Chapter 05 the derivative in graphing and application
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Transcript of Chapter 05 the derivative in graphing and application
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153
CHAPTER 5
The Derivative in Graphing and Applications
EXERCISE SET 5.1
1. (a) f β² > 0 and f β²β² > 0
y
x
(b) f β² > 0 and f β²β² < 0y
x
(c) f β² < 0 and f β²β² > 0y
x
(d) f β² < 0 and f β²β² < 0y
x
2. (a) y
x
(b) y
x
(c) y
x
(d) y
x
3. A: dy/dx < 0, d2y/dx2 > 0B: dy/dx > 0, d2y/dx2 < 0C: dy/dx < 0, d2y/dx2 < 0
4. A: dy/dx < 0, d2y/dx2 < 0B: dy/dx < 0, d2y/dx2 > 0C: dy/dx > 0, d2y/dx2 < 0
5. An inflection point occurs when f β²β² changes sign: at x = β1, 0, 1 and 2.
6. (a) f(0) < f(1) since f β² > 0 on (0, 1). (b) f(1) > f(2) since f β² < 0 on (1, 2).(c) f β²(0) > 0 by inspection. (d) f β²(1) = 0 by inspection.(e) f β²β²(0) < 0 since f β² is decreasing there. (f) f β²β²(2) = 0 since f β² has a minimum there.
7. (a) [4, 6] (b) [1, 4] and [6, 7] (c) (1, 2) and (3, 5)(d) (2, 3) and (5, 7) (e) x = 2, 3, 5
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154 Chapter 5
8. (1, 2) (2, 3) (3, 4) (4, 5) (5, 6) (6, 7)f β² β β β + + βf β²β² + β + + β β
9. (a) f is increasing on [1, 3] (b) f is decreasing on (ββ, 1], [3,+β](c) f is concave up on (ββ, 2), (4,+β) (d) f is concave down on (2, 4)(e) points of inflection at x = 2, 4
10. (a) f is increasing on (ββ,+β) (b) f is nowhere decreasing(c) f is concave up on (ββ, 1), (3,+β) (d) f is concave down on (1, 3)(e) f has points of inflection at x = 1, 3
11. f β²(x) = 2(xβ 3/2)f β²β²(x) = 2
(a) [3/2,+β) (b) (ββ, 3/2](c) (ββ,+β) (d) nowhere(e) none
12. f β²(x) = β2(2 + x)f β²β²(x) = β2
(a) (ββ,β2] (b) [β2,+β)(c) nowhere (d) (ββ,+β)(e) none
13. f β²(x) = 6(2x+ 1)2
f β²β²(x) = 24(2x+ 1)(a) (ββ,+β) (b) nowhere(c) (β1/2,+β) (d) (ββ,β1/2)(e) β1/2
14. f β²(x) = 3(4β x2)f β²β²(x) = β6x
(a) [β2, 2] (b) (ββ,β2], [2,+β)(c) (ββ, 0) (d) (0,+β)(e) 0
15. f β²(x) = 12x2(xβ 1)f β²β²(x) = 36x(xβ 2/3)
(a) [1,+β) (b) (ββ, 1](c) (ββ, 0), (2/3,+β) (d) (0, 2/3)(e) 0, 2/3
16. f β²(x) = x(4x2 β 15x+ 18)f β²β²(x) = 6(xβ 1)(2xβ 3)
(a) [0,+β), (b) (ββ, 0](c) (ββ, 1), (3/2,+β) (d) (1, 3/2)(e) 1, 3/2
17. f β²(x) = β3(x2 β 3x+ 1)
(x2 β x+ 1)3
f β²β²(x) =6x(2x2 β 8x+ 5)(x2 β x+ 1)4
(a) [ 3ββ
52 , 3+
β5
2 ] (b) (ββ, 3ββ
52 ], [3+
β5
2 ,+β)
(c) (0, 2ββ
62 ), (2 +
β6
2 ,+β) (d) (ββ, 0), (2ββ
62 , 2 +
β6
2 )
(e) 0, 2ββ6/2, 2 +
β6/2
18. f β²(x) = β x2 β 2
(x+ 2)2
f β²β²(x) =2x(x2 β 6)(x+ 2)3
(a) (ββ,ββ2), (β2,+β) (b) (β
β2,β2)
(c) (ββ,ββ6), (0,
β6) (d) (β
β6, 0), (
β6,+β)
(e) none
19. f β²(x) =2x+ 1
3(x2 + x+ 1)2/3
f β²β²(x) = β 2(x+ 2)(xβ 1)9(x2 + x+ 1)5/3
(a) [β1/2,+β) (b) (ββ,β1/2](c) (β2, 1) (d) (ββ,β2), (1,+β)(e) β2, 1
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Exercise Set 5.1 155
20. f β²(x) =4(xβ 1/4)3x2/3
f β²β²(x) =4(x+ 1/2)9x5/3
(a) [1/4,+β) (b) (ββ, 1/4](c) (ββ,β1/2), (0,+β) (d) (β1/2, 0)(e) β1/2, 0
21. f β²(x) =4(x2/3 β 1)3x1/3
f β²β²(x) =4(x5/3 + x)9x7/3
(a) [β1, 0], [1,+β) (b) (ββ,β1], [0, 1](c) (ββ, 0), (0,+β) (d) nowhere(e) none
22. f β²(x) =23xβ1/3 β 1
f β²β²(x) = β 29x4/3
(a) [β1, 0], [1,+β) (b) (ββ,β1], [0, 1](c) (ββ, 0), (0,+β) (d) nowhere(e) none
23. f β²(x) = βxeβx2/2
f β²β²(x) = (β1 + x2)eβx2/2
(a) (ββ, 0] (b) [0,+β)(c) (ββ,β1), (1,+β) (d) (β1, 1)(e) β1, 1
24. f β²(x) = (2x2 + 1)ex2
f β²β²(x) = 2x(2x2 + 3)ex2
(a) (ββ,+β) (b) none(c) (0,+β) (d) (ββ, 0)(e) 0
25. f β²(x) =x
x2 + 4
f β²β²(x) = β x2 β 4(x2 + 4)2
(a) [0,+β) (b) (ββ, 0](c) (β2,+2) (d) (ββ,β2), (2,+β)(e) β2,+2
26. f β²(x) = x2(1 + 3 lnx)f β²β²(x) = x(5 + 6 lnx)
(a) [eβ1/3,+β) (b) (0, eβ1/3]
(c) (eβ5/6,+β) (d) (0, eβ5/6)
(e) eβ5/6
27. f β²(x) =2x
1 + (x2 β 1)2
f β²β²(x) = β2 3x4 β 2x2 β 2
[1 + (x2 β 1)2]2
(a) [0 +β) (b) (ββ, 0]
(c)(ββ1+
β7β
3, ββ
1ββ
7β3
),(β1β
β7β
3,
β1+β
7β3
)(d)
(ββ,β
β1+β
7β3
),(ββ1β
β7β
3,
β1ββ
7β3
),(β1+
β7β
3,+β
)(e) four: Β±
β1Β±β7
3
28. f β²(x) =2
3x1/3β1β x4/3
f β²β²(x) =2(β1 + 3x4/3
)9x4/3
(1β x4/3
)3/2(a) [0, 1](b) [β1, 0](c) (β1,β3β3/4), (0, 3β3/4, 1)
(d) (3β3/4, 0), (3β3/4
(e) Β±3β3/4
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156 Chapter 5
29. f β²(x) = cosx+ sinxf β²β²(x) = β sinx+ cosx(a) [βΟ/4, 3Ο/4] (b) (βΟ,βΟ/4], [3Ο/4, Ο)(c) (β3Ο/4, Ο/4) (d) (βΟ,β3Ο/4), (Ο/4, Ο)(e) β3Ο/4, Ο/4
1.5
β1.5
^ 6
30. f β²(x) = (2 tan2 x+ 1) secxf β²β²(x) = secx tanx(6 tan2 x+ 5)
(a) [βΟ/2, Ο/2] (b) nowhere(c) (0, Ο/2) (d) (βΟ/2, 0)(e) 0
10
β10
^ 6
31. f β²(x) = β12sec2(x/2)
f β²β²(x) = β12tan(x/2) sec2(x/2))
(a) nowhere (b) (βΟ, Ο)(c) (βΟ, 0) (d) (0, Ο)(e) 0
10
β10
C c
32. f β²(x) = 2β csc2 xf β²β²(x) = 2 csc2 x cotx = 2
cosxsin3 x
(a) [Ο/4, 3Ο/4] (b) (0, Ο/4], [3Ο/4, Ο)(c) (0, Ο/2) (d) (Ο/2, Ο)(e) Ο/2
8
β2
0 p
33. f(x) = 1 + sin 2xf β²(x) = 2 cos 2xf β²β²(x) = β4 sin 2x(a) [βΟ,β3Ο/4], [βΟ/4, Ο/4], [3Ο/4, Ο](b) [β3Ο/4,βΟ/4], [Ο/4, 3Ο/4](c) (βΟ/2, 0), (Ο/2, Ο)(d) (βΟ,βΟ/2), (0, Ο/2)(e) βΟ/2, 0, Ο/2
2
0C c
34. f β²(x) = 2 sin 4xf β²β²(x) = 8 cos 4x
(a) (0, Ο/4], [Ο/2, 3Ο/4](b) [Ο/4, Ο/2], [3Ο/4, Ο](c) (0, Ο/8), (3Ο/8, 5Ο/8), (7Ο/8, Ο)(d) (Ο/8, 3Ο/8), (5Ο/8, 7Ο/8)(e) Ο/8, 3Ο/8, 5Ο/8, 7Ο/8
1
00 p
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Exercise Set 5.1 157
35. (a)
2
4
x
y (b)
2
4
x
y (c)
2
4
x
y
36. (a)
2
4
x
y (b)
2
4
x
y (c)
2
4
x
y
37. (a) g(x) has no zeros:There can be no zero of g(x) on the interval ββ < x < 0 because if there were, sayg(x0) = 0 where x0 < 0, then gβ²(x) would have to be positive between x = x0 and x = 0, saygβ²(x1) > 0 where x0 < x1 < 0. But then gβ²(x) cannot be concave up on the interval (x1, 0),a contradiction.There can be no zero of g(x) on 0 < x < 4 because g(x) is concave up for 0 < x < 4 and thus
the graph of g(x), for 0 < x < 4, must lie above the line y = β23x+ 2, which is the tangent
line to the curve at (0, 2), and above the line y = 3(x β 4) + 3 = 3x β 9 also for 0 < x < 4(see figure). The first condition says that g(x) could only be zero for x > 3 and the secondcondition says that g(x) could only be zero for x < 3, thus g(x) has no zeros for 0 < x < 4.Finally, if 4 < x < +β, g(x) could only have a zero if gβ²(x) were negative somewherefor x > 4, and since gβ²(x) is decreasing there we would ultimately have g(x) < β10, acontradiction.
(b) one, between 0 and 4(c) We must have lim
xβ+βgβ²(x) = 0; if the limit were β5
then g(x) would at some time cross the line g(x) =β10; if the limit were 5 then, since g is concave downfor x > 4 and gβ²(4) = 3, gβ² must decrease for x > 4and thus the limit would be β€ 4.
x1
4
4
y
slope 4slope β2/3
38. (a) f β²(x) = 3(xβ a)2, f β²β²(x) = 6(xβ a); inflection point is (a, 0)(b) f β²(x) = 4(xβ a)3, f β²β²(x) = 12(xβ a)2; no inflection points
39. For n β₯ 2, f β²β²(x) = n(nβ 1)(xβ a)nβ2; there is a sign change of f β²β² (point of inflection) at (a, 0)if and only if n is odd. For n = 1, y = xβ a, so there is no point of inflection.
40. If t is in the interval (a, b) and t < x0 then, because f is increasing,f(t)β f(x)tβ x β₯ 0. If x0 < t
thenf(t)β f(x)tβ x β₯ 0. Thus f β²(x0) = lim
tβx
f(t)β f(x)tβ x β₯ 0.
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158 Chapter 5
41. f β²(x) = 1/3β 1/[3(1 + x)2/3] so f is increasing on [0,+β)thus if x > 0, then f(x) > f(0) = 0, 1 + x/3β 3
β1 + x > 0,
3β1 + x < 1 + x/3.
2.5
00 10
42. f β²(x) = sec2 xβ 1 so f is increasing on [0, Ο/2)thus if 0 < x < Ο/2, then f(x) > f(0) = 0,tanxβ x > 0, x < tanx.
10
00 6
43. x β₯ sinx on [0,+β): let f(x) = xβ sinx.Then f(0) = 0 and f β²(x) = 1β cosx β₯ 0,so f(x) is increasing on [0,+β).
4
β1
0 4
44. Let f(x) = 1 β x2/2 β cosx for x β₯ 0. Then f(0) = 0 and f β²(x) = βx + sinx. By Exercise 43,f β²(x) β€ 0 for x β₯ 0, so f(x) β€ 0 for all x β₯ 0, that is, cosx β₯ 1β x2/2.
45. (a) Let f(x) = xβ ln(x+ 1) for x β₯ 0. Then f(0) = 0 and f β²(x) = 1β 1/(x+ 1) β₯ 0 for x β₯ 0,so f is increasing for x β₯ 0 and thus ln(x+ 1) β€ x for x β₯ 0.
(b) Let g(x) = xβ 12x
2 β ln(x+ 1). Then g(0) = 0 and gβ²(x) = 1β xβ 1/(x+ 1) β€ 0 for x β₯ 0since 1β x2 β€ 1. Thus g is decreasing and thus ln(x+ 1) β₯ xβ 1
2x2 for x β₯ 0.
(c) 2
00 2
1.2
00 2
46. (a) Let h(x) = ex β 1β x for x β₯ 0. Then h(0) = 0 and hβ²(x) = ex β 1 β₯ 0 for x β₯ 0, so h(x) isincreasing.
(b) Let h(x) = exβ1βxβ 12x
2. Then h(0) = 0 and hβ²(x) = exβ1βx. By Part (a), exβ1βx β₯ 0for x β₯ 0, so h(x) is increasing.
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Exercise Set 5.1 159
(c) 6
00 2
6
00 2
47. Points of inflection at x = β2,+2. Concave upon (β5,β2) and (2, 5); concave down on (β2, 2).Increasing on [β3.5829, 0.2513] and [3.3316, 5],and decreasing on [β5,β3.5829] and [0.2513, 3.3316].
250
β250
β5 5
48. Points of inflection at x = Β±1/β3. Concave up
on [β5,β1/β3] and [1/
β3, 5], and concave down
on [β1/β3, 1/β3]. Increasing on [β5, 0]
and decreasing on [0, 5].
1
β2
β5 5
49. f β²β²(x) = 290x3 β 81x2 β 585x+ 397
(3x2 β 5x+ 8)3 . The denominator has complex roots, so is always positive;
hence the x-coordinates of the points of inflection of f(x) are the roots of the numerator (if itchanges sign). A plot of the numerator over [β5, 5] shows roots lying in [β3,β2], [0, 1], and [2, 3].To six decimal places the roots are x = β2.464202, 0.662597, 2.701605.
50. f β²β²(x) =2x5 + 5x3 + 14x2 + 30xβ 7
(x2 + 1)5/2. Points of inflection will occur when the numerator changes
sign, since the denominator is always positive. A plot of y = 2x5+5x3+14x2+30xβ7 shows thatthere is only one root and it lies in [0, 1]. To six decimal place the point of inflection is located atx = 0.210970.
51. f(x1)βf(x2) = x21βx2
2 = (x1+x2)(x1βx2) < 0 if x1 < x2 for x1, x2 in [0,+β), so f(x1) < f(x2)and f is thus increasing.
52. f(x1) β f(x2) =1x1β 1x2=x2 β x1
x1x2> 0 if x1 < x2 for x1, x2 in (0,+β), so f(x1) > f(x2) and
thus f is decreasing.
53. (a) If x1 < x2 where x1 and x2 are in I, then f(x1) < f(x2) and g(x1) < g(x2), sof(x1) + g(x1) < f(x2) + g(x2), (f + g)(x1) < (f + g)(x2). Thus f + g is increasing on I.
(b) Case I: If f and g are β₯ 0 on I, and if x1 < x2 where x1 and x2 are in I, then0 < f(x1) < f(x2) and 0 < g(x1) < g(x2), so f(x1)g(x1) < f(x2)g(x2),(f Β· g)(x1) < (f Β· g)(x2). Thus f Β· g is increasing on I.
Case II: If f and g are not necessarily positive on I then no conclusion can be drawn: forexample, f(x) = g(x) = x are both increasing on (ββ, 0), but (f Β· g)(x) = x2 isdecreasing there.
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160 Chapter 5
54. (a) f β² and gβ² are increasing functions on the interval. By Exercise 53, f β² + gβ² is increasing.(b) f β² and gβ² are increasing functions on the interval. If f β² β₯ 0 and gβ² β₯ 0 on the interval then
by Exercise 53, f Β· g is concave up on the interval.If the requirement of nonnegativity is removed, then the conclusion may not be true: letf(x) = g(x) = x2 on (ββ, 0). Each is concave up,
55. (a) f(x) = x, g(x) = 2x (b) f(x) = x, g(x) = x+ 6 (c) f(x) = 2x, g(x) = x
56. (a) f(x) = ex, g(x) = e2x (b) f(x) = g(x) = x (c) f(x) = e2x, g(x) = ex
57. (a) f β²β²(x) = 6ax + 2b = 6a(x+
b
3a
), f β²β²(x) = 0 when x = β b
3a. f changes its direction of
concavity at x = β b3aso β b
3ais an inflection point.
(b) If f(x) = ax3 + bx2 + cx + d has three x-intercepts, then it has three roots, say x1, x2 andx3, so we can write f(x) = a(x β x1)(x β x2)(x β x3) = ax3 + bx2 + cx + d, from which it
follows that b = βa(x1 + x2 + x3). Thus βb
3a=13(x1 + x2 + x3), which is the average.
(c) f(x) = x(x2 β 3x+ 2) = x(xβ 1)(xβ 2) so the intercepts are 0, 1, and 2 and the average is1. f β²β²(x) = 6xβ 6 = 6(xβ 1) changes sign at x = 1.
58. f β²β²(x) = 6x+ 2b, so the point of inflection is at x = β b3. Thus an increase in b moves the point of
inflection to the left.
59. (a) Let x1 < x2 belong to (a, b). If both belong to (a, c] or both belong to [c, b) then we havef(x1) < f(x2) by hypothesis. So assume x1 < c < x2. We know by hypothesis thatf(x1) < f(c), and f(c) < f(x2). We conclude that f(x1) < f(x2).
(b) Use the same argument as in Part (a), but with inequalities reversed.
60. By Theorem 5.1.2, f is increasing on any interval [(2nβ1)Ο, 2(n+1)Ο] (n = 0,Β±1,Β±2, Β· Β· Β·), becausef β²(x) = 1 + cosx > 0 on ((2n β 1)Ο, (2n + 1)Ο). By Exercise 59 (a) we can piece these intervalstogether to show that f(x) is increasing on (ββ,+β).
61. By Theorem 5.1.2, f is decreasing on any interval [(2nΟ+Ο/2, 2(n+1)Ο+Ο/2] (n = 0,Β±1,Β±2, Β· Β· Β·),because f β²(x) = β sinx+ 1 < 0 on (2nΟ + Ο/2, 2(n+ 1)Ο + Ο/2). By Exercise 59 (b) we can piecethese intervals together to show that f(x) is decreasing on (ββ,+β).
62. By zooming on the graph of yβ²(t), maximum increase is at x = β0.577 and maximum decrease isat x = 0.577.
63.
t
1
2
y
infl pt
64.
t
1
2
y
no infl pts
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Exercise Set 5.2 161
65. y
x
infl pts
1
2
3
466. y
x
infl pts
1
2
3
4
67. (a) yβ²(t) =LAkeβkt
(1 +Aeβkt)2S, so yβ²(0) =
LAk
(1 +A)2
(b) The rate of growth increases to its maximum, which occurs when y is halfway between 0 and
L, or when t =1klnA; it then decreases back towards zero.
(c) From (2) one sees thatdy
dtis maximized when y lies half way between 0 and L, i.e. y = L/2.
This follows since the right side of (2) is a parabola (with y as independent variable) with
y-intercepts y = 0, L. The value y = L/2 corresponds to t =1klnA, from (4).
68. Find t so that N β²(t) is maximum. The size of the population is increasing most rapidly whent = 8.4 years.
69. t = 7.67 1000
00 15
70. Since 0 < y < L the right-hand side of (3) of Example 9 can change sign only if the factor Lβ 2ychanges sign, which it does when y = L/2, at which point we have
L
2=
L
1 +Aeβkt, 1 = Aeβkt,
t =1klnA.
