

J Vx^ - 8a: + r

{x^ -8x+ l)'/2
1/2

+ C

= V';c2 - 8;c + 1 + C

(b) V- = ^ cos x^, (0, 1)

y = I ;i: cos x^ dx = —\ cos(j:2)2x dx

1

sin (x^) + C

(0, 1): 1 = - sm(0) + C => C = 1

y = - sinU^) + 1

j 4^:^ sin a:^ ^ = I :

42. 4^3 sin;i^^ = sin;d(4x3) dx = -cosx^ + C

I cos 6x dx = -\ {<

1 .

44. I cos 6xdx = -\ (cos 6j:)(6) dx = — sin 6;t + C

5. J X sin x^ ctr = - I (

46. I X sin x^ ctr = - 1 (sin x^){2x) dx = -- cos x^ + C

/^

-/'

48. sec(l - x) tan(l - x) dx = - \ [sec(l - jc) tan(l - x)]{- I) dx = -sec(l - x) + C

»■/■

50. I Vtanlc sec^ ;c dt = ™^^ h C = -(tanx^/^ + C

52. ^'V ^ = - {cosx)-\-smx)dx = - ^ + C = t r- + C = -scc^a: + C

Jcos^x J -2 2cos2x 2

54.Jcsc2(f)^ = 2jcsc^

;c\/l

2/V2

dx = -2 cot - + C

(f)

-I""

56. /(j:) = I 77 sec 77X tan nxdx = sec 77X + C

Since /(1/3) = 1 = sec(77/3) + C, C = - 1. Thus

f(x) = sec 77X - 1.

Section 4.5 Integration by Substitution 475

5%. u = 2x+ \,x = -(u- 1), dx = ]rdu

I xjlx + 1 dx = \hu - \)^]-du
= \\^{u'''-u''')du

„3/2

= 3^(2r + l)V2[3(2x + 1) - 5] + C

2
30

(2;t + \y'\6x - 2) + C

60. M = 2— X, x = 2-M, iit= —du

\{x+ \)J2 - xdx= - (3 -m)v^

= - (3u ' - u

du

3/2

)rftt

= -(2tt3/2-|„5/2)+c

2m3/2

{5 - m) + C

= -j(2-xm5-(2-x)] + C

= -|(2 - ;c)3/2(;c + 3) + C

= ^{2x + l)3/2(3;c - 1) + C

62. Let u = X + 4, X = u - 4, du = dx.

= |(2«'/'-7M-'/2)d«

= |m3/2 - 14„l/2 + C

: jm'/2(2« - 21) + C

64. u = r - 4, r = M + 4, rff = rfu

I t^t- 4 dt= (« + 4)m'/3 dtt

/•

= (« + 4)m

,4/3 +4„l/3)d„

= ^«'/3 + 3„4/3 + C

3«^/3

iu + 1) + C

= ->A + 4[2(-T + 4) - 21] + C

= =jit- 4)''/3[(r - 4) + 7] + C

;-VFT4(2x - 13) + c

= ^r - 4)^/3(f + 3) + C

66. Let « = x^ + 8, rfM = 3^^ dx.

r

c

x2(x3 + 8)2 ate = ^ I (x^ + 8)2(3^2) die

3 3 J--

= ^[(64 + 8)3 - (-8 + 8)3] = 41,472

68. Let M = 1 - .r^, du = -Ixdx.

f x^V^^dx = -||" (1 - .t2)'/2(-2v) cfe = r-|(l - .X'V

:>-r;

70. Let M = 1 + 2x-, c/m = 4x dx.

C X ^^1 p

Jo v-m^ ' 4jo

(1 + lx-)-'/-(4.r)d:r = ^71 + li

[i-

^-i=l

476 Chapter 4 Integration

72. Let « = 4 + .x\ du = 2xdx.

xi/4 + x'^dx = ^\
Jo ^Jo

xi/4Tl(^ dx = i-\ (4 + x^y/^{2x) dx=\^{4 + x^Y'^

= ^(S''/^ - 4*/3) = 6 - ^^ - 3.619

0 5 2

74. Let H = 2x - L rfu = 2 A, x = -(« + 1).

When ;c = 1, M = L When x = 5,u = 9.

Ji 72F^M J, v4 2 4ji

^[f^(27) + 2(3)) - (I + 2

4LV3

26

3

,6. £;v . cos,)* = [f . si„,]:;; . (f . ,) - (g . f ) . f

'tt^ , 73\ _ 517^ _^ 2-73

78. u = x + 2, x = u — 2, dx = du

Whenx = -2, m = 0. When a: = 6, m = 8.

Area

( ;C-3/^rT2 die = I (m - 2)2 34rfM = ( (m'/3 _ 4„4/3 + 4„l/3) J„
J-2 Jo Jo

3 ,0/3 _ il„7/3 + 3„4/3

4752
35

Jo

80. A = (sin j: + cos 2x) dx =

- cos j: + - sin 2x I =2

•r

82. Let « = 2x, dM = 2 iic.

rnM J p/4 r J "1^/4 J

Area = esc 2Ar cot 2x dr = - esc 2jccot 2jr(2) dx = -— esc 2;c =-

J7r/12 2J^/i2 L 2 J,r/12 2

Jo

84. x^^x + 2(ic = 7.581

^/:

86. xVx - 1 otc = 67.505

90. I sin X cos X ate = I (sin x)' (cos x ate) = — r— + Q

sin X cos X otc = I
sinxcosxdr = ~

sinxcosxiic= - (cosx)'(— sinxatc) =

+ C,

(1 - sin^x)

^sin^_l

^ (-2 2 2

J-ir/2
si
0

88. sin2xatt = LO

.^

They differ by a constant: €2= C, +

1

Section 4.5 Integration by Substitution 477

92. f{x) = sin^ X cos x is even.

■w/Z

rnn

.

sin^

^ccos j:

dx

= sin^ j:(cos x) dx

-7r/2

Jo
fsin^
L 3

_2
3

-0

94. /(a:) = sin j: cos x is odd.

rir/2

sin X cos j: A: = 0

-ir/2

J-%

rir/4

96. (a) s

J-ir/4

sin X dtc = 0 since sin x is symmetric to the origin.

(b)

rn/4 r-w/i r -|Tr/4

COS xdx = 2] cos j: A = 2 sin X = v^ since cos jr is symmetric to the y-axis.

J-T/i Jo I Jo

r-^/2 r-r/2 r -1^/2

(c) I cos xdx = 2\ cos xatc=2sinjc =2
J-V2 Jo L Jo

p/2

(d) sin X cos .r tic = 0 since sin(— j:) cos(— j:) = — sin j: cos x and hence, is symmetric to the origin.

J-Tr/2

98. j (sin 3x + cos 3x) dx = I sin 3j: lic + cos 3xdx = 0 + 21 cos 3j; tic =
J-w J-TT J-TT Jo L

cos 3x dx = I — sin 3x

= 0

'/'

100. If« = 5 - x^,thendu = -2x&and \x{5 - x^f dx = "T (5 - x2)3(-2x) A = --|«3(i„.

^1'

= 4P

102. ^ = /t(100 - r)^

f

Q{t) = I A:(100 - tf dt = --(100 - f)' + C
(2(100) = C = 0

<2(r) = -jdOO - tf

Q(0) = -j(100)3 = 2,000,000 => /t = -6

Thus, Q{t) = 2(100 - tf. When t = 50, 2(50) =
$250,000.

106. (a) 70

104. R = 3.121 + 2.399 sin(0.524f + 1.377)
(a) ^

Relative minimum: (6.4, 0.7) or June
Relative maximum: (0.4, 5.5) or January

(b) Volume

Jo

if" 1

I R{i)dt-

(b) I R{{) dt = 37.47 inches

(c) 7 1 y?(f) ^f = T-(13) = 4.33 inches

1272 (5 thousand of gallons)

Maximum flow: /?== 61.713 at r = 9.36.
[(18.861, 61.178) is a relative maximum.]

478 Chapter 4 Integration

108. (a) "

(b) g is normegative because the graph of/ is positive at the
beginning, and generally has more positive sections than
negative ones.

(c) The points on g that correspond to the extrema of/ are
points of inflection of g.

(e) 4

^^-^

The graph of h is that of g shifted 2 units downward.

g{t) = I /U) dx
Jo

I f(x)dx+ I f

Jo Jtt/I

fix) dx+ \ f(x)dx = 2 + hit).

(d) No, some zeros of/, like x = tt/2, do not correspond to
an extrema of g. The graph of g continues to increase
after x = ir/l because /remains above the j:-axis.

110. False

Lix^ + l)2(fa = 11 (;c2 + l)(2;c) dx = ^ix^ + l)^ + C

112,

True

•*
sin j:

Ja

dx =

114.

False

i

— cos x\ = - cos b + cos a = - cos(fe + 2

tt) + cos a = I

sin j: lie

I sLn^ 2x cos 2xdx = -\ (sin

1 I 1 (sin2x)' 1

:cos2xdx = -\ (sin 2x)2(2 cos 2x) dx = - - — r—'- + C = - sin^ 2x + C

21 I i o

116. Because /is odd, /(-At) = -/(jc). Then

ra ro fa

fix) dx = fix) dx + fix) dx

J-a J-a Jo

= -( fix)dx+ \fix)dx.
Jo Jo

Let x = —u,dx= — rfM in the first integral.
When X = Q,u = 0. When x = -a,u = a.

I fix)dx= - \fi-u)i'du) + \fix)dx

J~a Jo Jo

= -\ fiu)du+ \fix)dx==0
Jo Jo

Section 4.6 Numerical Integration 479

Section 4.6 Numerical Integration

2. Exact:

f(f

+ 1 Uc =

x^

+ X

1.1667

Simpson's:

4. Exact:
Trapezoidal:
Simpson's:

6. Exact:

i.4|M:.i).2(M)!,,),4(l«.i).g.i

75
= £-1.1719
64

= 1.1667

= 0.5000

l+4|i)= + 2(i)" + 4(f + i].0.5004

12.0000

Trapezoidal: iG dx = ^[0 + 2 + 2^/2 + 23/3 + 23/4 + 2^5 + 2^6 + 2^7 + 2] « 11.7296

Jo 2

Simpson's: .^ otc == :^[0 + 4 + 2^2 + 4^ + 2i/4 + 4^5 + 2i/6 + 4i/l + 2] « 11.8632

Jo 3

8. Exact:

(4 - jc2) dx -■
Trapezoidal: (4 - x^)dx ^ ^3 + 2 4

Simpson's: I (4 - x^) a[r = - 3 + 4(4

1 i

2 ^
"3

+ 2(0) + 2

-0.6667

4
4, + 0 + 4(4-f

iJ]

5 I - -0,

5 = -0.7500

6667

10. Exact:

I

x^x^ + \dx = -

(.^ + 1)3/2

' = i(53/2 - 1) ^ 3.393

0 3

Trapezoidal:

+ 1 A - Ifo + 2(^)^(1/2)- + 1 + 2(i)yFTT + 2(|)v(3/2)^ + 1 + 27FTTJ -

3.457

xjx- + 1 A = ^1 0
Simpson's: j xV^lH'aLc = | 0 + 4[ijV(l/2)2 + 1 + 2(l)VPin' + 4f|jV(3/2)- + 1 + 2v'2^"+T j - 3.39

12. Trapezoidal: J- dx = 7 1 + 21 , ^ =) + 2(^=1=

jov/m? 4 \vi + (1/2)3/ \jY+

Simpson's: , etc = - 1 -I- 4 , =

JoTrn? 6L vvi + (1/2)

vTTTs
1

+ 2'

+ 4

1

1

Vl + (3/2)3/ 3

1

ym3/ \vi + (3/2)3/ 3 J

1.397

= 1.405

Graphing utility: 1.402

480 Chapter 4 Integration

14.
16.

Trapezoidal:

Simpson's:
Graphing utility

Trapezoidal:
Simpson's:

^xsmxdx = —

n/2 16

"tt

Jx sin xdx'=^ -—

.12 24

: 1.458

V
V

77/4

'^nw, f^-i^A^-, f^-C^AA.-, f^-PA^r.

«= 1.430
1.458

0 ^
« 0.271

tanCv2^ Wv V^V4

V 4 y \ 2 / V 4 / \\J A)
tan 0 . 4 tan(^)% 2 tanf^^)^ . 4 tanf^^//^)^ . tanf ^V

V 4 / V 2 ; V 4 y W 4/.

0

12

■ Q.l'il

Graphing utility: 0.256

J-ir/2
Vl + zos^xdx'- -^v/^ + 2Vl + cos^dr/S) + 2Vl + cos2(Tr/4) + 2Vl + cos2(37r/8) + l] = 1.910
0 16

J-7r/2
Vl + cos2A:dx«:^v/2 +4Vl + cos2(7r/8) +2^1 + cos2(tt/4) + 4V1 + cos2(37r/8) + l] = 1.910
0 24

Graphing utility: 1.910

,„ ^ .J, r sinj: , ttF, 2sm(7r/4) 2 sin(7r/2) 2 sm(37r/4) 1 , „.,^
20. Trapezoidal: dx « - 1 + 7;^ + 7^^-^ + — J^ ' + 0 =» 1.836

f ^gJM ^ ^ 4 1 ^ iji'Ml) ^ 2sin(^ ^ 4 sin(3,r/4) ^ 1 ^
Jo ^ I2L 7^/4 7r/2 3Tr/4 J

Simpson's:

Graphing utility: 1.852

22. Trapezoidal: Linear polynomials
Simpson's: Quadratic polynomials

24. f(x) =
fix) =

/'W =
f"U) =

. f'Kx) =

X

+ 1

-1

{x

+ 1)^

2

(x

+ 1)3

-6

ix

+ 1)"

24

(x + ly

(a) Trapezoidal: Error < ,.,-^. (2) = — = 0.01 since
1 2(4 } 96

f"{x) is maximum in [0, 1] when x = 0.
(h) Simpson's: Error < ^^^^(24) = ^^ » 0.0005
since /**"(x) is maximum in [0, 1] when x = 0.

Section 4.6 Numerical Integration 481

26. ru

in [0, 1].

(1 + xY
(a) |/'tx)| is maximum whenx = 0 and |/"(0)| = 2.

12/1

Trapezoidal: Error < 7:^3(2) < 0.00001, n- > 16,666.67, n > 129.10; let n = 130.

f'\x)

24

in [0, 1]

(1 + xf
(b) \f'^*\x)\ is maximum when a: = 0 and |/''"(0)| = 24.

Simpson's: Error < ^(24) < 0.00001. n-* > 13,333.33, « > 10.75; let n= 12. (In Simpson's Rule /i must be even.)

28. fix) = (x + 1)2/3
(a) /'W =

9{x + l)'*/3

in [0, 2].

l/'lt) I is maximum when x = Q and |/'tO) |

Trapezoidal: Error < Ty^l^l < 0.00001, w^ > 14,814.81, n > 121.72; let n = 122.

(b) /WW = -

56

81U + l)i»/3

. [0, 2]

\f''\x)\ is maximum when .r = 0 and [/("'(O)! = — .

ol

Simpson's: Error < t^^(^] < 0.00001, /i"* > 12,290.81, n > 10.53; let « = 12. (In Simpson's Rule n must
, . 1 0O/7 \ 8 1 /

be even.)

30. fix) = sin(.t^)

(a) fix) = 2[-2x= sinU-) + cosix^)] in [0. 1].

\f"ix)\ is maximum when.r = 1 and |/"(1)| = 2.2853.

Trapezoidal: Error < ^' ~ , (2.2853) < 0.00001, n- > 19,044.17, n > 138.00; let/! = 139.
' (b) /W(x) = (16x^ - 12) sin(x2) - 48x^ cos(a^) in [0, 1]

[/(■"(x)! is maximum when x == 0.852 and |/<'"(0.852)| == 28.4285.
(1 - 0)5

Simpson's: Error <

180«''

-(28.4285) < 0.00001, «■» > 15,793.61, « > 11.21;let/i= 12.

32. The program will vary depending upon the computer or programmable calculator that you use.

34. fix) = Vl - .r= on [0, 1].

n

Un)

Min)

Rin)

7t")

Sin)

4

0.8739

0.7960

0.6239

0.7489

0.7709

8

0.8350

0.7892

0.7100

0.7725

0.7803

10

0.8261

0.7881

0.7261

0.7761

0.7818

12

0.8200

0.7875

0.7367

0.7783

0.7826

16

0.8121

0.7867

0.7496

0.7808

0.7836

20

0.8071

0.7864

0.7571

0.7821

0.7841

482 Chapter 4 Integration

36. /W = ^ on [1,2].

n

L{n)

M{n)

R{n)

Tin)

Sin)

4

Q.IQIQ

0.6597

0.6103

0.6586

0.6593

8

0.6833

0.6594

0.6350

0.6592

0.6593

10

0.6786

0.6594

0.6399

0.6592

0.6593

12

0.6754

0.6594

0.6431

0.6593

0.6593

16

0.6714

0.6594

0.6472

0.6593

0.6593

20

0.6690

0.6593

0.6496

0.6593

0.6593

38. Simpson's Rule; « = 8

l-lsm^dde^' ^ V ^ ~ f ^'"' 0 + 4^]

6
^ 17.476

sin2 0 + 4.^/l -|sin2:^ + 2
3 16

>yrri^+...+^

1 2 . , TT

1 — r sm'^ —
3 2

40. (a) Trapezoidal:

I fix)dx^ Wf'^-^^ "^ ^'"^-^^^ "^ ^^"^-^^^ "^ ^^^-"^^^ "^ ^^^-^"^^ "^ ^^''■^^^ "*" ^(^-^^ "^ ^(^-^^^ "^ ^•^''■] " ^^-^^^

Simpson's:

2

/:

fix) dx - ^[4.32 + 4(4.36) + 2(4.58) + 4(5.79) + 2(6.14) + 4(7.25) + 2(7.64) + 4(8.08) + 8.14] = 12.592
(b) Using a graphing utility,

V = - 1.3727x3 + 4.0092x2 - 0.6202x + 4.2844

Integrating,

ydx'='n.
Jo

53

42. Simpson's Rule: n = 6

77 = 4

Jo 1 + ^-

4
3(6)

4 2 4 2 4 1"

1 + . . .. ,,x. + . . .^,... + ■ . ..,„..+ . . ...,..+ . . . +■

1 + (1/6)2 1 + (2/6)2 1 + (3/6)2 1 + (4/6)2 j + (5/5)2 2J

^ 3.14159

120
44. Area = ^7J^[75 + 2(81) + 2(84) + 2(76) + 2(67) + 2(68) + 2(69) + 2(72) + 2(68) + 2(56) + 2(42) + 2(23) + 0]

= 7435 sq m

46. The quadratic polynomial

p(^^ ^ Jl^LJ^imhL I (^ - x,)ix - x^) I jx - x,)ix - X,)

(x, - X2)(Xi - X3) ' (X2 - Xi)(x2 - X3) 2 (jj^ - x^)iXi - Xj) ^

passes through the three points.

