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Review Exercises for Chapter 3
43. r
v02 sin 2 32
45. Let f x x, x 100, dx 0.6. f x x f x f x dx
v0 2200 ftsec
changes from 10 to 11 dr
x
22002 cos 2 d 16
10
f x x 99.4
180
d 11 10
1 dx 2x
100
1 0.6 9.97 2100
Using a calculator: 99.4 9.96995
180
r dr
22002 20 cos 16 180
180 4961 feet
4961 feet 49. Let f x x, x 4, dx 0.02, f x 1 2x .
4 x, x 625, dx 1. 47. Let f x 4 x f x x f x f x d x
Then
1 dx 44x3
f 4.02 f 4 f 4 dx
1 4 624 4 625 f x x 1 4 625 3 4 5
4.02 4
1 1 0.02 2 0.02. 4 24
1 4.998 500
4 624 4.9980. Using a calculator,
51. In general, when x → 0, dy approaches y. 53. True
55. True
Review Exercises for Chapter 3 1. A number c in the domain of f is a critical number if f c 0 or f is undefined at c.
y 4
f ′(c) is 3 undefined.
f ′(c) = 0
x −4 −3
−1 −2 −3 −4
3. gx 2x 5 cos x, 0, 2
18
(6.28, 17.57)
g x 2 5 sin x 2
0 when sin x 5 . 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: 2, 17.57 Maximum
(2.73, 0.88) − 4
2 −1
1
2
4
163
164
Chapter 3
Applications of Differentiation
5. Yes. f 3 f 2 0. f is continuous on 3, 2 , differentiable on 3, 2.
7. f x 3 x 4 y
(a)
f x x 33x 1 0 for x 13.
6 4
c 13 satisfies f c 0.
2 x
−2
2
4
6
10
−4 −6
f 1 f 7 0 (b) f is not differentiable at x 4.
9.
f x x23, 1 ≤ x ≤ 8
f x x cos x,
11.
2 f x x13 3
f x 1 sin x f b f a 2 2 1 ba 2 2
f b f a 4 1 3 ba 81 7
f c 1 sin c 1
2 3 f c c13 3 7 c
13.
149
3
≤ x ≤ 2 2
c0
2744 3.764 729
f x Ax2 Bx C f x 2Ax B f x2 f x1 Ax22 x12 Bx2 x1 x2 x1 x2 x1 Ax1 x2 B f c 2Ac B Ax1 x2 B 2Ac Ax1 x2 c
x1 x2 Midpoint of x1, x2
2
15. f x x 12x 3 f x x 1 1 x 32x 1 2
x 13x 7 7 Critical numbers: x 1 and x 3
17. hx xx 3 x32 3x12 Domain: 0, 3 3 h x x12 x12 2 2 3 3x 1 x12x 1 2 2x Critical number: x 1
Interval:
< x < 1
1 < x <
7 3
7 3
< x <
Sign of f x:
f x > 0
f x < 0
f x > 0
Conclusion:
Increasing
Decreasing
Increasing
Interval:
0 < x < 1
Sign of h x:
h x < 0
h x > 0
Conclusion:
Decreasing
Increasing
1 < x <
Review Exercises for Chapter 3 19. ht 14t 4 8t
Test Interval: < t < 2
h t t 3 8 0 when t 2. Relative minimum: 2, 12
165
2 < t <
Sign of h t:
h t < 0
h t > 0
Conclusion:
Decreasing
Increasing
1 1 cos12t sin12t 3 4
21. y
v y 4 sin12t 3 cos12t (a) When t
1 , y inch and v y 4 inches/second. 8 4
(b) y 4 sin12t 3 cos12t 0 when
sin12t 3 3 ⇒ tan12t . cos12t 4 4
3 4 Therefore, sin12t and cos12t . The maximum displacement is 5 5 y (c) Period:
1345 41 53 125 inch. 2 12 6 1 6 6
Frequency:
23. f x x cos x, 0 ≤ x ≤ 2 f x 1 sin x f x cos x 0 when x Points of inflection:
3 , . 2 2
Test Interval: Sign of f x: Conclusion:
3 < x < 2 2
3 < x < 2 2
f x < 0
f x > 0
f x < 0
Concave downward
Concave upward
Concave downward
0 < x <
2
2 , 2 , 32, 32
25. gx 2x21 x2
y
g x 4x2x2 1 Critical numbers: x 0, ±
(−
1 2
1, 1 2 2
)
1
−2
g x 4 24x
(
1, 1 2 2
)
(0, 0)
2
x
−1
2
−2
g 0 4 > 0
Relative minimum at 0, 0
1 1 Relative maximums at ± , 2 2
y
27. 6
(5, f(5))
5 4
(3, f(3))