EXERCISE SET 5.2
1. (a)f (x)
x
y (b)
f (x)
x
y
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162 Chapter 5
(c)f (x)
x
y (d)
f (x)x
y
2. (a) y
x
(b) y
x
(c) y
x
(d) y
x
3. (a) f β²(x) = 6xβ 6 and f β²β²(x) = 6, with f β²(1) = 0. For the first derivative test, f β² < 0 for x < 1and f β² > 0 for x > 1. For the second derivative test, f β²β²(1) > 0.
(b) f β²(x) = 3x2 β 3 and f β²β²(x) = 6x. f β²(x) = 0 at x = Β±1. First derivative test: f β² > 0 forx < β1 and x > 1, and f β² < 0 for β1 < x < 1, so there is a relative maximum at x = β1,and a relative minimum at x = 1. Second derivative test: f β²β² < 0 at x = β1, a relativemaximum; and f β²β² > 0 at x = 1, a relative minimum.
4. (a) f β²(x) = 2 sinx cosx = sin 2x (so f β²(0) = 0) and f β²β²(x) = 2 cos 2x. First derivative test: if x isnear 0 then f β² < 0 for x < 0 and f β² > 0 for x > 0, so a relative minimum at x = 0. Secondderivative test: f β²β²(0) = 2 > 0, so relative minimum at x = 0.
(b) gβ²(x) = 2 tanx sec2 x (so gβ²(0) = 0) and gβ²β²(x) = 2 sec2 x(sec2 x + 2 tan2 x). First derivativetest: gβ² < 0 for x < 0 and gβ² > 0 for x > 0, so a relative minimum at x = 0. Second derivativetest: gβ²β²(0) = 2 > 0, relative minimum at x = 0.
(c) Both functions are squares, and so are positive for values of x near zero; both functions arezero at x = 0, so that must be a relative minimum.
5. (a) f β²(x) = 4(xβ 1)3, gβ²(x) = 3x2 β 6x+ 3 so f β²(1) = gβ²(1) = 0.(b) f β²β²(x) = 12(xβ 1)2, gβ²β²(x) = 6xβ 6, so f β²β²(1) = gβ²β²(1) = 0, which yields no information.(c) f β² < 0 for x < 1 and f β² > 0 for x > 1, so there is a relative minimum at x = 1;
gβ²(x) = 3(xβ 1)2 > 0 on both sides of x = 1, so there is no relative extremum at x = 1.
6. (a) f β²(x) = β5x4, gβ²(x) = 12x3 β 24x2 so f β²(0) = gβ²(0) = 0.(b) f β²β²(x) = β20x3, gβ²β²(x) = 36x2 β 48x, so f β²β²(0) = gβ²β²(0) = 0, which yields no information.(c) f β² < 0 on both sides of x = 0, so there is no relative extremum there; gβ²(x) = 12x2(xβ2) < 0
on both sides of x = 0 (for x near 0), so again there is no relative extremum there.
7. f β²(x) = 16x3 β 32x = 16x(x2 β 2), so x = 0,Β±β2 are stationary points.
8. f β²(x) = 12x3 + 12 = 12(x+ 1)(x2 β x+ 1), so x = β1 is the stationary point.
9. f β²(x) =βx2 β 2x+ 3(x2 + 3)2
, so x = β3, 1 are the stationary points.
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Exercise Set 5.2 163
10. f β²(x) = βx(x3 β 16)
(x3 + 8)2, so stationary points at x = 0, 24/3.
11. f β²(x) =2x
3(x2 β 25)2/3 ; so x = 0 is the stationary point.
12. f β²(x) =2x(4xβ 3)3(xβ 1)1/3 , so x = 0, 3/4 are the stationary points.
13. f(x) = | sinx| ={
sinx, sinx β₯ 0β sinx, sinx < 0 so f β²(x) =
{cosx, sinx > 0β cosx, sinx < 0 and f β²(x) does not exist
when x = nΟ, n = 0,Β±1,Β±2, Β· Β· Β· (the points where sinx = 0) becauselim
xβnΟβf β²(x) = lim
xβnΟ+f β²(x) (see Theorem preceding Exercise 61, Section 3.3). Now f β²(x) = 0 when
Β± cosx = 0 provided sinx = 0 so x = Ο/2 + nΟ, n = 0,Β±1,Β±2, Β· Β· Β· are stationary points.
14. When x > 0, f β²(x) = cosx, so x = (n+ 12 )Ο, n = 0, 1, 2, . . . are stationary points.
When x < 0, f β²(x) = β cosx, so x = (n+ 12 )Ο, n = β1,β2,β3, . . . are stationary points.
f is not differentiable at x = 0, so the latter is a critical point but not a stationary point.
15. (a) none(b) x = 1 because f β² changes sign from + to β there(c) none because f β²β² = 0 (never changes sign)
16. (a) x = 1 because f β²(x) changes sign from β to + there(b) x = 3 because f β²(x) changes sign from + to β there(c) x = 2 because f β²β²(x) changes sign there
17. (a) x = 2 because f β²(x) changes sign from β to + there.(b) x = 0 because f β²(x) changes sign from + to β there.(c) x = 1, 3 because f β²β²(x) changes sign at these points.
18. (a) x = 1 (b) x = 5 (c) x = β1, 0, 3
19. critical points x = 0, 51/3:f β²: 00
51/30
β ββ ββ ++ +β
x = 0: neitherx = 51/3: relative minimum
20. critical points x = β3/2, 0, 3/2: f β²: 00 0
0 3/2β3/2
+ β+ β+β ββ β + ++
x = β3/2: relative minimum;x = 0: relative maximum;x = 3/2: relative minimum
21. critical points x = 2/3: f β²: 0
2/3
+ β+ β+ β
x = 2/3: relative maximum;
22. critical points x = Β±β7: f β²: 00+ β+ β+ ++ +β
β 7 7x = ββ7: relative maximum;
x =β7: relative minimum
23. critical points x = 0: f β²: 0
0
β +β +β +
x = 0: relative minimum;
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164 Chapter 5
24. critical points x = 0, ln 3: f β²: 00
ln 30
β ββ ββ ++ +βx = 0: neither;x = ln 3: relative minimum
25. critical points x = β1, 1:f β²: 00
1β1
β +β +β ββ β+x = β1: relative minimum;x = 1: relative maximum
26. critical points x = ln 2, ln 3: f β²: 00
ln 3ln 2
+ β+ β+ ++ +βx = ln 2: relative maximum;x = ln 3: relative minimum
27. f β²(x) = 8β 6x: critical point x = 4/3f β²β²(4/3) = β6 : f has a maximum of 19/3 at x = 4/3
28. f β²(x) = 4x3 β 36x2: critical points at x = 0, 9f β²β²(0) = 0: Theorem 5.2.5 with m = 3: f has an inflection point at x = 0f β²β²(9) > 0: f has a minimum of β2187 at x = 9
29. f β²(x) = 2 cos 2x: critical points at x = Ο/4, 3Ο/4f β²β²(Ο/4) = β4: f has a maximum of 1 at x = Ο/4f β²β²(3Ο/4) = 4 : f has a minimum of 1 at x = 3Ο/4
30. f β²(x) = (xβ 2)ex: critical point at x = 2f β²β²(2) = e2 : f has a minimum of βe2 at x = 2
31. f β²(x) = 4x3 β 12x2 + 8x: critical points at x = 0, 1, 2
00 0
1 20
+ β+ β+β ββ β + ++relative minimum of 0 at x = 0relative maximum of 1 at x = 1relative minimum of 0 at x = 2
32. f β²(x) = 4x3 β 36x2 + 96xβ 64; critical points at x = 1, 4f β²β²(1) = 36: f has a relative minimum of β27 at x = 1f β²β²(4) = 0: Theorem 5.2.5 with m = 3: f has an inflection point at x = 4
33. f β²(x) = 5x4 + 8x3 + 3x2: critical points at x = β3/5,β1, 0f β²β²(β3/5) = 18/25 : f has a relative minimum of β108/3125 at x = β3/5f β²β²(β1) = β2 : f has a relative maximum of 0 at x = β1f β²β²(0) = 0: Theorem 5.2.5 with m = 3: f has an inflection point at x = 0
34. f β²(x) = 5x4 + 12x3 + 9x2 + 2x: critical points at x = β2/5,β1, 0f β²β²(β2/5) = β18/25 : f has a relative maximum of 108/3125 at x = β2/5f β²β²(β1) = 0: Theorem 5.2.5 with m = 3: f has an inflection point at x = β1f β²β²(0) = 2 : f has a relative maximum of 0 at x = 0
35. f β²(x) =2(x1/3 + 1)x1/3 : critical point at x = β1, 0
f β²β²(β1) = β23: f has a relative maximum of 1 at x = β1
f β² does not exist at x = 0. By inspection it is a relative minimum of 0.
36. f β²(x) =2x2/3 + 1x2/3 : no critical point except x = 0; since f is an odd function, x = 0 is an inflection
point for f .
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Exercise Set 5.2 165
37. f β²(x) = β 5(xβ 2)2 ; no extrema
38. f β²(x) = β2x(x4 β 16)
(x4 + 16)2; critical points x = β2, 0, 2
f β²β²(β2) = β18; f has a relative maximum of
18at x = β2
f β²β²(0) =18; f has a relative minimum of 0 at x = 0
f β²β²(2) = β18; f has a relative maximum of
18at x = 2
39. f β²(x) =2x
2 + x2 ; critical point at x = 0
f β²β²(0) = 1; f has a relative minimum of ln 2 at x = 0
40. f β²(x) =3x2
2 + x2 ; critical points at x = 0,β21/3 : f β² :
f β²β²(0) = 0 no relative extrema; f has no limit at x = β21/3
00
0
β +β +β ++ ++
β 23
41. f β²(x) = 2e2x β ex; critical point x = β ln 2f β²β²(β ln 2) = 1/2; relative minimum of β1/4 at x = β ln 2
42. f β²(x) = 2x(1 + x)e2x: critical point x = β1, 0f β²β²(β1) = β2/e2; relative maximum of 1/e2 at x = β1f β²β²(0) = 2: relative minimum of 0 at x = 0
43. f β²(x) = Β±(3β 2x) : critical points at x = 3/2, 0, 3f β²β²(3/2) = β2, relative maximum of 9/4 at x = 3/2x = 0, 3 are not stationary points:by first derivative test, x = 0 and x = 3 are each a minimum of 0
44. On each of the intervals (ββ,β1), (β1,+β) the derivative is of the form y = Β± 13x2/3 hence it is
clear that the only critical points are possibly β1 or 0. Near x = 0, x = 0, yβ² =1
3x2/3 > 0 so y
has an inflection point at x = 0. At x = β1, yβ² changes sign, thus the only extremum is a relativeminimum of 0 at x = β1.
45.
β2 1 3 5
β6
β4
2
x
y
( , β )32
254
46.
2 4 6 8
10
20
x
y
(4 β β17, 0)
(4 + β17, 0)
(4, 17)
(0, 1)
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166 Chapter 5
47.
β80
2 4
x
y(β2, 49)
(0, 5)
(3, β76)
( , β )12
272
( , 0)β7 β β574
( , 0)β7 + β574
48.
β1 0.5 1.5
β4
β2
4
6
x
y
13
5027
, ( )(0, 2)
(2, 0)
49.
β1 1
1
3
5y
x
( , )1 β β32
9 β 6β34
( , )1 + β32
9 + 6β34
(β , β β2 )1
β254
( , + β2 )1
β254
50.
β1.5 0.5 1.5
β4
β2
1
3
x
y
(ββ5, 0)
(ββ3, β4) (β3, β4)
(β5, 0)
(1, 0)
(0, 5)
(β1, 0)
51.
β1 1β1
0.5
1.5
x
y
(β , β )12
2716
52.
β0.4 0.6
β0.5
β0.3
β0.1
xy
29
16729
, ( )
13
127
, ( )
49
, 0( )
(0, 0)
53.
1
0.2
0.4
x
y
( , )β55
16β5125
( , )β155
4β5125
( , )ββ155
β4β5125
( , )ββ55
β16β5125
54.
β0.6 0.4 1.2
β0.4
0.4
x
y
β216β72401( , )ββ21
7
( , )β217
216β72401
(0, 0)
(1, 0)
(β1, 0)
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Exercise Set 5.2 167
55. (a) limxβββ
y = ββ, limxβ+β
y = +β;curve crosses x-axis at x = 0, 1,β1
y
x
β6
β4
β2
2
4
β2 β1 1
(b) limxβΒ±β
y = +β;curve never crosses x-axis
y
x
0.2
β1 1
(c) limxβββ
y = ββ, limxβ+β
y = +β;curve crosses x-axis at x = β1
y
x
β0.2
0.2
0.4
β1 1
(d) limxβΒ±β
y = +β;curve crosses x-axis at x = 0, 1
y
x
0.4
β1 1
56. (a) y
xa b
y
x
a b
y
xa b
(b) y
x
a b
y
x
a b
(c) y
x
a b
57. f β²(x) = 2 cos 2x if sin 2x > 0,f β²(x) = β2 cos 2x if sin 2x < 0,f β²(x) does not exist when x = Ο/2, Ο, 3Ο/2;critical numbers x = Ο/4, 3Ο/4, 5Ο/4, 7Ο/4, Ο/2, Ο, 3Ο/2relative minimum of 0 at x = Ο/2, Ο, 3Ο/2;relative maximum of 1 at x = Ο/4, 3Ο/4, 5Ο/4, 7Ο/4
1
00 o
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168 Chapter 5
58. f β²(x) =β3 + 2 cosx;
critical numbers x = 5Ο/6, 7Ο/6relative minimum of 7
β3Ο/6β 1 at x = 7Ο/6;
relative maximum of 5β3Ο/6 + 1 at x = 5Ο/6
12
00 o
59. f β²(x) = β sin 2x;critical numbers x = Ο/2, Ο, 3Ο/2relative minimum of 0 at x = Ο/2, 3Ο/2;relative maximum of 1 at x = Ο
1
00 o
60. f β²(x) = (2 cosxβ 1)/(2β cosx)2;critical numbers x = Ο/3, 5Ο/3relative maximum of
β3/3 at x = Ο/3,
relative minimum of ββ3/3 at x = 5Ο/3
0.8
0 o
β0.8
61. f β²(x) = lnx+ 1, f β²β²(x) = 1/x; f β²(1/e) = 0, f β²β²(1/e) > 0;relative minimum of β1/e at x = 1/e
2.5
β0.5
0 2.5
62. f β²(x) = β2 ex β eβx
(ex + eβx)2= 0 when x = 0.
By the first derivative test f β²(x) > 0 for x < 0and f β²(x) < 0 for x > 0; relative maximum of 1 at x = 0
1
0β2 2
63. f β²(x) = 2x(1β x)eβ2x = 0 at x = 0, 1.f β²β²(x) = (4x2 β 8x+ 2)eβ2x;f β²β²(0) > 0 and f β²β²(1) < 0, so a relative minimum of 0 at x = 0and a relative maximum of 1/e2 at x = 1.
0.14
0β0.3 4
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Exercise Set 5.2 169
64. f β²(x) = 10/xβ 1 = 0 at x = 10; f β²β²(x) = β10/x2 < 0;relative maximum of 10(ln(10)β 1) β 13.03 at x = 10
14
β4
0 20
65. Relative minima at x = β3.58, 3.33;relative maximum at x = 0.25
250
β250
β5 5
66. Relative minimum at x = β0.84;relative maximum at x = 0.84
1.2
β1.2
-6 6
67. Relative maximum at x = β0.27,relative minimum at x = 0.22
β4 2
2
4
6
8
x
f '(x)
f "(x)
y
68. Relative maximum at x = β1.11,relative minimum at x = 0.47relative maximum at x = 2.04
β5 β3 1 5
0.2
0.4
y
xf '(x)
f "(x)
69. f β²(x) =4x3 β sin 2x2βx4 + cos2 x
,
f β²β²(x) =6x2 β cos 2xβx4 + cos2 x
β (4x3 β sin 2x)(4x3 β sin 2x)4(x4 + cos2 x)3/2
Relative minima at x β Β±0.62, relative maximum at x = 0
2
β2
β2 1
x
y
f''(x)
f '(x)
70. Point of inflection at x = 0
β0.1 0.1
β0.2
0.1
0.2
x
y
f '(x)
f "(x)
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170 Chapter 5
71. (a) Let f(x) = x2 +k
x, then f β²(x) = 2x β k
x2 =2x3 β kx2 . f has a relative extremum when
2x3 β k = 0, so k = 2x3 = 2(3)3 = 54.
(b) Let f(x) =x
x2 + k, then f β²(x) =
k β x2
(x2 + k)2. f has a relative extremum when k β x2 = 0, so
k = x2 = 32 = 9.
72. (a) relative minima at x β Β±0.6436,relative maximum at x = 0
y
x1
1.2
1.4
1.6
1.8
2
β1.5 β0.5 0.5 1.5
(b) x β Β±0.6436, 0
73. (a) (β2.2, 4), (2, 1.2), (4.2, 3)(b) f β² exists everywhere, so the critical numbers are when f β² = 0, i.e. when x = Β±2 or r(x) = 0,
so x β β5.1,β2, 0.2, 2. At x = β5.1 f β² changes sign from β to +, so minimum; at x = β2f β² changes sign from + to β, so maximum; at x = 0.2 f β² doesnβt change sign, so neither; atx = 2 f β² changes sign from β to +, so minimum.Finally, f β²β²(1) = (12 β 4)rβ²(1) + 2r(1) β β3(0.6) + 2(0.3) = β1.2.
74. gβ²(x) exists everywhere, so the critical points are the points when gβ²(x) = 0, or r(x) = x, so r(x)crosses the line y = x. From the graph it appears that this happens precisely when x = 0.
75. f β²(x) = 3ax2 + 2bx+ c and f β²(x) has roots at x = 0, 1, so f β²(x) must be of the formf β²(x) = 3ax(xβ 1); thus c = 0 and 2b = β3a, b = β3a/2. f β²β²(x) = 6ax+ 2b = 6axβ 3a, sof β²β²(0) > 0 and f β²β²(1) < 0 provided a < 0. Finally f(0) = d, so d = 0; andf(1) = a+ b+ c+ d = a+ b = βa/2 so a = β2. Thus f(x) = β2x3 + 3x2.
76. (a) one relative maximum, located at x = n 0.3
00 14
(b) f β²(x) = cxnβ1(βx+ n)eβx = 0 at x = n.Since f β²(x) > 0 for x < n and f β²(x) < 0 forx > n itβs a maximum.
77. (a) f β²(x) = βxf(x). Since f(x) is alwayspositive, f β²(x) = 0 at x = 0,f β²(x) > 0 for x < 0 andf β²(x) < 0 for x > 0,so x = 0 is a maximum.
(b)
Β΅
y
x
2c1 Β΅,( )2c
1
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Exercise Set 5.3 171
78. (a) Because h and g have relative maxima at x0, h(x) β€ h(x0) for all x in I1 and g(x) β€ g(x0)for all x in I2, where I1 and I2 are open intervals containing x0. If x is in both I1 and I2then both inequalities are true and by addition so is h(x)+g(x) β€ h(x0)+g(x0) which showsthat h+ g has a relative maximum at x0.
(b) By counterexample; both h(x) = βx2 and g(x) = β2x2 have relative maxima at x = 0 buth(x) β g(x) = x2 has a relative minimum at x = 0 so in general h β g does not necessarilyhave a relative maximum at x0.
79. (a)
x
y
x0( )
f(x0) is not anextreme value.
(b)
x
y
x0( )
f(x0) is a relativemaximum.
(c)
x
y
x0( )
f(x0) is a relativeminimum.