Review Exercises for Chapter 4 483

Review Exercises for Chapter 4

2.

4. M = 3x

du = 3,dx

2

/j|-f/

{3jc)-'/3(3)d:t=(3;c)2/3 + C

6- [^^ §~~^^ = f U - 2 + x-2) ^

-j:^ - 2x - - + C
2 j:

10. fix) = 6U - 1)

/'W = 6(;c - 1) dx = 3{x - 1)2 + C,

Since the slope of the tangent line at (2, 1) is 3, it follows
that/'(2) = 3 + Ci = 3 when Ci = 0.

fix) = 3(.t - 1)2

fix) = 3U - \)'dx = (x - 1)3 + Q

/(2) = 1 + C2 = 1 when Cj = 0.
/W = U - 1)3

../,

8. I (5 cos x - 2 sec^ x)dx = 5s\nx-2tanx + C

12. 45 mph = 66 ft/sec
30 mph = 44 ft/sec
a(f) = -a

at + 66 since v(0) = 66 ft/sec.

-t~ + 66r since s{0} = 0.

v(r)

Solving the system

v(f) = -at + 66 = 44

s(t) = -^- + 66r = 264

we obtain t = 24/5 and a = 55/12. We now solve
-(55/12)f + 66 = 0 and get r = 72/5. Thus,

72

■mm.eeiii

" 475.2 ft.

Stopping distance from 30 mph to rest is
475.2 - 264 = 211.2 ft.

14. a(t} = -9.8m/sec2

v(f) = -9.8t + Vq = -9.8f + 40
s{t) = -4.9?2 + 40t (5(0) = 0)

40

(a) v(t) = -9.8r + 40 = 0 when f = ^ ^ 4.08 sec.

(b) i(4.08) = 81.63 m

(c) v(t) = -9.8f + 40 = 20 when t

(d) i(2.04)«= 61.2 m

20
9.8

2.04 sec.

484 Chapter 4 Integration

16. Xj = 2, j:2 ~ — 1, ^3 = 5, ;c4 = 3, X5 = 7
mjinf-h- l+5 + 3 + 7)--!|

37_
5 ■ 3 ■ 7 210

(c) 2(2^,- - ^,') = [2(2) - (2)2] + [2(- 1) - (- 1)2] + [2(5) - (5)^] + [2(3) - (3)^] + [2(7) - (7)^] =
1 = 1

(d) 2(x, - x,_,) = (-1 - 2) + [5 - (- 1)] + (3 - 5) + (7 - 3) = 5

-56

18. y = 9 - -x\ Ax = 1, n = 4

5(4) = 1

9 - i(4)) + (9 - ^(9)) + (9 - ^(16)) + 9 - |(25)]

= 22.5

5(4) = 1
== 14.5

9-^(9)1 +

9 - ^(16)) + (9 - |(25)) + (9 - 9)]

2

20. y = X- + 3, Ax = - right endpoints

n

Area = lim y* f{ci) Ax

= lim ^X Ft + 3

"4 w(«+ 1)(2«+ 1) , , 1

V 1 — ~^H

r^(n+l)(2n+l)^ 1 8^^
_3 w^ J 3

= lim -

n— >oo n

— lim

26
3

1 2

22. >> = -X?, Ax = -

4 w

Area = lim V /(ci) Ax

,. 1 -^Tn 24i 24/2 s/n

lim — y 8 + — + — ^ + — r
n->oc2n-^,L n « n J

n->oo n j^i L " « « J

= 1- if I 3 w(« + 1) ^ 3 n{n + l)(2n + 1) , 1 w2(n + l)^"!
n-^co nL«2 «2 6 n'4j

= 4 + 6 + 4+1 = 15

12 3 4

Review Exercises for Chapter 4 485

- <». ^ ^ m - <m - «(f )(!) ^ "(t)© = if <■ — ' = ^

J = m(0)

(!)

-i!)(!) -(!)(!-»

4JU/ 16'

(1 + 2 + 3) =

8

0\ wfc^(« + 1)
2n

•« = lao = !"(!)© = -er % - """- ^ "-^ - ""--^^'

«^

2«

(c) Area = lim ^?^%1I) = u„, ^^^^n - 1) ^ 1 , ^ 1 ^^^^^^^ ^ ^ (base) (height)
n->oo 2n n-»oo 2n 2 2 2

f ri 1* 1

(d) I mxdx = \ -nvxP- = -wi^

26. lim V3c((9 - c/^) ^xi = 3;c(9 - x^) dx

28.

r.

Vl6 -jc^atc = x'^^^)-

Stt (semicircle)

30. (a) J /(;c)^ = j /(^)dx + \ f{x) dx = 4 + (-1) = 3

(b) J/(;c)(&= -J/(x)d:x= -(-1)= 1

(c) \f{x)dx = Q

(d) - \Qf(x) dx= -\o\ f(x)dx= - 10(- 1) = 10

„ Pl2 , ri2x-21' r-67 -6 , 16 „,

34.J_V + 2)..= [f.2r];_=f

£

36. I {x!^ + Ix- - 5) dx =

-{

L

32 16

+ 10

52
15

^«-f(i4)'^ = />-^-^-')^ = [-^2^

- /

' 1 1\

/ , 1\

1

-1 + -

I

^ 2 8^

V 2/

8

rw/4 r -1^4

I. sec^ r rf/ = tan n = 1 - (- 1) = 2

J-ir/4 L J-7r/4

486 Chapter 4 Integration

i

42. {x + A)dx =

V

+ Ax

10

-2-1 12 3 4 5

44.

/>

x^ + X + 2) dx =

x^ x^

■I— )-(i4-)

= 12 7 ^9
3 6 2

46. V^(

Jo

\-x)dx^ (;c'/2 - ^3/2) dx

2^/2 _

.3

2 T
5 Jo

2 2

3 5 "

4
" 15

48. Area

rV3
Jo

= tanjc

sec^ j: etc

= V3

;c = 3/2

50. :r^ \ x'dx =

x^ = 2

y

= 2

52. F'ix) = -
x^

54. F'(x) = csc2;c

56. (^ + ;^) dx^ I (jc2 + 2 + ;c-2) die

j;3 1

3 X

Review Exercises for Chapter 4 487

58. u = jc' + 2,, du = 2>x^ dx

jx^VJ^TI^ = \U^ + 3)'^' 3j;2alr = |(x3 + 3)3/2 + c

60. M = jr^ + 6j: - 5, (ftt = (2x + 6) ^

\{x4V-5f = ij(.2?6x-5P 'i^ = ^U^ + 6.-5)-' + C= 2(;,2 + ^ _ 5) + C

62. .r sin 3x^ dx = -\ (sin 3Ar2)(6.t) iir = -7 cos 3x- + C

64. M^^£it= (sin;c)-'/2j,os^^ = 2(sinjc)'/2 + c = 27ii^ + C
J Vsinjc J

66. I sec 2.r tan 2r ^ = - I (sec 2x tan 2x)(2) (ic = - sec Ir + C
68. I cor' a CSC- ada = - (cot a)''(-csc- a) da = --cot^a + C

f ;cV+ l)3A = ^f I
Jo -^Jo

70. I ;cV + 1)3 A = 11 (x3 + l)3(3.r:)d:x = ^

Jo 12

(x3+ 1)^1 =^16-1) ^

72.

f ^::#^'^ = C^-^ ~ 8)-./2(2x)^ = [f(^ - 8)-]; = 1(277 - 1)

74. m = j:+1,x = m— l,cir = rf«

When x= - I, m = 0. When .r = 0, m = 1.

76.

27r| xVa: + 1 dx = 27r (« - D^Judu

= 2tt\ (tt5/2 _ 2„3/2 + „l/2) ^„ = 27Tf|«'/2 _ l„5/2 + ^ 3/2I' = HE

Jq L' 5 3 Jo 1U5

r-n/4

sin 2r

J--7r/4

It otr = 0 since sin 2a: is an odd function.

78. « = 1 — x,x = 1 — u, dx = —du

When .X = a, M = 1 -a. When .v = fc, « = 1 - fc.

^0= I

Ja

''^155-.v3(l-.)3/2^.= l'55'—

32

32

Jl-n

(1 -m)3m3/2^„

1155

32

(„9/2 _ 3„7/2 + 3„5/2 _ „3/2) ^„ = 1122 ^„112 _ r„9/2 + £„,,,, _ £^^,
Jl-a 32 [11 3 7 5 Jl-n

= -^ 7777(105"' - 385m= + 495« - 231) = -r^(105u3 - 385m- -i- 495m - 231)
32 L1155 Jl-n L lo Ji-o

(a) P,
(b)

mV2
16

(lOStt' - 385m= + 495m - 231)

"10.75

- 0.025 =

0. 0.25

r,,5/2 no

^oj 1 = Tr(105M3 - 385m- + 495m - 231) = 0.736 = 73.6%
L lo Jo.5

488 Chapter 4 Integration

f

Jo

TTt 2

80. I 1.75 sin ^;- A =

2 77

1.75 cos

f]'

--(1.75)(- 1 - 1) = - - 2.2282 liters

IT IT

Increase is

2 _ 5J_^ L9

77 TT TT

= 0.6048 liters.

J"' y3/2

Simpson's Rule (n = 4):

Jo

Grapliing utility: 0.166

£& =

■ 2(1/4)3/^ 2(1/2)3/^ 2(3/4)3/^ r

_ 3 - (1/4)2 + 3 _ (1/2)2 -^ 3 _ (3/4)2 "^ 2.

4(1/4)V2 2(1/2PA 4(3/4)V2 f
3 - (1/4)2 -^ 3 _ (1/2)2 3 _ (3/4)2 2.

== 0.172
= 0.166

Jo

84. Trapezoidal Rule (n = 4): J\ + %\v?xdx =» 3.820

Jo

Simpson's Rule (« = 4): 3.820
Grapliing utility: 3.820

Problem Solving for Chapter 4

2. (a) F(x) = sin r^ dt

r

X

0

1.0

1.5

1.9

2.0

2.1

2.5

3.0

4.0

5.0

F(x)

-0.8048

-0.4945

-0.0265

0.061 1

0

-0.0867

-0.3743

-0.0312

-0.0576

-0.2769

(b) G{x) =

x-2

i

sin f2 dt

X

1.9

1.95

1.99

2.01

2.05

2.1

G(x)

-0.6106

-0.6873

-0.7436

-0.7697

-0.8174

-0.8671

limG(x) = -0.75

(c)F'(2) = lim^^W^

i-»2 X — 2

= lim —

Jr-»2X

-2J2

sin t^ dt

= lim G{x)

i->2

Since F'{x) = sinx^, F'{2) = sin4 = lim G(x).

x^2

(Note: sin 4= -0.7568)

Problem Solving for Chapter 4 489

4. Let d be the distance traversed and a be the uniform acceleration.
We can assume that v(0) = 0 and s{0) = 0. Then

a{t) = a

v(r) = at

s{t) = -ai^.

s(t) = d when t = -./ — .
V a

The highest speed is v = a

/2d_
/ a

Jlad.

The lowest speed is v = 0.

The mean speed is -(j2ad + o) = .

f7d
V 2

The time necessary to traverse the distance d at the mean speed is
d fid

t =

Jadjl V a
which is the same as the time calculated above.

6. (a)

I I I I I I t I I I* <

0.2 0.4 0.6 0.8 1.0

(b) V is increasing (positive acceleration) on (0, 0.4) and (0.7, 1.0).
v(0.4) - v(0) _ 60 - 0

(c) Average acceleration =

0.4 - 0 0.4

(d) This integral is the total distance traveled in miles.

= 150mi/hr-

1

385

v(i) dt = -^[0 + 2(20) + 2(60) + 2(40) + 2(40) + 65] = ^ = 38.5 miles
Jo 1" 10

(e) One approximation is

,- -, _ v(0.9) - v(0.8) 50-40 ,^ ...2

(other answers possible)

490 Chapter 4 Integration

8. I mix -t)dt= \ xf{t) dt-\ tfit) dt = x\ fit) dt- \ tfit) dt
h Jo Jo Jo Jo

Thus, £ j f(t)(x -t)dt = xfix) + j fit) dt - xfix) = J fit) dt
Differentiating the other integral.

Thus, the two original integrals have equal derivatives.

\fit)ix -')dt=\ M /(v) dv\dt + C

Letting Jt = 0, we see that C = 0.

J"' 2 T 2

Vx dx = -x'''^ = -. The corresponding
0 3 Jo 3

Riemann Sum usmg right-hand endpoints is

5(n) =

'1

Thus, lim

= -^[^^ + 72 + • • • + v^]

yi + V2 + ■ • ■ + >/« 2

-,3/2

3'

12. (a) Area = ( (9 - x^) & = 2 I (9 - jc^) ate

= 2[27 - 9] = 36

(b) Base = 6, height = 9. Area = \bh = |(6)(9) = 36.

(c) Let the parabola be given by y = b^ — a^x^, a, b > 0.

rb/n

Area

rb,

= 2 it^- aV) dx
Jo

b^x - a^j^^

= 2

= 2

a 3 a

4^
3 a

2b

Base = — , height = b^
a

. .• . ,^ ■ . 2/'2fo\,,„ 4fc3

Archimedes Formula: Area = - — Hb^) = — —

3\ a / i a

4 -2-1 12 4 5

Problem Solving for Chapter 4 491

14. (a) (1 + /)' = 1 + 3( + 3r + P => (1 + i^ - i^ = 3P + 3i + 1

(b) 3r + 3; + 1 = (; + 1)^ - (3

J;(3/2 + 3,+ l)= 2[('+iP-'']

1=1 1 = 1

= (23 - V) + (33 - 23) + ■ • ■ + [((n + 1)3 - «3)]

= (« + 1)3 - 1
Hence, (n + 1)3 = J^(3r + 3/ + 1) + 1

i=l

(c) (n + 1)3 - 1 = 2(3,-^ + 3/ + 1) = 23,-2 + 3Mk±i) + „

1 = 1 1=1 -^

£3/2 = „3 + 3„2 + 3„ _

3n(« + 1)

2«3 + 6«- + 6« - 3n2 - 3/j - 2n

2«3 + 3?!- + n

n(/i + l)(2n + 1)

A., ^ w(^; + \)(2n + 1)
,^/ 6

16. (a) C

(b) C

/•20

= »'J.

12 sin

12 sin

vit - 8)

12

7T<r-8)
12

r 14.4 iT{t-

dt= cos —

L 77 12

-8)

-14.4

(-1 - 1) ^$9.17

A. \ 14.4 77<f-8) ^,,

j\dt = cos -r 0.6f

J L TT 12 Jio

14.4/ -V^

10.

H^

14.4/ 73

= $3.14

Savings = 9.17 - 3.14 = $6.03.

18. (a) Let A

Jo/U) +

fix)

f(b - x) ■
Let u = b - x, du = - dx.

, r f{b-u) . ^-
^ f /«^-«) ■

^ f /(fc-.x) ,

Jo/(^--^)+/W

Then,

2A

= f fix) , . f /(fc--v) ■

Jo/W + /(i - x) ^^ "^ Jo/(fc - .t) + fix) '^

Jo

\dx = b.

Thus, A = -.

(b)

Jo sin(l — j:) + sm.x 2

CHAPTER 5

Logarithmic, Exponential,

and Other Transcendental Functions

Section 5.1 The Natural Logarithmic Function: Differentiation .... 493

Section 5.2 The Natural Logarithmic Function: Integration 498

Section 53 hiverse Functions 503

Section 5.4 Exponential Functions: Differentiation and Integration . . 509

Section 5.5 Bases Other than e and Applications 516

Section 5.6 Differential Equations: Growth and Decay 522

Section 5.7 Differential Equations: Separation of Variables 527

Section 5.8 Inverse Trigonometric Functions: Differentiation 535

Section 5.9 Inverse Trigonometric Functions: Integration 539

Section 5.10 Hyperbolic Functions 543

Review Exercises 548

Problem Solving 554

CHAPTER 5

Logarithmic, Exponential, and Other Transcendental Functions

Section 5.1 The Natural Logarithmic Function: Differentiation

Solutions to Even-Numbered Exercises

2. (a)

(b) 3

The graphs are identical.

4. (a) In 8.3 = 2.1163
(b) -A = 2.1163

6. (a) In 0.6= -0.5108
(b) -df -0.5108

8. /(.x)= -Inx

Reflection in the .i-axis
Matches (d)

10. /W = -\n(-x)

Reflection in the v-axis and the .v-axis
Matches (c)

12. /(.v) = -2\nx
Domain: .ir > 0

14. fix) = \n\x\
Domain: x i^ Q

16. g(x) = 2 + In.T
Domain: .r > 0

18. (a) In 0.25 = In 5 = In 1 - 2 In 2 = - 1.3862

(b) In 24 = 3 In 2 + In 3 = 3.1779

(c) In ^12 = 5(2 In 2 + In 3) = 0.8283

(d) In^ = Inl - (3In2 + 21n3) = -4.2765

22. \nxyz = In.v + In v + In c

20. ln^^= In 2'/- = ?ln2

24. Inv a - 1 = ln(a - 1)"= = (;) ln(a - 1)

26. In 3e= = In 3 + 2 In e = 2 + In 3

28. In - = In 1 - In e = - 1
e

493

494 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

30. 3 Inx + 21ny - 41nz = \i\x^ + Iny^ - Inz!*

= ln

x'y^

32. 2[\nx - ln(jc + 1) - In^ - 1)] = 2 In

In

{x + l){x - 1) W - 1

X V

34. |[ln(;c2 + 1) - InU + 1) - \n(x - 1)] = |ln. /' t ^ ,.

In

V(i^

36.

3

/

38. lim ln(6 — ;c) = — oo

a:->6"

40. lim In— 7==

^-►5* Vx - 4

ln5 = 1.6094

42.