2 1 −1
29. The first derivative is positive and the second derivative is negative. The graph is increasing and is concave down.
7
3
−3
1 g ± 8 < 0 2
(6, 0) (0, 0) 2 3 4 5
x 7
166
Chapter 3
Applications of Differentiation
31. (a) D 0.0034t4 0.2352t3 4.9423t2 20.8641t 94.4025 (b)
369
0
29 0
(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
x →
2x2 2 2 lim 5 x → 3 5x2 3
35. lim
2x 3 x4
39. f x
3x2
37. hx
Discontinuity: x 4 lim
x →
x →
5 cos x 0, since 5 cos x ≤ 5. x
3 2 x
Discontinuity: x 0
2x 3 2 3x lim 2 x → 1 4x x4
lim
x →
3x 2 2
Vertical asymptote: x 4
Vertical asymptote: x 0
Horizontal asymptote: y 2
Horizontal asymptote: y 2
41. f x x3
243 x
43. f x
x1 1 3x2
Relative minimum: 3, 108
Relative minimum: 0.155, 1.077
Relative maximum: 3, 108
Relative maximum: 2.155, 0.077 0.2
200
−2
−5
5
5
− 1.4
− 200
Vertical asymptote: x 0
Horizontal asymptote: y 0
45. f x 4x x2 x4 x Domain: , ; Range: , 4 f x 4 2x 0 when x 2. f x 2 Therefore, 2, 4 is a relative maximum. Intercepts: 0, 0, 4, 0
y
5
)2, 4)
4 3 2 1
x 1
2
3
5
Review Exercises for Chapter 3 47. f x x16 x2, Domain: 4, 4 , Range: 8, 8
y
2
2, 8
8
Domain: 4, 4 ; Range: 8, 8
6 4
16 2x2 f x 0 when x ± 22 and undefined when x ± 4. 16 x2 f x
2xx2 24 16 x232
2
(− 4, 0)
(4, 0) x
8
6
2
2
4
6
8
(0, 0)
8
2
2,
8
f 22 > 0
Therefore, 22, 8 is a relative minimum. f 22 < 0
Therefore, 22, 8 is a relative maximum. Point of inflection: 0, 0 Intercepts: 4, 0, 0, 0, 4, 0 Symmetry with respect to origin 49. f x x 13x 32
y
Domain: , ; Range: ,
4
f x x 12x 35x 11 0 when x 1,
11 , 3. 5
f x 4x 15x2 22x 23 0 when x 1,
( 115 , 1.11(
(1, 0)
(2.69, 0.46) (3, 0)
2
x
−2
4 −2
11± 6 . 5
6
(1.71, 0.60)
−4
f 3 > 0 Therefore, 3, 0 is a relative minimum. f
115 < 0
Therefore,
is a relative maximum. 115, 3456 3125
Points of inflection: 1, 0,
11 5
6
11 5
, 0.60 ,
6
, 0.46
Intercepts: 0, 9, 1, 0, 3, 0 51. f x x13x 323
y
Domain: , ; Range: ,
4 3
x1 f x 0 when x 1 and undefined when x 3, 0. x 313x23 2 f x 53 is undefined when x 0, 3. x x 343 3 4 is By the First Derivative Test 3, 0 is a relative maximum and 1, a relative minimum. 0, 0 is a point of inflection.
Intercepts: 3, 0, 0, 0
2 1
) 3, 0)
)0, 0) x
5
4
2
) 1,
1
1
1.59) 3
2
167
168
Chapter 3
Applications of Differentiation
x1 x1
53. f x
x
1
y
Domain: , 1, 1, ; Range: , 1, 1, f x
2 < 0 if x 1. x 12
f x
4 x 13
4
y
1 2
x 2
2
4
2
Horizontal asymptote: y 1 Vertical asymptote: x 1 Intercepts: 1, 0, 0, 1 55. f x
4 1 x2
y 5
Domain: , ; Range: 0, 4
8x 0 when x 0. f x 1 x22 3 81 3x2 . 0 when x ± f x 1 x23 3
(0, 4)
4
(−
3,3 3
(
(
3,3 3
1
2
(
2 1 −3
−2 −1
x −1
3
f 0 < 0 Therefore, 0, 4 is a relative maximum. Points of inflection: ± 33, 3 Intercept: 0, 4 Symmetric to the y-axis Horizontal asymptote: y 0
57. f x x3 x
y
4 x
10
Domain: , 0, 0, ; Range: , 6 , 6, f x 3x2 1 f x 6x
4 x2
3x4
4 0 when x ± 1. x2 x2
8 6x4 8 0 x3 x3
f 1 < 0 Therefore, 1, 6 is a relative maximum. f 1 > 0 Therefore, 1, 6 is a relative minimum. Vertical asymptote: x 0 Symmetric with respect to origin
5
(1, 6) x
2
1
(−1, −6) −5
1
x
2
0
Review Exercises for Chapter 3
59. f x x2 9
y
Domain: , ; Range: 0, f x
2xx2 9 0 when x 0 and is undefined when x ± 3. x2 9
10
5
2x2 9 is undefined at x ± 3. f x 2 x 9
)0, 9)
) 3, 0)
)3, 0) x
4
f 0 < 0
2
2
4
Therefore, 0, 9 is a relative maximum. Relative minima: ± 3, 0 Points of inflection: ± 3, 0 Intercepts: ± 3, 0, 0, 9 Symmetric to the y-axis 61. f x x cos x
y
)2 , 2
Domain: 0, 2 ; Range: 1, 1 2
3 3 , 2 2
f x 1 sin x ≥ 0, f is increasing. f x cos x 0 when x Points of inflection:
1)
2
3 , . 2 2
3 3 , , , 2 2 2 2
)0, 1)
, 2 2 x
2
Intercept: 0, 1 63. x2 4y2 2x 16y 13 0 (a) x 2 2x 1 4y 2 4y 4 13 1 16
y
x 1 4y 2 4 x 12 y 22 1 4 1 The graph is an ellipse: 2
2
4
(1, 3) 3 2 1
Maximum: 1, 3
(1, 1) x −1
Minimum: 1, 1
1
2
3
(b) x2 4y2 2x 16y 13 0 2x 8y
dy dy 2 16 0 dx dx dy 8y 16 2 2x dx dy 2 2x 1x dx 8y 16 4y 8
The critical numbers are x 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.