EXERCISE SET 5.3
1. Vertical asymptote x = 4horizontal asymptote y = β2
6 8
β6
β4
2
x
x = 4
y = β2
y
2. Vertical asymptotes x = Β±2horizontal asymptote y = 0
β4 4
β8
β6
β4
2
4
6
8
x
x = 2x = β2
y
3. Vertical asymptotes x = Β±2horizontal asymptote y = 0
β44
β4
β2
2
4
x
x = 2
x = β2
y
4. Vertical asymptotes x = Β±2horizontal asymptote y = 1
β4 4
β6
β2
4
8
x
y
y = 1
x = 2x = β2
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172 Chapter 5
5. No vertical asymptoteshorizontal asymptote y = 1
β4
0.75
1.25
x
y
y = 1
(β , )2
β3
14 ( , )2
β3
14
6. No vertical asymptoteshorizontal asymptote y = 1
β4 β2 2 4
0.8
x
y
y = 1
(1, 0)(β1, 0)
(0, 1)
(( ) )β β18 β 3β333 β β33/3
13
β18 β 3β33 β 1 21
3( , )
( , )(( ) )β β18 + 3β33
3 + β33/313
β18 β 3β33 β 1 21
3
7. Vertical asymptote x = 1horizontal asymptote y = 1
2 4
β2
2
4
x
y
x = 1
y = 1
(β , β )1
β2133
8. Vertical asymptote x = 0,β3horizontal asymptote y = 2
β6 β4 β2 2
β6
β2
8
x
x = β3
y
y = 2(β2, )74
9. Vertical asymptote x = 0horizontal asymptote y = 3
β5 5
4
8
x
y = 3
y
(6, )259
(4, )114
(2, 3)
10. Vertical asymptote x = 1horizontal asymptote y = 3
3 6
6
12
18
x
x = 1
y
y = 3(β2, )1
3
(β1, 0)
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Exercise Set 5.3 173
11. Vertical asymptote x = 1horizontal asymptote y = 9
β10 10
20
30
x
y
y = 10
x = 1
( , 9)13
(β , 0)13
(β1, 1)
12. Vertical asymptote x = 1horizontal asymptote y = 3
β10 β6 β2 4 8 12
6
9
12
15
xx = 1
y
y = 3
(β , )53
61172048(β , )7
374732500
13. Vertical asymptote x = 1horizontal asymptote y = β1
2 4
β4
β2
x
x = 1
y = β1
y
(β1, β )12
14. Vertical asymptote x = 1horizontal asymptote y = 0
β4 β2 4 6
β4
β2
3
5
x
x = 1
y
(β21/3, )22/3
3
( , )(28 + 12β5)1/3
2
(28 + 12β5)2/3
18 + 6β5
15. (a) horizontal asymptote y = 3 asxβ Β±β, vertical asymptotes of x = Β±2
y
x
β5
5
10
β5 5
(b) horizontal asymptote of y = 1 asxβ Β±β, vertical asymptotes at x = Β±1
y
x
β10
10
β5 5
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174 Chapter 5
(c) horizontal asymptote of y = β1 asxβ Β±β, vertical asymptotes at x = β2, 1
y
x
β10
10
β5 5
(d) horizontal asymptote of y = 1 asxβ Β±β, vertical asymptote at x = β1, 2
y
x
β10
10
β5 5
16. (a) y
x
a b
(b) Symmetric about the line x =a+ b2
means f(a+ b2
+ x)= f
(a+ b2β x)for any x.
Note thata+ b2
+ xβ a = xβ aβ b2
anda+ b2
+ xβ b = x+ aβ b2, and the same equations
are true with x replaced by βx. Hence(a+ b2
+ xβ a)(
a+ b2
+ xβ b)=(xβ aβ b
2
)(x+
aβ b2
)= x2 β
(aβ b2
)2
The right hand side remains the same if we replace x with βx, and so the same is true ofthe left hand side, and the same is therefore true of the reciprocal of the left hand side. But
the reciprocal of the left hand side is equal to f(a+ b2
+ x). Since this quantity remains
unchanged if we replace x with βx, the condition of symmetry about the line x = a+ b2
hasbeen met.
17. limxβΒ±β
β£β£β£β£ x2
xβ 3 β (x+ 3)β£β£β£β£ = lim
xβΒ±β
β£β£β£β£ 9xβ 3
β£β£β£β£ = 0
10
β5
10
x
x = 3
y = x + 3
y
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Exercise Set 5.3 175
18.2 + 3xβ x3
xβ (3β x2) =
2xβ 0 as xβ Β±β
β5 β3 β1 1 3 5
β20
β10
5
15
25
x
y
y = 3 β x2
19. y = x2 β 1x=x3 β 1x
;
yβ² =2x3 + 1x2 ,
yβ² = 0 when
x = β 3
β12β β0.8;
yβ²β² =2(x3 β 1)x3
x
y
β(β0.8, 1.9)
(1, 0)
20. y =x2 β 2x
= xβ 2xso
y = x is an oblique asymptote;
yβ² =x2 + 2x2 ,
yβ²β² = β 4x3
x
yy = x
4
4
21. y =(xβ 2)3x2 = xβ 6 + 12xβ 8
x2 so
y = xβ 6 is an oblique asymptote;
yβ² =(xβ 2)2(x+ 4)
x3 ,
yβ²β² =24(xβ 2)x4
x
y
β10
(β4, β13.5)
y = x β 6
(2, 0)10
22. y = xβ 1xβ 1x2 = xβ
(1x+1x2
)so
y = x is an oblique asymptote;
yβ² = 1 +1x2 +
1x3 ,
yβ²β² = β2 1x3 β 6
1x4
β6 β4 β2 2 4 6
4
6
x
y
y = x
(β3, β )259
(β1, 1)
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176 Chapter 5
23. y =x3 β 4xβ 8x+ 2
= x2 β 2xβ 8x+ 2
so
y = x2 β 2x is a curvilinear asymptote as xβ Β±β
β4 4
10
30
x
y
y = x2 β 2x
(β3, 23)
(0, β4)x = β2
24. y =x5
x2 + 1= x3 β x+ x
x2 + 1so
y = x3 β x is a curvilinear asymptote
β2 β1 1 2
β10
β5
5
10
x
y
y = x3 β x
25. (a) VI (b) I (c) III (d) V (e) IV (f) II
26. (a) When n is even the function is defined only for x β₯ 0; as n increases the graph approachesthe line y = 1 for x > 0.
y
x
(b) When n is odd the graph is symmetric with respect to the origin; as n increases the graphapproaches the line y = 1 for x > 0 and the line y = β1 for x < 0.
y
x
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Exercise Set 5.3 177
27. y =β4x2 β 1
yβ² =4xβ4x2 β 1
yβ²β² = β 4(4x2 β 1)3/2 so
extrema when x = Β±12,
no inflection points
β1 1
1
2
3
4
x
y
28. y = 3βx2 β 4;
yβ² =2x
3(x2 β 4)2/3 ;
yβ²β² = β 2(3x2 + 4)9(x2 β 4)5/3
β2
(β2, 0) (2, 0)
(0, β2)
2
x
y3
29. y = 2x+ 3x2/3;
yβ² = 2 + 2xβ1/3;
yβ²β² = β23xβ4/3
x
y
4
5
(0, 0)
30. y = 2x2 β 3x4/3
yβ² = 4xβ 4x1/3
yβ²β² = 4β 43xβ2/3
β4 β2 2 4
2
4
6
8
x
y
(β1, β1) (1, β1)
31. y = x1/3(4β x);
yβ² =4(1β x)3x2/3 ;
yβ²β² = β4(x+ 2)9x5/3
β10
10
x
y
(1, 3)
(β2, β6 )23
32. y = 5x2/3 + x5/3
yβ² =5(x+ 2)3x1/3
yβ²β² =10(xβ 1)9x4/3
β6 β4 β2 2 4
β2
4
6
8
x
y
(1, 6)
(β2, 3 Γ 22/3)
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178 Chapter 5
33. y = x2/3 β 2x1/3 + 4
yβ² =2(x1/3 β 1)3x2/3
yβ²β² = β2(x1/3 β 2)9x5/3
10 30β10
2
6
x
y
(β8, 12)
(0, 4)
(1, 3) (8, 4)
34. y =8(βxβ 1)x
;
yβ² =4(2β
βx)
x2 ;
yβ²β² =2(3βxβ 8)x3
15
x
y
(4, 2)649
158
,
4
( )
35. y = x+ sinx;yβ² = 1 + cosx, yβ² = 0 when x = Ο + 2nΟ;yβ²β² = β sinx; yβ²β² = 0 when x = nΟn = 0,Β±1,Β±2, . . .
c
cx
y
36. y = xβ tanx;yβ² = 1β sec2 x; yβ² = 0 when x = 2nΟyβ²β² = β2 sec2 x tanx = 0 when x = nΟn = 0,Β±1,Β±2, . . .
β5 β3 3 5
β8
β4
2
6
10
x
x = β1.5 x = 1.5
x = β4.5 x = 4.5y
37. y =β3 cosx+ sinx;
yβ² = ββ3 sinx+ cosx;
yβ² = 0 when x = Ο/6 + nΟ;yβ²β² = β
β3 cosxβ sinx;
yβ²β² = 0 when x = 2Ο/3 + nΟ
oβ2
2
x
y
38. y = sinx+ cosx;yβ² = cosxβ sinx;yβ² = 0 when x = Ο/4 + nΟ;yβ²β² = β sinxβ cosx;yβ²β² = 0 when x = 3Ο/4 + nΟ
βo o
β2
2
x
y
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Exercise Set 5.3 179
39. y = sin2 xβ cosx;yβ² = sinx(2 cosx+ 1);yβ² = 0 when x = Β±Ο, 2Ο, 3Ο and when
x = β23Ο,23Ο,43Ο,83Ο;
yβ²β² = 4 cos2 x+ cosxβ 2; yβ²β² = 0 whenx β Β±2.57, Β±0.94, 3.71, 5.35, 7.22, 8.86
β4 6 10
β1
1.5
x
y
(5.35, 0.06)
(β2.57, 1.13)
(2.57, 1.13) (3.71, 1.13)
(8.86, 1.13)
(7.22, 0.06)(β0.94, 0.06) (0.94, 0.06)
(o, β1)
(Γ₯, 1)(C, 1) (c, 1)
(8, )54(*, )5
454(g, ) 5
4(w, )
40. y =βtanx;
yβ² =sec2 x2βtanx
; so yβ² > 0 always;
yβ²β² = sec2 x3 tan2 xβ 14(tanx)3/2
,
yβ²β² = 0 when x =Ο
6
2
4
6
8
10
x
x = 6
y
3
41. (a) limxβ+β
xex = +β, limxβββ
xex = 0
β1
β3β51
x
y
(β2, β0.27)(β1, β0.37)
(b) y = xex;yβ² = (x+ 1)ex;yβ²β² = (x+ 2)ex
y
x
β0.8
0.2
1 2
(1, ) 1e (2, )2
e2
42. limxβ+β
f(x) = 0, limxβββ
f(x) = ββ
f β²(x) = (1β x)eβx, f β²β²(x) = (xβ 2)eβxcritical point at x = 1;relative maximum at x = 1point of inflection at x = 2horizontal asymptote y = 0 as xβ +β
43. (a) limxβ+β
x2
e2x= 0, lim
xβββ
x2
e2x= +β
1 2 3
0.3
x
y
(0, 0)
(0.29, 0.05)
(1, 0.14)(1.71, 0.10)
(b) y = x2/e2x = x2eβ2x;
yβ² = 2x(1β x)eβ2x;
yβ²β² = 2(2x2 β 4x+ 1)eβ2x;
yβ²β² = 0 if 2x2 β 4x+ 1 = 0, when
x =4Β±β16β 84
= 1Β±β2/2 β 0.29, 1.71
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180 Chapter 5
44. (a) limxβ+β
x2e2x = +β, limxβββ
x2e2x = 0.
β3 β2 β1
y
x
0.2
0.3
(β1.71, 0.10)(β1, 0.14)
(0, 0)
(β0.29, 0.05)
(b) y = x2e2x;
yβ² = 2x(x+ 1)e2x;
yβ²β² = 2(2x2 + 4x+ 1)e2x;
yβ²β² = 0 if 2x2 + 4x+ 1 = 0, when
x =β4Β±
β16β 84
= β1Β±β2/2 β β0.29,β1.71
45. (a) limxβΒ±β
x2eβx2= 0
(b)
β1β3 1 3
0.1
x
y
1e(β1, ) 1
e(1, )
(β1.51, 0.23) (1.51, 0.23)
(β0.47, 0.18) (0.47, 0.18)
y = x2eβx2;
yβ² = 2x(1β x2)eβx2;
yβ² = 0 if x = 0,Β±1;yβ²β² = 2(1β 5x2 + 2x4)eβx
2
yβ²β² = 0 if 2x4 β 5x2 + 1 = 0,
x2 =5Β±β17
4,
x = Β± 12
β5 +β17 β Β±1.51,
x = Β± 12
β5ββ17 β Β±0.47
46. (a) limxβΒ±β
f(x) = 1 y
x
0.4
1
β10 β5 5 10
(0, 0)
(ββ2/3, eβ3/2) (β2/3, eβ3/2)
(b) f β²(x) = 2xβ3eβ1/x2so f β²(x) < 0 for x < 0 and f β²(x) > 0
for x > 0. Set u = x2 and use the given result to findlimxβ0
f β²(x) = 0, so (by the First Derivative Test) f(x)
has a minimum at x = 0. f β²β²(x) = (β6xβ4 + 4xβ6)eβ1/x2,
so f(x) has points of inflection at x = Β±β2/3.
47. (a) limxβββ
f(x) = 0, limxβ+β
f(x) = ββ
β1 2 3 4
β20
β10
10
xx = 1
y
(2, βe2)
(b) f β²(x) = βex(xβ 2)(xβ 1)2 so
f β²(x) = 0 when x = 2
f β²β²(x) = βex(x2 β 4x+ 5)(xβ 1)3 so
f β²β²(x) = 0 alwaysrelative maximum when x = 2, no point of inflectionasymptote x = 1
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Exercise Set 5.3 181
48. (a) limxβββ
f(x) = 0, limxβ+β
f(x) = +β
β2 β1 1
1
2
3
x
y
(β2/3, β2/32/3e2/3)
(β1.48, 0.30) (0.15, 0.33)
(b) f β²(x) =ex(3x+ 2)3x1/3 so
f β²(x) = 0 when x = β23
f β²β²(x) =ex(9x2 + 12xβ 2)
9x4/3 so
points of inflection when f β²β²(x) = 0 at
x = β2ββ6
3,β2 +
β6
3,
relative maximum at
(β23, eβ2/3
(β23
)2/3)
absolute minimum at (0, 0)
y
x
0.6
1
1.8
1 2 3 4
(2, )
(3.41, 1.04)
(0.59, 0.52)
(0, 0)
4e
49. limxβ+β
f(x) = 0,
limxβββ
f(x) = +β
f β²(x) = x(2β x)e1βx, f β²β²(x) = (x2 β 4x+ 2)e1βxcritical points at x = 0, 2;relative minimum at x = 0,relative maximum at x = 2points of inflection at x = 2Β±
β2
horizontal asymptote y = 0 as xβ +β
50. limxβ+β
f(x) = +β, limxβββ
f(x) = 0
f β²(x) = x2(3 + x)exβ1, f β²β²(x) = x(x2 + 6x+ 6)exβ1
critical points at x = β3, 0;relative minimum at x = β3points of inflection at x = 0,β3Β±
β3 β 0,β4.7,β1.27
horizontal asymptote y = 0 as xβ ββ
y
x
β0.4
0.4
0.8
β4 β2
(β4.7, β0.35)
(0, 0)
(β1.27, β0.21)
(β3, β0.49)
1
1
x
y
(eβ1, βeβ1)
51. (a) limxβ0+
y = limxβ0+
x lnx = limxβ0+
lnx1/x
= limxβ0+
1/xβ1/x2 = 0;
limxβ+β
y = +β
(b) y = x lnx,yβ² = 1 + lnx,yβ²β² = 1/x,yβ² = 0 when x = eβ1
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182 Chapter 5
52. (a) limxβ0+
y = limxβ0+
lnx1/x2 = lim
xβ0+
1/xβ2/x3 = 0,
limxβ+β
y = +β
1
β0.2β0.1
0.10.2
x
y
(eβ1/2, β eβ1) 12
(eβ3/2, β eβ3) 32
(b) y = x2 lnx, yβ² = x(1 + 2 lnx),yβ²β² = 3 + 2 lnx,yβ² = 0 if x = eβ1/2,yβ²β² = 0 if x = eβ3/2,limxβ0+
yβ² = 0
53. (a) limxβ0+
x2 lnx = 0 by the rule given, limxβ+β
x2 lnx = +β by inspection, and f(x) not defined
for x < 0y
x
38e3
12e3/2( ), β
( )18e
12
, β12 βe
(b) y = x2 ln 2x, yβ² = 2x ln 2x+ xyβ²β² = 2 ln 2x+ 3yβ² = 0 if x = 1/(2
βe),
yβ²β² = 0 if x = 1/(2e3/2)
y
x
1
2
β2 2(0, 0)
(1, ln 2)(β1, ln 2)
54. (a) limxβ+β
f(x) = +β; limxβ0
f(x) = 0
(b) y = ln(x2 + 1), yβ² = 2x/(x2 + 1)
yβ²β² = β2 x2 β 1
(x2 + 1)2
yβ² = 0 if x = 0yβ²β² = 0 if x = Β±1
4 5β1
1
3
5
x
(e3/2, )3e2
(eβ3/2, β )32e
y55. (a) limxβ+β
f(x) = +β, limxβ0+
f(x) = 0
(b) y = x2/3 lnx
yβ² =2 lnx+ 33x1/3
yβ² = 0 when lnx = β32, x = eβ3/2
yβ²β² =β3 + 2 lnx9x4/3 ,
yβ²β² = 0 when lnx =32, x = e3/2
8 16 24 32 40
0.5
1
x
y (e15/4 , )15
4e5/4
(e3, )3e
56. (a) limxβ0+
f(x) = ββ, limxβ+β
f(x) = 0
(b) y = xβ1/3 lnx
yβ² =3β lnx3x4/3
yβ² = 0 when x = e3;
yβ²β² =4 lnxβ 159x7/3
yβ²β² = 0 when x = e15/4
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Exercise Set 5.3 183
57. (a) 0.4
β0.2
β0.5 3
(b) yβ² = (1β bx)eβbx, yβ²β² = b2(xβ 2/b)eβbx;relative maximum at x = 1/b, y = 1/be;point of inflection at x = 2/b, y = 2/be2.Increasing b moves the relative maximumand the point of inflection to the left anddown, i.e. towards the origin.
58. (a) 1
0β2 2
(b) yβ² = β2bxeβbx2,
yβ²β² = 2b(β1 + 2bx2)eβbx2;
relative maximum at x = 0, y = 1; pointsof inflection at x = Β±
β1/2b, y = 1/
βe.
Increasing bmoves the points of inflectiontowards the y-axis; the relative maximumdoesnβt move.
59. (a) The oscillations of ex cosx about zero in-crease as x β +β so the limit does notexist, and lim
xβββex cosx = 0.
(b) y
x
β2
46
β2 β1 1 2
(0,1)
(1.52, 0.22)
(c) The curve y = eax cos bx oscillates between y = eax and y = βeax. The frequency ofoscillation increases when b increases.
y
x
β5
5
β1 2
a = 1
b = 1
b = 2
b = 3y
x
5
10
β1 0.5 1
b = 1
a = 1a = 2
a = 3
60. (a) 10
00 6
(b) yβ² =n2 β 2x2
nxnβ1eβx
2/n,
yβ²β² =n4 β n3 β 4x2n2 β 2x2n+ 4x4
n2 xnβ2eβx2/n,
relative maximum when x =nβ2, y =
3e;
point of inflection when
x =
β(2n+ 1Β±
β8n+ 1
4
)n.
As n increases the maxima move up andto the right; the points of inflection moveto the right.
61. (a) x = 1, 2.5, 4 and x = 3, the latter being a cusp(b) (ββ, 1], [2.5, 3)(c) relative maxima for x = 1, 3; relative minima for x = 2.5(d) x = 0.8, 1.9, 4
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184 Chapter 5
62. (a) f β²(x) = β2h(x) + (1β 2x)hβ²(x), f β²(5) = β2h(5)β 9hβ²(5). But from the graphhβ²(5) β β0.2 and f β²(5) = 0, so h(5) = β(9/2)hβ²(5) β 0.9
(b) f β²β²(x) = β4hβ²(x) + (1 β 2x)hβ²β²(x), f β²β²(5) β 0.8 β 9hβ²β²(5) and since hβ²β²(5) is clearly negative,f β²β²(5) > 0 and thus f has a minimum at x = 5.
63. Let y be the length of the other side of the rectangle,then L = 2x+2y and xy = 400 so y = 400/x and henceL = 2x+ 800/x. L = 2x is an oblique asymptote
L = 2x+800x=2(x2 + 400)
x,
Lβ² = 2β 800x2 =
2(x2 β 400)x2 ,
Lβ²β² =1600x3 ,
Lβ² = 0 when x = 20, L = 80
20
100
x
L
64. Let y be the height of the box, then S = x2+4xy andx2y = 500 so y = 500/x2 and hence S = x2 + 2000/x.The graph approaches the curve S = x2 asymptotically
S = x2 +2000x
=x3 + 2000
x,
Sβ² = 2xβ 2000x2 =
2(x3 β 1000)x2 ,
Sβ²β² = 2 +4000x3 =
2(x3 + 2000)x3 ,
Sβ²β² = 0 when x = 10, S = 300
30
1000
x
S
65. yβ² = 0.1x4(6xβ 5);critical numbers: x = 0, x = 5/6;relative minimum at x = 5/6,y β β6.7Γ 10β3
β1 1
0.01x
y
66. yβ² = 0.1x4(x+ 1)(7x+ 5);critical numbers: x = 0, x = β1, x = β5/7,relative maximum at x = β1, y = 0;relative minimum at x = β5/7, y β β1.5Γ 10β3
1
0.001 x
y
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Exercise Set 5.4 185
EXERCISE SET 5.4
1. relative maxima at x = 2, 6; absolute maximum at x = 6; relative and absolute minima at x = 0, 4
2. relative maximum at x = 3; absolute maximum at x = 7; relative minima at x = 1, 5; absoluteminima at x = 1, 5
3. (a) y
x
10
(b) y
x
2 7
(c) y
x
53 7
4. (a) y
x
(b) y
x
(c) y
x
β5 5
5. x = 1 is a point of discontinuity of f .