>> = Inx
, 3

3/2

= -Inx

At (1, 0),

y'

_ 3
~ 2-

44. >- = lnx'/2 = iln^:

•' 2x
At(l,0),>'' = i

46. /iW = ln(2x2 + 1)

1 \x

2x2 + r ^ 2jc2 + 1

48. y = ;c In X

^ = 4-) + Inx = 1 + Inx
dx \xJ

50. >- = lnVx2 - 4 = - InU^ - 4)
rfy _ 1/ 2r

a[r 2\jc2 - 4/ x^ - 4

52. f(x) = In
/'(x) =

2x

,x + 3

1 1

In 2x - ln(x + 3)
3

X x + 3 x{x + 3)

54. h(t)

h'(t)

\nt

ti\/t) -\nt \ -Int

56. y = In(lnx)
dy _ lA _ 1

lie \nx a: In x

58. y = In 3/^ = h^i^ " D " ln(^ + D]

X + 1 3'-

, ^ ir_! !_

' 3U - 1 x+ \

1 2 2

3 ^2 - 1 " 3(a:2 - 1)

60. /W = \n{x + V4 + x^)
1

/'W =

x+ V4 + x2 V x/4T

1 +

1

-.--c ^^5

-i.^5

> ^ -

^' '"

c-Jc-f

-.;c<ef<

Section 5. 1 The Natural Logarithmic Function: Differentiation 495

62. > = ^ 4ln( ) = —^^ -\ril + JJT^) + -\nx

1

H

^ ^ -lx'^{x/Jx- + 4) + 4;cV;c^ + 4 _ 1/ 1

dx 4x* 4^2 + Vxn^J\V^^^T4] ' 4x

Note that:

1

1

2 - y;c- + 4 2 - Vx- + 4

2 + Va;2 + 4 2 + 7.x:2 + 4 2 - Va:= + 4

-;c2

Hence, -i- =

-1

y^M^ 1 (2 - Jx^ + 4)
dx " IxJx^ + 4 x^ 4 -.t2

-1 + (l/2)(2 - V?T4) ^ y/jc^ + 4 ^ 1
Ixjx^ + 4 .t^ 4j:

Vx^ + 4/ ' 4x

- V.t- + 4 Vj:' + 4 J_ ^ Jx^ + 4
4;c7;c- + 4 -T^ 4j: .x^

64. >! = ln|csc;<:|

— CSCJ: • cot AT

y

cscj:

= -cotx

66. 3; = Inlsecx + tan.i;|

dy _ sec .r tan .r + sec'.r
dx sec .r + tan -T

sec .i:(sec x + tan x)

sec a: + tan x

secx

68. y = InVl + sin^.x: = -ln(l + sin^x)
dy /'1\2 sin.i:cos;c sin j: cos x

dx \ll 1 + sin-x 1 + sin-.r

70. ^(.r)

/•Ihj:

+ 3)rff

g [x] = L(ln.r)- + 3J— (ln.r) =

dx X

(Second Fundamental Theorem of Calculus)

72. (a) y = 4 - x2 - Inl -X + 1 ), (0. 4)

^=-2;, \ i

dx (l/2).r + 1 V2

= -2jc-

1

x + 2

When.x: = 0,^= -\.
dx 2

Tangent line: >• - 4 = - -[x - 0)

y = --.r + 4

74.

ln(.xT) + 5.r = 30

In.T

+ In y + 5.r = 30

1_
x

+ 1^ + 5 = 0
ydx

\dy _ 1
y dx X

- 5

dy_ y

dx X '

- 5y = -

-F^

(b)

496 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

76.

y = j:(ln x) — Ax

y' = x\-\ + \nx — A = — 3 + lnj:
(x + y) — xy' = X + x\nx — Ax — x(—?) + \r\x) = Q

1%. y = x - \nx
Domain: x > Q

V ' = 1 = 0 when x = I.

X

/'=-> 0
X-

Relative minimum: (1, 1)

80. v =

Injc

Domain: x > Q

, _ x{\lx) -\nx _ 1 — Inj:

= 0 when x = e.

u.-')w.¥^)

„ ^ ^{-\lx) - (1 - ln;c)(2x) ^ 2(lnx) - 3

Relative maximum: (e, e~ ')
Point of inflection: {e'^l^,\e''^l'^

0 when x = e^l'^.

82. y = x^ln -. Domain j: > 0

y = j:'I-I + 2j:ln- = jcI 1 + 2in-l = 0 when

-1 = 21n =

In-

x = 4e-i/2

/'=l+21n^ + 2x(y = 3 + 21n|

y"= 0 when;c = 4€-3/2 ' '

Relative minimum: (4e" '^^, - 8e" ')
Point of inflection: (Ae'^^^, -lAe'^)

4(4f-3'2.-24e-^

84. /(x)=xln^ /(I) = 0
f'(x) = 1 +ln;c, /(1)= 1

fix) = ^. /"(I) = 1

P,W =/(l) +/'(1)U - 1) = ;c - 1, P,(l) = 0

P^ix) = /(I) + f'{\)(x - 1) + |/"(1)U - 1)^

= ix- 1)+|U- 1)^ ^2(1) = 0

/•,'« = 1, P,'(l) = l

Fz'W = 1 + U - 1) = ;c, PjXl) = 1

The values off, Pj, Pj, and their first derivatives agree at
X = 1. The values of the second derivatives of/ and Pj
agree at;c = 1.

\P2

/

V

Section 5.1 The Natural Logarithmic function: Differentiation 497

86. Find x such that In x = 3 - x.

fix) = X + (In a:) -3 = 0

n

1

2

3

Xn

2

2.2046

2.2079

fUn)

-0.3069

-0.0049

0.0000

Approximate root: x = 2.208
90. y

\ X- +

- 1^
1

\r

lny = ^lnU2- I) - \n{x^ + 1)]

}_dy ^ I
y dx 2

2x 2x 1

.r= - 1 .v2 + ij

^ = A^ - 1 r 2x 1
ate " V.)c2+ iLx^- ij

U" - l)'/-2x

(.X^ + l)'/2(;c2 - 1)(.^2 + 1)

(x^ + iy'~{x- - 1)

1/2

88. y = yU - \){x - 2)(;c - 3)

\ny = ^[\n(x - 1) + ln(;c - 2) + ln(jc - 3j]

i(^uir_L_ + _^ + _L_i

y\dxl llx - \ X - 2 x-3]

^ If 3j:- - 12x + 11 1
2|_U- l)U-2)(x-3)J

dy ^ 3x' - lit + 11
dx 2v

3x- - 12x + 11
2VU - 1)U - 2)U - 3)

92. y

{x +\){x + 2)

ix - l){x - 2)
Iny = InU + D + InU + 2) - ln(.t - 1) - Hx - 2)

1 1

+

1

1

Udy

y\dxj X + \ X + 2 x — \ X — 2

dy

dx

XT — I X — 4

-6x- + 12 ]

■^ - DCr^ - 4) J

(x-

-6(x^ - 2)

(x + l)(x + 2)

ix - \)ix - 2) ix+ l)(x - l){x + 2)U - 2)

6(x^ - 2)
U - D^Ct - 2)2

94. The base of the natural logarithmic function is e.

96. gix) = ln/(x),/(x) > 0
.fix)

g'(x)

fix)

(a) Yes. If the graph of g is increasing, then ^'(.v) > 0.
Since/(x) > 0, you know that/'U) = g'ix)fix) and
thus, fix) > 0. Therefore, the graph of/ is increasing

(b) No. Let/(.r) = .r- + 1 (positive and concave up), gix)
Intxr + 1) is not concave up.

98. t

5.315

6.7968 + In.v'
(a) 50

1000 < x

(b) f(1167.41) = 20 years

T= (1167.41)(20)(12) = $280,178.40

(c) f(1068.45) ^ 30 years

T = (1068.45)(30)(12) = $384,642.00

(d) ^= -5.315(-6.7968 +
dx

5.315

.,,.(1)

.v(- 6.7968 + ln.v)=

Whenx = 1167.41. dt/dx « -0.0645. When.r = 1068.45,
dt/dx=- -0.1585.

(e) There are two obvious benefits to paying a higher monthly
payment;

1. The term is lower

2. The total amount paid is lower.

498 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

100. (a) 350

_, ,„ , 34.96 ,3.955

(b) T (p) = + — 7=^

P Jp

r'(10) = 4.75 deg/lb/in^

r'(70) = 0.97 deg/lb/in^

(C) 30

lim r'(p) = 0

102. >> = lOlnf^^"^ ^^°° — ^j - VlOO - x^ = 10[ln(l0 + VlOO - ;c2) - Inx] - VlOO - x^
(a) ?2 (c) lim ^ = 0

j:->10- dX

dy

Vioo - xKio + Vioo - x^)

-a

yioo"

-10

yioo"

10 + Vioo - x^.

-10

-10 +

Vioo - jc'

Vioo - r

10+ Vioo - x'^
Vioo - x^

+ 1

10 + Vioo - x^

10

X Vioo - x^

10

x

w

x

10 + Vioo - x2 Jc

;c(lO - VTOO^^ - 10 ^ Vioo - x^

When x = 5, dy/dx = - Vs. When x = 9, (fy/dc = - Vl9/9.

104. y = in X

y ' = - > 0 for jc > 0.

Since In x is increasing on its entire domain (0, oo), it is a
strictly monotonia function and therefore, is one-to-one.

106. False

TT is a constant.
d,

dx

[In tt] = 0

Section 5.2 The Natural Logarithmic Function: Integration

\ — dx= \o\-dx= lOlnj

x\ + C

4. u = X — 5, du = dx
1

/.

:& = Inx - 51 + C

\z^2'' = \h

2'' = -2kT2^'^'^

'x- 5
S. u = 3 - x^, du = — 3x'^ dx

= -In |3x + 2| + C

Js^'^^-l/l^^-^x^)^

--ln|3 -x^\ + C

Section 5.2 The Natural Logarithmic Function: Integration 499

10. M = 9 - x\du = -2xdx

J V 9 - x^ 2J

r xU + 2) if 3;c- + 6^:

J ^ + 3x2 - 4 3J ^ + 3^2 _

dx (u=x^ + 3x' - 4)

^ln|x3 + 3x2-4| + C

,^ r2;r + 7A-3^ f

7a- - 3 ^ I /^ , , 19 , _,
-dx = 2x + 1 1 + dx

x-2)
= x^ + lU + 191nU - 2| + C

18.

fx^ - 3;c2 + 4;c - 9 , f

J — ^^n — '^ = J

-3 + .1: +

a:- + 3

a[t

a;2 1

= -3;t + y + -lnU= + 3) + C

-1^^^- = /

^-^K-— "-^i-

3 2

+ 19;c- n51nU + 5| + C

JxlnCr^) 3 Jinx x

J ln|ln|x|| +C

22. M = 1 + x'/3 du

3x2/3

a^

jx2/3(l +x'/3)'^ Y

/3)'^ = 3jY^(^j^

31n|l +x'''3| + c

= Inlx - ll +

1

2(x - 1)

- + c

26. M = 1 + V3x, rfM = -A=dx ^ (& = -(«- 1) rf«
2v'3x 3

J 1 + V3I J«3

l)d«

rfM

^M - ln|M|] + C

= 1^1 + V3^ - ln(l + v/3l)] + C

= |v/3x-|ln(l + V3^) + C,

500 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

/:

'x- 1

28. u = x'/3 -\,du= T-^ dx => dx = Z{M + Xfdu

dx = 2{u+\)^du

= 3 f ^^^(«' + 2« + 1) du
= 3 ( M^ + 3« + 3 + -j rf«
= 3rj + ^ + 3M + lnlM|l + C

3^

3 2

1/3 - 1)3 , SU'/J - 1)2

+ ■

+ 3(;c'/3 - 1) + ln|x'/3 - l|

+ C

3x2/3

= 3 ln|;c'/3 - 1 1 + =^ + 3;c'/3 + x + Q

9. \tan5ede = -

30. |tan5erfe = Tl^^rffl
cos 56

= --ln|cos5e| + C

32. I sec - (fe = 2 I sec -( - j <ic

= 21n

x , X

sec - + tan -

2 2

+ C

34. M = cot t, du = — csc^ f A

rcsc-^ t

f-

cot f

-rf; = -In cot f + C

/'

36. (sec f + tan r) A = Inlsec r + tan fl - Inlcos t\ + C

= ln

sec r + tan f

cos t

+ C

= ln|sec t(sec f + tan r)| + C

38. y

h

2x

dx

= ln|;c2 _ 9| + c

(0,4): 4 = ln|0- 9| + C
)' = ln|;c2_ 9| +4- in9

C = 4 - ln9

)

■^^ (0, 4)

^

7

V

40. , = f-J££il-

J tan f + 1

dt

= ln|tanf + 1| + C
(77,4): 4 = ln|0+ 1| + C =» C = 4
r = Inltant + ll + 4

w

^

<T.'4)'

42.^ = ^,(1,-2)

(a) y

X 2

(In 1)2

+ C ^ C= -2

Hence, y = - 2.

Section 5.2 The Natural Logarithmic Function: Integration 501

^•£7T3'^=K^2|]-

In 3 - In 1 = In 3

46. u = \n X, du = — dx

X

J^ x\nx j^ \lnxjx

In InU

= ln2

48. f^— |-A= I 1^+ |— TT^
Jo j: + 1 Jo Jo ^ + 1

[

= U - 21nl;c + ll

1 - 2 In 2

J -0.2 ro.2

(esc 20 - cot ley de = (csc^ 20-2 esc 20 eot 20 + cot^ 20) dd
0.1 Jo.i

ro.2

•20- 2ese20cot20 - I) dO

J -0.2
(2 esc2
0.1

= -cot 20 + esc 20 - 0 = 0.1

0024

52. ln|sinx| + C = In

+ C = -ln|eseA:| + C

54. -Inlesej: + cotJ:| + C = —In

-In

(escj: + cot x)(ese x - eot x)

(esc ;c - cot x)

+ C = -In

csc-.t - eot"^.x

esc j: — eot X

+ C

1

cscx - eotx

+ C = In [esc x - cotxl + C

56.

\- ^dx = -(l + V^f + 6(l + V^) - 41n(l + v^) + Ci

J 1 + V x

= 4v^ - X - 4 ln(l + v^) + C where C = C, + 5.

58. — dx = -rflnlsec Zx + ian2x\ — sin 2x] + C

ps

J-7r/4

^. , sin"^x - eos'j: ,
60. dx

[ini

secx + tanx - 2 sinx

ir/4

-ir/4

Inl'^^'l - 2v^ - - 1.066

Note: In Exercises 62 and 64, you can use the Second Fundamental Theorem of Calculus or integrate the function.

62. F{x)

F'{x) = tanx

= I tantdt

Jo

64. F{.x) = j^^

dt

Zx 2

F\x) = 4 = -

.r- .V

502 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

66.

A = 3

Matches (a)

68. A

r^-i

X + 4 In ;

>.]'

4 + 4 In 4 - 1

3 + 4 In 4 » 8.5452

••I

70. {2x - tan(0.3x)) dx

x'- + ■

10

ln|cos(0.3x)|

10,

16 + Ylncos(1.2)

]-hf

In cos(0.3)

« 11.7686

72. Substitution: (m = x^ + 4) and Power Rule

74. Substitution: (« = tan ;c) and Log Rule

76. Answers will vary.

78. Average value =

1 r4(x-

4 - 2J2 ^2

4., J

+ 1)

dx

+ -z\dx
X x-^'

re

82. f

I

= 2

1

In 4-;^- In 2 + 1

ln2 + ^l = ln4 + ^« 1.

8863

10

In 2 ,„ r - 100

dT

80. Average value

10

In 2

ln(r - 100)

= i^[.n200-lnl50] = i^[ln(|;

2-oJo

TTX ,

sec-— dx
6

l/'6

In

TTJC . TTX

sec -^ + tan — -
6 6

L2\Tr,
= -[ln(2 + 73) - ln(l + 0)]

= -ln(2+ 73)

=» 4.1504 units of time

84.

dS^k

dt ~ t

S{t) = \ -dt = k\n\t\ + C = klnt + C since r > 1.

5(2) = A: In 2 + C = 200
S(4) = k\n4 + C = 300
Solving this system yields k = 100/ln 2 and C = 100. Thus,

5W - -!5^ + 100 - 1001

[^-1

Section 5.3 Inverse Functions 503

86. k= I: /iW = X- \

Vx- 1

k = 0.5: /0.5W =

A: = 0.1:/o.,W

lim L (x) = \nx

;t-.0*

0.5
0.1

= 2(v^ - 1)

= io('^- 1)

88. False

— [Inx] = -
ax or

90. False; the integrand has a nonremovable discontinuity at
x = 0.

Section 5.3 Inverse Functions

2. (a) f(x) = 3 - 4a:

/(.W)=/f^)-3-4i^Ux

'g(/W) = g(3 - 4x) = ^ ^l ^-"^ = X

(b)

/\..

4. (a) /W = 1 - x' •

f(g{x)) = /( yp^) = 1 - ( yr^^)

= 1 - (1 - x) = JC
g(/W) = gil - x')

= Vi - (1 - ;c^) = l^ = .t

6. (a) /W = \6 - x\ x > 0

g(x) = 716^^
/(gW) =/(yT6^^) = 16 - (VT6^^)2

= 16 - (16 - x) = X
g{f(x)) = g(16 - ;c2) = Vl6 - (16 - x^)

(b)

8 12 16 20

= vj: = JC

8. (a) f(x) = — — , .X > 0
1 + X

g(x) = ^-^^, 0 < .X < 1

f{g{x))=f

gifix)) = gl

x

1 -.V

1 1

X X

1

1 + X

1 + -Y _ X 1 + -V

1 ~ 1 + .X- ' 1

1 +.r

(b)

504 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

10. Matches (b)

14. /U) = 5.r- 3

One-to-one; has an inverse

12. Matches (d)

16. F(x) =

x^ + A

Not one-to-one; does not have an
inverse

18. git)

1

Jt^ + l
Not one-to-one; does not have an
inverse

20. fix) = 5x^x - 1

One-to-one; has an inverse

22. hix) =\x + 4\ - \x-4\

Not one-to-one; does not have an inverse

24. fix) = cos-

3x

, . 3 . 3j: Itt 4tt

fix) = —— sin—- = 0 when x = 0, -r-, ^r-, ...
•' 22 33

/is not strictly monotonic on (— oo, oo). Therefore, /does not have an inverse.