169
170
Chapter 3
Applications of Differentiation
65. Let t 0 at noon.
(100 − 12t, 0) (0, 0)
L d 2 100 12t2 10t2 10,000 2400t 244t 2
A
(100, 0)
d
300 dL 2400 488t 0 when t 4.92 hr. dt 61
B (0, −10t)
Ship A at 40.98, 0; Ship B at 0, 49.18 d 2 10,000 2400t 244t 2 4098.36 when t 4.92 4:55 P.M.. d 64 km 67. We have points 0, y, x, 0, and 1, 8. Thus,
y
08 8x y8 or y . m 01 x1 x1
(0, y)
10
(1, 8)
8 6
Let f x L 2 x 2
x 8x 1 . 2
4 2
x f x 2x 128 x1 x
(x, 0)
x 1 x 0 x 12
x 2
4
6
8
10
64x 0 x 13
x x 13 64 0 when x 0, 5 minimum. Vertices of triangle: 0, 0, 5, 0, 0, 10 69.
A Average of basesHeight
x 2 s
3s2 2sx x2
2
s
see figure
s
dA 1 s xs x 3s2 2sx x2 dx 4 3s2 2sx x2
x−s 2
22s xs x 0 when x 2s. 43s2 2sx x2 A is a maximum when x 2s. 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. 60 6x y6 or y 04 4x x4
Let f x L2 x2 y2 x 2 f x 2x 72
x 6x 4 . 2
0 x x 4 x 4 4 2
3 x x 43 144 0 when x 0 or x 4 144.
L 14.05 feet
s
x−s 2 x
m
3s 2 + 2sx − x 2 2
Review Exercises for Chapter 3 csc
73. csc
L1 or L1 6 csc 6
2 9 or L L2
2
9 csc
see figure
L1 θ
2
L2
θ 9
L L1 L2 6 csc 9 csc
171
6
(π2 − θ(
2 6 csc 9 sec
dL 6 csc cot 9 sec tan 0 d tan3
3 2 2 ⇒ tan 3 3 3
sec 1 tan2 csc L6
1 23
23
323 223
313
sec 323 223 tan 213
323 22312 323 22312 9 3323 22332 ft 21.07 ft Compare to Exercise 72 using a 9 and b 6. 13 2 313
75. Total cost Cost per hourNumber of hours T
v 110 11v 550 5 600 v 60 v 2
dT 11 550 11v 2 33,000 2 dv 60 v 60v 2 0 when v 3000 1030 54.8 mph. d 2T 1100 3 > 0 when v 1030 so this value yields a minimum. dv 2 v 77. f x x3 3x 1 From the graph you can see that f x has three real zeros. f x 3x2 3 f xn
f xn
f xn f xn
1.5000
0.1250
3.7500
0.0333
1.5333
2
1.5333
0.0049
4.0530
0.0012
1.5321
n
xn
f xn
f xn
f xn f xn
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
f xn
f xn
f xn f xn
1
1.9000
0.1590
7.8300
0.0203
1.8797
2
1.8797
0.0024
7.5998
0.0003
1.8794
n
xn
1
xn
f xn f xn
xn
xn
f xn f xn
f xn f xn
The three real zeros of f x are x 1.532, x 0.347, and x 1.879.
172
Chapter 3
Applications of Differentiation
79. Find the zeros of f x x4 x 3. fx 4x3 1 From the graph you can see that f x has two real zeros. f changes sign in 2, 1.
n
xn
f xn
f xn
f xn fxn
1
1.2000
0.2736
7.9120
0.0346
1.1654
2
1.1654
0.0100
7.3312
0.0014
1.1640
xn
f xn fxn
On the interval 2, 1: x 1.164. f changes sign in 1, 2. f xn
f xn fxn
0.5625
12.5000
0.0450
1.4550
1.4550
0.0268
11.3211
0.0024
1.4526
1.4526
0.0003
11.2602
0.0000
1.4526
n
xn
1
1.5000
2 3
f xn
xn
f xn fxn
On the interval 1, 2; x 1.453. 81.
y x1 cos x x x cos x dy 1 x sin x cos x dx dy 1 x sin x cos x dx
83.