6. Since f is monotonically increasing on (0, 1), one might expect a minimum at x = 0 and a maximumat x = 1. But both points are discontinuities of f .
7. f β²(x) = 8x β 12, f β²(x) = 0 when x = 3/2; f(1) = 2, f(3/2) = 1, f(2) = 2 so the maximum valueis 2 at x = 1, 2 and the minimum value is 1 at x = 3/2.
8. f β²(x) = 8β 2x, f β²(x) = 0 when x = 4; f(0) = 0, f(4) = 16, f(6) = 12 so the maximum value is 16at x = 4 and the minimum value is 0 at x = 0.
9. f β²(x) = 3(xβ 2)2, f β²(x) = 0 when x = 2; f(1) = β1, f(2) = 0, f(4) = 8 so the minimum is β1 atx = 1 and the maximum is 8 at x = 4.
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186 Chapter 5
10. f β²(x) = 6x2 + 6xβ 12, f β²(x) = 0 when x = β2, 1; f(β3) = 9, f(β2) = 20, f(1) = β7, f(2) = 4, sothe minimum is β7 at x = 1 and the maximum is 20 at x = β2.
11. f β²(x) = 3/(4x2+1)3/2, no critical points; f(β1) = β3/β5, f(1) = 3/
β5 so the maximum value is
3/β5 at x = 1 and the minimum value is β3/
β5 at x = β1.
12. f β²(x) =2(2x+ 1)3(x2 + x)1/3
, f β²(x) = 0 when x = β1/2 and f β²(x) does not exist when x = β1, 0;
f(β2) = 22/3, f(β1) = 0, f(β1/2) = 4β2/3, f(0) = 0, f(3) = 122/3 so the maximum value is 122/3
at x = 3 and the minimum value is 0 at x = β1, 0.
13. f β²(x) = 1 β 2 cosx, f β²(x) = 0 when x = Ο/3; then f(βΟ/4) = βΟ/4 +β2; f(Ο/3) = Ο/3 ββ
3; f(Ο/2) = Ο/2β2, so f has a minimum of Ο/3ββ3 at x = Ο/3 and a maximum of βΟ/4+
β2
at x = βΟ/4.
14. f β²(x) = cosx+sinx, f β²(x) = 0 for x in (0, Ο) when x = 3Ο/4; f(0) = β1, f(3Ο/4) =β2, f(Ο) = 1
so the maximum value isβ2 at x = 3Ο/4 and the minimum value is β1 at x = 0.
15. f(x) = 1 + |9 β x2| ={
10β x2, |x| β€ 3β8 + x2, |x| > 3 , f
β²(x) ={β2x, |x| < 32x, |x| > 3 thus f β²(x) = 0 when
x = 0, f β²(x) does not exist for x in (β5, 1) when x = β3 because limxββ3β
f β²(x) = limxββ3+
f β²(x) (see
Theorem preceding Exercise 61, Section 3.3); f(β5) = 17, f(β3) = 1, f(0) = 10, f(1) = 9 so themaximum value is 17 at x = β5 and the minimum value is 1 at x = β3.
16. f(x) = |6 β 4x| ={
6β 4x, x β€ 3/2β6 + 4x, x > 3/2 , f
β²(x) ={β4, x < 3/24, x > 3/2 , f
β²(x) does not exist when
x = 3/2 thus 3/2 is the only critical point in (β3, 3); f(β3) = 18, f(3/2) = 0, f(3) = 6 so themaximum value is 18 at x = β3 and the minimum value is 0 at x = 3/2.
17. f β²(x) = 2x β 1, f β²(x) = 0 when x = 1/2; f(1/2) = β9/4 and limxβΒ±β
f(x) = +β. Thus f has aminimum of β9/4 at x = 1/2 and no maximum.
18. f β²(x) = β4(x+ 1); critical point x = β1. Maximum value f(β1) = 5, no minimum.
19. f β²(x) = 12x2(1 β x); critical points x = 0, 1. Maximum value f(1) = 1, no minimum becauselim
xβ+βf(x) = ββ.
20. f β²(x) = 4(x3 + 1); critical point x = β1. Minimum value f(β1) = β3, no maximum.
21. No maximum or minimum because limxβ+β
f(x) = +β and limxβββ
f(x) = ββ.
22. No maximum or minimum because limxβ+β
f(x) = +β and limxβββ
f(x) = ββ.
23. limxββ1β
f(x) = ββ, so there is no absolute minimum on the interval;
f β²(x) = 0 at x β β2.414213562, for which y β β4.828427125. Also f(β5) = β13/2, so theabsolute maximum of f on the interval is y β β4.828427125 taken at x β β2.414213562.
24. limxββ1+
f(x) = ββ, so there is no absolute minimum on the interval.
f β²(x) = 3/(x+ 1)2 > 0, so f is increasing on the interval (β1, 5] and the maximum must occur atthe endpoint x = 5 where f(5) = 1/2.
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Exercise Set 5.4 187
25. limxβΒ±β
= +β so there is no absolute maximum.
f β²(x) = 4x(x β 2)(x β 1), f β²(x) = 0 when x = 0, 1, 2,and f(0) = 0, f(1) = 1, f(2) = 0 so f has an absoluteminimum of 0 at x = 0, 2.
8
0β2 4
26. (xβ1)2(x+2)2 can never be less than zero because it isthe product of two squares; the minimum value is 0 forx = 1 or β2, no maximum because lim
xβ+βf(x) = +β.
15
0β3 2
27. f β²(x) =5(8β x)3x1/3 , f β²(x) = 0 when x = 8 and f β²(x)
does not exist when x = 0; f(β1) = 21, f(0) = 0,f(8) = 48, f(20) = 0 so the maximum value is 48 atx = 8 and the minimum value is 0 at x = 0, 20.
50
0β1 20
28. f β²(x) = (2 β x2)/(x2 + 2)2, f β²(x) = 0 for x in theinterval (β1, 4) when x =
β2; f(β1) = β1/3,
f(β2) =
β2/4, f(4) = 2/9 so the maximum value isβ
2/4 at x =β2 and the minimum value is β1/3 at
x = β1.
0.4
β0.4
β1 4
29. f β²(x) = β1/x2; no maximum or minimum becausethere are no critical points in (0,+β).
25
00 10
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188 Chapter 5
30. f β²(x) = β x(xβ 2)(x2 β 2x+ 2)2 , and for 1 β€ x < +β, f
β²(x) = 0
when x = 2. Also limxβ+β
f(x) = 2 and f(2) = 5/2 and
f(1) = 2, hence f has an absolute minimum value of 2 atx = 1 and an absolute maximum value of 5/2 at x = 2.
3
01 8
31. f β²(x) =1β 2 cosxsin2 x
; f β²(x) = 0 on [Ο/4, 3Ο/4] only when
x = Ο/3. Then f(Ο/4) = 2β2 β 1, f(Ο/3) =
β3 and
f(3Ο/4) = 2β2 + 1, so f has an absolute maximum value
of 2β2 + 1 at x = 3Ο/4 and an absolute minimum value ofβ
3 at x = Ο/3.
3
03 9
32. f β²(x) = 2 sinx cosxβsinx = sinx(2 cosxβ1), f β²(x) = 0 forx in (βΟ, Ο) when x = 0, Β±Ο/3; f(βΟ) = β1, f(βΟ/3) =5/4, f(0) = 1, f(Ο/3) = 5/4, f(Ο) = β1 so the maximumvalue is 5/4 at x = Β±Ο/3 and the minimum value is β1 atx = Β±Ο.
1.5
β1.5
C c
0.2
01 4
33. f β²(x) = x2(3 β 2x)eβ2x, f β²(x) = 0 for x in [1, 4] when
x = 3/2; if x = 1, 3/2, 4, then f(x) = eβ2,278eβ3, 64eβ8;
critical point at x = 3/2; absolute maximum of278eβ3 at
x = 3/2, absolute minimum of 64eβ8 at x = 4
0.76
0.641 2.7
34. f β²(x) = (1β ln 2x)/x2, f β²(x) = 0 on [1, e] for x = e/2;
if x = 1, e/2, e then f(x) = ln 2, 2/e, (ln 2 + 1)/e;
absolute minimum of1 + ln 2e
at x = e,
absolute maximum of 2/e at x = e/2
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Exercise Set 5.4 189
3.0
β2.5
0 4
35. f β²(x) = β3x2 β 10x+ 3x2 + 1
, f β²(x) = 0 when x =13, 3.
Then f(0) = 0, f(13
)= 5 ln
(109
)β 1,
f(3) = 5 ln 10 β 9, f(4) = 5 ln 17 β 12 and thus f has anabsolute minimum of 5(ln 10β ln 9)β 1 at x = 1/3 and anabsolute maximum of 5 ln 10β 9 at x = 3.
20
β20
β2 2
36. f β²(x) = (x2 + 2x β 1)ex, f β²(x) = 0 at x = β2 +β2 and
x = β1ββ2 (discard), f(β1+
β2) = (2β2
β2)e(β1+
β2) β
β1.25, absolute maximum at x = 2, f(2) = 3e2 β 22.17,absolute minimum at x = β1 +
β2
37. f β²(x) = β[cos(cosx)] sinx; f β²(x) = 0 if sinx = 0 or ifcos(cosx) = 0. If sinx = 0, then x = Ο is the criti-cal point in (0, 2Ο); cos(cosx) = 0 has no solutions be-cause β1 β€ cosx β€ 1. Thus f(0) = sin(1), f(Ο) =sin(β1) = β sin(1), and f(2Ο) = sin(1) so the maxi-mum value is sin(1) β 0.84147 and the minimum valueis β sin(1) β β0.84147.
1
β1
0 o
38. f β²(x) = β[sin(sinx)] cosx; f β²(x) = 0 if cosx = 0 or ifsin(sinx) = 0. If cosx = 0, then x = Ο/2 is the criticalpoint in (0, Ο); sin(sinx) = 0 if sinx = 0, which gives nocritical points in (0, Ο). Thus f(0) = 1, f(Ο/2) = cos(1),and f(Ο) = 1 so the maximum value is 1 and the minimumvalue is cos(1) β 0.54030.
1.5
00 c
39. f β²(x) ={
4, x < 12xβ 5, x > 1 so f β²(x) = 0 when x = 5/2, and f β²(x) does not exist when x = 1
because limxβ1β
f β²(x) = limxβ1+
f β²(x) (see Theorem preceding Exercise 61, Section 3.3); f(1/2) = 0,
f(1) = 2, f(5/2) = β1/4, f(7/2) = 3/4 so the maximum value is 2 and the minimum value isβ1/4.
40. f β²(x) = 2x + p which exists throughout the interval (0, 2) for all values of p so f β²(1) = 0 becausef(1) is an extreme value, thus 2 + p = 0, p = β2. f(1) = 3 so 12 + (β2)(1) + q = 3, q = 4 thusf(x) = x2 β 2x+ 4 and f(0) = 4, f(2) = 4 so f(1) is the minimum value.
41. The period of f(x) is 2Ο, so check f(0) = 3, f(2Ο) = 3 and the critical points.f β²(x) = β2 sinxβ 2 sin 2x = β2 sinx(1 + 2 cosx) = 0 on [0, 2Ο] at x = 0, Ο, 2Ο andx = 2Ο/3, 4Ο/3. Check f(Ο) = β1, f(2Ο/3) = β3/2, f(4Ο/3) = β3/2.Thus f has an absolute maximum on (ββ,+β) of 3 at x = 2kΟ, k = 0,Β±1,Β±2, . . .and an absolute minimum of β3/2 at x = 2kΟ Β± 2Ο/3, k = 0,Β±1,Β±2, . . ..
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190 Chapter 5
42. cosx
3has a period of 6Ο, and cos
x
2a period of 4Ο, so f(x) has a period of 12Ο. Consider the
interval [0, 12Ο]. f β²(x) = β sin x3β sin x
2, f β²(x) = 0 when sin
x
3+ sin
x
2= 0 thus, by use of
the trig identity sin a + sin b = 2 sina+ b2
cosaβ b2, 2 sin
(5x12
)cos(β x12
)= 0 so sin
5x12
= 0 or
cosx
12= 0. Solve sin
5x12
= 0 to get x = 12Ο/5, 24Ο/5, 36Ο/5, 48Ο/5 and then solve cosx
12= 0
to get x = 6Ο. The corresponding values of f(x) are β4.0450, 1.5450, 1.5450,β4.0450, 1, 5, 5 so themaximum value is 5 and the minimum value is β4.0450 (approximately).
43. Let f(x) = xβ sinx, then f β²(x) = 1β cosx and so f β²(x) = 0 when cosx = 1 which has no solutionfor 0 < x < 2Ο thus the minimum value of f must occur at 0 or 2Ο. f(0) = 0, f(2Ο) = 2Ο so 0 isthe minimum value on [0, 2Ο] thus xβ sinx β₯ 0, sinx β€ x for all x in [0, 2Ο].
44. Let h(x) = cosx β 1 + x2/2. Then h(0) = 0, and it is sufficient to show that hβ²(x) β₯ 0 for0 < x < 2Ο. But hβ²(x) = β sinx+ x β₯ 0 by Exercise 43.
45. Let m = slope at x, then m = f β²(x) = 3x2β 6x+5, dm/dx = 6xβ 6; critical point for m is x = 1,minimum value of m is f β²(1) = 2
46. (a) limxβ0+
f(x) = +β, limxβ(Ο/2)β
f(x) = +β, so f has no maximum value on the interval. By
table 5.4.3 f must have a minimum value.(b) According to table 5.4.3, there is an absolute minimum value of f on (0, Ο/2). To find the
absolute minimum value, we examine the critical points (Theorem 5.4.3).f β²(x) = secx tanx β cscx cotx = 0 at x = Ο/4, where f(Ο/4) = 2
β2, which must be the
absolute minimum value of f on the interval (0, Ο/2).
47. limxβ+β
f(x) = +β, limxβ8β
f(x) = +β, so there is no absolute maximum value of f for x > 8. By
table 5.4.3 there must be a minimum. Since f β²(x) =2x(β520 + 192xβ 24x2 + x3)
(xβ 8)3 , we must solve
a quartic equation to find the critical points. But it is easy to see that x = 0 and x = 10 are realroots, and the other two are complex. Since x = 0 is not in the interval in question, we must havean absolute minimum of f on (8,+β) of 125 at x = 10.
48. (a)dC
dt=
K
aβ b(aeβat β beβbt
)sodC
dt= 0 at t =
ln(a/b)aβ b . This is the only stationary point and
C(0) = 0, limtβ+β
C(t) = 0, C(t) > 0 for 0 < t < +β, so it is an absolute maximum.
(b) 0.7
00 10
49. The slope of the line is β1, and the slope of the tangent to y = βx2 is β2x so β2x = β1, x = 1/2.The line lies above the curve so the vertical distance is given by F (x) = 2 β x + x2; F (β1) = 4,F (1/2) = 7/4, F (3/2) = 11/4. The point (1/2,β1/4) is closest, the point (β1,β1) farthest.
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Exercise Set 5.5 191
50. The slope of the line is 4/3; and the slope of the tangent to y = x3 is 3x2 so 3x2 = 4/3, x2 = 4/9,x = Β±2/3. The line lies below the curve so the vertical distance is given by F (x) = x3 β 4x/3+ 1;F (β1) = 4/3, F (β2/3) = 43/27, F (2/3) = 11/27, F (1) = 2/3. The closest point is (2/3, 8/27),the farthest is (β2/3,β8/27).
51. The absolute extrema of y(t) can occur at the endpoints t = 0, 12 or when dy/dt = 2 sin t = 0, i.e.t = 0, 12, kΟ, k = 1, 2, 3; the absolute maximum is y = 4 at t = Ο, 3Ο; the absolute minimum isy = 0 at t = 0, 2Ο.
52. (a) The absolute extrema of y(t) can occur at the endpoints t = 0, 2Ο or whendy/dt = 2 cos 2tβ 4 sin t cos t = 2 cos 2tβ 2 sin 2t = 0, t = 0, 2Ο, Ο/8, 5Ο/8, 9Ο/8, 13Ο/8;the absolute maximum is y = 3.4142 at t = Ο/8, 9Ο/8; the absolute minimum is y = 0.5859at t = 5Ο/8, 13Ο/8.
(b) The absolute extrema of x(t) occur at the endpoints t = 0, 2Ο or when
dx
dt= β 2 sin t+ 1
(2 + sin t)2= 0, t = 7Ο/6, 11Ο/6. The absolute maximum is x = 0.5774 at t = 11Ο/6
and the absolute minimum is x = β0.5774 at t = 7Ο/6.
53. f β²(x) = 2ax+ b; critical point is x = β b2a
f β²β²(x) = 2a > 0 so f(β b2a
)is the minimum value of f , but
f
(β b2a
)= a
(β b2a
)2+ b
(β b2a
)+ c =
βb2 + 4ac4a
thus f(x) β₯ 0 if and only if
f
(β b2a
)β₯ 0, βb
2 + 4ac4a
β₯ 0, βb2 + 4ac β₯ 0, b2 β 4ac β€ 0
54. Use the proof given in the text, replacing βmaximumβ by βminimumβ and βlargestβ by βsmallestβand reversing the order of all inequality symbols.
EXERCISE SET 5.5
1. If y = x+ 1/x for 1/2 β€ x β€ 3/2 then dy/dx = 1β 1/x2 = (x2 β 1)/x2, dy/dx = 0 when x = 1. Ifx = 1/2, 1, 3/2 then y = 5/2, 2, 13/6 so(a) y is as small as possible when x = 1.(b) y is as large as possible when x = 1/2.
2. Let x and y be nonnegative numbers and z the sum of their squares, then z = x2 + y2. Butx+ y = 1, y = 1βx so z = x2+(1βx)2 = 2x2β 2x+1 for 0 β€ x β€ 1. dz/dx = 4xβ 2, dz/dx = 0when x = 1/2. If x = 0, 1/2, 1 then z = 1, 1/2, 1 so
(a) z is as large as possible when one number is 0 and the other is 1.(b) z is as small as possible when both numbers are 1/2.
3. A = xy where x + 2y = 1000 so y = 500 β x/2 andA = 500x β x2/2 for x in [0, 1000]; dA/dx = 500 β x,dA/dx = 0 when x = 500. If x = 0 or 1000 then A = 0,if x = 500 then A = 125, 000 so the area is maximumwhen x = 500 ft and y = 500β 500/2 = 250 ft.
x
y
Stream
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192 Chapter 5
4. The triangle in the figure is determined by the two sides oflength a and b and the angle ΞΈ. Suppose that a + b = 1000.Then h = a sin ΞΈ and area = Q = hb/2 = ab(sin ΞΈ)/2. Thisexpression is maximized by choosing ΞΈ = Ο/2 and thus Q =ab/2 = a(1000 β a)/2, which is maximized by a = 500. Thusthe maximal area is obtained by a right isosceles triangle, wherethe angle between two sides of 500 is the right angle.
A B
C
ha
b
c
ΞΈ
5. Let x and y be the dimensions shown in the figure and A thearea, then A = xy subject to the cost condition 3(2x)+2(2y) =6000, or y = 1500β 3x/2. Thus A = x(1500β 3x/2) = 1500xβ3x2/2 for x in [0, 1000]. dA/dx = 1500 β 3x, dA/dx = 0 whenx = 500. If x = 0 or 1000 then A = 0, if x = 500 thenA = 375, 000 so the area is greatest when x = 500 ft and (fromy = 1500β 3x/2) when y = 750 ft.
Heavy-duty
Standard
x
y
6. Let x and y be the dimensions shown in the figure and A thearea of the rectangle, then A = xy and, by similar triangles,x/6 = (8β y)/8, y = 8β 4x/3 so A = x(8β 4x/3) = 8xβ 4x2/3for x in [0, 6]. dA/dx = 8 β 8x/3, dA/dx = 0 when x = 3. Ifx = 0, 3, 6 then A = 0, 12, 0 so the area is greatest when x = 3in and (from y = 8β 4x/3) y = 4 in.
8
y
x
6
10
7. Let x, y, and z be as shown in the figure and A the area of therectangle, then A = xy and, by similar triangles, z/10 = y/6,z = 5y/3; also x/10 = (8 β z)/8 = (8 β 5y/3)/8 thus y =24/5β 12x/25 so A = x(24/5β 12x/25) = 24x/5β 12x2/25 forx in [0, 10]. dA/dx = 24/5β 24x/25, dA/dx = 0 when x = 5. Ifx = 0, 5, 10 then A = 0, 12, 0 so the area is greatest when x = 5in. and y = 12/5 in.