26. fix) = j^ - 6^2 -t- 12x

/'W = 3;c2 - 12t -K 12 = 3(;c - 2)^ > 0 for all x.

/is increasing on (-oo, oo). Therefore, /is strictly
monotonic and has an inverse.

28. fix) = ln(x- 3),;c > 3
1

fix)

x-3

> OfoTx > 3.

/is increasing on (3, oo). Therefore, /is strictly monotonic
and has an inverse.

30. fix) = 3x = y

"I .

X

/-w = f

32. fix) =x^ - 1

x = ^y + 1
y = ^x + 1

r'W = ^^TTT

34. fix) =x^ = y, 0 < X

X = Vy

y = Vx

f-'ix) = ^

Section 5.3 Inverse Functions 505

36. fix) = Jx^- A = y, x>2
X = ^y^ + 4
y = Jx^ + 4

/-'W = V.r^ + 4, jc > 0

38.

fix) = 3 V2x - 1

y5 + 243
^ 486

.t^ + 243
^ 486

j^ + ''43

486

h^

C^

1

The graphs of/and/~'
are reflections of each
other across the line v =

40. fix) = a:V5 = y

V = ^5/3

/-'W = Ar5/3

/^

^

The graphs of f and/" ' are
reflections of each other across
the line v = x.

X + 2
42. fix) = ^-^ = V
.r

/-'W =

y- 1

2
j: - 1

2
;c - 1

■ — : —

^

1

\

44.

X

0

2

4

/-'W

6

2

0

The graphs of /and/"' are
reflections of each other across
- .. • the line y = x.

46. fix) =ki2- X- X?) is one-to-one for all k ^ 0. Since/-'(3) = -2,/(-2) = 3 = ^2 - (-2) - (-2)3) = 12A.- => k = \.

48. fix) = \x + 2| on [-2,00)

/'W=^^l) = l >0on(-2,oo)

X + 2

/is increasing on [-2, 00). Therefore, /is strictly
monotonic and has an inverse.

52. fix) = secjcon

0.-

50. fix) = cot -t on (0, tt)

fix) = -csc-.t < Oon (0, tt)

/is decreasing on {0, tt). Therefore, /is strictly monotonic
and has an inverse.

fix) = sec X tan x > 0 on I 0, — I

/is increasing on [0, it/2). Therefore, /is strictly monotonic and has an inverse.

506 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

54. /(.r) = 2-^ = yon(0, 10)

2^2 _ 3 = y.Zy

xK2 - y) = 3

f-'{x)

56. (a), (b)

= -V2

3

~ y

= *V2-.

2

K'

/^.

12

The graphs of/ and/"' are
reflections of each other
across the line y = x.

58. (a), (b)

(c) h is not one-to-one and does not have an inverse.
The inverse relation is not an inverse function.

(c) Yes, /is one-to-one and has an inverse. The inverse
relation is an inverse function.

60. /(x) = -3

Not one-to-one; does not have an inverse

64. f(x) = 16 — j:^ is one-to-one for j: > 0.
16-;c^ = y
16 -y = x^
4/16^ = ^:
Vl6 - x = y

f-\x) = yi6 - X, X < \6

62.

f{x) =

ax + b

/is one-to-one; has an inverse

ax

+ b

= y

X

y-b

a

y

X - b
a

f-

-'W

-'-\a^Q
a

66.

/w =

1^-

3| is one-to-one for X > 3.

x

- 3 =

= y

X =

--y + 3

y

-x + 3

f-

Kx)-

= X -1- 3, X > 0

68. No, there could be two times t^ i= t-^^ for which

hit,) = h{t^.

70. Yes, the area function is increasing and hence one-to-one.
The inverse function gives the radius r corresponding to
the area A.

72. fix) = ^ix? + 2x3);/(-3) = ^("243 - 54) = - 11 = a.
1

fix) = ^5x^ + 6x2)

(/-')'(- n)

1

1

27

^ ^ ^\_

7'(/-'(-ll)) /'(-3) 5(-3)^ + 6(-3)2"l7

Section 5.3 Inverse Functions 507

74. f{x) = cos 2x,/(0) = 1 = a

f'{x) = -2sin2x

^•^"'^'^'^ =7lrW) =7W = ^i^ = ^ which is undefined.

76. /W = v^r^^,/(8) = 2 = a

1

Z'U)

(/-')'(2)

2jx- 4

■/'{/- '(2)) /'(8) 1/(278^^) 1/4

78. (a) Domain/ = Domain/ ' = (-00,00)
(b) Range/ = Range/"' = (-00,00)
(c)

80. (a) Domain / = [0, 00), Domain /-' = (0, 4]
(b) Range / = (0, 4], Range /"' = [0, 00)
(c)

(d) /(.r) = 3 - 4x, (1, - 1)

fix) = -4

/'(I) = -4

f-\x)=^. (-1,1)

(r')w = -^

(/-')'(-!) =4

(d) /w=i',.

f'(,A ~<iX „ , ^

/(->) (^ + „r /(I) 2

rw=7^

— 9

1

2

1 2 3

82. x = 2 ln{y^ - 3)

1 = 2-

1

.rfy

^^2^"^

dtc 4y ■A'^"''^^'^^ 16 16-

In Exercises 84 and 86, use the following.

fix) = ix-3aQdg{x) = x-^
J^\x) = S{x + 3) andg-Hx) = ^

84. (g-' o/-')(-3) = r'(/-'(-3) = g-'(O) = 0

86. (g-' og-')(-4) = g-'(g-H-4)) = g-il-4)
= V^^= -^4

508 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

In Exercises 88 and 90, use the following.

fix) = X + 4 and g(x) = 2x - 5

J^\x) =x-4 andg-\x) = ^^

90. (g "/)(x) = gifix))
= gix + 4)
= 2(x + 4) - 5
= 2a: + 3

88. if-'

'g-')(x)=f-'(g-'ix))

■ -rm

=^-

x-3

Hence, {g »/)"'«

x-3

(Note:(go/)-'=/-'og-')

92. The graphs of/ and / ' are mirror images with respect
to the line y = x.

94. Theorem 5.9: Let/be differentiable on an interval /.
If/ has an inverse g, then g is differentiable at any x for
which f'(g{x)) + 0. Moreover,

g\x)

1

/'(gW)

. f\g{x)) + 0

96. /is not one-to-one because different j:- values yield the
same y- value.

Example: /(3) = /

Not continuous at ±2.

98. If/ has an inverse, then /and/ ' are both one-to-one.

Let (/-')~'W = y then x = r\y) and/W = y.
Thus, (/-•)-' =/

100. If/has an inverse and/(xi) = f(x-^, then/"'(/Ui)) =/~'(/U2)) =^ ^i ~ ^2- Therefore,/is one-to-one. If/W is one-to-
one, then for every value b in the range, there corresponds exactly one value a in the domain. Define g{x) such that the
domain of g equals the range of/ and g(Z?) = a. By the reflexive property of inverses, g = /"'.

102. True; if/has ay-intercept.

104. False

Let/U) = xoxg{x) = \/x.

106. From Theorem 5.9, we have:

,. ,_fUx)m-f"{g{x))g\x)
ng{x)) ■ [i/r(gU)))]

\f'ig{x)W
figix))

\fUxm

If/is increasing and concave down, then/' > 0 and/" < 0 which implies that g is increasing and concave up.

Section 5.4 Exponential Functions: Differentiation and Integration 509

Section 5.4 Exponential Functions: Differentiation and Integration

2. e-2 = 0.1353.

lnO.1353. .. = -2

8. 4^^ = 83

83

e'

X = ln(f

- 3.033

12.

200e-

-4c —

15

e~

-4i =

15 3
200 40

^

4x =

■»(i)

x =

i'"(f)'

0.648

16.

ln4x

= 1

4a:

= e'

= e

a;

e

" 4

- 0.680

In 0.5 = -0.6931...

6.

gta2x= 12

0.6931. . . = i

2;c = 12
X = 6

10. -6 + 3^^ = 8

ie' = 14

-^

-

. = ln(^

) =

= 1.540

14. In;c2=10

^2 = glO

= ±e' ^±148.4132

18. Hx - 2)2 = 12

(x - IT' = e>2
X - 2 = e«

x = 2 + e^ == 405.429

20. V = W

22. V = e"^/2

24. (a)

(b)

"^

V.

Horizontal asymptotes: y = 0 and y = S

Horizontal asymptote: v = 4

510 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

26. y = Ce"""

Horizontal asymptote; y = 0
Reflection in the y-axis
Matches (d)

28. y =

1 +e-

lim

j^oo 1 + «-"

lim -z

x-*-ca I + e

= c

0

Horizontal asymptotes: y = C and y = 0
Matches (b)

30. fix) = e^/3

g(x} = In x^ = 3 In jc

32./W = e--'
g(x) = 1 + Inj:

34. In the same way.

lim

(-3-=

e^ for r > 0.

,^,,111 1 1 1

36. l + l+x + T + :rT + ttt + ^rr +

2.71825396

2 6 24 120 720 5040

e == 2.718281828

.>l + l+^ + ^ + :^ + 4. + a.+ 1

2 6 24 120 720 5040

38. (a) y = e^

y' = 2e^
At(0, l),y'= 2.

(h) y = e-^
y'= -2e-^
At (0,1),)''= -2.

40. /W = e'-^
fix) = -e'--

42. >> = e

= ^-^

^=-2x.-^

44. v = x^e

dy
dx

—x^e ^ + 2xe ^

= xe-'{2 - x)

46. g(/J = e-^'^

g'it) = e-V/^Cer-^)

S^S/r'

fe

48. >> = Ini

1 + e'

1 - e^,
ln(l + e') - ln(l - e')

+ ■

dr 1 + e' ■ 1 - e'

2e^

1 - e

2x

50. > = Inl

e^ + e"

In(e^ + g-^ - In 2

dy _ e^ — e ■
dx e^ + e~-

e^ - 1

e^ + 1

52. y =

e' — e

^~ 2

54. y = xe^ — e' = e'ix — 1)

dx

= e' + e'ix - \) = xe^

Section 5.4 Exponential Functions: Differentiation and Integration 511

56. f(x) = e^\nx

fix) = -

X

58. y = \ne' = X
dx

60. e^ + X? - f = 10

\x^ + y\e^ + 2x - ly^ = Q
\ dx dx

di
dx

(xe^ - 2y) = -ye^' - 2x

dy
dx

xe^ - ly

62. g{x) = -Jx + e'Xnx

1 e'

— P + — + e* In j:

,,, . 1 xe' — e' e'

1 , e^jlx-l) , ,,
- H :: h e' In jc

4xv^

;c2

64. y = ^^(3 cos 2x - 4 sin Ix)

y' = e^i-6smlx - 8 cos 2x) + f^(3 cos 2x - 4 sin 2x)
= e^{—\Qsix\lx - 5 cos Ix) = - 5e'^(2 sin Ix + cos 2x)

y" = - 5e'(4 cos 2x - 2 sin 2j:) — Se'il sin 2j: + cos Ix) = - 5e^(5 cos 2j:) = - 25^^ cos 2x
y"- ly' = -25e'cos2jc - 2(-5e')(2 sin 2j: + cos 2;c) = -5e*(3cos2x - 4 sin 2j:) = -5y
Therefore, y" - ly' = - Sy => y " - 2y ' + 5.v = 0.

/■

66. f{x) =

e' — e ^

/'W = ^-^ > 0

/W

oX — a X

= 0 when x = 0.

Point of inflection: (0,0)

(0.0)

/

A

68. g(x) = ^e-(--3)V2

^'W = -^(;c-3>-(-3)V2
^2^

5"W

1

/27r
Relative maximum

{x - l){x - 4)e-(^-3)V2

^ ('•*)

(3, 0.399)

Points of inflection: I 2, ^=«- '/^ |, ( 4, — L=^- '/:

- (2, 0.242), (4, 0.242)

70. fix) = xe-''

f'(x) = -A:e~-' + e~' = e"^(l -;«:) = 0 when.v = 1.

f"(x) = -e--^ + (-e-')(l - x) = e-'ix - 1) = Owhenx = 2.

Relative maximum: (l.e'')

Point of mflection: (2, 2e-2)

(I..--')

/

512 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

72. fix) = -2 + e3^(4 - 2x)

fix) = eH-2) + 3e3^(4 - 2x) = e'^ClO - 6x) = Owhen;c = f.
fix) = e^1-6) + Se^HlO - 6x) = £^^(24 - 18;c) = Owhen;c = f.
Relative maximum: (5, 96.942)
Point of inflection: (|, 70.798)

74. (a) fie) = fie + x)

lOce"'^ = 10(c + ;c)e-<''+^)

C_ _ C + X

ce"^" = ic + x)e''
ce' = c + X
ce^ — c = X

X

(c) Aix) = -J^^/»-^*
e^ — 1

(b) AW = jcf (c) = X

10'
10x2

e^- 1

e^ - 1

-(V(e'-i)l

e^/d-c')

(d) c =

e'- 1

lim c = 1
lim c = 0

The maximum area is 4.591 for a: = 2.118 and

fix) = 2.547.

76. Let (j;o> y^) be the desired point any = e ^.

-e " (Slope of tangent line)

1

-, = e' (Slope of normal line)

y — e~'^ = e^'ix — x^

We want (0, 0) to satisfy the equation;

f( 0.4263, ^-<'«")

1 = x^e^
XqC^ -1=0
Solving by Newton's Method or using a computer, the solution is Xq = 0.4263.
(0.4263, e-o"2«3)

Section 5.4 Exponential Functions: Differentiation and Integration 513

78. V= 15,0{)Oe-o«286, 0 < f < 10

(a) 20.000

dV

(b) ^ = -9429e-o«286'
at

Whenr =1,^= -5028.84.
dt

Whenr = 5,^- -406.89.
dt

(C) 20.000

80. 1.56e ''■^cos4.9r < 0.25 (3 inches equals one-fourth foot.) Using a graphing
utility or Newton's Method, we have f > 7.79 seconds.

'O'V

82. (a) V, = -1686.79f + 23,181.79

V^ = 109.52f2 - 3220.12t + 28,110.36

(b) The slope represents the rate of decrease in value of
the car.

(c) V3 = 31,450.77(0.8592)' = 31,450.77e^'"5i8t

dV,

(e) -r^ = -4774.2e-

dt

dV^

For f = 5, — ^ = -2235 dollars/year.

dV,
For f = 9, — ^ = - 1218 dollars/year.

(d) Horizontal asymptote: lim V3(r) = 0

As f — > 00, the value of the car approaches 0.

/"l

-^

\^'

/

\

84. fix) = e--^/2,/(0) = 1
fix) = -xe-^'\f'{Q) = 0

fix) = .t^e-^/2 - e-^/2 = e-^/2(;c2 - l),/t0) = - 1
P.W = 1 + 0(.t - 0) = 1, Pi(0) =1
P, '(.v) = 0, P, '(0) = 0

P^ix) = 1 + 0(.r - 0) - i(.r - 0)= = 1 - y, P^lO) = 1

P^'ix)^ -x,P2'{0) = 0

P.'\x) = -UP, 'to) = -1

The values of/, P,, P2 and their first derivatives agree at ;c = 0. The values of the second derivati\es of
/and P2 agree at jc = 0.

514 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

86. n* term is x"/n\ in polynomial:
r= x^ ;c^

X^ JC'

Conjecture: e^ = 1 + x + — + — +

»■/:

90. J e^-'dx

-e-i + 1 = 1-

>

88. Let M = -x^, du = -Ax^dx.
3^

e "\-Ax^)dx = e "' + C / ' vj

v.. ^~

^

c.

7

92.

[:t2^/2^ = ll

jt ->,' ' J ^

^^^(¥) '^ " \^"' "^ *^

J^d.= -ije

94. r^dx=-\\e^'4-^)dx=-\e^'^ + C

96. Let M = 1 + e^JT, rfw = le^dx.

+ e^A:) + c

98. Let u = ~z~. du = —x dx.

rV2 r.

xe-'^^-dx = -
Jo Jo

xe -''-dx = - f %--=/2(-x) a[x = [-^-^2!^-^=

1 -e-' =

e - 1

100. Let M = e^ + e^^ ^m = (e' - e-^)a[r.

/

e' + e'"

dx = ln(e^ + g-^) + C

102. Let M = e^ + e-", du = {e^ - e'") dx

- le

-2

e^ + e"

+ C

104. ^ ^ alt = (e^ + 2 + e-'^)dx

= e* + 2a: - e"^ + C

/'

106. le'^^^seclxtanlxdx = -e'^^ + C
(u = sec 2x,du = 2 sec 2j: tan 2x)

ln(e2»-i)a[x= I (

108. \\nie^-'^)dx = {2x - \)dx
x^ - X + C

112. f'{x) = I (sin;c + e^ a^ = -cosx + ^e^ + Q
/'(O) = -1+1+ C, =1^ C, = 1

e-^Pa[r

(e^ - 2 + e^2')a[«

= |e^ - 2;c - ^e~^ + C

f'(x) = -cosx + -e^+ 1

/W

= /

- cos x + -e^ + I ]dx

= - sin ;c + -e^ + x + C,
4 '^

/fO) = - + Q = -^Q = 0

/(x) == X - sinx + -e^
4

Section 5.4 Exponential Functions: Differentiation and Integration 515

114. (a)

(b)

di
dx

-0.21^

y = \xe-°^dx = —- \e-°^ {-OAx)dx

-0.2^

0.4

+ C = -2.5e-

+ C

0, -|]: -|= -2.5e'> + C= -2.5 + C =^ C=l

y = -2.5e-°-'' + 1

116.

f

Jn

e ' dx =

118.

f

Jo

(e-^ + 2)dj:

1

1 _.

+ 4 + 1 = 4.491

Jo

120. (a) Jxe" dx,n= 12

Midpoint Rule: 92.1898
Trapezoidal Rule: 93.8371
Simpson's Rule: 92.7385
Graphing Utility: 92.7437

/:

(b) 2x6-" dx,n = 12
Jo

Midpoint Rule: 1.1906

Trapezoidal Rule: 1.1827

Simpson's Rule: 1.1880

Graphing Utility: 1.18799

122.

\ 0.3-°^' dt = \
Jo

+ 1 =

-0.3 J = ±

2

-0.3a: = ln-= -In 2

ln2 ,,, .
X = —I- ~ 2.31 mmutes

124.

t

0

1

2

3

4

R

425

240

118

71

36

In/?