S 4 r 2. dr r ± 0.025 dS 8r dr 89± 0.025 ± 1.8 square cm dS 8 r dr 2 dr 100 100 100 S 4 r 2 r
2± 0.025 100 ± 0.56% 9
4 V r3 3 dV 4 r 2 dr 492± 0.025 ± 8.1 cubic cm dV 4 r 2 dr 3 dr 100 100 100 V 43r 3 r
3± 0.025 100 ± 0.83% 9
Problem Solving for Chapter 3 1. Assume y1 < d < y2. Let gx f x dx a. g is continuous on a, b and therefore has a minimum c, gc on a, b. The point c cannot be an endpoint of a, b because ga fa d y1 d < 0 gb fb d y2 d > 0 Hence, a < c < b and gc 0 ⇒ fc d.
Problem Solving for Chapter 3 3. (a) For a 3, 2, 1, 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) px 4ax3 12x 4xax2 3 0 ⇒ x 0, ±
3a
p x 12ax2 12 12ax2 1 For x 0, p 0 12 < 0 ⇒ p has a relative maximum at 0, 0. (c) If a > 0, x ±
3a are the remaining critical numbers.
3a 12 3a 12 24 > 0 ⇒ p has relative minima for a > 0.
p ±
(d) 0, 0 lies on y 3x2. Let x ±
3a. Then
px a
3a
2
6
3 a
2
a=1 a=3 y
2ax 2x
2x2
8 7 6 5 4 3 2
a
For a ≥ 0, there is one relative minimum at 0, 0.
(c) For a < 0, there are two relative minima at x ±
−2
2a.
(d) There are either 1 or 3 critical points. The above analysis shows that there cannot be exactly two relative extrema. c x2 x c c c 2x 0 ⇒ 2 2x ⇒ x3 ⇒ x x2 x 2
cx 3
2c 2 x3
If c 0, f x x2 has a relative minimum, but no relative maximum.
2c is a relative minimum, because f 2c > 0. c If c < 0, x is a relative minimum too. 2 If c > 0, x
3
3
Answer: all c.
a=2
a=0
a = −1 a = −2 a = −3
(b) For a < 0, there is a relative maximum at 0, 1.
f x
a = −1 a = −3
3x2 is satisfied by all the relative extrema of p.
p x 16x2 2a
fx
1
2
3
3a a9 18a a9.
5. px x 4 ax2 1
7. f x
a=1 x
−3
(a) px
a=3
2
9 Thus, y 3 ± a
4x4
y
a=2
3
x −1 −2
2
−8
a = −2 a=0
173
174
Chapter 3
9. Set
Applications of Differentiation
f b f a f ab a k. b a2
Define Fx f x f a fax a kx a2. Fa 0, Fb f b f a fab a kb a2 0 F is continuous on a, b and differentiable on a, b. There exists c1, a < c1 < b, satisfying Fc1 0. Fx fx fa 2kx a satisfies the hypothesis of Rolle’s Theorem on a, c1: Fa 0, Fc1 0. There exists c2, a < c2 < c1 satisfying F c2 0. Finally, F x f x 2k and F c2 0 implies that k
f c2 . 2
Thus, k
11. E E
f b f a f ab a f c2 1 ⇒f b f a fab a f c2b a2. b a2 2 2
tan 1 0.1 tan 10 tan tan2 0.1 tan 1 10 tan
1 10 tan 10 sec2 2 tan sec2 10 tan tan2 10 sec2 0 1 10 tan
⇒ 1 10 tan 10 sec2 2 tan sec2 10 tan tan2 10 sec2 ⇒ 10 sec2 2 tan sec2 100 tan sec2 20 tan2 sec2 100 tan sec2 10 tan2 sec2 ⇒ 10 2 tan 10 tan2 ⇒ 10 tan2 2 tan 10 0 tan
2 ± 4 400 0.90499, 1.10499 20
Using the positive value, 0.7356, or 42.14 . 13. v 2400 sin v 2400 cos 0
3 2n, 2n, n an integer 2 2
Problem Solving for Chapter 3 x y 4 1 or y x 4. 3 4 3
15. The line has equation Rectangle:
4 4 Area A xy x x 4 x2 4x. 3 3 8 8 3 Ax x 4 0 ⇒ x 4 ⇒ x 3 3 2 Dimensions:
3 2 2
Calculus was helpful.
Circle: The distance from the center r, r to the line
r
r r 1 3 4
x y 1 0 must be r: 3 4
12 7r 12 7r 12 5 12 5
1 1 9 16 5r 7r 12 ⇒ r 1 or r 6.
Clearly, r 1. x y 1 and satisfies x y r. 3 4
Semicircle: The center lies on the line
7 12 r r 1 ⇒ r 1 ⇒ r . No calculus necessary. 3 4 12 7
Thus
17. y 1 x21 y
2x 1 x22
y
3 23x2 1 1 0 ⇒ x± ± 3 x2 13 3
y′′:
+++ −−−− −−−− +++ −
3 3
0
3 3
The tangent line has greatest slope at
19. (a)
x sin x
3 3
3
,
4
and least slope at 33, 34 .