8
z
x
y
6
10
8. A = (2x)y = 2xy where y = 16 β x2 so A = 32x β 2x3 for0 β€ x β€ 4; dA/dx = 32 β 6x2, dA/dx = 0 when x = 4/
β3. If
x = 0, 4/β3, 4 then A = 0, 256/(3
β3), 0 so the area is largest
when x = 4/β3 and y = 32/3. The dimensions of the rectangle
with largest area are 8/β3 by 32/3.
β4 4
16
x
y
y
x
9. A = xy where x2 + y2 = 202 = 400 so y =β400β x2 and
A = xβ400β x2 for 0 β€ x β€ 20;
dA/dx = 2(200β x2)/β400β x2, dA/dx = 0 when
x =β200 = 10
β2. If x = 0, 10
β2, 20 then A = 0, 200, 0 so the
area is maximum when x = 10β2 and y =
β400β 200 = 10
β2.
x
y10
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January 27, 2005 11:44 L24-CH05 Sheet number 41 Page number 193 black
Exercise Set 5.5 193
10. Let R denote the rectangle with corners (Β±8,Β±10). Sup-pose the lower left corner of the square S is at the point(x0, y0) = (x0,β4x0). Let Q denote the desired region.
It is clear that Q will be a rectangle, and that its left andbottom sides are subsets of the left and bottom sides of S. Theright and top edges of S will be subsets of the sides of R (if Sis big enough).
From the drawing it is evident that the area of Q is (8βx0)(10+4x0) = β4x2
0 + 22x0 + 80. This function is maximized whenx0 = 11/4, for which the area of Q is 441/4.
The maximum possible area for Q is 441/4, taken when x0 = 11/4.
β6 2
β6
y = β4x
y = β4x
(x0, y0)
R
S
11. Let x= length of each side that uses the $1 per foot fencing,y= length of each side that uses the $2 per foot fencing.
The cost is C = (1)(2x) + (2)(2y) = 2x+ 4y, but A = xy = 3200 thus y = 3200/x so
C = 2x+ 12800/x for x > 0,dC/dx= 2β 12800/x2, dC/dx = 0 when x = 80, d2C/dx2 > 0 so
C is least when x = 80, y = 40.
12. A = xy where 2x+2y = p so y = p/2βx and A = px/2βx2 forx in [0, p/2]; dA/dx = p/2 β 2x, dA/dx = 0 when x = p/4. Ifx = 0 or p/2 then A = 0, if x = p/4 then A = p2/16 so the areais maximum when x = p/4 and y = p/2β p/4 = p/4, which is asquare. x
y
13. Let x and y be the dimensions of a rectangle; the perimeter is p = 2x + 2y. But A = xy thusy = A/x so p = 2x + 2A/x for x > 0, dp/dx = 2 β 2A/x2 = 2(x2 β A)/x2, dp/dx = 0 whenx =βA, d2p/dx2 = 4A/x3 > 0 if x > 0 so p is a minimum when x =
βA and y =
βA and thus
the rectangle is a square.
14. With x, y, r, and s as shown in the figure, the sum of the
enclosed areas is A = Οr2+s2 where r =x
2Οand s =
y
4because
x is the circumference of the circle and y is the perimeter of the
square, thus A =x2
4Ο+y2
16. But x+ y = 12, so y = 12β x and
A =x2
4Ο+(12β x)2
16=Ο + 416Ο
x2 β 32x+ 9 for 0 β€ x β€ 12.
dA
dx=Ο + 48Ο
xβ 32,dA
dx= 0 when x =
12ΟΟ + 4
. If x = 0,12ΟΟ + 4
, 12
then A = 9,36Ο + 4
,36Οso the sum of the enclosed areas is
x y12
r s
cut
(a) a maximum when x = 12 in. (when all of the wire is used for the circle)
(b) a minimum when x = 12Ο/(Ο + 4) in.
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January 27, 2005 11:44 L24-CH05 Sheet number 42 Page number 194 black
194 Chapter 5
15. (a)dN
dt= 250(20β t)eβt/20 = 0 at t = 20, N(0) = 125,000, N(20) β 161,788, and
N(100) β 128,369; the absolute maximum is N = 161788 at t = 20, the absoluteminimum is N = 125,000 at t = 0.
(b) The absolute minimum ofdN
dtoccurs when
d2N
dt2= 12.5(tβ 40)eβt/20 = 0, t = 40.
16. The area of the window is A = 2rh+ Οr2/2, the perimeter
is p = 2r + 2h+ Οr thus h =12[pβ (2 + Ο)r] so
A= r[pβ (2 + Ο)r] + Οr2/2= pr β (2 + Ο/2)r2 for 0 β€ r β€ p/(2 + Ο),
dA/dr = pβ (4 + Ο)r, dA/dr = 0 when r = p/(4 + Ο) andd2A/dr2 < 0, so A is maximum when r = p/(4 + Ο). 2r
r
h
17. Let the box have dimensions x, x, y, with y β₯ x. The constraint is 4x+ y β€ 108, and the volumeV = x2y. If we take y = 108β 4x then V = x2(108β 4x) and dV/dx = 12x(βx+ 18) with rootsx = 0, 18. The maximum value of V occurs at x = 18, y = 36 with V = 11, 664 in2.
18. Let the box have dimensions x, x, y with x β₯ y. The constraint is x+2(x+y) β€ 108, and the volumeV = x2y. Take x = (108β 2y)/3 = 36β 2y/3, V = y(36β 2y/3)2, dV/dy = (4/3)y2 β 96y + 1296with roots y = 18, 54. Then d2V/dy2 = (8/3)y β 96 is negative for y = 18, so by the secondderivative test, V has a maximum of 10, 368 in2 at y = 18, x = 24.
19. Let x be the length of each side of a square, then V = x(3 β 2x)(8 β 2x) = 4x3 β 22x2 + 24xfor 0 β€ x β€ 3/2; dV/dx = 12x2 β 44x + 24 = 4(3x β 2)(x β 3), dV/dx = 0 when x = 2/3 for0 < x < 3/2. If x = 0, 2/3, 3/2 then V = 0, 200/27, 0 so the maximum volume is 200/27 ft3.
20. Let x = length of each edge of base, y = height. The cost isC = (cost of top and bottom) + (cost of sides) = (2)(2x2) + (3)(4xy) = 4x2 + 12xy, butV = x2y = 2250 thus y = 2250/x2 so C = 4x2 + 27000/x for x > 0, dC/dx = 8xβ 27000/x2,dC/dx = 0 when x = 3
β3375 = 15, d2C/dx2 > 0 so C is least when x = 15, y = 10.
21. Let x = length of each edge of base, y = height, k = $/cm2 for the sides. The cost isC = (2k)(2x2) + (k)(4xy) = 4k(x2 + xy), but V = x2y = 2000 thus y = 2000/x2 soC = 4k(x2 + 2000/x) for x > 0, dC/dx = 4k(2xβ 2000/x2), dC/dx = 0 whenx = 3β1000 = 10, d2C/dx2 > 0 so C is least when x = 10, y = 20.
22. Let x and y be the dimensions shown in the figure and V the vol-ume, then V = x2y. The amount of material is to be 1000 ft2,thus (area of base) + (area of sides) = 1000, x2 + 4xy = 1000,
y =1000β x2
4xso V = x2 1000β x2
4x=
14(1000x β x3) for
0 < x β€ 10β10.
dV
dx=14(1000β 3x2),
dV
dx= 0
when x =β1000/3 = 10
β10/3.
If x = 0, 10β10/3, 10
β10 then V = 0,
50003
β10/3, 0;
the volume is greatest for x = 10β10/3 ft and y = 5
β10/3 ft.
xx
y
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January 27, 2005 11:44 L24-CH05 Sheet number 43 Page number 195 black
Exercise Set 5.5 195
23. Let x = height and width, y = length. The surface area is S = 2x2 + 3xy where x2y = V , soy = V/x2 and S = 2x2 + 3V/x for x > 0; dS/dx = 4x β 3V/x2, dS/dx = 0 when x = 3
β3V/4,
d2S/dx2 > 0 so S is minimum when x = 3
β3V4, y =
43
3
β3V4.
24. Let r and h be the dimensions shown in the figure, thenthe volume of the inscribed cylinder is V = Οr2h. But
r2 +(h
2
)2= R2 thus r2 = R2 β h
2
4
so V = Ο(R2 β h
2
4
)h = Ο
(R2hβ h
3
4
)
for 0 β€ h β€ 2R. dVdh
= Ο(R2 β 3
4h2),dV
dh= 0 h
2
h
r
R
when h = 2R/β3. If h = 0, 2R/
β3, 2R
then V = 0,4Ο3β3R3, 0 so the volume is largest
when h = 2R/β3 and r =
β2/3R.
25. Let r and h be the dimensions shown in the figure,then the surface area is S = 2Οrh+ 2Οr2.
But r2 +(h
2
)2= R2 thus h = 2
βR2 β r2 so
S = 4ΟrβR2 β r2 + 2Οr2 for 0 β€ r β€ R,
dS
dr=4Ο(R2 β 2r2)βR2 β r2
+ 4Οr;dS
dr= 0 when
R2 β 2r2βR2 β r2
= βr (1)
R2 β 2r2 = βrβR2 β r2
R4 β 4R2r2 + 4r4 = r2(R2 β r2)5r4 β 5R2r2 +R4 = 0
h2
h
r
R
and using the quadratic formula r2 =5R2 Β±
β25R4 β 20R4
10=5Β±β5
10R2, r =
β5Β±β5
10R, of
which only r =
β5 +β5
10R satisfies (1). If r = 0,
β5 +β5
10R, 0 then S = 0, (5+
β5)ΟR2, 2ΟR2 so
the surface area is greatest when r =
β5 +β5
10R and, from h = 2
βR2 β r2, h = 2
β5ββ5
10R.
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196 Chapter 5
26. Let R and H be the radius and height of the cone, andr and h the radius and height of the cylinder (see fig-ure), then the volume of the cylinder is V = Οr2h.
By similar triangles (see figure)H β hH
=r
Rthus
h =H
R(Rβ r) so V = ΟH
R(Rβ r)r2 = ΟH
R(Rr2 β r3)
for 0 β€ r β€ R. dVdr
= ΟH
R(2Rrβ3r2) = ΟH
Rr(2Rβ3r),
dV
dr= 0 for 0 < r < R when r = 2R/3. If
r = 0, 2R/3, R then V = 0, 4ΟR2H/27, 0 so the maxi-
mum volume is4ΟR2H
27=4913ΟR2H =
49Β· (volume of
cone).
R
r
H
h
27. From (13), S = 2Οr2 + 2Οrh. But V = Οr2h thus h = V/(Οr2) and so S = 2Οr2 + 2V/r for r > 0.dS/dr = 4Οrβ 2V/r2, dS/dr = 0 if r = 3
βV/(2Ο). Since d2S/dr2 = 4Ο+4V/r3 > 0, the minimum
surface area is achieved when r = 3βV/2Ο and so h = V/(Οr2) = [V/(Οr3)]r = 2r.
28. V = Οr2h where S = 2Οr2 + 2Οrh so h =S β 2Οr22Οr
, V =12(Sr β 2Οr3) for r > 0.
dV
dr=12(S β 6Οr2) = 0 if r =
βS/(6Ο),
d2V
dr2= β6Οr < 0 so V is maximum when
r =βS/(6Ο) and h =
S β 2Οr22Οr
=S β 2Οr22Οr2
r =S β S/3S/3
r = 2r, thus the height is equal to the
diameter of the base.
29. The surface area is S = Οr2 + 2Οrhwhere V = Οr2h = 500 so h = 500/(Οr2)and S = Οr2 + 1000/r for r > 0;dS/dr = 2Οr β 1000/r2 = (2Οr3 β 1000)/r2,dS/dr = 0 when r = 3
β500/Ο, d2S/dr2 > 0
for r > 0 so S is minimum when r = 3β500/Ο cm and
h =500Οr2
=500Ο
( Ο500
)2/3
= 3β500/Ο cm
r
h
30. The total area of material used isA = Atop +Abottom +Aside = (2r)2 + (2r)2 + 2Οrh = 8r2 + 2Οrh.The volume is V = Οr2h thus h = V/(Οr2) so A = 8r2 + 2V/r for r > 0,dA/dr = 16rβ2V/r2 = 2(8r3βV )/r2, dA/dr = 0 when r = 3
βV /2. This is the only critical point,
d2A/dr2 > 0 there so the least material is used when r = 3βV /2,
r
h=
r
V/(Οr2)=Ο
Vr3 and, for
r = 3βV /2,
r
h=Ο
V
V
8=Ο
8.
31. Let x be the length of each side of the squares and y the height of the frame, then the volume isV = x2y. The total length of the wire is L thus 8x+ 4y = L, y = (Lβ 8x)/4 soV = x2(L β 8x)/4 = (Lx2 β 8x3)/4 for 0 β€ x β€ L/8. dV/dx = (2Lx β 24x2)/4, dV/dx = 0 for0 < x < L/8 when x = L/12. If x = 0, L/12, L/8 then V = 0, L3/1728, 0 so the volume is greatestwhen x = L/12 and y = L/12.
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January 27, 2005 11:44 L24-CH05 Sheet number 45 Page number 197 black
Exercise Set 5.5 197
32. (a) Let x = diameter of the sphere, y = length of an edge of the cube. The combined volume is
V =16Οx3 + y3 and the surface area is S = Οx2 + 6y2 = constant. Thus y =
(S β Οx2)1/2
61/2
and V =Ο
6x3 +
(S β Οx2)3/2
63/2 for 0 β€ x β€βS
Ο;
dV
dx=Ο
2x2 β 3Ο
63/2x(S β Οx2)1/2 =
Ο
2β6x(β6xβ
βS β Οx2).
dV
dx= 0 when x = 0, or when
β6x =
βS β Οx2, 6x2 = S β Οx2, x2 =
S
6 + Ο, x =
βS
6 + Ο. If x = 0,
βS
6 + Ο,
βS
Ο,
then V =S3/2
63/2 ,S3/2
6β6 + Ο
,S3/2
6βΟso that V is smallest when x =
βS
6 + Ο, and hence when
y =
βS
6 + Ο, thus x = y.
(b) From Part (a), the sum of the volumes is greatest when there is no cube.
33. Let h and r be the dimensions shown in the figure, then the
volume is V =13Οr2h. But r2+h2 = L2 thus r2 = L2βh2
so V =13Ο(L2βh2)h =
13Ο(L2hβh3) for 0 β€ h β€ L. dV
dh=
13Ο(L2 β 3h2).
dV
dh= 0 when h = L/
β3. If h = 0, L/
β3, 0
then V = 0,2Ο9β3L3, 0 so the volume is as large as possible
when h = L/β3 and r =
β2/3L.
h L
r
34. Let r and h be the radius and height of the cone (see figure).The slant height of any such cone will be R, the radius ofthe circular sheet. Refer to the solution of Exercise 33 to
find that the largest volume is2Ο9β3R3. h R
r
35. The area of the paper is A = ΟrL = Οrβr2 + h2, but
V =13Οr2h = 10 thus h = 30/(Οr2)
so A = Οrβr2 + 900/(Ο2r4).
To simplify the computations let S = A2,
S = Ο2r2(r2 +
900Ο2r4
)= Ο2r4 +
900r2
for r > 0,
dS
dr= 4Ο2r3β 1800
r3=4(Ο2r6 β 450)
r3, dS/dr = 0 when
r = 6β450/Ο2, d2S/dr2 > 0, so S and hence A is least
when r = 6β450/Ο2 cm, h =
30Ο
3βΟ2/450 cm.
h L
r
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198 Chapter 5
36. The area of the triangle is A =12hb. By similar trian-
gles (see figure)b/2h=
Rβh2 β 2Rh
, b =2Rhβh2 β 2Rh
so A =Rh2
βh2 β 2Rh
for h > 2R,dA
dh=Rh2(hβ 3R)(h2 β 2Rh)3/2 ,
dA
dh= 0 for h > 2R when h = 3R, by the first deriva-
tive test A is minimum when h = 3R. If h = 3R thenb = 2
β3R (the triangle is equilateral).
R
h β R
R
h
bb/2
h2 β 2Rh
37. The volume of the cone is V =13Οr2h. By similar tri-
angles (see figure)r
h=
Rβh2 β 2Rh
, r =Rhβ
h2 β 2Rh
so V =13ΟR2 h3
h2 β 2Rh =13ΟR2 h2
hβ 2R for h > 2R,
dV
dh=13ΟR2h(hβ 4R)
(hβ 2R)2 ,dV
dh= 0 for h > 2R when
h = 4R, by the first derivative test V is minimumwhen h = 4R. If h = 4R then r =
β2R.
r
R
h β R
R
h
h2 β 2Rh
38. The area is (see figure)
A=12(2 sin ΞΈ)(4 + 4 cos ΞΈ)
= 4(sin ΞΈ + sin ΞΈ cos ΞΈ)for 0 β€ ΞΈ β€ Ο/2;dA/dΞΈ= 4(cos ΞΈ β sin2 ΞΈ + cos2 ΞΈ)
= 4(cos ΞΈ β [1β cos2 ΞΈ] + cos2 ΞΈ)= 4(2 cos2 ΞΈ + cos ΞΈ β 1)= 4(2 cos ΞΈ β 1)(cos ΞΈ + 1)
dA/dΞΈ = 0 when ΞΈ = Ο/3 for 0 < ΞΈ < Ο/2. If ΞΈ = 0, Ο/3, Ο/2 thenA = 0, 3
β3, 4 so the maximum area is 3
β3.
42 cos ΞΈ
ΞΈ
4 cos ΞΈ
2 sin ΞΈ2
39. Let b and h be the dimensions shown in the figure,
then the cross-sectional area is A =12h(5 + b). But
h = 5 sin ΞΈ and b = 5 + 2(5 cos ΞΈ) = 5 + 10 cos ΞΈ so
A =52sin ΞΈ(10 + 10 cos ΞΈ) = 25 sin ΞΈ(1 + cos ΞΈ) for
0 β€ ΞΈ β€ Ο/2.
dA/dΞΈ= β25 sin2 ΞΈ + 25 cos ΞΈ(1 + cos ΞΈ)= 25(β sin2 ΞΈ + cos ΞΈ + cos2 ΞΈ)= 25(β1 + cos2 ΞΈ + cos ΞΈ + cos2 ΞΈ)= 25(2 cos2 ΞΈ + cos ΞΈ β 1) = 25(2 cos ΞΈ β 1)(cos ΞΈ + 1).
dA/dΞΈ = 0 for 0 < ΞΈ < Ο/2 when cos ΞΈ = 1/2, ΞΈ = Ο/3.If ΞΈ = 0, Ο/3, Ο/2 then A = 0, 75
β3/4, 25 so the cross-
sectional area is greatest when ΞΈ = Ο/3.
b5 cos ΞΈ
ΞΈ
55
5
h = 5 sin ΞΈ
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Exercise Set 5.5 199
40. I = kcosΟ*2
, k the constant of proportionality. If h is the height of the lamp above the table then
cosΟ = h/* and * =βh2 + r2 so I = k
h
*3= k
h
(h2 + r2)3/2for h > 0,
dI
dh= k
r2 β 2h2
(h2 + r2)5/2,dI
dh= 0
when h = r/β2, by the first derivative test I is maximum when h = r/
β2.
41. Let L, L1, and L2 be as shown in the figure, thenL = L1 + L2 = 8 csc ΞΈ + sec ΞΈ,
dL
dΞΈ= β8 csc ΞΈ cot ΞΈ + sec ΞΈ tan ΞΈ, 0 < ΞΈ < Ο/2
= β8 cos ΞΈsin2 ΞΈ
+sin ΞΈcos2 ΞΈ
=β8 cos3 ΞΈ + sin3 ΞΈ
sin2 ΞΈ cos2 ΞΈ;
dL
dΞΈ= 0 if sin3 ΞΈ = 8 cos3 ΞΈ, tan3 ΞΈ = 8, tan ΞΈ = 2 which gives
the absolute minimum for L because limΞΈβ0+
L = limΞΈβΟ/2β
L = +β.
If tan ΞΈ = 2, then csc ΞΈ =β5/2 and sec ΞΈ =
β5 so L = 8(
β5/2) +
β5 = 5
β5 ft.
1
8
ΞΈ
L2
L1
L
42. Let x= number of steers per acrew= average market weight per steerT = total market weight per acre
then T = xw where w = 2000β 50(xβ 20) = 3000β 50xso T = x(3000β 50x) = 3000xβ 50x2 for 0 β€ x β€ 60,dT/dx = 3000 β 100x and dT/dx = 0 when x = 30. If x = 0, 30, 60 then T = 0, 45000, 0 so thetotal market weight per acre is largest when 30 steers per acre are allowed.