6.052

5.481

4.771

4.263

3.584

(a) InR = -0.6155f + 6.0609

K = g-0.6155; + 6.0«)9 = 428.78^""°*"^'

(b)

(c)

R(t)dt =
Jo Jo

428.78€

-0.61551 J, -.

dt = 637.2 liters

516 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

126. The graphs of f{x) = In j: and g{x) = e^ are mirror
images across the line y = x.

128. (a) Log Rule: (m = e^ + 1)
(b) Substitution: (m = x^)

130. \n-r = ]xie' -\ne^ = a- b
Ine""* = a - b

Therefore, In -r = In e" * and since y = In x is one-to-one, we have -r = e"

Section 5.5 Bases Other than e and Applications

Atro=16,y = y =-

-=(r

Atto = 2,y=(|)'^'-0.7579

6. log2,9 = log2,272/3=|

8. log„ - = log„ 1 - log„ a = - 1

. (a) 272/3 =

-- 9

10g27 9 =

2
" 3

(b) 163/" =

= 8

10gl6 8 =

I

A

12. (a) logs I = -2

3-2 = 1

(b) 49'/2 = 7
log49 7 = 5

14. y = 3^

16. y = 2^

j:

-1

0

1

2

3

y

1
9

1
3

1

3

9

18. y = 3-W

x

-2

-1

0

1

2

j:

0

±1

±2

y

16

2

1

2

16

y

1

I

3

1
9

12 3 4

20.

(a) log3

81 ^

= X

y--

1

" 81

X -

= -4

(b) logs

36 =

= X

6^ =

-- 36

X -

= 2

22. (a) log, 27 = 3
If' = 11
b = 3
(b) log, 125 = 3
fo3 = 125
b = 5

Section 5.5 Bases Other than e and Applications 517

24. (a) log3JC + log3(A:-2) = 1

logjW^c - 2)] = 1

x(x - 2) = 3'

;c2 - 2;t - 3 = 0

U + l)(x - 3) = 0

x= -10Rx = 3

j: = 3 is the only solution since the domain of the
logarithmic function is the set of all positive real
numbers.

(b) logjoU + 3) - logioj: = 1

logi

x + 3
' X

x + 3

= 10'

a: + 3 = 10a:

3 = 9jt

1
^ = 3

26. 5^ = 8320

6;cln 5 = In 8320

^ hi 8320
^~ 6 In 5

- 0.935

28. 3(5^-') = 86

5.-1 = 86
3

ix - l)ln 5 = ln(y)

^86

Ini

X- I =

In 5

;c= 1 +

In 5

= 3.085

30.

(-IT

365r ln( 1 + |^ j = In 2

t =

1

ta2

365

'"(-I)

6.932

32. log,o(f - 3) = 2.6

f - 3 = 1026

r = 3 + 102« = 401.107

34. logsVx - 4 = 3.2

4 = 53-2

;C - 4 = (53.2)2 = 56.4 ^

jc = 4 + 5*" = 29,748.593

36. /(f) = 300(1.0075'^) - 735.41
Zero: t = 10

(10.0)

38. g(x)= 1 -21og,o[xU-3)]
Zeros: x = -0.826,3.826

518 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

40. fix) = 3^

g(x) = l0g3 X

X

-2

-1

0

1

2

fix)

I

9

1

3

1

3

9

X

I
9

1
3

1

3

9

six)

-2

-1

0

1

2

42. gix) = 2--

g'ix) = -i\n2)2-

44. y = jcCe-^')

^ = 4-2(ln6)6-2') + 6-2^
dx

= 6-^-2;c(ln6) + l]

= 6-2^(1 - 2;c In 6)

32/
46. fit) = y

/'(/)

t(2 In 3) 3^ - 3^

3^(2fln3 - 1)

48. gia) = S""/- sin 2a

g'(a) = 5-"/2 2 cos 2a - lOn 5) S-"/^ sin 2a

50. y = logio (2j:) = logm 2 + logmX
dx ;clnlO a: In 10

52. /iW = log3

tV^c - 1

= logj x + - log3 (jt - 1) - logj 2

h'ix)=^^ + -

jc In 3 2 (jc - 1) In 3

-0

1

In 3

1
In 3

1_

X ^ 2{;t

3x- 2

2xix - 1)

]

54. y = logi

x^- 1

= logloU-- 1) - lOgioX

1

(iy 2x

dx ~ ix^ - l)lnlO X In 10

= 1 r 2x _ n

~ In loL^c- - 1 x\

1 r ;c^+ 1 ]
In loUU^ - 1)J

56. /(f) = r3/2 log, jm = I

3/2

1 ln(f + 1)

2 In 2

fit) =

1

21n2

[^rl

- + |/'/Mn(f + 1) I

58. )' = x'-'

Iny = (jc - l)(lnx)

rJ:-2 C„ _

ix - 1 + X In x)

60.

y = (1 + xf'^
Iny = -In(l + ;c)
1 I 1

KS=KTi7)-»»-f7

dx X

zM
xL^c +

(1 + x)^/^

InU + 1)1

"_J InU + 1)"

x+ \ X

^■\

62. 5-^^

In 5

+ C

64.

5^) A = I (27 - 25) dx

Section 5.5 Bases Other than e and Applications 519

66.

J(3-.)7'-^>^^=4j-

2(3 - x)!^^'"^' dx

2In7

£7(3 -.P] + c

S. 2

68. I 2 cos jc dx, « = sin jc, du = cos a; olx
1

In 2

2suijr + (;;

70. (a) y

,'-SWs-—- v////v\\v.

(b) "T = e^^-^cosx (-77,2)

ox

=/'

(it, 2): 2 = e^""' + C = 1 + C ^ C = 1

72. \og^x

\nx _ logio^t
In b logio b

74. /W = logioJT

(a) Domain: x > 0

(b) y = logioJ:
10>' = a:

/-•W =10'

(c) log,o 1000 = log.o 1(P = 3
Iog,olO,000 = log,o 10* = 4

If 1000 < .V < 10,000, then 3 < f{x) < 4.

(d) If/W < 0, thenO < .t < 1.
(e)/U) + 1 = log,oJt + log,ol0
= log,o(10;c)
X must have been increased by a factor of 10.

(f) 10glo(^-j = log,o-^| - log,o-t2

= in — n = 2n
Thus, jtiAj = 10^ = 100^.

76. f(x) = a^

(a) fiu + v) = a" + '' = a" a'' =f{u)f(v)

(b) f{Zx) = a^= {a^y = [f{x)Y

78. V{t) = 20,0001
(a)

3V

V{2) = 20,0001 |r = $11,250

(b)f=20,00o(ln|)(f

Whenf = 1: ^« -4315.23
When f = 4: ^ = - 1820.49

(c)

Horizontal asymptote: v ' = 0

As the car ages, it is worth less
each year and depreciates less
each year, but the value of the
car will never reach $0.

520 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

80. P = $2500, r = 6% = 0.06, t = 20

/ 0.06 N^O"
A = 2500(l +^^^j

A = 2500e(°'^**^* = 8300.29

n

1

2

4

12

365

Continuous

A

8017.84

8155.09

8226.66

8275.51

8299.47

8300.29

82. P = $5000, r = 7% = 0.07, t = 25
0.07^25"

A = 50001 1 + 1

A = 5000eO'"<25)

84. 100,000 = Pe°

P = 100,OOOe-oo«'

n

1

2

4

12

365

Continuous

A

27,137.16

27,924.63

28,340.78

28,627.09

28,768.19

28,773.01

f

1

10

20

30

40

50

P

94,176.45

54,881.16

30,119.42

16,529.89

9071.80

4978.71

0 07\3i
86. 100,000 = P\\+ ^1

P = ,00,0001 1 + H)-

t

1

10

20

30

40

50

p

93,240.01

49,661.86

24,663.01

12,248.11

6082.64

3020.75

88. LetP = $100, 0 < r < 20.

400

(a) A = lOOfiOo^r
A(20) = 182.21

(b) A = lOOe""^'
A(20)« 271.83

(c) A = lOOgOo*'
A(20) « 332.01

92. (a) 12.000

(b) Limiting size: 10,000 fish

, , 10,000

(c) p(t) =

p\t) =

~ (\ + 19e-'/5)2

p'{\) = 113.5 fish/month

p'(10) » 403.2 fisii/month

1 + 19e-'/5

e-'/i /19

(1 + 19e'/5)2l 5

38,000e-"'5

: 10,000)

90. (a) lim

0.86

= 0.86 or

(b)

-0.86(-0.25)(e-°-^") 0.2156-"-^"

(1 + e-o-zs")

(1 + e-o-25")

P'(3) - 0.069
F'(IO) - 0.016

,. , 38,000, ,,,
(d) p'tr) = -r-{e-''^)

1 - 19e-'/^
(1 + 19e-'/=)3

h

19e-'/5 = 1
t

In 19

f = 5 In 19 = 14.72

Section 5.5 Bases Other than e and Applications 521

94. (a) ^'i = 6.0536;c + 97.5571

y^ = 100.0751 + 17.8148 In X
Vj = 99.4557(1.0506)^
y^ = 101.2875xO"'-'i

(b) 150

100

>3 seems best.

(c) The slope of 6.0536 is the annual rate of change in
the amount given to philanthropy.

(d) For 1996, .« = 6 and y,' = 6.0536, >',' = 2.9691,
j3' = 6.6015,^4' -3.2321.

^3 is increasing at the greatest rate in 1 996.

96. A = S-'dx = [-^1 = -^ - 23.666
Jo Lin 3 Jo In 3

98.

X

1

10-'

10-2

10-^

io-«

(1 + x)'"^

2

2.594

2.705

2.718

2.718

100.

f

0

1

2

3

4

y

600

630

661.50

694.58

729.30

y = ciJd)

When f = 0, >- = 600
y = 600{kf)

C = 600.

630^^05 66yO 69i58»,io5 ^^^ «= 1 05

600 ' 630 ' 661.50 ' 694.58

Let/t = 1.05.
y = 600(1.05)'

102. True.

/(e" + ') -f{e") = Ine"*' - In e"
= n + \ - n
= 1

104. True.

cPy _
dx" '

Ce'

yforn = 1,2,3,

106. True.

f(x) = g(x)e' = 0 ^

g(x) = 0 since e' > 0 for all .r.

522 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions
108. y = x^^''

y' l^\

— = sinx — + cos;c • Injc

y ' = -t^"-' — - + cos X • In X

Tangent line: y — — = IIjc — —

y^x

Section 5.6 Differential Equations: Growth and Decay

1 '^y A A dy , Jx

2

dy _
dx

= 4 -

- y

dy

A-y

-dx

■

'^ dv

= -

-dx

, 4

-/>

ln|4-

- y\dy -

= —x

+ c.

A - y -

= «-•>

■+c, =

Ce

y =

= 4-

- Ce"^

/3yy^ = |,

/x dx
2 3 '

8. y'-

= j:(i + y)

y'

1 +.V

= ;c

f/' dx-

= X dx

Ji+y

= xdx

ln(l + y) =

=f-.

l+y-

= g(rV2) + C,

y --

= gC, ^/2 _ 1

= C^/2 - 1

9y2 _ 4^/2 = c

10. xy + y' = 100;c

y' = 100;c + xy = ^100 - y)

y' ^
100 - > "^

jTo^'^^^r'^"

-ln(100-y)=y + C,

ln(100 - y) = -| - C,
100 - >- = e-0^/2)-c,

-y = e-c. e-^/2 - 100
>- = 100 - Ce-^/^

Section 5.6 Differential Equations: Growth and Decay 523

12. ^ = /t(10 - t)
dt

dp ^

—-dt =
dt

-

= k(\Q -

- t)dt

dP--

->■

-ty- + c

P =

^>-

-tY + c

14.

dx'

= kx{L - y)

1

L -

dy

ydx

= kx

1

L-y

dx

= fccdx

1

, L -

— rfv =

V

2 ^^'

-ln(L

-y)-

L - y = e-(*^/2)-c,

-y = -L + g-C, g-tar'/2

y = L - Ce-'=^/2

16. (a)

"^^iiii

1 ! ; ' I s J '■ \ I

:oi

^ = xdx

inbl = y + c

y = e^/2+c = c,^/2

1 v.

18.

dy
dt

hi-

Vt, (0, 10) 15

■Jtdt

y = -^^1^ + C

10 = -|(0)3/2 + C ^ C = 10

^ = -2^'''' "^ '°

rfv 3
20. ^- = 4v, (0, 10)

dt 4'

In V = 7 r + Ci
4

10 = Ce° => C = 10
V = lOe^'/*

(o, 10)

22. f = ^

N = Ce'^ (Theorem 5.16)

(0, 250): C = 250

400 S

(1,400): 400 = 250e* =^ k = In-^ = In ^

When t = A,N= 250e'"=(8/5' = 250e'°<«/5>'
^25o(8V 8192

24.^ = ^

P = Ce*' (Theorem 5.16)
(0, 5000): C = 5000

(1.4750): 4750 = 5000£>* => A: = Inf^j

When f = 5. P = 5000e'"*'9/=o>(5>

= 500o(^j' - 3868.905

524 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

26. y = Ce^, (0, 4), [s, |

C = 4

y = 4e>'

, = MiM._o.4159

y = 4g -0.41591

28.

C- (3,1), (4.

5)

^ = Ce-
5 = Ce"*

2Ce« = |ce'"--
10^3* = £«
10 = e*
fc = In 10 = 2.3026

y = (7^2.3026;
5 = (7^2.3026(4)

C - 0.0005

y = O.OOOSe^^o^*'

30. y'

dy
dt

ky

-r- > 0 when v > 0. Quadrants I and II.
dx

34. Since y = Cef'^<'/2)/i62o];_ .^^,e have 1.5 = Ce['"<'''2)/i62o](iooo) => c == 2.30 which implies that the initial quantity is 2.30 grams.
When t = 10,000, we have^' = 2.30e['"'<'/2)/i62o](io,ooo) ^ o.03 gram.

36. Since y = Ce^\^(Ui)/5no\<^ we have 2.0 = Ce['"(i/2)/573o](io,ooo) =^ c = 6.70 which implies that the initial quantity is 6.70
grams. When t = 1000, we have y = 6.70et'°<i/2)/5730]{iooo) ^ 5 94 ^^^^

38. Since y = Ce^^(U2)/5iio\t^ ^e have 3.2 = Cef'-C/^'/^sojiooo =^ c = 3.61.
Initial quantity: 3.61 grams.
When t = 10,000, we have >- = 1.08 grams.

40. Since y = Ce['"<'/2)/24.360],^ ^e have 0.4 = Ce^^imvi^imm.om) =^ c = 0.53 which implies that the initial quantity is 0.53
gram. When t = 1000, we have y = o.53ef'°<'/^'/^-3«'l<"»o) = 0.52 gram.

dy fc, w

42. Since — - ky, y - Ce*^ or >- = yQe^.

po = ^0^""*

k =

In 2

5730

0.15>'o = Joe'-"" 2/5730),
(In 2)r

In 0.15 =

t = —

5730

5730 In 0.15
In 2

15,682.8 years.

44. Since A = 20,0006°"^^', the time to double is given by
40,000 = 20,000^0 055r ^^ ^e have

2 = gO.055,

In 2 = 0.055r
In 2

t =

0.055

12.6 years.

Amount after 10 years:

A = 20,000e«'055'"0' = $34,665.06

Section 5.6 Differential Equations: Growth and Decay 525

46. Since A = lO.OOOe" andA = 20,000 when ? = 5, we
have the following.

20,000 = lO.OOOeS-^

In 2

Amount after 10 years: A = lO.OOOet""^'/']"'" = $40,000

0.1386 = 13.8

48. Since A = 2000e" andA = 5436.56 when t= 10. we
have the following.

5436.56 = 2000e'O'

ln(5436.56/2000)

'■ = To

The time to double is given by
4000 = 2000e"""

In 2 ^ „,

'=aTo'^^-^^y"^'-

= 0.10 = 10%

50. 500,000 = P\l +

0.06Y'a(-'o)
12

P = 500,000(1.005)-''8o « $45,631.04

52. 500,000 = Pll +

P = 500,000 1 +

== $53,143.92

0.09 V '^'<^'
12

0.09
12

54. (a) 2000 = 1000(1 + 0.6)'
2 = 1.06'
ln2 = rlnl.06
In 2

t =

In 1.06

= 11.90 years

(b) 2000 = 1000( 1 +

2 = I 1 +

0.06
12

0.06^ '2'

12

ln2 = 12rln 1 +

0.06
12

t =

1

In 2

^^nfl.^f)

1 1 .58 years

(c) 2000 = lOOOl

i'^m

^ /, 0.06\365'

In 2 = 365r In 1 +

0.06
365

In 2

365

1 (^ ^ 0.06

= 11.55 years

(d) 2000 = lOOOgOo*'

2 = g0.06l

In 2 = 0.06r
In 2

' = 006°="-^^^'^^

56. (a) 2000 = 1000(1 + 0.055)

2 = 1.055'

ln2 = rlnl.055

In 2

In 1.055

= 12.95 years

(b) 2000 = 1000( 1 + ^^j

2=1 +

ln2 = 12fln 1 +

0.055 y 2'
12 J

0.055

12

t =

In 2

'H'^'-W)

= 12.63 years

/ 0 055\2''''
(c) 2000 = 1000(^1 + ^j

/, 0.055

.055 \3«'

In 2 = 365r Inl 1 +

0.055
365

t =

In 2

365

, /, 0.055

= 12.60 years

(d) 2000 = lOOOeOO"'

In 2 = 0.055r

'^-Me^'--^^^'"^

526 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

58. P= Ce'^ = Cgoo^i'

P(-\) = 11.6 = CgOo^K-i) => c = 11.9652

P= 11.9652e003"

P(10)== 16.31 or 16,310,000 people in 2010

60. P = Ce'^ = Ce-°-^'

P{-\) = 3.6 = Ce-oo*"-" => C = 3.5856

P = S.SSSee-ooo^'

P(10) == 3.45 or 3,450,000 people in 2010

62. (a) Af= 100.1596(1.2455)'

(b) N = 400 when t = 6.3 hours (graphing utility)
Analytically,

400 = 100.1596(1.2455)'
400

'■''''' = 100.1596
fin 1.2455 = In 3.9936
hi 3.9936

t =

In 1.2455

= 3.9936

~ 6.3 hours.