0.1
0.2
0.3
0.4
0.5
1.0
0.09983
0.19867
0.29552
0.38942
0.47943
0.84147
sin x ≤ x (b) Let f x sin x. Then fx cos x and on 0, x you have by the Mean Value Theorem, fc cosc Hence,
f x f 0 , 0 < c < x x0 sin x x
sin x cosc ≤ 1 x
⇒ sin x ≤ x ⇒ sin x ≤ x
175
43. r
v02 sin 2 32
45. Let f x x, x 100, dx 0.6. f x x f x f x dx
v0 2200 ftsec
changes from 10 to 11 dr
x
22002 cos 2 d 16
10
f x x 99.4
180
d 11 10
1 dx 2x
100
1 0.6 9.97 2100
Using a calculator: 99.4 9.96995
180
r dr
22002 20 cos 16 180
180 4961 feet
4961 feet 49. Let f x x, x 4, dx 0.02, f x 1 2x .
4 x, x 625, dx 1. 47. Let f x 4 x f x x f x f x d x
Then
1 dx 44x3
f 4.02 f 4 f 4 dx
1 4 624 4 625 f x x 1 4 625 3 4 5
4.02 4
1 1 0.02 2 0.02. 4 24
1 4.998 500
4 624 4.9980. Using a calculator,
51. In general, when x → 0, dy approaches y. 53. True
55. True
Review Exercises for Chapter 3 1. A number c in the domain of f is a critical number if f c 0 or f is undefined at c.
y 4
f ′(c) is 3 undefined.
f ′(c) = 0
x −4 −3
−1 −2 −3 −4
3. gx 2x 5 cos x, 0, 2
18
(6.28, 17.57)
g x 2 5 sin x 2
0 when sin x 5 . 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: 2, 17.57 Maximum
(2.73, 0.88) − 4
2 −1
1
2
4
163
164
Chapter 3
Applications of Differentiation
5. Yes. f 3 f 2 0. f is continuous on 3, 2 , differentiable on 3, 2.
7. f x 3 x 4 y
(a)
f x x 33x 1 0 for x 13.
6 4
c 13 satisfies f c 0.
2 x
−2
2
4
6
10
−4 −6
f 1 f 7 0 (b) f is not differentiable at x 4.
9.
f x x23, 1 ≤ x ≤ 8
f x x cos x,
11.
2 f x x13 3
f x 1 sin x f b f a 2 2 1 ba 2 2
f b f a 4 1 3 ba 81 7
f c 1 sin c 1
2 3 f c c13 3 7 c
13.
149
3
≤ x ≤ 2 2
c0
2744 3.764 729
f x Ax2 Bx C f x 2Ax B f x2 f x1 Ax22 x12 Bx2 x1 x2 x1 x2 x1 Ax1 x2 B f c 2Ac B Ax1 x2 B 2Ac Ax1 x2 c
x1 x2 Midpoint of x1, x2
2
15. f x x 12x 3 f x x 1 1 x 32x 1 2
x 13x 7 7 Critical numbers: x 1 and x 3
17. hx xx 3 x32 3x12 Domain: 0, 3 3 h x x12 x12 2 2 3 3x 1 x12x 1 2 2x Critical number: x 1
Interval:
< x < 1
1 < x <
7 3
7 3
< x <
Sign of f x:
f x > 0
f x < 0
f x > 0
Conclusion:
Increasing
Decreasing
Increasing
Interval:
0 < x < 1
Sign of h x:
h x < 0
h x > 0
Conclusion:
Decreasing
Increasing
1 < x <
Review Exercises for Chapter 3 19. ht 14t 4 8t
Test Interval: < t < 2
h t t 3 8 0 when t 2. Relative minimum: 2, 12
165
2 < t <
Sign of h t:
h t < 0
h t > 0
Conclusion:
Decreasing
Increasing
1 1 cos12t sin12t 3 4
21. y
v y 4 sin12t 3 cos12t (a) When t
1 , y inch and v y 4 inches/second. 8 4
(b) y 4 sin12t 3 cos12t 0 when
sin12t 3 3 ⇒ tan12t . cos12t 4 4
3 4 Therefore, sin12t and cos12t . The maximum displacement is 5 5 y (c) Period:
1345 41 53 125 inch. 2 12 6 1 6 6
Frequency:
23. f x x cos x, 0 ≤ x ≤ 2 f x 1 sin x f x cos x 0 when x Points of inflection:
3 , . 2 2
Test Interval: Sign of f x: Conclusion:
3 < x < 2 2
3 < x < 2 2
f x < 0
f x > 0
f x < 0
Concave downward
Concave upward
Concave downward
0 < x <
2
2 , 2 , 32, 32
25. gx 2x21 x2
y
g x 4x2x2 1 Critical numbers: x 0, ±
(−
1 2
1, 1 2 2
)
1
−2
g x 4 24x
(
1, 1 2 2
)
(0, 0)
2
x
−1
2
−2
g 0 4 > 0
Relative minimum at 0, 0
1 1 Relative maximums at ± , 2 2
y
27. 6
(5, f(5))
5 4
(3, f(3))