43. (a) The daily profit is
P = (revenue) β (production cost) = 100xβ (100, 000 + 50x+ 0.0025x2)= β100, 000 + 50xβ 0.0025x2
for 0 β€ x β€ 7000, so dP/dx = 50β 0.005x and dP/dx = 0 when x = 10, 000. Because 10,000is not in the interval [0, 7000], the maximum profit must occur at an endpoint. When x = 0,P = β100, 000; when x = 7000, P = 127, 500 so 7000 units should be manufactured and solddaily.
(b) Yes, because dP/dx > 0 when x = 7000 so profit is increasing at this production level.(c) dP/dx = 15 when x = 7000, so P (7001)β P (7000) β 15, and the marginal profit is $15.
44. (a) R(x) = px but p = 1000β x so R(x) = (1000β x)x(b) P (x) = R(x)β C(x) = (1000β x)xβ (3000 + 20x) = β3000 + 980xβ x2
(c) P β²(x) = 980β 2x, P β²(x) = 0 for 0 < x < 500 when x = 490; test the points 0, 490, 500 to findthat the profit is a maximum when x = 490.
(d) P (490) = 237,100(e) p = 1000β x = 1000β 490 = 510.
45. The profit is
P = (profit on nondefective)β (loss on defective) = 100(xβ y)β 20y = 100xβ 120y
but y = 0.01x + 0.00003x2 so P = 100x β 120(0.01x + 0.00003x2) = 98.8x β 0.0036x2 for x > 0,dP/dx = 98.8β 0.0072x, dP/dx = 0 when x = 98.8/0.0072 β 13, 722, d2P/dx2 < 0 so the profit ismaximum at a production level of about 13,722 pounds.
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200 Chapter 5
46. To cover 1 mile requires 1/v hours, and 1/(10 β 0.07v) gallons of diesel fuel, so the total cost to
the client is C =15v+
1.5010β 0.07v ,
dC
dv=0.0315v2 + 21v β 1500v2(0.07v β 10)2 . By the second derivative test, C
has a minimum of 50.6 cents/mile at v = 65.08 miles per hour.
47. The distance between the particles is D =β(1β tβ t)2 + (tβ 2t)2 =
β5t2 β 4t+ 1 for t β₯ 0. For
convenience, we minimize D2 instead, so D2 = 5t2 β 4t + 1, dD2/dt = 10t β 4, which is 0 whent = 2/5. d2D2/dt2 > 0 so D2 and hence D is minimum when t = 2/5. The minimum distance isD = 1/
β5.
48. The distance between the particles is D =β(2tβ t)2 + (2β t2)2 =
βt4 β 3t2 + 4 for t β₯ 0. For
convenience we minimize D2 instead so D2 = t4β 3t2+4, dD2/dt = 4t3β 6t = 4t(t2β 3/2), whichis 0 for t > 0 when t =
β3/2. d2D2/dt2 = 12t2 β 6 > 0 when t =
β3/2 so D2 and hence D is
minimum there. The minimum distance is D =β7/2.
49. Let P (x, y) be a point on the curve x2 + y2 = 1. The distance between P (x, y) and P0(2, 0) isD =
β(xβ 2)2 + y2, but y2 = 1 β x2 so D =
β(xβ 2)2 + 1β x2 =
β5β 4x for β1 β€ x β€ 1,
dD
dx= β 2β
5β 4xwhich has no critical points for β1 < x < 1. If x = β1, 1 then D = 3, 1 so the
closest point occurs when x = 1 and y = 0.
50. Let P (x, y) be a point on y =βx, then the distance D between P and (2, 0) is
D =β(xβ 2)2 + y2 =
β(xβ 2)2 + x =
βx2 β 3x+ 4, for 0 β€ x β€ 3. For convenience we find the
extrema for D2 instead, so D2 = x2 β 3x+4, dD2/dx = 2xβ 3 = 0 when x = 3/2. If x = 0, 3/2, 3then D2 = 4, 7/4, 4 so D = 2,
β7/2, 2. The points (0, 0) and (3,
β3) are at the greatest distance,
and (3/2,β3/2) the shortest distance from (2, 0).
51. (a) Draw the line perpendicular to the given line that passes through Q.(b) Let Q : (x0, y0). If (x,mx + b) is the point on the graph closest to Q, then the distance
squared d2 = D = (xβ x0)2 + (mx+ bβ y0)2 is minimized. Then
dD/dx = 2(xβ x0) + 2m(mx+ bβ y0) = 0
at the point (x,mx+ b) which minimizes the distance.
On the other hand, the line connecting the pointQ with the line y = mx+b is perpendicular tothis line provided the connecting line has slope β1/m. But this condition is (yβy0)/(xβx0) =β1/m, or m(yβ y0) = β(xβx0). Since y = mx+ b the condition becomes m(mx+ bβ y0) =β(xβ x0), which is the equivalent to the expression above which minimizes the distance.
52. (a) Draw the line from the point Q through the center P of the circle. The nearer and fartherpoints are the points on C that are, respectively closest to and farthest from Q.
(b) Let Q : (xQ, yQ) and P : (xP , yP ), and let the equation of the circle be (x β xP )2 + (y βyP )2 = c2 where c is constant. Let T be the square of the distance from (x, y) to Q:T = (x β xQ)2 + (y β yQ)2. It is desired to find the minimum (and maximum) value of Twhen (x, y) lies on the circle.This occurs when the derivative of T is zero, i.e. 2(x β xQ) + 2(y β yQ)(dy/dx) = 0. Notethat dy/dx here expresses the slope of the line tangent to the circle at the point (x, y). Anequivalent expression is dy/dx = β(xβ xQ)/(yβ yQ), which is the negative reciprocal of theslope of the line connecting (x, y) with Q. Thus the minimum and maximum distances areachieved on this line.
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Exercise Set 5.5 201
53. (a) The line through (x, y) and the origin has slope y/x, and the negative reciprocal is βx/y. If(x, y) is a point on the ellipse, then x2 β xy + y2 = 4 and, differentiating,2xβ x(dy/dx)β y + 2y(dy/dx) = 0.For the desired points we have dy/dx = βx/y, and inserting that into the previous equationresults in 2x + x2/y β y β 2x = 0, 2xy + x2 β y2 β 2xy = 0, x2 β y2 = 0, y = Β±x. Insertingthis into the equation of the ellipse we have x2 β x2 + x2 = 4 with solutions y = x = Β±2or y = βx = Β±2/
β3. Thus there are four solutions, (2, 2), (β2,β2), (2/
β3,β2/
β3) and
(β2/β3, 2/β3).
(b) In general, the shortest/longest distance from a point to a curve is taken on the line connectingthe point to the curve which is perpendicular to the tangent line at the point in question.
54. The tangent line to the ellipse at (x, y) has slope dy/dx,where x2 β xy + y2 = 4. This yields2xβyβxdy/dx+2ydy/dx = 0, dy/dx = (2xβy)/(xβ2y). The line through (x, y) that also passes throughthe origin has slope y/x. Check to see if the twoslopes are negative reciprocals: (2x β y)/(x β 2y) =βx/y; y(2x β y) = βx(x β 2y); x = Β±y, so thepoints lie on the line y = x or on y = βx.
y
x
β2 β1 1 2
β2
β1
1
2
55. If P (x0, y0) is on the curve y = 1/x2, then y0 = 1/x20. At P the slope of the tangent line is β2/x3
0
so its equation is y β 1x2
0= β 2
x30(xβ x0), or y = β
2x3
0x+
3x2
0. The tangent line crosses the y-axis
at3x2
0, and the x-axis at
32x0. The length of the segment then is L =
β9x4
0+94x2
0 for x0 > 0. For
convenience, we minimize L2 instead, so L2 =9x4
0+94x2
0,dL2
dx0= β36
x50+92x0 =
9(x60 β 8)2x5
0, which
is 0 when x60 = 8, x0 =
β2.d2L2
dx20> 0 so L2 and hence L is minimum when x0 =
β2, y0 = 1/2.
56. If P (x0, y0) is on the curve y = 1 β x2, then y0 = 1 β x20. At P the slope of the tangent line is
β2x0 so its equation is y β (1 β x20) = β2x0(x β x0), or y = β2x0x + x2
0 + 1. The y-intercept is
x20 + 1 and the x-intercept is
12(x0 + 1/x0) so the area A of the triangle is
A =14(x2
0 + 1)(x0 + 1/x0) =14(x3
0 + 2x0 + 1/x0) for 0 β€ x0 β€ 1.
dA/dx0 =14(3x2
0 + 2 β 1/x20) =
14(3x4
0 + 2x20 β 1)/x2
0 which is 0 when x20 = β1 (reject), or when
x20 = 1/3 so x0 = 1/
β3. d2A/dx2
0 =14(6x0 + 2/x3
0) > 0 at x0 = 1/β3 so a relative minimum and
hence the absolute minimum occurs there.
57. At each point (x, y) on the curve the slope of the tangent line is m =dy
dx= β 2x
(1 + x2)2for any
x,dm
dx=2(3x2 β 1)(1 + x2)3
,dm
dx= 0 when x = Β±1/
β3, by the first derivative test the only relative
maximum occurs at x = β1/β3, which is the absolute maximum because lim
xβΒ±βm = 0. The
tangent line has greatest slope at the point (β1/β3, 3/4).
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202 Chapter 5
58. Let x be how far P is upstream from where the manstarts (see figure), then the total time to reach T is
t= (time from M to P ) + (time from P to T )
=βx2 + 1rR
+1β xrW
for 0 β€ x β€ 1,
where rR and rW are the rates at which he can rowand walk, respectively.
M
TP
1
1x
(a) t =βx2 + 13
+1β x5
,dt
dx=
x
3βx2 + 1
β 15sodt
dx= 0 when 5x = 3
βx2 + 1,
25x2 = 9(x2 + 1), x2 = 9/16, x = 3/4. If x = 0, 3/4, 1 then t = 8/15, 7/15,β2/3 so the time
is a minimum when x = 3/4 mile.
(b) t =βx2 + 14
+1β x5
,dt
dx=
x
4βx2 + 1
β 15sodt
dx= 0 when x = 4/3 which is not in the
interval [0, 1]. Check the endpoints to find that the time is a minimum when x = 1 (he shouldrow directly to the town).
59. With x and y as shown in the figure, the maximumlength of pipe will be the smallest value of L = x+ y.By similar triangles
y
8=
xβx2 β 16
, y =8xβx2 β 16
so
L = x+8xβx2 β 16
for x > 4,dL
dx= 1β 128
(x2 β 16)3/2 ,dL
dx= 0 when
(x2 β 16)3/2= 128x2 β 16= 1282/3 = 16(22/3)
x2= 16(1 + 22/3)x= 4(1 + 22/3)1/2,
d2L/dx2 = 384x/(x2 β 16)5/2 > 0 if x > 4 so L is smallest when x = 4(1 + 22/3)1/2.For this value of x, L = 4(1 + 22/3)3/2 ft.
8
4
x
y
x2 β 16
60. s= (x1 β xΜ)2 + (x2 β xΜ)2 + Β· Β· Β·+ (xn β xΜ)2,ds/dxΜ= β2(x1 β xΜ)β 2(x2 β xΜ)β Β· Β· Β· β 2(xn β xΜ),ds/dxΜ= 0 when
(x1 β xΜ) + (x2 β xΜ) + Β· Β· Β·+ (xn β xΜ)= 0(x1 + x2 + Β· Β· Β·xn)β (xΜ+ xΜ+ Β· Β· Β·+ xΜ)= 0
(x1 + x2 + Β· Β· Β·+ xn)β nxΜ= 0
xΜ=1n(x1 + x2 + Β· Β· Β·+ xn),
d2s/dxΜ2 = 2 + 2 + Β· Β· Β·+ 2 = 2n > 0, so s is minimum when xΜ =1n(x1 + x2 + Β· Β· Β·+ xn).
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Exercise Set 5.5 203
61. Let x = distance from the weaker light source, I = the intensity at that point, and k the constantof proportionality. Then
I =kS
x2 +8kS
(90β x)2 if 0 < x < 90;
dI
dx= β2kS
x3 +16kS
(90β x)3 =2kS[8x3 β (90β x)3]
x3(90β x)3 = 18kS(xβ 30)(x2 + 2700)
x3(xβ 90)3 ,
which is 0 when x = 30;dI
dx< 0 if x < 30, and
dI
dx> 0 if x > 30, so the intensity is minimum at a
distance of 30 cm from the weaker source.
62. If f(x0) is a maximum then f(x) β€ f(x0) for all x in some open interval containing x0 thusβf(x) β€
βf(x0) because
βx is an increasing function, so
βf(x0) is a maximum of
βf(x) at x0.
The proof is similar for a minimum value, simply replace β€ by β₯.
x
y
A(2, 1)
B(5, 4)
P(x, 0)
52x β 2 5 β x
οΏ½ οΏ½οΏ½
63. ΞΈ= Ο β (Ξ±+ Ξ²)= Ο β cotβ1(xβ 2)β cotβ1 5β x
4,
dΞΈ
dx=
11 + (xβ 2)2 +
β1/41 + (5β x)2/16
= β 3(x2 β 2xβ 7)[1 + (xβ 2)2][16 + (5β x)2]
dΞΈ/dx = 0 when x =2Β±β4 + 282
= 1Β± 2β2,
only 1 + 2β2 is in [2, 5]; dΞΈ/dx > 0 for x in [2, 1 + 2
β2),
dΞΈ/dx < 0 for x in (1 + 2β2, 5], ΞΈ is maximum when x = 1 + 2
β2.
οΏ½οΏ½
οΏ½
x
10
2
64. ΞΈ= Ξ±β Ξ²
= cotβ1(x/12)β cotβ1(x/2)
dΞΈ
dx= β 12
144 + x2 +2
4 + x2
=10(24β x2)
(144 + x2)(4 + x2)
dΞΈ/dx = 0 when x =β24 = 2
β6, by
the first derivative test ΞΈ ismaximum there.
65. Let v = speed of light in the medium. The total time required for the light to travel from A to Pto B is
t = (total distance from A to P to B)/v =1v(β(cβ x)2 + a2 +
βx2 + b2),
dt
dx=1v
[β cβ xβ
(cβ x)2 + a2+
xβx2 + b2
]
anddt
dx= 0 when
xβx2 + b2
=cβ xβ
(cβ x)2 + a2. But x/
βx2 + b2 = sin ΞΈ2 and
(cβ x)/β(cβ x)2 + a2 = sin ΞΈ1 thus dt/dx = 0 when sin ΞΈ2 = sin ΞΈ1 so ΞΈ2 = ΞΈ1.
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204 Chapter 5
66. The total time required for the light to travel from A to P to B is
t = (time from A to P ) + (time from P to B) =βx2 + a2
v1+
β(cβ x)2 + b2
v2,
dt
dx=
x
v1βx2 + a2
β cβ xv2β(cβ x)2 + b2
but x/βx2 + a2 = sin ΞΈ1 and
(cβ x)/β(cβ x)2 + b2 = sin ΞΈ2 thus
dt
dx=sin ΞΈ1v1β sin ΞΈ2
v2sodt
dx= 0 when
sin ΞΈ1v1
=sin ΞΈ2v2
.
67. (a) The rate at which the farmer walks is analogous to the speed of light in Fermatβs principle.
(b) the best path occurs when ΞΈ1 = ΞΈ2(see figure).
x 1 β x
34
14
ΞΈ2
ΞΈ1House
Barn
(c) by similar triangles,x/(1/4)= (1β x)/(3/4)
3x= 1β x4x= 1x= 1/4 mi.
EXERCISE SET 5.6
1. f(x) = x2 β 2, f β²(x) = 2x, xn+1 = xn βx2n β 22xn
x1 = 1, x2 = 1.5, x3 = 1.416666667, . . . , x5 = x6 = 1.414213562
2. f(x) = x2 β 5, f β²(x) = 2x, xn+1 = xn βx2n β 52xn
x1 = 2, x2 = 2.25, x3 = 2.236111111, x4 = 2.2360679779, x4 = x5 = 2.2360679775
3. f(x) = x3 β 6, f β²(x) = 3x2, xn+1 = xn βx3n β 63x2
n
x1 = 2, x2 = 1.833333333, x3 = 1.817263545, . . . , x5 = x6 = 1.817120593
4. xn β a = 0
5. f(x) = x3 β 2xβ 2, f β²(x) = 3x2 β 2, xn+1 = xn βx3n β 2xn β 23x2
n β 2
x1 = 2, x2 = 1.8, x3 = 1.7699481865, x4 = 1.7692926629, x5 = x6 = 1.7692923542
6. f(x) = x3 + xβ 1, f β²(x) = 3x2 + 1, xn+1 = xn βx3n + xn β 13x2
n + 1x1 = 1, x2 = 0.75, x3 = 0.686046512, . . . , x5 = x6 = 0.682327804
7. f(x) = x5 + x4 β 5, f β²(x) = 5x4 + 4x3, xn+1 = xn βx5n + x
4n β 5
5x4n + 4x3
n
x1 = 1, x2 = 1.333333333, x3 = 1.239420573, . . . , x6 = x7 = 1.224439550
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Exercise Set 5.6 205
8. f(x) = x5 β 3x+ 3, f β²(x) = 5x4 β 3, xn+1 = xn βx5n β 3xn + 35x4
n β 3
x1 = β1.5, x2 = β1.49579832, x3 = β1.49577135 = x4 = x5
9. f(x) = x4 + x2 β 4, f β²(x) = 4x3 + 2x,
xn+1 = xn βx4n + x
2n β 4
4x3n + 2xn
x1 = 1, x2 = 1.3333, x3 = 1.2561, x4 = 1.24966, . . . ,x7 = x8 = 1.249621068;by symmetry, x = β1.249621068 is the other solution.
16
β5
β2.2 2.2
10. f(x) = x5 β 5x3 β 2, f β²(x) = 5x4 β 15x2,
xn+1 = xn βx5n β 5x3
n β 25x4
n β 15x2n
x1 = 2, x2 = 2.5,x3 = 2.327384615, . . . , x7 = x8 = 2.273791732
10
β20
β2.5 2.5
11. f(x) = 2 cosxβ x, f β²(x) = β2 sinxβ 1
xn+1 = xn β2 cosxβ xβ2 sinxβ 1
x1 = 1, x2 = 1.03004337, x3 = 1.02986654,x4 = x5 = 1.02986653
8
β6
O o
12. f(x) = sinxβ x2,f β²(x) = cosxβ 2x,
xn+1 = xn βsinxn β x2
n
cosxn β 2xn
x1 = 1, x2 = 0.891395995,x3 = 0.876984845, . . . , x5 = x6 = 0.876726215
0.3
β1.3
0 1.5
13. f(x) = xβ tanx,
f β²(x) = 1β sec2 x = β tan2 x, xn+1 = xn +xn β tanxntan2 xn
x1 = 4.5, x2 = 4.493613903, x3 = 4.493409655,x4 = x5 = 4.493409458
100
β100
6 i
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206 Chapter 5
14. f(x) = 1 + ex sinx, f β²(x) = ex(cosx+ sinx)
xn+1 = xn β1 + exn sinxn
exn(cosxn + sinxn)
x1 = 3, x2 = 3.2249, x3 = 3.1847,. . . ,x10 = x11 = 3.183063012
3
β22
0 c
15. The graphs of y = x3 and y = 1 β x intersect nearthe point x = 0.7. Let f(x) = x3 + x β 1, so thatf β²(x) = 3x2 + 1, and
xn+1 = xn βx3 + xβ 13x2 + 1
.
If x1 = 0.7 then x2 = 0.68259109,x3 = 0.68232786, x4 = x5 = 0.68232780.
4
β1
β1 2
16. The graphs of y = sinx and y = x3β 2x2+1 intersectat points near x = β0.8 and x = 0.6 and x = 2. Letf(x) = sinxβx3+2x2β1, then f β²(x) = cosxβ3x2+4x,so
xn+1 = xn βcosxβ 3x2 + 4x
sinxβ x3 + 2x2 + 1.
If x1 = β0.8, then x2 = β0.783124811,x3 = β0.782808234,x4 = x5 = β0.782808123; if x1 = 0.6, thenx2 = 0.568003853, x3 = x4 = 0.568025739 ; if x1 = 2, thenx2 = 1.979461151, x3 = 1.979019264, x4 = x5 = 1.979019061
β1.5 2.5
β1
2
17. The graphs of y = x2 and y =β2x+ 1 intersect at
points near x = β0.5 and x = 1; x2 =β2x+ 1,
x4 β 2xβ 1 = 0. Let f(x) = x4 β 2xβ 1, thenf β²(x) = 4x3 β 2 so
xn+1 = xn βx4n β 2xn β 14x3
n β 2.
If x1 = β0.5, then x2 = β0.475,x3 = β0.474626695,x4 = x5 = β0.474626618; ifx1 = 1, then x2 = 2,x3 = 1.633333333, . . . , x8 = x9 = 1.395336994.
4
0β0.5 2
18. The graphs of y = 18x
3 β 1 and y = cosxβ 2 intersectwhen x = 0 and near the point x = β2. Let f(x) =18x
3 + 1 β cosx so that f β²(x) = 38x
2 + sinx. Then
xn+1 = xn βx3/8 + 1β cosx3x2/8 + sinx
.