66. (a) 20 = 30(1 - e"*)
30e3o* = 10

ln(l/3) -ln3_ ^^,..

A?= 30(1 - e-00366r)

64. >> = Ce'', (0, 742,000), (2, 632,000)

C = 742,000
632,000 = 742,000e2*

. = iB(63|m2)_oo,o2

y - 742,OOOe-o"802'

Whenf = 4, >> = $538,372.

(b) 25 = 30(1 - e-o-0366,)

-ln6 ,„^
^ = ^00366 ^'^'^^y^

(a) 4 = 25(1 - e«") ==> 1 -

(b) 25,000 units (lim S = 25)

25

2i

25

(c) When r = 5, 5 = 14.545 which is 14,545 units.

k = Inl

-0.1744

(d) =

70. (a) R = 979.3993(1.0694)' = 979.3993eOo«^"

/ = -0.1385r* + 2.1770f3 - 9.9755f2 + 23.8513f + 266.4923

(b) 2<X»

Rate of growth = R'(t) = 65.7eOo«^"

(C) 500

(d) Pit) = ^

Section 5.7 Differential Equations: Separation of Variables 527

72. 93 = lOlog.o^^ = lOdog.o/ + 16) 74. Since ^ = % - 80)

6.7 = log.o/ => /= 10

-6.7

kdt

I A< VI I ' I

10~i6 -v'-^io- • "/ ln(y-80) = fa + C.

80= 101og,o77^ni=10(log,o/+ 16)

/7^* = l

= log,o 1^1= IQ-* When f = 0, >> = 1500. Thus, C = In 1420.

;nf = \,y = 1120. Thus,
k(\) + In 1420 = ln(1120- 80)

Percentage decrease: ( '»';;:J°")(100) ==95% When r = 1, , = 1 120. Thus

104

k = In 1040- In 1420 = In 7^.
142

Thus, .V = i420e['"<'*'/"»2)]' + 80.
When t = 5,y-= 379.2°.

76. True 78. True

Section 5.7 Differential Equations: Separation of Variables

Ixy
2. Differential equation: y = ^-^ — ^

X y

Solution: x^ + y^ = Cy
Check: 2x + lyy' = Cy' • '
-2x

y

(2y - C)

, _ -~2xy _ —Ixy _ —2xy Ixy

^ " 2>'2 - Cy ~ 2y2 - U^ + y^) " y^ - ^2 - ^2 J ^2

4. Differential equation: y" + 2y ' + 2y = 0
Solution: y = C[e~'cosx + C-,e~^sinjc

Check: y' = -(Cj + Cj)e-^sinA: + (-C, + C2)e"^cosx

y" = 2 Cif"^ sin x — IC-^e'^ cos .x
y"+ 2y' + 2y = 2C,<'--"^ sin a- - 2C2e--' cos .« +

2(-(Ci + C.)e-^^mx+ (-Ci + C^-'^cosx) + 2(C,e-"^cos.x + Qe""^ sin .r)
= (2C, - 2Ci - 2C2 + 2C2)e--'sin.»: + (-2C2 - 2Ci + 2C2 + 2Ci)f "'^cos.v = 0

6. y = |(e^2A- _,_ ^)
y' = f(-2e-^ + e")
y"=f(4e-^ + e')
Substituting, y" + 2y' = f(4e--t + ^t) + 2(5)(-2e"^ + e') = 2e■^

528 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

In Exercises 8-12, the differential equation is ^'''* — 16y = 0.

8. y = 3 COS Zx

;y(4) = 48 COS 2x
■y(4) _ i6y = 48 cos It - 48 cos 2x = 0, Yes.

10.

«(4)

>> = 5 In X
30

v('') = -:

30

I6y = — T- 801n;c^ 0, No.

12. y = 3e^ — 4 sin 2jc

yW = 48e^ - Msinlx
yW - i6>. = (24e^ - 64 sin 2x) - 16(3e2» - 4 sin 2r) = 0, Yes

In 14-18, the differentia] equation isxy' — 2y = x^e^.

14. V = ;cV, y' = xV + Ixe' = e^(;c2 + 2x)

xy' - 2y = x{e^{x^ + 2x)) - 2{x^e^) = x^e^, Yes.

16. >> = sin X, y ' = cos j:

xy ' — 2^ = x(cos x) — 2{sin jc) ¥" x'e^. No.

18. y = xV - ix-,y' = ;cV + 2xe^ - 10a:

xy' -ly = ;c[a:V + Ixe - lOx] - 2[.tV - S.t^] = jt^e^, Yes.

20. v = A siB (Jit

d^ , , .

-TT = —Aar sin oi/

Since (d^/dt^) + 16>' = O, we have

—Aui^ sin cot + 16A sin a)t = 0.
Thus, or = 16 and oj = ±4.

22. 2x^ - y^ = C passes through (3, 4)
2(9) -16 = C=^C = 2
Particular solution: 2x' — y^ = 2

24. Differential equation: yy' + x = 0
General solution: x^ + y~ = C
Particular solutions; C = 0, Point

C= 1,C = 4, Circles

26. Differential equation: 3a: + 2yy ' = 0

General solution: 3x^ + 2y^ = C

6x + 4yy' = 0

2(3x + 2yy') = 0

3x + 2yy' = 0

Initial condition:

yd) -= 3: 3fl)2 + 2(3)^ = 3 + 18 = 21 = C

Particular solution: 3x^ + 2y- = 21

Section 5.7 Differential Equations: Separation of Variables 529

28. Differential equation: xy" + y' = 0
General solution: >> = C, + Cj In j:

y'=C,ny"=-Cn

xy"+y' = x[-C,^] + C^ = 0

Initial conditions: y(2) = 0, >> '(2) = -

0 = C, + C2 In 2

2 2

C, = 1, C, = -In 2

30. Differential equation: 9y" - 12y' + 4>' = 0
General solution: y = e-^/^(Ci + C,j:)

Particular solution: y = — In 2 + In jc = In -

y"= l^^/^ff C, + Q + Iq.t) + ^^4C2 = \eM\c, + 2C, + \c.x

3 V3

3 V3

<^"

9y"- 12y' + 4y = 9(^e^/3 j/|q + IC. + 1^.^] - 12(e^''3)(|Ci + Q + jQa:) + 4{e^/3)(C| + Qj) = 0

Initial conditions: y(0) = 4, y(3) = 0
0 = e\C^ + 3C2)
4 = (1)(C, + 0) ^ C, = 4

0 = e-(4 + 3C2) => C2 = -J

Particular solution: y = e^l\ 4 — — x

32. ^ = .x3 - 4;c
dx

V

y = I (.x3 - 4.t) dr = — - 2x2 + C

34.

dx: ~ 1 + e'

Jt^

e"

dx = ln(l + e^) + C

1/: "^

36. -— = xcosx-

dx

y = \x cos(.r-) dx = — sin(.r-) + C

(u = x~, du = 2xdx)

38. -7- = tan- x = sec- .r — 1
dx

I (sec- X —

1) (ix = tan .r — .r -1- C

dv

40. -j- = xV5 - X. Let « = V5 - x, «= = 5 - x. dx = - 2u dw
dx

y = X

= xVS^^dr = (5 - u-)u(-2u

(-10«= + 2«'')du

3 5

-j(5 - x)V2 + |{5 - x)^/- + C

530 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

42.

dx

y = '■

5e-V2Qrj = 5(-

'2)je-

-i

= - lOe-^/2

+ c

46,

dr
' ds

dr =

r =

0.055

0.055 ds
0.0255^ + C

48. xy'^y

If-f

dx

X

44.

dy _x^ + 2
dx 2>y^

|3y2rfy = J

{x^ + 2)dx

f = ^ + 2x + C

\ny = \a X + \n C = \n Cx
y = Cx

dy
50. y-r- = 6 cos vx
dx

h^\

6 cos TTX dx

/ 6 .

-r = — sin TTX + C,

2 77 '

>^ = — smTix + C

TT

,dy

52. V^?^^^ = 5jc
dx

/-/

5;c

■ dx

Jx^- 9

54. 4y^ = 3e^

Wydy = pe^dx
2y2 = 3e^ + C

56. Jx+ Jyy' = 0

Ll/2^y = - Ll/2^
|y3/2 = _|^/2 + c,
y/2 + ^/2 = c

Initial condition: ^(1) = 4,

(4)3/2 + (1)3/2 = 8 + 1 = 9 = C

Particular solution: y'l'^ + x'l'^ = 9

58. Ixy' - \nx^ = 0

l<

60. yVl -a:^T- = -^v1 -y^

ox

(1 _y2)-l/2y^y = \{l-X^)-'^^xdx
-(1 -/)l/2= _(1 -;c2)l/2+ c

y(0) = 1: 0=-l + C=>C=l

yr^= yn^

1

dx

= 21nx

dy

■ln;c

X

^

y '■

(lnx)2

2

+ c

yd)

= 1:1--

= C

y =

)^\^xY-

4- 2

62.

ds

-2s

H^l

= e-2^rf5

-g-'' = --e"^ + C

^(0) = 0: -1 = -| + C^C=-|

1,1

2 2

g-r = i -2r + i

2 2

-r=ln|.- + i| = lnl

r = lnl

1 + e-^\

2 ;

1 +e-

Section 5.7 Differential Equations: Separation of Variables 531

64. dT + k(T - 70) dt = 0
dT

ln(r- 70) = -kt + Ci

r - 70 = Ce-*'

Initial condition: T(0) = 140;

140 - 70 = 70 = Ce" = C

Particular solution: T- 70 = 70e-*', T= 70(1 + e~*')

66.

ir 3jt

In y = in x^ + In C
Initial condition: y{S) = 2, 2' = CCS^), C =

Particular solution: Sv^ = x^, y

±rV3

68. m - — - -

dx X — 0 X

n-\

dx

X

In >> = In JT + C, = Tn jr + In C = In Cc
y = Cx

70. /U, y) = .r3 + BxV - 2y2
/(rx, ty) = t^x^ + Ifx^ - Ifr
Not homogeneous

72. /U,y) =

xy

r;if fy
r-xy

xy

tjx- + y2 7.^2 + y2
Homogeneous of degree 1

74. /U, y) = tanU + y)

f(tx, fy) = tan(fx + ty) = tan[f(.x + y)]
Not homogeneous

76. /(.r,y) = tan^ ■

ty V

f(tx, rv) = tan — = tan -

tx X

Homogeneous of degree 0

78. y

.ty

xy2 dy = {:c' -\- y") dx

y = vx, dy = X d\) -\- V dx
x{vxY{x dv + V dx) = {x^ + (vxY) dx
X* v^ dv + x^ v^ dx = x^ dx + v^ x^ dx
xv^ dv = dx
"l

Jv2.v = /i

dx

- = ln|x| + C
(^y = 3 ln|.t| + C

80.

X- + y-
2xy

, y = vx

dv X' + v-x-

V + X — =

dx 2x- V
2v dx + 2x dv = dx

1:3^- -/f

ln(v2- 1) = -Inx + lnC= In-

v2- 1 =

4-1 = ^

X- X

y- — -ir = Cx

532 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

, 2y + 3y
82. y' = -,y = vx

dv 2x + 3vx

V + x-r = = 2 + 3v

dx x

dx

/if. -If

ln|l + v| = \nx^ + \nC = \nx'^C
l+v=x^C

X

X

y = Cy? — X

84. — y^ etc + j:U + y) rfy = 0, >> = vx

-x'^v^dx + {x'^ + x^v){vdx + xdv) = 0

/^— If

V + In V = - In ;c + In Ci = In

Ci

1 ^1
V = In —

XV
VX

y

y = Ce-y^''
Initial condition: ^(l) =1,1 = Ce~' =^ C = e
Particular solution: y = e'~^/^

86. (2jc2 + y2) dx + xydy = 0
Let y = v;c, ofy = j: dv + V tic.
(2x' + vV) dx + x(vx){x dv + V dx) = Q
(2^2 + 2^2 V 2) dx + x^v dv = Q
(2 + lv^)dx = -j:vrfv

;c 1 + v'

1

rfv

-21nx = -ln(l + v2) + C;

ln;c-2 = ln(l + v^y^ + In C
x-2 = C(l + v2)'/2

- = C(x2 + /)l/2
X

y{\) = 0: 1 = C(l + 0) => C = 1

-= y?T7

88- T

ax

+ Q

3/2 + ^2

1 = xjx^ + f

Section 5.7 Dijferential Equations: Separation of Variables 533

90. ^ = 0.25x(4 - y)

ax

dy

A-y

025x dx

\^>-h

25xdx
1

\jxdx

1

In b - 4| = --x^ + C,

;y = 4 + Ce-^um

92. £ = 2-j.,>.(0) = 4

I I I I

\ \ \ \V\

\ \ \ \

\ \ \ \

(111

\ \ \ \ \

\ \ \ \ \

\ \ \ \ \

N S N V

/./,/./ /././././.

I I I I l\l I I I I

94. ^ = 0.2jc(2 - y), y(0) = 9

96. ^ = Ay, :y = Ce*'
dt

Initial conditions: ^(O) = 20, y(l) = 16
20 ^ Ce° = C
16 = 20e*

A: = ln|

Particular solution; y = 20e"°<''/5'

When 75% has been changed:

5 = 20e"°(^/5'

1 = grln(4/5)

ln(l/4)
ln(4/5)

98. ^ = /t(x - 4)
dx

The direction field satisfies {dy/dx) = 0 along x = 4:
Matches (b).

6.2 hr

100. ^ = ky""
dx

The direction field satisfies (dy/dx) = 0 along y = 0, and
grows more positive as y increases. Matches (d).

102. From Exercise 101,

w = 1200 - Ce-^ k = 1

w = 1200 - Ce-'

w{0) = H'o = 1200 - C ^ C= 1200 - Ho

w = 1200 - (1200 - Wo)e-'

534 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

104. Let the radio receiver be located at {xq, 0).
The tangent line toy = x - j:^ joins (- 1, 1)
and (xq, 0).

(a) If [x, y) is the point of tangency on the >> = x - x^,
then

1 - 2;c =

y— l_x-x^— 1

X + I X + 1

X — 2x^ +1— 2jc = .r — j:-— 1
a:2 + It - 2 = 0

-2 ± 74 + 8

= -1 + 73

Then

y = x-x"- = 373 - 5
1-0 1 - 373 + 5 6 - 373

l--to -1 + 1-73 -73

73 = (1 + Xo){6 - 373)

= 6 - 373 + xg(6 - 373)
473 - 6

° 6-373

= 1.155

(c) 10

(b) Now let the transmitter be located at (- 1, h).

1 o y- h _ X- x^ - h
'-^ = rri- x+1

x-2x^+\-2x = x-x^-h

x^ + 2x-h-l=0

(-2 + 74 + Mh + 1))

1 + 72 + ;z

y = j: — x^

Then,

= 372 + h- h- 4
0 /; - (372 + h~ h- A)

■1--X0 -1 - (-1 + 72 + /;)

2/! + 4 - 372 + /i

-72 + /i

Xo+ 1

72 + /!

2/! + 4 - 372 + h

hjl + h

2h+ A- 372 + h

- 1

There is a vertical asymptote at /i =5, which is the
height of the mountain.

106. Given family (hyperbolas): x^ — 2)r = C

Ix - Ayy' = 0

X

y =

2y

Orthogonal trajectory: y' =

-ly

n-\i

dx

\ny = -21nA: + \nk
y = foc-2 =

108. Given family (parabolas): y^ = 2Cx

2yy'=2C

' = £ = yl(l\=y.

^ y 2x\yj 2x

Orthogonal trajectory (ellipse):

2x

Jyrfy=-J:

2xdx

= -x^ + Ki

2x'^ + y^ = K

J^

^

Section 5.8 Inverse Trigonometric Functions: Differentiation 535

110. Given family (exponential functions): y = Ce^

y' = Ce' = y

Orthogonal trajectory (parabolas): y' = —

\ydy= -\dx

y- = -2x + K

112. The number of initial conditions matches the number of
constants in the general solution.

114. TWo families of curves are mutually orthogonal if each
curve in the first family intersects each curve in the
second family at right angles.

116. Tnie

dy

^=U-2)Cv+l)

118. True

X' + y^ = 2Cy

dy X

dx C - y

X K - X Kx

-x^

x2 + y2 = 2Kx

dy ^K- X
dx V

2Kx - Ix- _ X- +y^- 2x^ _ y^

C — y y Cy — y^ 2Cy — ly'^ x''- + y — 2y x- — _v'

= -1

Section 5.8 Inverse Trigonometric Functions: Differentiation

2. y = arccos x
(a)

x

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

y

3.142

2.499

2.214

1.982

1.772

1.571

1.369

1.159

0.927

0.634

0

(c)

(d) Intercepts: I 0, — ) and (1, 0)
No symmetry

^-f

3 '

-f)

(-^._). -v^.-f

6. arcsin 0 = 0

536 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

8. arccos 0 = —

2

10. arc cot(- J3) = -^
6

12. arccos(^-— j = -

14. arcsin(-0.39) = -0.40

16. arctan(-3) = -1.25

18. (a) tanl arccos ^r- j = tanl — 1 = 1

(b) cos arcsin

5\ 12

13/ 13

20. (a) sec arctani -7) =

5

vS

-3

(b) tan

syiT

arcsinl-- = ^^

22. y = sec(arc tan 4x)
8 = arctan 4x

■y = sec e = Vl + I6x^

24. y = cos(arccotx)
6 - arccot x

y = cos 6 =

vO^+T

26. y = sec[arcsin(;c ~ 1)]
9 = arcsin(.r — 1)
1

>> = sec 6 =

Jlx - x^

( . X- h\
28. y = cosi arcsin I

6 = arcsin

X- h

y = cos 0 =

Vr^ - U - /z)-

Vr^-U-Zl)^

30.

v^

^,

32. arctan(2x - 5) = - 1

2x - 5 = tan(-l)
1,

X = -(tan(-l) + 5) = 1.721

Asymptote: x = 0

arccos r = 0

cos 6 =

tan e =

74^-^

Section 5.8 Inverse Trigonometric Functions: Differentiation 537

. arccosx

—

arcsecj:

X

=

cos(arcsec x)

X

=

x

x^

=

1

X

^

±1

V?^\

36. (a) arcsm(— x) = — arcsin^, \x\ < 1.