2 1 −1
29. The first derivative is positive and the second derivative is negative. The graph is increasing and is concave down.
7
3
−3
1 g ± 8 < 0 2
(6, 0) (0, 0) 2 3 4 5
x 7
166
Chapter 3
Applications of Differentiation
31. (a) D 0.0034t4 0.2352t3 4.9423t2 20.8641t 94.4025 (b)
369
0
29 0
(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
x →
2x2 2 2 lim 5 x → 3 5x2 3
35. lim
2x 3 x4
39. f x
3x2
37. hx
Discontinuity: x 4 lim
x →
x →
5 cos x 0, since 5 cos x ≤ 5. x
3 2 x
Discontinuity: x 0
2x 3 2 3x lim 2 x → 1 4x x4
lim
x →
3x 2 2
Vertical asymptote: x 4
Vertical asymptote: x 0
Horizontal asymptote: y 2
Horizontal asymptote: y 2
41. f x x3
243 x
43. f x
x1 1 3x2
Relative minimum: 3, 108
Relative minimum: 0.155, 1.077
Relative maximum: 3, 108
Relative maximum: 2.155, 0.077 0.2
200
−2
−5
5
5
− 1.4
− 200
Vertical asymptote: x 0
Horizontal asymptote: y 0
45. f x 4x x2 x4 x Domain: , ; Range: , 4 f x 4 2x 0 when x 2. f x 2 Therefore, 2, 4 is a relative maximum. Intercepts: 0, 0, 4, 0
y
5
)2, 4)
4 3 2 1
x 1
2
3
5
Review Exercises for Chapter 3 47. f x x16 x2, Domain: 4, 4 , Range: 8, 8
y
2
2, 8
8
Domain: 4, 4 ; Range: 8, 8
6 4
16 2x2 f x 0 when x ± 22 and undefined when x ± 4. 16 x2 f x
2xx2 24 16 x232
2
(− 4, 0)
(4, 0) x
8
6
2
2
4
6
8
(0, 0)
8
2
2,
8
f 22 > 0
Therefore, 22, 8 is a relative minimum. f 22 < 0
Therefore, 22, 8 is a relative maximum. Point of inflection: 0, 0 Intercepts: 4, 0, 0, 0, 4, 0 Symmetry with respect to origin 49. f x x 13x 32
y
Domain: , ; Range: ,
4
f x x 12x 35x 11 0 when x 1,
11 , 3. 5
f x 4x 15x2 22x 23 0 when x 1,
( 115 , 1.11(
(1, 0)
(2.69, 0.46) (3, 0)
2
x
−2
4 −2
11± 6 . 5
6
(1.71, 0.60)
−4
f 3 > 0 Therefore, 3, 0 is a relative minimum. f
115 < 0
Therefore,
is a relative maximum. 115, 3456 3125
Points of inflection: 1, 0,
11 5
6
11 5
, 0.60 ,
6
, 0.46
Intercepts: 0, 9, 1, 0, 3, 0 51. f x x13x 323
y
Domain: , ; Range: ,
4 3
x1 f x 0 when x 1 and undefined when x 3, 0. x 313x23 2 f x 53 is undefined when x 0, 3. x x 343 3 4 is By the First Derivative Test 3, 0 is a relative maximum and 1, a relative minimum. 0, 0 is a point of inflection.
Intercepts: 3, 0, 0, 0
2 1
) 3, 0)
)0, 0) x
5
4
2
) 1,
1
1
1.59) 3
2
167
168
Chapter 3
Applications of Differentiation
x1 x1
53. f x
x
1
y
Domain: , 1, 1, ; Range: , 1, 1, f x
2 < 0 if x 1. x 12
f x
4 x 13
4
y
1 2
x 2
2
4
2
Horizontal asymptote: y 1 Vertical asymptote: x 1 Intercepts: 1, 0, 0, 1 55. f x
4 1 x2
y 5
Domain: , ; Range: 0, 4
8x 0 when x 0. f x 1 x22 3 81 3x2 . 0 when x ± f x 1 x23 3
(0, 4)
4
(−
3,3 3
(
(
3,3 3
1
2
(
2 1 −3
−2 −1
x −1
3
f 0 < 0 Therefore, 0, 4 is a relative maximum. Points of inflection: ± 33, 3 Intercept: 0, 4 Symmetric to the y-axis Horizontal asymptote: y 0
57. f x x3 x
y
4 x
10
Domain: , 0, 0, ; Range: , 6 , 6, f x 3x2 1 f x 6x
4 x2
3x4
4 0 when x ± 1. x2 x2
8 6x4 8 0 x3 x3
f 1 < 0 Therefore, 1, 6 is a relative maximum. f 1 > 0 Therefore, 1, 6 is a relative minimum. Vertical asymptote: x 0 Symmetric with respect to origin
5
(1, 6) x
2
1
(−1, −6) −5
1
x
2
0
Review Exercises for Chapter 3
59. f x x2 9
y
Domain: , ; Range: 0, f x
2xx2 9 0 when x 0 and is undefined when x ± 3. x2 9
10
5
2x2 9 is undefined at x ± 3. f x 2 x 9
)0, 9)
) 3, 0)
)3, 0) x
4
f 0 < 0
2
2
4
Therefore, 0, 9 is a relative maximum. Relative minima: ± 3, 0 Points of inflection: ± 3, 0 Intercepts: ± 3, 0, 0, 9 Symmetric to the y-axis 61. f x x cos x
y
)2 , 2
Domain: 0, 2 ; Range: 1, 1 2
3 3 , 2 2
f x 1 sin x ≥ 0, f is increasing. f x cos x 0 when x Points of inflection:
1)
2
3 , . 2 2
3 3 , , , 2 2 2 2
)0, 1)
, 2 2 x
2
Intercept: 0, 1 63. x2 4y2 2x 16y 13 0 (a) x 2 2x 1 4y 2 4y 4 13 1 16
y
x 1 4y 2 4 x 12 y 22 1 4 1 The graph is an ellipse: 2
2
4
(1, 3) 3 2 1
Maximum: 1, 3
(1, 1) x −1
Minimum: 1, 1
1
2
3
(b) x2 4y2 2x 16y 13 0 2x 8y
dy dy 2 16 0 dx dx dy 8y 16 2 2x dx dy 2 2x 1x dx 8y 16 4y 8
The critical numbers are x 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.