If x1 = β2 then x2 = β2.70449471,x3 = β2.46018026, . . . , x6 = x7 = β2.40629382
0
β3
β3 2
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Exercise Set 5.6 207
19. The graphs of y = 1 and y = ex sinx intersect near thepoints x = 1 and x = 3. Let f(x) = 1βex sinx, f β²(x) =βex(cosx+ sinx), and
xn+1 = xn +1β ex sinx
ex(cosx+ sinx). If x1 = 1 then
x2 = 0.65725814, x3 = 0.59118311, . . . , x5 = x6 =0.58853274, and if x1 = 3 then x2 = 3.10759324, x3 =3.09649396, . . . , x5 = x6 = 3.09636393.
8
00 3.2
20. The graphs of y = eβx and y = lnx intersect near x = 1.3; let
f(x) = eβx β lnx, f β²(x) = βeβx β 1/x, x1 = 1.3,
xn+1 = xn +eβxn β lnxneβxn + 1/xn
, x2 = 1.309759929,
x4 = x5 = 1.309799586
1
β4
0 2
21. (a) f(x) = x2 β a, f β²(x) = 2x, xn+1 = xn βx2n β a2xn
=12
(xn +
a
xn
)
(b) a = 10; x1 = 3, x2 = 3.166666667, x3 = 3.162280702, x4 = x5 = 3.162277660
22. (a) f(x) =1xβ a, f β²(x) = β 1
x2 , xn+1 = xn(2β axn)
(b) a = 17; x1 = 0.05, x2 = 0.0575, x3 = 0.058793750, x5 = x6 = 0.058823529
23. f β²(x) = x3 + 2xβ 5; solve f β²(x) = 0 to find the critical points. Graph y = x3 and y = β2x+ 5 to
see that they intersect at a point near x = 1.25; f β²β²(x) = 3x2 + 2 so xn+1 = xn βx3n + 2xn β 53x2
n + 2.
x1 = 1.25, x2 = 1.3317757009, x3 = 1.3282755613, x4 = 1.3282688557, x5 = 1.3282688557 so theminimum value of f(x) occurs at x β 1.3282688557 because f β²β²(x) > 0; its value is approximatelyβ4.098859123.
24. From a rough sketch of y = x sinx we see that the maximum occurs at a point near x = 2, whichwill be a point where f β²(x) = x cosx+ sinx = 0. f β²β²(x) = 2 cosxβ x sinx so
xn+1 = xn βxn cosxn + sinxn2 cosxn β xn sinxn
= xn βxn + tanxn2β xn tanxn
.
x1 = 2, x2 = 2.029048281, x3 = 2.028757866, x4 = x5 = 2.028757838; the maximum value isapproximately 1.819705741.
25. A graphing utility shows that there are two inflection points at x β 0.25,β1.25. These points
are the zeros of f β²β²(x) = (x4 + 4x3 + 8x2 + 4x β 1) eβx
(x2 + 1)3. It is equivalent to find the zeros of
g(x) = x4+4x3+8x2+4xβ1. One root is x = β1 by inspection. Since gβ²(x) = 4x3+12x2+16x+4,
Newtonβs Method becomes
xn+1 = xn βx4n + 4x
3n + 8x
2n + 4xn β 1
4x3n + 12x2
n + 16xn + 4
With x0 = 0.25, x1 = 0.18572695, x2 = 0.179563312, x3 = 0.179509029,x4 = x5 = 0.179509025. So the points of inflection are at x β 0.18, x = β1.
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208 Chapter 5
26. f β²(x) = β2 tanβ1 x+1β 2xx2 + 1
= 0 for x = x1 β 0.2451467013, f(x1) β 0.1225363521
27. Let f(x) be the square of the distance between (1, 0) and any point (x, x2) on the parabola, thenf(x) = (xβ 1)2 + (x2 β 0)2 = x4 + x2 β 2x+ 1 and f β²(x) = 4x3 + 2xβ 2. Solve f β²(x) = 0 to find
the critical points; f β²β²(x) = 12x2 + 2 so xn+1 = xn β4x3
n + 2xn β 212x2
n + 2= xn β
2x3n + xn β 16x2
n + 1.
x1 = 1, x2 = 0.714285714, x3 = 0.605168701, . . . , x6 = x7 = 0.589754512; the coordinates areapproximately (0.589754512, 0.347810385).
28. The area is A = xy = x cosx so dA/dx = cosxβ x sinx. Find x so that dA/dx = 0;
d2A/dx2 = β2 sinxβ x cosx so xn+1 = xn +cosxn β xn sinxn2 sinxn + xn cosxn
= xn +1β xn tanxn2 tanxn + xn
.
x1 = 1, x2 = 0.864536397, x3 = 0.860339078, x4 = x5 = 0.860333589; y β 0.652184624.
29. (a) Let s be the arc length, and L the length of the chord, then s = 1.5L. But s = rΞΈ andL = 2r sin(ΞΈ/2) so rΞΈ = 3r sin(ΞΈ/2), ΞΈ β 3 sin(ΞΈ/2) = 0.
(b) Let f(ΞΈ) = ΞΈ β 3 sin(ΞΈ/2), then f β²(ΞΈ) = 1β 1.5 cos(ΞΈ/2) so ΞΈn+1 = ΞΈn βΞΈn β 3 sin(ΞΈn/2)1β 1.5 cos(ΞΈn/2)
.
ΞΈ1 = 3, ΞΈ2 = 2.991592920, ΞΈ3 = 2.991563137, ΞΈ4 = ΞΈ5 = 2.991563136 rad so ΞΈ β 171β¦.
30. r2(ΞΈ β sin ΞΈ)/2 = Οr2/4 so ΞΈ β sin ΞΈ β Ο/2 = 0. Let f(ΞΈ) = ΞΈ β sin ΞΈ β Ο/2, then f β²(ΞΈ) = 1β cos ΞΈ
so ΞΈn+1 =ΞΈn β sin ΞΈn β Ο/2
1β cos ΞΈn.
ΞΈ1 = 2, ΞΈ2 = 2.339014106, ΞΈ3 = 2.310063197, . . . , ΞΈ5 = ΞΈ6 = 2.309881460 rad; ΞΈ β 132β¦.
31. If x = 1, then y4 + y = 1, y4 + y β 1 = 0. Graph z = y4 and z = 1β y to see that they intersect
near y = β1 and y = 1. Let f(y) = y4 + y β 1, then f β²(y) = 4y3 + 1 so yn+1 = yn βy4n + yn β 14y3n + 1
.
If y1 = β1, then y2 = β1.333333333, y3 = β1.235807860, . . . , y6 = y7 = β1.220744085;if y1 = 1, then y2 = 0.8, y3 = 0.731233596, . . . , y6 = y7 = 0.724491959.
32. If x = 1, then 2y β cos y = 0. Graph z = 2y and z = cos y to see that they intersect near y = 0.5.
Let f(y) = 2y β cos y, then f β²(y) = 2 + sin y so yn+1 = yn β2yn β cos yn2 + sin yn
.
y1 = 0.5, y2 = 0.450626693, y3 = 0.450183648, y4 = y5 = 0.450183611.
33. S(25) = 250,000 =5000i
[(1 + i)25 β 1
]; set f(i) = 50iβ (1+ i)25+1, f β²(i) = 50β25(1+ i)24; solve
f(i) = 0. Set i0 = .06 and ik+1 = ikβ[50iβ (1 + i)25 + 1
]/[50β 25(1 + i)24
]. Then i1 = 0.05430,
i2 = 0.05338, i3 = 0.05336, . . . , i = 0.053362.
34. (a) x1 = 2, x2 = 5.3333,x3 = 11.055, x4 = 22.293,x5 = 44.676
0.5
00 15
(b) x1 = 0.5, x2 = β0.3333, x3 = 0.0833, x4 = β0.0012, x5 = 0.0000 (and xn = 0 for n β₯ 6)
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Exercise Set 5.7 209
35. (a) x1 x2 x3 x4 x5 x6 x7 x8 x9 x10
0.5000 β0.7500 0.2917 β1.5685 β0.4654 0.8415 β0.1734 2.7970 1.2197 0.1999
(b) The sequence xn must diverge, since if it did converge then f(x) = x2 + 1 = 0 would have asolution. It seems the xn are oscillating back and forth in a quasi-cyclical fashion.
36. (a) xn+1 = xn, i.e. the constant sequence xn is generated.
(b) This is equivalent to f(xn) = 0 as in part (a).
(c) xn = xn+2 = xn+1 βf(xn+1)f β²(xn+1)
= xn βf(xn)f β²(xn)
β f(xn+1)f β²(xn+1)
, so f(xn+1) = βf β²(xn+1)f β²(xn)
f(xn)
37. (a) |xn+1 β xn| β€ |xn+1 β c|+ |cβ xn| < 1/n+ 1/n = 2/n(b) The closed interval [c β 1, c + 1] contains all of the xn, since |xn β c| < 1/n. Let M be an
upper bound for |f β²(x)| on [c β 1, c + 1]. Since xn+1 = xn β f(xn)/f β²(xn) it follows that|f(xn)| β€ |f β²(xn)||xn+1 β xn| < M |xn+1 β xn| < 2M/n.
(c) Assume that f(c) = 0. The sequence xn converges to c, since |xn β c| < 1/n. By thecontinuity of f , f(c) = f( lim
nβ+βxn) = lim
nβ+βf(xn).
Let Ξ΅ = |f(c)|/2. Choose N such that |f(xn)β f(c)| < Ξ΅/2 for n > N . Then|f(xn)β f(c)| < |f(c)|/2 for n > N , so β|f(c)|/2 < f(xn)β f(c) < |f(c)/2|.If f(c) > 0 then f(xn) > f(c)β |f(c)|/2 = f(c)/2.If f(c) < 0, then f(xn) < f(c) + |f(c)|/2 = β|f(c)|/2, or |f(xn)| > |f(c)|/2.
(d) From (b) it follows that limnβ+β
f(xn) = 0. From (c) it follows that if f(c) = 0 then
limnβ+β
f(xn) = 0, a contradiction. The conclusion, then, is that f(c) = 0.
EXERCISE SET 5.7
1. f(3) = f(5) = 0; f β²(x) = 2xβ 8, 2cβ 8 = 0, c = 4, f β²(4) = 0
2. f(0) = f(2) = 0, f β²(x) = 3x2 β 6x+ 2, 3c2 β 6c+ 2 = 0; c = 6Β±β36β 246
= 1Β±β3/3
3. f(Ο/2) = f(3Ο/2) = 0, f β²(x) = β sinx, β sin c = 0, c = Ο
4. f(β1) = f(3) = 0; f β²(1) = 0; f β²(x) = 2(1β x)/(4 + 2xβ x2); 2(1β c) = 0, c = 1
5. f(0) = f(4) = 0, f β²(x) =12β 12βx,12β 12βc= 0, c = 1
6. f(1) = f(3) = 0, f β²(x) = β 2x3 +
43x2 , β
2c3+
43c2
= 0, β6 + 4c = 0, c = 3/2
7. (f(5)β f(β3))/(5β (β3)) = 1; f β²(x) = 2xβ 1; 2cβ 1 = 1, c = 1
8. f(β1) = β6, f(2) = 6, f β²(x) = 3x2 + 1, 3c2 + 1 =6β (β6)2β (β1) = 4, c2 = 1, c = Β±1 of which only
c = 1 is in (β1, 2)
9. f(0) = 1, f(3) = 2, f β²(x) =1
2βx+ 1
,1
2βc+ 1
=2β 13β 0 =
13,βc+ 1 = 3/2, c+ 1 = 9/4, c = 5/4
10. f(4) = 15/4, f(3) = 8/3, solve f β²(c) = (15/4β8/3)/1 = 13/12; f β²(x) = 1+1/x2, f β²(c) = 1+1/c2 =13/12, c2 = 12, c = Β±2
β3, but β2
β3 is not in the interval, so c = 2
β3.
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210 Chapter 5
11. f(β5) = 0, f(3) = 4, f β²(x) = β xβ25β x2
, β cβ25β c2
=4β 0
3β (β5) =12, β2c =
β25β c2,
4c2 = 25β c2, c2 = 5, c = ββ5
(we reject c =β5 because it does not satisfy the equation β2c =
β25β c2)
12. f(2) = 1, f(5) = 1/4, f β²(x) = β1/(xβ 1)2, β 1(cβ 1)2 =
1/4β 15β 2 = β1
4, (cβ 1)2 = 4, cβ 1 = Β±2,
c = β1 (reject), or c = 3
13. (a) f(β2) = f(1) = 0The interval is [β2, 1]
(b) c = β1.296
β2
β2 1
(c) x0 = β1, x1 = β1.5, x2 = β1.32, x3 = β1.290, x4 = β1.2885843
14. (a) m =f(β2)β f(1)β2β 1 =
0 + 3β3 = β1 so y + 3 = β(xβ 1), y = βxβ 2
(b) f β²(x) = 3x2 β 4 = β1 has solutions x = Β±1; discard x = 1, so c = β1(c) y β (3) = β(xβ (β1)) or y = βx+ 2(d) 4
β4
β3 2
15. (a) f β²(x) = sec2 x, sec2 c = 0 has no solution (b) tanx is not continuous on [0, Ο]
16. (a) f(β1) = 1, f(8) = 4, f β²(x) = 23xβ1/3
23cβ1/3 =
4β 18β (β1) =
13, c1/3 = 2, c = 8 which is not in (β1, 8).
(b) x2/3 is not differentiable at x = 0, which is in (β1, 8).
17. (a) Two x-intercepts of f determine two solutions a and b of f(x) = 0; by Rolleβs Theorem thereexists a point c between a and b such that f β²(c) = 0, i.e. c is an x-intercept for f β².
(b) f(x) = sinx = 0 at x = nΟ, and f β²(x) = cosx = 0 at x = nΟ + Ο/2, which lies between nΟand (n+ 1)Ο, (n = 0,Β±1,Β±2, . . .)
18.f(x1)β f(x0)x1 β x0
is the average rate of change of y with respect to x on the interval [x0, x1]. By
the Mean-Value Theorem there is a value c in (x0, x1) such that the instantaneous rate of change
f β²(c) =f(x1)β f(x0)x1 β x0
.
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Exercise Set 5.7 211
19. Let s(t) be the position function of the automobile for 0 β€ t β€ 5, then by the Mean-Value Theoremthere is at least one point c in (0, 5) wheresβ²(c) = v(c) = [s(5)β s(0)]/(5β 0) = 4/5 = 0.8 mi/min = 48 mi/h.
20. Let T (t) denote the temperature at time with t = 0 denoting 11 AM, then T (0) = 76 andT (12) = 52.
(a) By the Mean-Value Theorem there is a value c between 0 and 12 such thatT β²(c) = [T (12)β T (0)]/(12β 0) = (52β 76)/(12) = β2β¦ F/h.
(b) Assume that T (t1) = 88β¦F where 0 < t1 < 12, then there is at least one point c in (t1, 12)where T β²(c) = [T (12)βT (t1)]/(12βt1) = (52β88)/(12βt1) = β36/(12βt1). But 12βt1 < 12so T β²(c) < β3β¦F.
21. Let f(t) and g(t) denote the distances from the first and second runners to the starting point,and let h(t) = f(t) β g(t). Since they start (at t = 0) and finish (at t = t1) at the same time,h(0) = h(t1) = 0, so by Rolleβs Theorem there is a time t2 for which hβ²(t2) = 0, i.e. f β²(t2) = gβ²(t2);so they have the same velocity at time t2.
22. f(x) = x β (2 β x) ln(2 β x); solve f(x) = 0 in (0, 1). f(0) = β2 ln 2 < 0; f(1) = 1 > 0. By theIntermediate-Value Theorem (Theorem 2.5.7) there exists x in (0, 1) such that f(x) = 0.
23. (a) By the Constant Difference Theorem f(x)β g(x) = k for some k; since f(x0) = g(x0), k = 0,so f(x) = g(x) for all x.
(b) Set f(x) = sin2 x + cos2 x, g(x) = 1; then f β²(x) = 2 sinx cosx β 2 cosx sinx = 0 = gβ²(x).Since f(0) = 1 = g(0), f(x) = g(x) for all x.
24. (a) By the Constant Difference Theorem f(x) β g(x) = k for some k; since f(x0) β g(x0) = c,k = c, so f(x)β g(x) = c for all x.
(b) Set f(x) = (xβ 1)3, g(x) = (x2 + 3)(xβ 3). Thenf β²(x) = 3(xβ 1)2, gβ²(x) = (x2 + 3) + 2x(xβ 3) = 3x2 β 6x+ 3 = 3(x2 β 2x+ 1) = 3(xβ 1)2,so f β²(x) = gβ²(x) and hence f(x)β g(x) = k. Expand f(x) and g(x) to geth(x) = f(x)β g(x) = (x3 β 3x2 + 3xβ 1)β (x3 β 3x2 + 3xβ 9) = 8.
(c) h(x) = x3 β 3x2 + 3xβ 1β (x3 β 3x2 + 3xβ 9) = 8
25. By the Constant Difference Theorem it follows that f(x) = g(x) + c; since g(1) = 0 and f(1) = 2we get c = 2; f(x) = xex β ex + 2.
26. By the Constant Difference Theorem f(x) = tanβ1 x+ C and
2 = f(1) = tanβ1(1) + C = Ο/4 + C,C = 2β Ο/4, f(x) = tanβ1 x+ 2β Ο/4.
27. (a) If x, y belong to I and x < y then for some c in I,f(y)β f(x)y β x = f β²(c),
so |f(x)βf(y)| = |f β²(c)||xβy| β€M |xβy|; if x > y exchange x and y; if x = y the inequalityalso holds.
(b) f(x) = sinx, f β²(x) = cosx, |f β²(x)| β€ 1 =M , so |f(x)β f(y)| β€ |xβ y| or| sinxβ sin y| β€ |xβ y|.
28. (a) If x, y belong to I and x < y then for some c in I,f(y)β f(x)y β x = f β²(c),
so |f(x)β f(y)| = |f β²(c)||xβ y| β₯M |xβ y|; if x > y exchange x and y; ifx = y the inequality also holds.
(b) If x and y belong to (βΟ/2, Ο/2) and f(x) = tanx, then |f β²(x)| = sec2 x β₯ 1 and| tanxβ tan y| β₯ |xβ y|
(c) y lies in (βΟ/2, Ο/2) if and only if βy does; use Part (b) and replace y with βy
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212 Chapter 5
29. (a) Let f(x) =βx. By the Mean-Value Theorem there is a number c between x and y such thatβ
y ββx
y β x =12βc<
12βxfor c in (x, y), thus
βy ββx <
y β x2βx
(b) multiply through and rearrange to getβxy <
12(x+ y).
30. Suppose that f(x) has at least two distinct real solutions r1 and r2 in I. Thenf(r1) = f(r2) = 0 so by Rolleβs Theorem there is at least one number between r1 and r2 wheref β²(x) = 0, but this contradicts the assumption that f β²(x) = 0, so f(x) = 0 must have fewer thantwo distinct solutions in I.
31. (a) If f(x) = x3+4xβ 1 then f β²(x) = 3x2+4 is never zero, so by Exercise 30 f has at most onereal root; since f is a cubic polynomial it has at least one real root, so it has exactly one realroot.
(b) Let f(x) = ax3 + bx2 + cx+ d. If f(x) = 0 has at least two distinct real solutions r1 and r2,then f(r1) = f(r2) = 0 and by Rolleβs Theorem there is at least one number between r1 andr2 where f β²(x) = 0. But f β²(x) = 3ax2 + 2bx+ c = 0 forx = (β2bΒ±
β4b2 β 12ac)/(6a) = (βbΒ±
βb2 β 3ac)/(3a), which are not real if b2 β 3ac < 0
so f(x) = 0 must have fewer than two distinct real solutions.
32. f β²(x) =12βx,12βc=β4ββ3
4β 3 = 2ββ3. But
14<
12βc<
12β3for c in (3, 4) so
14< 2β
β3 <
12β3, 0.25 < 2β
β3 < 0.29, β1.75 < β
β3 < β1.71, 1.71 <
β3 < 1.75.
33. By the Mean-Value Theorem on the interval [0, x],
tanβ1 xβ tanβ1 0xβ 0 =
tanβ1 x
x=
11 + c2
for c in (0, x), but
11 + x2 <
11 + c2
< 1 for c in (0, x) so1
1 + x2 <tanβ1 x
x< 1,
x
1 + x2 < tanβ1 x < x.
34. (a)d
dx[f2(x) β g2(x)] = 2f(x)f β²(x) β 2g(x)gβ²(x) = 2f(x)g(x) β 2g(x)f(x) = 0, so f2 β g2 is
constant.