Let>' = arcsin(— j:). Then,

— ^=sin3' => x = — siny => x = sin(— y).

Thus, —y = sicsmx =^ y = — arcsin j:. Therefore,
arcsin(-j:) = -arcsinx.

(b) arccos(-.r) = tt - arccos x, |j:| < 1.

Lety = arccos(-.r). Then,

—X = cosy =^ X = — cosv => ;c = cos(7r — y).

Thus, TT — y = arccos x =^ y = tt - arccos x.
Therefore, arccos(-.i:) = tt — arccos x.

38. f(x) = arctan x +

X = tanly —

Domain: (— oo, oo)

Range: (0, tt)

f(x) is the graph of arctan x shifted ir/l units upward.

42. f(t) — arcsin t-
2t

f\t)

40. fix) = arccosl —

— = cos y
4

X = A cos y

Domain: [-4,4]

Range: [0, tt]

44. f(x) = arcsec 2x
2

VT^

fix) =

\2x\ V4x- - 1 \x\j4x~- 1

c^^-_

o*

^^r /,..-]

<y:f

d

/i-t^+<«i^''7

A/' '

-V,

46. /W = arctan Vx

fix)

1 + xj\2j'xl 2v/x(l + ;c)

50. f(x) = arcsin a: + arccos x =

/'W = 0

48. h{x) = x^ arctan x

h \x) = 2x arctan x +

1 +x'

52. y = ]n{t- + 4) - - arctan -

It 1

f + A 2

'* 5

TV©

2t

1 2f- 1

ri + 4 r^ + 4 r + 4

i-C.rc..^7.

54. y = -

.rV'4 - .IT + 4 i

56. V = .V arctan Iv — r ln(l + \yr)
4

,_(4-^)-./.(-^),VT^,2-;,====

rfv

iv 1 + 4.T-

+ arctan(lr)

1/ 8.V

4V 1 + 4.V-

arctan(lv)

JiT^x

+ Ja^^- +

vT

74"

538 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

. X

58. y = 25 arcsin - - a:V25 - x^

60. y = arctan

1

j>'=5

-^=i_- 721-1^ -.-(25 -.r'/^(-2.)

25

(25 - x^) ^ x^

V25 - x- 725 - X- 725 - x^

2x2
725 - x^-

2 2(^2 + 4)

1 1

2 1 + (x/2)2 2
2 x

+ -(.2 + 4)-2(2x)

jc^ + 4 (;c2 + 4)2

2x^ + S + X
(x^ + 4)2

62. f(x) = arctan .r, a = 1
1

fXx) =

1 +.2

, , _ —2x

^^''' " (1 + .2)2

P,{x)=f{\)+f'{\){x-\) = ^ + \{x-\)

AW

P,(X) =/(l) +/'(1)(X - 1) + |/"(1)(X - 1)2 = J + \{X - 1) - ^(X - 1)2

64. /(;c) = arcsin X — 2x
1

/'(;c) =

7r

-2 = 0 when 7l - x^ =

1

x = ±

fix)

75
2 ■

X

(1 - .2)3/2

/"I'^'l > 0

73

Relative minimum: I ^r— , - 0.68

r(-f)<o

Relative maximum: ( — r-, 0.68

73

66. f(x) = arcsin x — 2 arctan x
1 2

fix)

71^

1 +.2

= 0

1 + 2.2 + ;c4 = 4(J _ ^2)

;d + 6.2 - 3 = 0

. = ±0.681

By the First Derivative Test, (-0.681, 0.447) is a relative
maximum and (0.681, —0.447) is a relative minimum.

68. arctan 0 = 0. tt is not in the range of >> = arctan ..

70. The derivatives are algebraic. See Theorem 5.18.

72. (a) cot e =

(b)^ =

= arccotl-
-3 dx

dt .2 + 9 dt

If.= 10,^- 11.001 rad/hr.
dt

74. cos d =

750

6 = arccos

V s

de^Mds^^ -1 (~'^^^\ ^

dt ds dt 71 - (750A)2V s^ I dt

750 ds

sjs^ - 7502 dt

If. = 3,— =66.667 rad/hr.
dt

A lower altitude results in a greater rate of change of 6.

Section 5.9 Inverse Trigonometric Functions: Integration 539

76. (a) Let y = arcsin u. Then
smy = u
cos y • y ' = u'
dy _ u'

dr cos y J\ - u^'

(c) Let y = arcsec «. Then '

secy = u

dy
sec >■ tan y — = «
ax

dy _ u' _ u'

dx secy tan y \u\Ju- - V

Note: The absolute value sign in the fomiula for
the derivative of arcsec u is necessary because the
inverse secant function has a positive slope at every
value in its domain.

(e) Let y = arccot u. Then
cot y = M

2 dy
— esc' y -— = M
dx

di
dx

- esc- y

1 + m2

(b) Lety = arctan u. Then
tany = tt

2 dy
sec'y— - = u
dx

dy _ u'

dx sec^y i + u}

(d) Let y = arccos u. Then

cosy = u

. dy
— sin y -— = u
dx

dy _ m' _

dx sin y

vr

(f ) Let y = arccsc m. Then

cscy = u

dv
— esc V cot V -p = u
' dx

dy-_

dx

-cscycoty |«|Vm- - 1

Note: The absolute value sign in the formula for the
derivative of arccsc u is necessary because the inverse
cosecant function has a negative slope at every value in
its domain.

78. f(x) = sin X

g{x) = arcsin(sinx)

(a) The range of y = arcsin x is - it/2 S y ^ tt/I.

(b) Maximum: v/l
Minimum: - n/l

80. False

The range of y = arcsin x

[IT tt]

82. False

arcsin^ 0 + arccos- 0 = 0 + | y ) ^1

Section 5.9 Inverse Trigonometric Functions: Integration

r 3 3 r 2 3 r'

, dx = -\ , dx = - arcsin(2jc) + C 4.

J VI - 4x2 2} VI - 4x- 2 I

V4

-,dx =

arcsm -

TT

6

r 4 4 r 3 4 r I r i t^

^- JtT^'^ = sJlTa?^'^ = 3 arc.an{3..) + C 8. J^^^J-^^t = [l^<^tan-J _ =

^"•/4 + (.v-l)=''"^ = I^^V")

+ C

TT

36

^•/FTT'^^ = /^--')'^^ = i-

1-3 - X + C

540 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

14. Let M = r^ du = 2t dt.

16. Let u — x^,du = 2x dx.

lAre"' = lJ(ipTW^''^* ^arctanf + C l^^J^,^ = \\

V^^^^^ 2j xVM^ - 22

(2j:)dc

1 ^^ ^

= — arcsec y + C

18. Let M = arccos x, du =

yn^

dx.

n/V2
Jo

arccos x

dx

ri/72_
Jo

yr^

-dx

—- arccos^ j:

1/72 3,

32

« 0.925

20. Let M = 1 + j:^, Jm = 2x die.
J-/3l+^' 2j_^l+.

-J2x)dx

= |-ln(l+;c2)

->/3

-In 2

^r

;dx

1(73)2

3 + U - 2)2 Ji (V3)2 + U - 2)2

-d:x:

1 x-2

— ;= arctan — 1=-

73 V 73

7377

18

J-V2
g 1 + Sin-;

dx = arctan(sin x)

7r/2
0

2L
4

26.

— dx. u = ^Jx, du = — -1=
x) 2vJc

J 27^(1 +

3 f 2m du _ (_du__ _
2J m(1 + tt2) - -^J 1 + j<2 -

ate, dx = ludu

= 3 arctan m + C
3 arctanTc + C

r 4x + 3 , , ^x f -2x , ,r

28. \-^=J=dx^ {-2)\-j=^=dx + ■i\-j^==dx = -471 -x2 + 3arcsinA:+ C

^"•Ju+l)2 + 4'^ = 2j(.+ l)2 + 4^-J(;c+l)2 + 4^

= - ln(;c2 + 2x + 5) - - arctan! —r— I + C

32.

34.

{^ dx r- dx ri /x + 2\12 1 (4

L.2 + 4.+ 13 = L(x + 2)2 + 9 = [3 =^''H~5-JJ_2 = 3 ^'^^ll

/.2?2x + 2^ = l^T^l'^ - ^/l+U+l)2'^ = ln|.2 + 2. + 2| - 7 arctan(x + 1) + C

36. , ^ ^r =

J V-;c2 + 4;c J

V4 - (x2 - 4j: + 4)
2

a!x

74 - (;c - 2):

A

. ■ /^ - 2\
= 2 arcsin — r — +

38. Let tt = j:2 - 2x, dw = (2a; - 2) dx.
= 7.x2 - 2x + C

40.

f ^* = f ,L=^- „sec|. - ,| . C

J U - l)7;c2 - 2x Jix- 1)7U -1)2-1

Section 5.9 Inverse Trigonometric Functions: Integration 541

42. ha u = x^ - A, du = 2x dx.

\

79 + 8p^r?

-.dx

^w

2x , 1 /;c2 _ 4\

')

44. Let u = Jx--1, u^ + 2 = X, 2u du = dx

I :~ dx = I -r ^M = I ; — du = l\du — 6] —:.

J;c+1 Jm2 + 3J„2 + 3 J J„2 +

c/m

= 2m ;= arctan ^

V3 ^

^ + C = 2Vj: - 2 - 2V3 arctan .^'^ + C

/3\2 9 9 9 / 3\^ 9

46. The term is ( ^ 1 = ^: x^ + 3a: = ;c2 + 3;c + ^ - ;^ = ( x + 1 1 - ^

48. (a) J e^ die cannot be evaluated using the basic integration rules.

(b) \xerdx = \e^ + C,u = ,? ^ fl

J 2 (c)Jp

^*' ^^^ J dx cannot be evaluated using the basic integration rules.

J yf^ dx = Ij :p^ & = i arctan(x^) + C, « = x^

(c) -re'/^dx = -ei/^+ Cm =

1

(b)
(c)

4^3 . 1

1 + X

^dx = - ln(l + x") + C, M = 1 + x"*

52. (a)

dy

(b) ^ = xyi6^, (0, -2)

dj

= xdx

arcsinl jl = — + C

(0, -2): arcsini --1 = C =^ C
. (y\ x' ir

2L
6

V . /at 7r\

4 = nT~?J

V = 4 sin

2 6

-f = 7>fe-<°"^

7 1 /■ / /
/ / / / /
^ / / yjf

//III

y / / / /
y y y y y

^^ y y y

N N -^ ^ ^

^ ^ ^ X X

542 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

56.

A = . dx = arcsin — = —

Jo 74^^ L 2jo 6

58.

arcsin xdtc ~ 0.571
Jo

1 P*^
60.FW=-| ^

+ 1

dt

(a) F(a:) represents the average value off{x) over the interval [x, x + 2]. Maximum at ;c = — 1, since the graph is
greatest on [— 1, 1].

-[=

(b) F(x) = I arctan f
1

F'{x) =

= arctan(x + 2) — arctan x

1 (1 + x^) - (;c^ + 4x + 5) _ -4(x+ 1)

1 + (x + 2)2 1 + x^ (x2 + l)(x2 + 4x + 5) (x^ + l)(x2 + 4x + 5)

= 0 whenx = — 1.

^' J V6x - x^ '

62. ^ ,dx

J Vox - X-

(a) 6x - x2 = 9 - (x2 - 6x + 9) = 9 - (x - 3)^

dx . /x — 3

1

J V6x — x^ J .

X- J 79 - (x - 3)2

(b) u = Vx, M~ = X, 2u du = dx

2

— arcsin — - — + C

/ti^^'"''"^^/

. du = 2 arcsin —p= ] + C = 2 arcsini

76^^ V76,

(i)

+ c

(c)

The antiderivatives differ by
a constant, 7r/2.

Domain: [0, 6]

64. Let/(x) = arctan X

1 +x2

,rY N 1 1 -x2 2x2

1 + x"' (1 + x^r (1 + X*-)

Since/(0) = 0 and/is increasing for x > 0, arctan x - ^ > 0 for x > 0. Thus,

X

arctan X >

1 +x2'
Let g{x) = x — arctan x

Since g{0) = 0 and g is increasing for x > 0, x — arctan x > 0 for x > 0. Thus, x > arctan x.
Therefore,

X

2 4 6 8 10

1 +X2

- < arctan X < x.

Section 5.10 Hyperbolic Functions 543

Section 5.10 Hyperbolic Functions

pO + pO

2. (a) cosh(0) = ^ = 1

4. (a) siiih-'(O) = 0
(b) tanh-'(O) = 0

(b) sech(l)

e + e

= 0.648

6. (a) csch-'(2) = Inl ^ "^„ | - 0.481

(b) coth->(3)=^ln(|j = 0.347

„ 1 + cosh 2x 1 + (e=^ + e"2x)/2 g2x + 2 + e'^ fe' + C'^Y

8. = = = — T— =cosh^^

fe' — e~^\le' + e~-^\ e" — e^
10. 2 sinh x cosh jc = 2 = = sinh 2x

12. 2 coshi — :H- 1 cosh( — -^ 1 = 2

= 2[^^±^

e-y + e-

gU-y)h + g-(j:-y)/2"|

e"" + e~'' e^ + e~y

cosh X + cosh y

14. tanh x = —

2

- + sech^jc = 1 ^ sech^j: = - => sech^: = -r—
2/ 4 2

cosh X =

1 273

/3/2

coth;c = :f^ = 2

sinh X = tanh x cosh j:

1V2V3\ v^

2/V 3

Putting these in order:

73

sinh X = —— csch x = Jl>

cosh jt = -^i; — sech x = — ;-

3 2

tanhj:

1

coth X = 2

csch ;c

73/3

73

16. y = coth(3x)
y' = -3csch2(3jc)

18. g(x) = In(coshjc)
1

gW

cosh X

(sinh.v) = tanh.v

20. y = X cosh X — sinh x

y' = X sinh x + cosh x — cosh .r = x sinh .v

22. /i(r) = r - coth r

/( '(0 = 1 + csch- 1 = coth- f

544 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

24. g{x) = sech- 3j:

g'{x) = -2 s&Mix) sech(3j:) tanh(3x)(3)
= - 6 sech^ 3x tanh 3x

26. fix) = 6"^"

fix) = (cosh;c)(e=''=^^)

28. >- = sechU + 1)

>> ' = — sechU + 1) tanh(jc + 1)

30. fix) = X sinhCr - 1) - coshU - 1)

fix) = j:cosh(x - 1) + sinhU - 1) - sinh(jr - 1) = xcosh(x - 1)

fix) = 0 for X = 0. By the First Derivative

Test,(0, -cosh(- 1)) = (0, - 1.543) is a relative minimum.

32. hix) = 2 tanh x - x

2

\

(0.88. 0.53)

(-0.88,-0.531

\

34. y = a cosh j:
y' = a sinh .t
>>"= a cosh.x
Therefore, y" — y = 0.

Relative maximum: (0.88, 0.53)
Relative minimum: (—0.88, —0.53)

36. fix) = cosh a; /(I) = cosh(O) = 1

fix) = sinhx /'(I) = sinh(O) == 0

fix) = coshx /"(I) = cosh(O) « 1

/'i(^)=/(0)+/'(0)(x-0)= 1
P2(;c) = 1+5^2

V

/

/>,

38. (a) >- = 18 + 25 cosh—, -25 < x < 25

(b) At X = ±25, y = 18 + 25 cosh(l) = 56.577.
At.t = 0,y = 18 + 25 = 43.

X

(c) y' ~ sinh— .At .c = 25, y' = sinh(l) = 1.175

40. Let M = VJ, d«

Let u = Vj:, d« = — 7= ^. 42. Let u = cosh x, du = sinh j; <

ijx

^= — dx = l\ cosh V^l — ^ j (it = 2 sinh VA + C J ' + sinh^ a: J cosh-^ j:

(ic =

1

1 + sinh-^ X J cosh-^ x cosh .x

= — sechx + C

+ C

44. Let u = 2x - I, du = 2 dx.

\ sech^ilx - l)dx = \\ sech^dx - 1)(2) dx

/sech2(2.-l)<i. = i/s

^tanh(2x- 1) + C

46. Let u = sech x, du =
sech' X tanh .r dlr =

/'

sech X tanh jr tic.
sech^ xi - sech j: tanh x) dx

■/

--sech^ JT + C

Section 5.10 Hyperbolic Functions 545

„ f ,, , fl + coshZx
5. I cosh'' xdx = \ c

ir sinhlr]

50.

r

Jo

V25~

:dx =

3 Jo

. 4
arcsin —

-jc + - sinh 2x + C
2 4

52.

— , ^ dx = 2 / (2) aLc = -W-

J xVl + 4jt2 J (2x)Vl + (2;c)2 ^ \

+ Vl + 4x-
|2x|

+ C

54. Let u = sinh x, du = cosh jt (ic.

I

cosh a: , . /sinhjc\ „

, ^= ate = arcsin — - — + C

V9 - sinh-jc V 3 '

56. y = tanh-'l-
^ 1

©

1 - (;c/2)2V2/ 4 - X

arcsinl I + C

58. y = sech^Hcos 2j:), 0 < .r < —

y =

-1

cos 2xVl - cos- 2a:
since sin 2x: > 0 for 0 < x < 17/4.

( — 2 sin 2r) =

2sin2x

cos2r|sin2x| cos 2x

= 2 sec 2x,

60. >- = (csch-'A:P
y' = 2 csch" ' j:|

- 2 csch ' .V

x\jr+7-/ \x\^l + x^

62. y = ;c tanh"' x + InVl - x^ = x tanh"' .r + x ln(l - x^) 64. See page 401, Theorem 5.22.

y = x\

1 -x^

=)

1 \ _ -X

+ tanh ' JC + r = tanh ' x

1 -x2

66. Equation of tangent line through P = {xg, yg)

y

- a sech-' ^ + Va- - Xg^ =

HX - Xg)

When x = 0,

y = a sech" ' '- Va- - jCn" + Va^ - JTn^ = a sech~ ' — .

a a

Hence, Q is the point [0, a sech~'(.*o/'')]-

68.

istance from P lo Q: d = y/x

o^ + (-Va^-.o^r =

■' J. 'f "'^

- = 4©'"

3-.r

9-X* 2}9-(x^r-

3 +.t2

= -±,„

3-.r

+ c

3+:r

+ c

546 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

70. Let tt = x^'\ du = -Jxdx.

72.

f ^ =f.