169
170
Chapter 3
Applications of Differentiation
65. Let t 0 at noon.
(100 − 12t, 0) (0, 0)
L d 2 100 12t2 10t2 10,000 2400t 244t 2
A
(100, 0)
d
300 dL 2400 488t 0 when t 4.92 hr. dt 61
B (0, −10t)
Ship A at 40.98, 0; Ship B at 0, 49.18 d 2 10,000 2400t 244t 2 4098.36 when t 4.92 4:55 P.M.. d 64 km 67. We have points 0, y, x, 0, and 1, 8. Thus,
y
08 8x y8 or y . m 01 x1 x1
(0, y)
10
(1, 8)
8 6
Let f x L 2 x 2
x 8x 1 . 2
4 2
x f x 2x 128 x1 x
(x, 0)
x 1 x 0 x 12
x 2
4
6
8
10
64x 0 x 13
x x 13 64 0 when x 0, 5 minimum. Vertices of triangle: 0, 0, 5, 0, 0, 10 69.
A Average of basesHeight
x 2 s
3s2 2sx x2
2
s
see figure
s
dA 1 s xs x 3s2 2sx x2 dx 4 3s2 2sx x2
x−s 2
22s xs x 0 when x 2s. 43s2 2sx x2 A is a maximum when x 2s. 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. 60 6x y6 or y 04 4x x4
Let f x L2 x2 y2 x 2 f x 2x 72
x 6x 4 . 2
0 x x 4 x 4 4 2
3 x x 43 144 0 when x 0 or x 4 144.
L 14.05 feet
s
x−s 2 x
m
3s 2 + 2sx − x 2 2
Review Exercises for Chapter 3 csc
73. csc
L1 or L1 6 csc 6
2 9 or L L2
2
9 csc
see figure
L1 θ
2
L2
θ 9
L L1 L2 6 csc 9 csc
171
6
(π2 − θ(
2 6 csc 9 sec
dL 6 csc cot 9 sec tan 0 d tan3
3 2 2 ⇒ tan 3 3 3
sec 1 tan2 csc L6
1 23
23
323 223
313
sec 323 223 tan 213
323 22312 323 22312 9 3323 22332 ft 21.07 ft Compare to Exercise 72 using a 9 and b 6. 13 2 313
75. Total cost Cost per hourNumber of hours T
v 110 11v 550 5 600 v 60 v 2
dT 11 550 11v 2 33,000 2 dv 60 v 60v 2 0 when v 3000 1030 54.8 mph. d 2T 1100 3 > 0 when v 1030 so this value yields a minimum. dv 2 v 77. f x x3 3x 1 From the graph you can see that f x has three real zeros. f x 3x2 3 f xn
f xn
f xn f xn
1.5000
0.1250
3.7500
0.0333
1.5333
2
1.5333
0.0049
4.0530
0.0012
1.5321
n
xn
f xn
f xn
f xn f xn
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
f xn
f xn
f xn f xn
1
1.9000
0.1590
7.8300
0.0203
1.8797
2
1.8797
0.0024
7.5998
0.0003
1.8794
n
xn
1
xn
f xn f xn
xn
xn
f xn f xn
f xn f xn
The three real zeros of f x are x 1.532, x 0.347, and x 1.879.
172
Chapter 3
Applications of Differentiation
79. Find the zeros of f x x4 x 3. fx 4x3 1 From the graph you can see that f x has two real zeros. f changes sign in 2, 1.
n
xn
f xn
f xn
f xn fxn
1
1.2000
0.2736
7.9120
0.0346
1.1654
2
1.1654
0.0100
7.3312
0.0014
1.1640
xn
f xn fxn
On the interval 2, 1: x 1.164. f changes sign in 1, 2. f xn
f xn fxn
0.5625
12.5000
0.0450
1.4550
1.4550
0.0268
11.3211
0.0024
1.4526
1.4526
0.0003
11.2602
0.0000
1.4526
n
xn
1
1.5000
2 3
f xn
xn
f xn fxn
On the interval 1, 2; x 1.453. 81.
y x1 cos x x x cos x dy 1 x sin x cos x dx dy 1 x sin x cos x dx
83.