]{x+ 2)V.t2 + 4;c + 8 J

ate

U + 2)V(^ + 2)- + 4

= 4-P4?l^'-

74. f—

l)V2? + 4r+8

otc

= f L

J (x + \)j2{x +1)2 + 6

^J U + \)J(x + \Y +

(;c + 1)V(^ + IP + (V3)'

.^=_J_,„fVI+7U+i)i+i^^^

j: + 1

V6

76. Let tt = 2(a: - 1), du = 2dx.

f ^^f

J (x - 1)7-4x2 + 8x - 1 J

1" J 2(x - 1)V(^^-[2(x-1)?

(&= -

^In

73

73 + 7-4^2 + 8x - 1

2(x - 1)

+ C

78. y = 1:1 ^dt = \^ —,dx + 3 It -'r:^ -dx

^ }Ax-x? J 4x - x^ J (jc - 2)2 - 4

= In 4x - x^ +=T In

(x - 2) - 2

(;c - 2) + 2

+ C = lnl4x - x2| + - In

X -4

+ C

80. A = tanh 2x dx

2g2x _ g-Zx
g2i + g-2i

ip 1

2J0 £2^ + e-2^'
= [|ln{e2' + e-2-)|

= |ln(e^ + e-*)-|ln2

(it

(2)(e2^ - e-^) dx

= ln,/ 7 °= 1.654

84. (a) v(;) = -32r

82. A

I

^(fe

3 v4^^^'
U ln(x + 7x2 _ 4)

= 6 ln(5 + 721) - 6 ln(3 + Js)

(b) 5(r)

Jv(f)rff=|

(-32f)dr= -16f + C

siO) = - 16(0)2 + C = 400 => C = 400
i(r) = - 16r2 + 400

—CONTINUED—

Section 5. JO Hyperbolic Functions 547

84. —CONTINUED—

(c)

dv
dt

-32 + kv^

J 32 - fcv2 J '

Let « = V^ V, then du = Vfc dv.

1

1

:ln

Vfc 2732
Since v(0) = 0, C = 0.

V32 + yitv

V32 - V^v

= -f + C

In

-2V32/tf

- ^-2^32*;

732- Jkv
732 + Jkv

732 + 7fcv = 6-2^^2* '(732 - 7tv)

i'(7t + 7ite-2v^') = 732(e-2v^' - 1)

■ . ^(e-2v^>_ i) e>/?af

7t(e-2>'^'+ l) ' e-^'

732

v^

zif

-Jm.

'ilkt J. „-v^2itf

gVJ2A, + g

= -^tanh(732fcf)
7fc

86. Lety = arcsin(tanh j:). Then,

e" — e""^

sin y = tanh x = :- and

e' + e

e^ — e '
tan y = = sinh x.

Thus, y = arctan(sinh jc). Therefore,
arctan(sinh ;c) = arcsin(tanh ;c).

(d) lim

732
7^

tanh(732fcr)

J22

Jk'

The velocity is bounded by - 732/7^-

(e) Since /tanh(cf) dt = (1/c) In cosh(cf) (which can be
verified by differentiation), then

/32

s(t)

/

7fc

732 1

tanh(732ytr)<if

ln[cosh(732itr)] + C

7fc 732fc
= — iln[cosh(732/:r)] + C.
When r = 0,

5(0) = C

= 400 => 400 - (l/k) ln[cosh(732fcf)].
When A: = 0.01,

izW = 400 - 100 In(cosh7o32 t)

Si(f) = - 16t2 + 400.

s^it) = 0 when ; = 5 seconds.

Sjit) = 0 when r ~ 8.3 seconds

When air resistance is not neglected, it takes
approximately 3.3 more seconds to reach the ground.

88. y = sech~' j:

sech y = X

— (sechy)(tanhy)y' = 1

-1
y =

90. y = sinh^'x

sinh y = JC
(cosh y)y' = 1

1

(sechy)(tanhy) (sechy)7l - sech-y xVT

coshy 7sinh-y + 1 Vx- + 1

548 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

Review Exercises for Chapter 5

2. /(jc) = ]n(x - 3)

Horizontal shift 3 units to

the right

Vertical asymptote: x = 3

4. \n[{x^ + l){x - 1)] = lnU2 + 1) + \n(x - 1)

r 25xr' ~\
6. 3[ln;t - 2 InU^ + 1)] + 2 In 5 = 3In;c - 6Mx'^ + 1) + In 5^ = Inx^ - \n{x^ + 1)« + In 25 = In , ^ ^ ^.^

8. In X + ln(x - 3) = 0

\nx{x - 3) = 0

xix -2) = e°

x2 - 3.T - I = 0

3 ± Vn

10. h{x) = In^^^^ = Inx + ]n{x - 1) - ln(x - 2)

X - 2

.,,,11 1 x2-4x + 2

h{x) = - +

X X— 1 X — 2 x' — 3x^ + 2x

3 + 713 , . 3-713 „
X = only since < 0.

12. /(x) = ln[x(x2 - 2)2/3] = Inx + |ln(x2 - 2)

„, , 1 2/ 2x
f'(x) = - + '

X 3\x2 — 2/ 3x3 — fix

7x2-6

14. y = -jfa + te - fl ln(a + bx)]

dy _ U ab

dx b^\ a + bx) a + bx

16. y = 1 — :; In

ax a' X

1 h

= + -?[ln(a + bx) - Inx]

ax a^

(fa

aV x2/ a\a + bx xj
ax2 a2Lx(a + bx)\

1

ax2 ax{a + bx)

(g + bx) - bx 1

ax\a + bx) x\a + bx)

20. M = In X, <iM = - <fa

X

/¥*4/<'""©-=>'>'-^

18. M = x2 — \,du = Ixdx

S^^dx = ^(^^dx = |lnlx2 - 11 + C
J x2 - 1 2j x2 - 1 2 ' '

2..f^^ = f„„.,.(l)^.[i<,n,p]; = l

-/Xf

X dx

In

77
COS| "T "■ JC

)ir

= 0-,.|J=)4..2

Review Exercises for Chapter 5 549

26.

(a)

fix)

= 5x-

1

y

= 5x-

1

y + l
5

= X

x + 1
5

= y

f-Uy\

x + 1

(b)

(c) /-'(/W) ^f-\5x - 7) = ^^"^ l^^'' = X

/</-«) =/m-m—'

28. (a) fix) = x3 + 2
y = x^ + 2

Vy - 2 = ;c

V:c - 2 = y

(b)

A

^'r^

-fl

^

(c) /-'(/W) =r '(x^ + 2) = y(;c3 + 2) - 2 = x
/(/-'W) =/{4/^r^^) = {V:^^^f + 2=x

30. (a) fix) =x'^- 5,x>0
y = x^-5
Vy + 5 = x
Vx + 5 = y
/-'W = Vx + 5

(b)

^

/

(c) /"'(/W) =/-'(x^ - 5) = VCr^ - 5) + 5 = .r forx > 0.
/{/-'W) =/(Vx + 5) = {Jx + Sf -5=x

32. fix) = xVx- 3

/(4) = 4

/'W = VT^^ + ^xix - 3)-'/^
/'(4) =1 + 2 = 3
(/-')'(4)=^-;^ = |

34. /(.r) = lnx

f-'ix) = e^

(/-')'W = ^

(/-')'(0) = e°= 1

36. (a) fix) = e'-

y = e'"-'
In y = 1 - X
X = 1 — In y
y = 1 - Inx
/-'(x) = 1 - Inx

(b)

\

\^/

/-'

(c)/-'(/(x)) =/-'(«'-') = 1 - He'-')
= 1 - (1 - x) = X
/(/"'W) =/(l - Inx) = fi-ii-'^-'* = ftox = J.

550 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

38. y = 4e-

-5-4-3-2-1 ..12345

-3--

-4
-5

40. g{x) = Inl

gXx) = 1

1 + e'j

In e* - ln(l + e^) = x - ln(l + e-')
e^ 1

\ + e" \ + e'

42. h{z) = e-^'^
h\z) = -ze-^'^

44. y = 3e^3/r

Qp-3/r

y' = 3e-V'(3r2) = ^

46. f(e) =|e=ta2e

/'(e) =005 266^*" 29

48. cos n?- = xe^'

-2xsmx^ = xey^ + «>■
dx

di _ 2x sin jc^ + e^
dx ~ xey

50. Let u = -, du = — j- dx.

dx = -e'/-^ + C

52. Let « = e^ + e'^', J« = (2e^ - e'^^) d^r.

fe^ - e-^ ^ 1 r2g^ - 2g-
Je^ + e-^ 2J e^ + e-2

■dx

54. Let M = jc' + L -i" = Sx^ alx.

Lv+'aLr = 1 e

1^(2^+1^= gx=+l(3^2)^ = ^^ + l + c

= - \n{e^ + e-^) + C

56.j^^dx='-j^2e^dx

= - ln(e^ + 1) + C

58. (a), (c) 'oooo

(b) V = SOOOe-ofi', 0 < r < 5
V'(t) = -4800e-o«'
V'(l) = -2634.3 dollars/year
V'(4) = -435.4 dollars/year

60. Area = 2e^' dx

Jo

-2e-

Jo

62. g{x) = 6(2--^)

64. y = log4j:2

Review Exercises for Chapter 5 551

66. fix) = A'e^

fix) = 4^e^ + (In 4)4^?^ = 4V(1 + In 4)

X

70. /i(jc) = logs ~rj ^ '°g5 ^ ~ logsC^ ~ 1)

'''«=i;^

1 1

x j: — 1

1
In 5

xix - 1).

68. >> = ;c(4-^)

y' = 4"-' — X • 4~' In 4

72.

f2-iA
j — dt-

1
In 2

2-1 A + c

74. ? = 50 log,o!

18,000

,18,000 - h)
(a) Domain: 0 < h < 18,000
(b)

100- •

Vertical asymptote: h = 18,000

76. 2P = Pe'O'-

2 = e'O'
ln2 = lOr
In 2

/■ = — - 6.93%

(c) f = 50 log

18,000
Hl8,000-/!

8,000

18,000 - h
18,000 -/! = 18,0O0(10-'/5O)

h = 18,000(1 - lO-'/^o)
As ;!->l 8,000, t-^oo.
(d) r = 50 logio 18,000 - 50 log,(,(18,000 - h)

dt

50

78.

dh (In

10)( 18,000 -

-h)

d-'t

50

dh'' (In

10)( 18.000 -

-h)-

No critical numbers

As r increases, the rate

of chan^

eof tlie

altitude is

increasing.

>- =

^2J

^

'

^-(600) =

^2J

= 3.868

grams

80. (a)

^= -0.012y,5 > 50
ds

^/M-

aoTl'""^'^^'

When s = 50, v = 28 = Ce-'^onm
y = 28e''*-'''»^, 5 > 50

C = 2%e°^

(b)

Speed(s)

50

55

60

65

70

Miles per Gallon iy)

28

26.4

24.8

23.4

22.0

82.

dx

e'

-2r

1

+

e"

-2i

f

e"

Zr

rfx

1 r -2e--'^
"2J 1 +«--

otc

y = -- ln(l + e---^) + C

552 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

84. V ' — e^' sin j: = 0

dy , .
—- = e' sin X
dx

e-y dy = si

sin j: (it

86.

-e >■= -

cos x + Ci

1

(C= -Q)

cos j: + C

y = \r

1

= -ln|cosx + C

cosjc + C

-j- = '— (homogeneous differential equa

3{x + y)dx - xdy = 0

Let y = vx, dy = X dv + V dx.

3ix + vx)dx - x(xdv + vdx) = 0

{3x + 2vx)dx- x^dv = 0

(3 + 2v)dx = xdv

fi . r 1

3 + 2v

dv

\n\x\ = - ln|3 + 2v| + C, = ln(3 + 2v)>/2 + In Cj

;c = C2(3 + 2v)'/2
;c2 = C(3 + 2v) = c( 3 + 2(^j

jr' = C{3x + 2y) = 3C;c + 2Cy

_^x^ -3Cx
^ IC

88. ^ = /tv - 9.8
dt

(a)

1^7^ =f^

-ln|/tv - 9.8| = r + Ci

ln|fcv - 9.8| = fa + Cj
Jtv - 9.8 = e*'+c, = (2^gh

1

9.8 + Cjg*'

At ? = 0,

Vo = -(9.8 + C3) => C3 = fcvo - 9.1

V = -{9.8 + (/b'o - 9.8)6'=']
Note that k < Q since the object is moving downward.

(b) limvW=^

(c) i(f) = J|[9.8 + (A:vo - 9.8)e*']rff

9.8? + |(/fcvo - 9.8)e^ I + C

= ^ + -^(H - 9.8)e*' + C

5(0) = ^(fcvo - 9.8) + C ^ C = So - ^(fcvo - 9.8)

^(') = ^ + ;^K - 9.8)e*' + 5o - -^^vo - 9.8)
Q Kr 1

Review Exercises for Chapter 5 553

90. hix) = -3arcsin(2x)

92. (a) Let e = arccot 2
cot 0 = 2

tan(arccot 2) = tan 6

(b) Let d = arcsec V5
sec 6 = v/5

slarcsec Vs ) = cos 8 = — ;= .

94. y = arctanlx' - 1)

2x

2x

1 + U- - 1)2 ;c^ - 2jc2 + 2

96. y = X arctan e^

^ 2U + e-'V 1 + e'

98. y = V;c2 - 4 - 2 arcsec -, 2 < j: < 4

y =

X-- 4 V^2T74

V?^^ (|x|/2)V(a:/2)2- 1 Vx^'^ \x\J^^^^ \x\V^^^^ X

100. Let M = 5;c, du = 5 dx.

Jr^'^ = i/(v^)2V(5./^^^

1 5x

r arctan — ^ + C

5^3 V3

102. J.

1.1 ■« , ^

— r cit = — arctan — + C

16 + a:^ 4 4

104. I ^ -^ (it = 4 I -7=L= 0^ + :1^ I '

ix: + - (4 - x^)-^'-(-2x)dx = 4arcsin^+ J4 - x- + C

106. Let u = arcsin x, du

jr=i^-

dx.

\

arcsin j: , 1, . ^^ ^
, dx = -(arcsin x)- + C

108.

(n

Since the area of region A is 11 I sin.vrf>).

the shaded area is I arcsin.riv = — — 1 == 0.571.
Jo

110. y = xtanh-'2jc

y = x\

2r

, , , + tanh"' It = -^ + tanh-' 2jt

1 - 4x^/ 1 - 4x-

112. Let tt = .r3. du = ?,x^ dx.

I ;r(sech x^)- dx = ^\ (sech .■^)^3jr) dr = | tanh .t^ + C

554 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

Problem Solving for Chapter 5

2. (a)

smx dx = — \ smxdx

I

arccos xdx = 2\ — ] = tt

Jo

smxdx = Q

(b)

-I 1 1 \—*^x

It It 'in In

(sin j:
Jo

+ 2)dx = 2(2 tt) = An

(d)

1

^ 1 + (tanx)-^

is symmetric with respect to the

_J ^ nl\\ ^77

(tanx)v^ 2\2/ 4

4. y = O.S-' and y = 1.2-' intersect >> = j:.
>' = 2' does not intersect y = x.

Suppose y = x\s tangent to >> = o^ at (x, y).

c^ = X =^ a= x^l'.
y'=a'lna=l =s>j:ln x^l-

\z=^\'!\x=\=>x = e,a = e^l"

For 0 < (3 < e^^' ~ 1.445, the curve y = of intersects y = x.

6. (a) y = /(x) = arcsin x
sin >> = X

J-nl

Area A = | sin >> • rfy = — cos y

tt/6

V2 , V3 V3 - 72

]"" = _ V2 ^ V3 ^

Jir/fi 2 2

= 0.1589

^--^^ i)(fh§-»'^"«

I

72/2

(bj I arcsin jccic = Area(C) = ("TJ

7r\/7^

A - B

8 2 12

72 1 \ , 72 - 73

^'X-T2J+ 2

= 0.1346

—CONTINUED—

Problem Solving for Chapter 5 555

6. —CONTINUED—

(c) Area A

e^dy

3

3-1=2

AreaB= | ln;ca[r = 3(ln 3) - A = 3 In 3 - 2 = ln27 - 2 = 1.2958
(d) isny = X

Area A

riT/3

Jit/ 4

izny dy

= —In cos vl

7r/3

7r/4

= -In^ + ln^ = ln72 = |ln2
Area C = I arctan x dx = (|^j( VI) - - In 2 - (j
= ^(4V3-3)-iin2»0.6818

(1)

8. y = e'

y' = e^
y — b = e"{x — a)

y = e"x — ae" + b Tangent line
Ify = 0,

e"x = ae" — b
bx = ab - b (b = e")
X = a — I
c = a — I
Thus, a - c = a — (a - I) = I.

10. Let u = tan X. du = sec- x dx

r-r/A

Area

r'r/4

Jo sin-

1

x + 4 cos- X

J"ir/4 1

sec-jc
0 tan2.r-^4

Jo «= + 4
= [^arctang

= farctan(|)

(it

12. (a) f = >'(1 ->').v(0)=|

ln|y| - ln|l - y\ = t + C

y

In

\-y

V

= t+ C

e'+c= c,e'

1 -V

V = C,e' - yC^e'

C,e' 1

Hence, v =

1 1

4 \ + C.
1

1 -(- C.e' 1 + C^e"

C, = 3

1 + 3e-
— CONTINUED—

(b)

= y(l - y) = y - .-v-^

dy

dt

cT-y „ , ^ , „ .,- 1

-pr = V = y — 2\T => V = 0 lor v = —
dt' ' ... -2

-^> OifO< V < ^and-^ < Oif^ < V < 1.
dt- ■ 2 dt- 2 ■

Thus, the rate of growth is maximum at y = -. the
point of inflection.

556 Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

12. —CONTINUED—

(c) y' = y(l -y),ym-2
1

As before, y

1 + Ce-

^(0) = 2 = :p^=>Q=4

Thus, y =

1

The graph is different:

14. (a) u = 985.93 - 985.93

(120,000)(0.095)
12

(-^r

(b) The larger part goes for interest. The curves intersect when t = 27.7 years.

(c) The slopes are negatives of each other. Analytically,

du dv

» = 985.93 -V =>^=-^

m'(15) = -v'(15) = -14.06.

(d) t = 12.7 years

Again, the larger part goes for interest.

3-3:^S7-^

Full text of "Calculus Complete Solutions Guide"

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