S 4 r 2. dr r ± 0.025 dS 8r dr 89± 0.025 ± 1.8 square cm dS 8 r dr 2 dr 100 100 100 S 4 r 2 r
2± 0.025 100 ± 0.56% 9
4 V r3 3 dV 4 r 2 dr 492± 0.025 ± 8.1 cubic cm dV 4 r 2 dr 3 dr 100 100 100 V 43r 3 r
3± 0.025 100 ± 0.83% 9
Problem Solving for Chapter 3 1. Assume y1 < d < y2. Let gx f x dx a. g is continuous on a, b and therefore has a minimum c, gc on a, b. The point c cannot be an endpoint of a, b because ga fa d y1 d < 0 gb fb d y2 d > 0 Hence, a < c < b and gc 0 ⇒ fc d.
Problem Solving for Chapter 3 3. (a) For a 3, 2, 1, 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) px 4ax3 12x 4xax2 3 0 ⇒ x 0, ±
3a
p x 12ax2 12 12ax2 1 For x 0, p 0 12 < 0 ⇒ p has a relative maximum at 0, 0. (c) If a > 0, x ±
3a are the remaining critical numbers.
3a 12 3a 12 24 > 0 ⇒ p has relative minima for a > 0.
p ±
(d) 0, 0 lies on y 3x2. Let x ±
3a. Then
px a
3a
2
6
3 a
2
a=1 a=3 y
2ax 2x
2x2
8 7 6 5 4 3 2
a
For a ≥ 0, there is one relative minimum at 0, 0.
(c) For a < 0, there are two relative minima at x ±
−2
2a.
(d) There are either 1 or 3 critical points. The above analysis shows that there cannot be exactly two relative extrema. c x2 x c c c 2x 0 ⇒ 2 2x ⇒ x3 ⇒ x x2 x 2
cx 3
2c 2 x3
If c 0, f x x2 has a relative minimum, but no relative maximum.
2c is a relative minimum, because f 2c > 0. c If c < 0, x is a relative minimum too. 2 If c > 0, x
3
3
Answer: all c.
a=2
a=0
a = −1 a = −2 a = −3
(b) For a < 0, there is a relative maximum at 0, 1.
f x
a = −1 a = −3
3x2 is satisfied by all the relative extrema of p.
p x 16x2 2a
fx
1
2
3
3a a9 18a a9.
5. px x 4 ax2 1
7. f x
a=1 x
−3
(a) px
a=3
2
9 Thus, y 3 ± a
4x4
y
a=2
3
x −1 −2
2
−8
a = −2 a=0
173
174
Chapter 3
9. Set
Applications of Differentiation
f b f a f ab a k. b a2
Define Fx f x f a fax a kx a2. Fa 0, Fb f b f a fab a kb a2 0 F is continuous on a, b and differentiable on a, b. There exists c1, a < c1 < b, satisfying Fc1 0. Fx fx fa 2kx a satisfies the hypothesis of Rolle’s Theorem on a, c1: Fa 0, Fc1 0. There exists c2, a < c2 < c1 satisfying F c2 0. Finally, F x f x 2k and F c2 0 implies that k
f c2 . 2
Thus, k
11. E E
f b f a f ab a f c2 1 ⇒f b f a fab a f c2b a2. b a2 2 2
tan 1 0.1 tan 10 tan tan2 0.1 tan 1 10 tan
1 10 tan 10 sec2 2 tan sec2 10 tan tan2 10 sec2 0 1 10 tan
⇒ 1 10 tan 10 sec2 2 tan sec2 10 tan tan2 10 sec2 ⇒ 10 sec2 2 tan sec2 100 tan sec2 20 tan2 sec2 100 tan sec2 10 tan2 sec2 ⇒ 10 2 tan 10 tan2 ⇒ 10 tan2 2 tan 10 0 tan
2 ± 4 400 0.90499, 1.10499 20
Using the positive value, 0.7356, or 42.14 . 13. v 2400 sin v 2400 cos 0
3 2n, 2n, n an integer 2 2
Problem Solving for Chapter 3 x y 4 1 or y x 4. 3 4 3
15. The line has equation Rectangle:
4 4 Area A xy x x 4 x2 4x. 3 3 8 8 3 Ax x 4 0 ⇒ x 4 ⇒ x 3 3 2 Dimensions:
3 2 2
Calculus was helpful.
Circle: The distance from the center r, r to the line
r
r r 1 3 4
x y 1 0 must be r: 3 4
12 7r 12 7r 12 5 12 5
1 1 9 16 5r 7r 12 ⇒ r 1 or r 6.
Clearly, r 1. x y 1 and satisfies x y r. 3 4
Semicircle: The center lies on the line
7 12 r r 1 ⇒ r 1 ⇒ r . No calculus necessary. 3 4 12 7
Thus
17. y 1 x21 y
2x 1 x22
y
3 23x2 1 1 0 ⇒ x± ± 3 x2 13 3
y′′:
+++ −−−− −−−− +++ −
3 3
0
3 3
The tangent line has greatest slope at
19. (a)
x sin x
3 3
3
,
4
and least slope at 33, 34 .
0.1
0.2
0.3
0.4
0.5
1.0
0.09983
0.19867
0.29552
0.38942
0.47943
0.84147
sin x ≤ x (b) Let f x sin x. Then fx cos x and on 0, x you have by the Mean Value Theorem, fc cosc Hence,
f x f 0 , 0 < c < x x0 sin x x
sin x cosc ≤ 1 x
⇒ sin x ≤ x ⇒ sin x ≤ x
175
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