(b) f β²(x) = 12 (e
x β eβx) = g(x), gβ²(x) = 12 (e
x + eβx) = f(x)
35. (a)d
dx[f2(x) + g2(x)] = 2f(x)f β²(x) + 2g(x)gβ²(x) = 2f(x)g(x) + 2g(x)[βf(x)] = 0,
so f2(x) + g2(x) is constant.
(b) f(x) = sinx and g(x) = cosx
36. Let h = f β g, then h is continuous on [a, b], differentiable on (a, b), and h(a) = f(a) β g(a) = 0,h(b) = f(b)β g(b) = 0. By Rolleβs Theorem there is some c in (a, b) where hβ²(c) = 0. Buthβ²(c) = f β²(c)β gβ²(c) so f β²(c)β gβ²(c) = 0, f β²(c) = gβ²(c).
37. y
xc
]
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Exercise Set 5.8 213
38. (a) Suppose f β²(x) = 0 more than once in (a, b), say at c1 and c2. Then f β²(c1) = f β²(c2) = 0 andby using Rolleβs Theorem on f β², there is some c between c1 and c2 where f β²β²(c) = 0, whichcontradicts the fact that f β²β²(x) > 0 so f β²(x) = 0 at most once in (a, b).
(b) If f β²β²(x) > 0 for all x in (a, b), then f is concave up on (a, b) and has at most one relativeextremum, which would be a relative minimum, on (a, b).
39. (a) similar to the proof of Part (a) with f β²(c) < 0
(b) similar to the proof of Part (a) with f β²(c) = 0
40. Let x = x0 be sufficiently near x0 so that there exists (by the Mean-Value Theorem) a number c(which depends on x) between x and x0, such that
f(x)β f(x0
xβ x0= f β²(c).
Since c is between x and x0 it follows that
f β²(x0)= limxβx0
f(x)β f(x0)xβ x0
(by definition of derivative)
= limxβx0
f β²(c) (by the Mean-Value Theorem)
= limxβx0
f β²(x) (since lim f β²(x) exists and c is between x and x0).
41. If f is differentiable at x = 1, then f is continuous there;
limxβ1+
f(x) = limxβ1β
f(x) = f(1) = 3, a+ b = 3; limxβ1+
f β²(x) = a and
limxβ1β
f β²(x) = 6 so a = 6 and b = 3β 6 = β3.
42. (a) limxβ0β
f β²(x) = limxβ0β
2x = 0 and limxβ0+
f β²(x) = limxβ0+
2x = 0; f β²(0) does not exist because f is
not continuous at x = 0.
(b) limxβ0β
f β²(x) = limxβ0+
f β²(x) = 0 and f is continuous at x = 0, so f β²(0) = 0;
limxβ0β
f β²β²(x) = limxβ0β
(2) = 2 and limxβ0+
f β²β²(x) = limxβ0+
6x = 0, so f β²β²(0) does not exist.
43. From Section 3.2 a function has a vertical tangent line at a point of its graph if the slopes ofsecant lines through the point approach +β or ββ. Suppose f is continuous at x = x0 andlimxβx+
0
f(x) = +β. Then a secant line through (x1, f(x1)) and (x0, f(x0)), assuming x1 > x0, will
have slopef(x1)β f(x0)x1 β x0
. By the Mean Value Theorem, this quotient is equal to f β²(c) for some
c between x0 and x1. But as x1 approaches x0, c must also approach x0, and it is given thatlimcβx+
0
f β²(c) = +β, so the slope of the secant line approaches +β. The argument can be altered
appropriately for x1 < x0, and/or for f β²(c) approaching ββ.
EXERCISE SET 5.8
1. (a) positive, negative, slowing down (b) positive, positive, speeding up(c) negative, positive, slowing down
2. (a) positive, slowing down (b) negative, slowing down(c) positive, speeding up
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214 Chapter 5
3. (a) left because v = ds/dt < 0 at t0(b) negative because a = d2s/dt2 and the curve is concave down at t0(d2s/dt2 < 0)(c) speeding up because v and a have the same sign(d) v < 0 and a > 0 at t1 so the particle is slowing down because v and a have opposite signs.
4. (a) III(b) I(c) II
5.
t (s)
s (m)
6. (a) when s β₯ 0, so 0 < t < 2 and 4 < t β€ 8 (b) when the slope is zero, at t = 3(c) when s is decreasing, so 0 β€ t < 3
7.
1 2 3 4 5 6
β10
β5
0
5
10
15
t
|v|
6
t
a
β15
15
8. (a) v β (30β 10)/(15β 10) = 20/5 = 4 m/s(b)
t
v
t
a
2525
(1) (2)
9. (a) At 60 mi/h the tangent line seems to pass through the points (0, 20) and (16, 100). Thus the
acceleration would bev1 β v0t1 β t0
8860=100β 2016β 0
8860β 7.3 ft/s2.
(b) The maximum acceleration occurs at maximum slope, so when t = 0.
10. (a) At 60 mi/h the tangent line seems to pass through the points (0, 20) and (17.75, 100). Thusthe acceleration would be (recalling that 60 mi/h is the same as 88 ft/s)v1 β v0t1 β t0
8860=100β 2017.75β 0
8860β 6.6 ft/s2.
(b) The maximum acceleration occurs at maximum slope, so when t = 0.
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Exercise Set 5.8 215
11. (a) t 1 2 3 4 5s 0.71 1.00 0.71 0.00 β0.71v 0.56 0.00 β0.56 β0.79 β0.56a β0.44 β0.62 β0.44 0.00 0.44
(b) to the right at t = 1, stopped at t = 2, otherwise to the left(c) speeding up at t = 3; slowing down at t = 1, 5; neither at t = 2, 4
12. (a) t 1 2 3 4 5s 0.37 2.16 4.03 4.68 4.21v 1.10 2.16 1.34 0 β0.84a 1.84 0 β1.34 β1.17 β0.51
(b) to the right at t = 1, 2, 3, stopped at t = 4, to the left at t = 5(c) speeding up at t = 1, 5; slowing down at t = 3; stopped at t = 4, neither at t = 2
13. (a) v(t) = 3t2 β 6t, a(t) = 6tβ 6(b) s(1) = β2 ft, v(1) = β3 ft/s, speed = 3 ft/s, a(1) = 0 ft/s2
(c) v = 0 at t = 0, 2
(d) for t β₯ 0, v(t) changes sign at t = 2, and a(t) changes sign at t = 1; so the particle is speedingup for 0 < t < 1 and 2 < t and is slowing down for 1 < t < 2
(e) total distance = |s(2)β s(0)|+ |s(5)β s(2)| = | β 4β 0|+ |50β (β4)| = 58 ft
14. (a) v(t) = 4t3 β 8t, a(t) = 12t2 β 8(b) s(1) = 1 ft, v(1) = β4 ft/s, speed = 4 ft/s, a(1) = 0 ft/s2
(c) v = 0 at t = 0,β2
(d) for t β₯ 0, v(t) changes sign at t =β2, and a(t) changes sign at t =
β6/3. The particle is
speeding up for 0 < t <β6/3 and
β2 < t and slowing down for
β6/3 < t <
β2.
(e) total distance = |s(β2)β s(0)|+ |s(5)β s(
β2)| = |0β 4|+ |529β 0| = 533 ft
15. (a) s(t) = 9β 9 cos(Οt/3), v(t) = 3Ο sin(Οt/3), a(t) = Ο2 cos(Οt/3)
(b) s(1) = 9/2 ft, v(1) = 3Οβ3/2 ft/s, speed = 3Ο
β3/2 ft/s, a(1) = Ο2/2 ft/s2
(c) v = 0 at t = 0, 3(d) for 0 < t < 5, v(t) changes sign at t = 3 and a(t) changes sign at t = 3/2, 9/2; so the particle
is speeding up for 0 < t < 3/2 and 3 < t < 9/2 and slowing down for 3/2 < t < 3 and9/2 < t < 5
(e) total distance = |s(3)β s(0)|+ |s(5)β s(3)| = |18β 0|+ |9/2β 18| = 18 + 27/2 = 63/2 ft
16. (a) v(t) =4β t2(t2 + 4)2
, a(t) =2t(t2 β 12)(t2 + 4)3
(b) s(1) = 1/5 ft, v(1) = 3/25 ft/s, speed = 3/25 ft/s, a(1) = β22/125 ft/s2
(c) v = 0 at t = 2
(d) a changes sign at t = 2β3, so the particle is speeding up for 2 < t < 2
β3 and it is slowing
down for 0 < t < 2 and for 2β3 < t
(e) total distance = |s(2)β s(0)|+ |s(5)β s(2)| =β£β£β£β£14 β 0
β£β£β£β£+β£β£β£β£ 529 β 14
β£β£β£β£ = 1958ft
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216 Chapter 5
17. (a) s(t) = (t2 + 8)eβt/3 ft, v(t) =(β 1
3 t2 + 2tβ 8
3
)eβt/3 ft/s, a(t) =
( 19 t
2 β 43 t+
269
)eβt/3 ft/s2
(b) s(1) = 9eβ1/3, v(1) = βeβ1/3, a(1) = 53eβ1/3
(c) v = 0 for t = 2, 4(d) v changes sign at t = 2, 4 and a changes sign at t = 6 Β±
β10, so the particle is speeding up
for 2 < t < 6ββ10 and 4 < t < 6 +
β10, and slowing down for 0 < t < 2, 6β
β10 < t < 4
and t > 6 +β10
(e) total distance = |s(2)β s(0)|+ |s(4)β s(2)|+ |s(4)|= |12eβ2/3 β 8eβ1/3|+ |24eβ4/3 β 12eβ2/3|+ |33eβ5/3 β 24eβ4/3| β 8.33 ft
18. (a) s(t) = 14 t
2 β ln(t+ 1), v(t) = t2 + tβ 22(t+ 1)
, a(t) =t2 + 2t+ 32(t+ 1)2
(b) s(1) = 14 β ln 2 ft, v(1) = 0 ft/s, speed = 0 ft/s, a(1) =
34 ft/s
2
(c) v = 0 for t = 1(d) v changes sign at t = 1 and a does not change sign, so the particle is slowing down for
0 < t < 1 and speeding up for t > 1(e) total distance = |s(5) β s(1)| + |s(1) β s(0)| = |25/4 β ln 6 β (1/4 β ln 2)| + |1/4 β ln 2| =
23/4 + ln(2/3) β 5.345 ft
19. v(t) =5β t2(t2 + 5)2
, a(t) =2t(t2 β 15)(t2 + 5)3
0.25
00 20
s(t)
0.2
β0.05
0 20
v(t)
0.01
β0.15
0 10
a(t)
(a) v = 0 at t =β5 (b) s =
β5/10 at t =
β5
(c) a changes sign at t =β15, so the particle is speeding up for
β5 < t <
β15 and slowing down
for 0 < t <β5 and
β15 < t
20. v(t) = (1β t)eβt, a(t) = (tβ 2)eβt
2
00 2
s(t)
1
β0.2
0 2
v(t)
0.5
β2
0 3
a(t)
(a) v = 0 at t = 1 (b) s = 1/e at t = 1(c) a changes sign at t = 2, so the particle is speeding up for 1 < t < 2 and slowing down for
0 < t < 1 and 2 < t
21. s = β4t+ 3v = β4a = 0
30
Not speeding up,not slowing down
t = 3/4
β3
t = 3/2 t = 0 s
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Exercise Set 5.8 217
22. s = 5t2 β 20v = 10ta = 10
200
Speeding up
β20
t = 0 s
23. s = t3 β 9t2 + 24tv = 3(tβ 2)(tβ 4)a = 6(tβ 3)
2018160
t = 2 (Stopped)t = 0
(Stopped) t = 4t = 3
s
Speeding up
Slowing down
Slowing down
24. s = t3 β 6t2 + 9t+ 1v = 3(tβ 1)(tβ 3)a = 6(tβ 2)
531
t = 1 (Stopped)
t = 0
(Stopped) t = 3 t = 2
s
Speeding up
Slowing down
25. s = 16teβt2/8
v = (β4t2 + 16)eβt2/8a = t(β12 + t2)eβt2/8
t = 2β3
0 10 20
Speeding up
Slowing down
t = +β
t = 0t = 2 (Stopped)
s
26. s = t+ 25/(t+ 2);v = (tβ 3)(t+ 7)/(t+ 2)2a = 50/(t+ 2)3
12.5108s
Slowing down
t = 3t = 0
Speeding up
27. s ={cos t, 0 β€ t β€ 2Ο1, t > 2Ο
v ={β sin t, 0 β€ t β€ 2Ο0, t > 2Ο
a ={β cos t, 0 β€ t < 2Ο0, t > 2Ο
10-1s
Speeding up
t = ct = c/2
t = 3c/2 t = 2ct = 0
(Stoppedpermanently)
Slowing down
28. s ={
2t(tβ 2)2, 0 β€ t β€ 313β 7(tβ 4)2, t > 3
v ={6t2 β 16t+ 8, 0 β€ t β€ 3β14t+ 56, t > 3
a ={12tβ 16, 0 β€ t < 3β14, t > 3 1364/27 60
t = 4 (Stopped)
t = 2/3 (Stopped)t = 0
t = 2(Stopped)
t = 3
t = 4/3
s
Speeding up
Slowing down
29. (a) v = 10t β 22, speed = |v| = |10t β 22|. d|v|/dt does not exist at t = 2.2 which is the onlycritical point. If t = 1, 2.2, 3 then |v| = 12, 0, 8. The maximum speed is 12 ft/s.
(b) the distance from the origin is |s| = |5t2 β 22t| = |t(5t β 22)|, but t(5t β 22) < 0 for1 β€ t β€ 3 so |s| = β(5t2 β 22t) = 22tβ 5t2, d|s|/dt = 22β 10t, thus the only critical point ist = 2.2. d2|s|/dt2 < 0 so the particle is farthest from the origin when t = 2.2. Its position iss = 5(2.2)2 β 22(2.2) = β24.2.
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218 Chapter 5
30. v = β 200t(t2 + 12)2
, speed = |v| = 200t(t2 + 12)2
for t β₯ 0. d|v|dt
=600(4β t2)(t2 + 12)3
= 0 when t = 2, which
is the only critical point in (0,+β). By the first derivative test there is a relative maximum, andhence an absolute maximum, at t = 2. The maximum speed is 25/16 ft/s to the left.
31. s = ln(3t2 β 12t+ 13)
v =6tβ 12
3t2 β 12t+ 13
a = β6(3t2 β 12t+ 11)
(3t2 β 12t+ 13)2
(a) a = 0 when t = 2Β±β3/3;
s(2ββ3/3) = ln 2; s(2 +
β3/3) = ln 2;
v(2ββ3/3 = β
β3; v(2 +
β3) =
β3
(b) v = 0 when t = 2s(2) = 0; a(2) = 6
32. s = t3 β 6t2 + 1, v = 3t2 β 12t, a = 6tβ 12.(a) a = 0 when t = 2; s = β15, v = β12.(b) v = 0 when 3t2 β 12t = 3t(t β 4) = 0, t = 0 or t = 4. If t = 0, then s = 1 and a = β12; if
t = 4, then s = β31 and a = 12.
33. (a) 1.5
00 5
(b) v =2tβ2t2 + 1
, limtβ+β
v =2β2=β2
34. (a) a =dv
dt=dv
ds
ds
dt= vdv
dsbecause v =
ds
dt
(b) v =3
2β3t+ 7
=32s;dv
ds= β 3
2s2; a = β 9
4s3= β9/500
35. (a) s1 = s2 if they collide, so 12 t
2 β t + 3 = β 14 t
2 + t + 1, 34 t
2 β 2t + 2 = 0 which has no realsolution.
(b) Find the minimum value of D = |s1 β s2| =β£β£ 3
4 t2 β 2t+ 2
β£β£. From Part (a), 34 t
2 β 2t + 2
is never zero, and for t = 0 it is positive, hence it is always positive, so D = 34 t
2 β 2t + 2.dD
dt= 3
2 tβ 2 = 0 when t =43 .d2D
dt2> 0 so D is minimum when t = 4
3 , D =23 .
(c) v1 = t β 1, v2 = β12t + 1. v1 < 0 if 0 β€ t < 1, v1 > 0 if t > 1; v2 < 0 if t > 2, v2 > 0 if
0 β€ t < 2. They are moving in opposite directions during the intervals 0 β€ t < 1 and t > 2.
36. (a) sA β sB = 20β 0 = 20 ft(b) sA = sB , 15t2 +10t+20 = 5t2 +40t, 10t2 β 30t+20 = 0, (tβ 2)(tβ 1) = 0, t = 1 or t = 2 s.(c) vA = vB , 30t+10 = 10t+40, 20t = 30, t = 3/2 s. When t = 3/2, sA = 275/4 and sB = 285/4
so car B is ahead of car A.
37. r(t) =βv2(t), rβ²(t) = 2v(t)vβ²(t)/[2
βv(t)] = v(t)a(t)/|v(t)| so rβ²(t) > 0 (speed is increasing) if v
and a have the same sign, and rβ²(t) < 0 (speed is decreasing) if v and a have opposite signs.If v(t) > 0 then r(t) = v(t) and rβ²(t) = a(t), so if a(t) > 0 then the particle is speeding up and aand v have the same sign; if a(t) < 0, then the particle is slowing down, and a and v have oppositesigns.
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Review Exercises, Chapter 5 219
If v(t) < 0 then r(t) = βv(t), rβ²(t) = βa(t), and if a(t) > 0 then the particle is speeding up and aand v have opposite signs; if a(t) < 0 then the particle is slowing down and a and v have the samesign.
REVIEW EXERCISES, CHAPTER 5
3. f β²(x) = 2xβ 5f β²β²(x) = 2
(a) [5/2,+β) (b) (ββ, 5/2](c) (ββ,+β) (d) none(e) none
4. f β²(x) = 4x(x2 β 4)f β²β²(x) = 12(x2 β 4/3)
(a) [β2, 0], [2,+β) (b) (ββ,β2], [0, 2](c) (ββ,β2/
β3), (2/
β3,+β) (d) (β2/
β3, 2/β3)
(e) β2/β3, 2/
β3
5. f β²(x) =4x
(x2 + 2)2
f β²β²(x) = β4 3x2 β 2
(x2 + 2)3
(a) [0,+β) (b) (ββ, 0](c) (β
β2/3,
β2/3) (d) (ββ,β
β2/3), (
β2/3,+β)
(e) ββ2/3,
β2/3
6. f β²(x) = 13 (x+ 2)
β2/3
f β²β²(x) = β 29 (x+ 2)
β5/3
(a) (ββ,+β) (b) none(c) (ββ,β2) (d) (β2,+β)(e) β2
7. f β²(x) =4(x+ 1)3x2/3
f β²β²(x) =4(xβ 2)9x5/3
(a) [β1,+β) (b) (ββ,β1](c) (ββ, 0), (2,+β) (d) (0, 2)(e) 0, 2
8. f β²(x) =4(xβ 1/4)3x2/3
f β²β²(x) =4(x+ 1/2)9x5/3
(a) [1/4,+β) (b) (ββ, 1/4](c) (ββ,β1/2), (0,+β) (d) (β1/2, 0)(e) β1/2, 0
9. f β²(x) = β 2xex2
f β²β²(x) =2(2x2 β 1)ex2
(a) (ββ, 0] (b) [0,+β)(c) (ββ,β
β2/2), (
β2/2,+β) (d) (β
β2/2,β2/2)
(e) ββ2/2,β2/2
10. f β²(x) =2x
1 + x4
f β²β²(x) = β2(β1 + 3x4)
(1 + x4)2
(a) [0,+β) (b) (ββ, 0](c)(β1/31/4, 1/31/4) (d)(ββ,β1/31/4), (1/31/4,+β)(e)β1/31/4, 1/31/4
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220 Chapter 5
11. f β²(x) = β sinxf β²β²(x) = β cosx
(a) [Ο, 2Ο] (b) [0, Ο](c) (Ο/2, 3Ο/2) (d) (0, Ο/2), (3Ο/2, 2Ο)(e) Ο/2, 3Ο/2
1
β1
0 o
12. f β²(x) = sec2 xf β²β²(x) = 2 sec2 x tanx
(a) (βΟ/2, Ο/2) (b) none(c) (0, Ο/2) (d) (βΟ/2, 0)(e) 0
10
β10
^ 6
13. f β²(x) = cos 2xf β²β²(x) = β2 sin 2x
(a) [0, Ο/4], [3Ο/4, Ο] (b) [Ο/4, 3Ο/4]
(c) (Ο/2, Ο) (d) (0, Ο/2)
(e) Ο/2
0.5
β0.5
0 p
14. f β²(x) = β2 cosx sinxβ 2 cosx = β2 cosx(1 + sinx)f β²β²(x) = 2 sinx (sinx+ 1)β 2 cos2 x = 2 sinx(sinx+ 1)β 2 + 2 sin2 x = 4(1 + sinx)(sinxβ 1/2)Note: 1 + sinx β₯ 0
(a) [Ο/2, 3Ο/2] (b) [0, Ο/2], [3Ο/2, 2Ο]
(c) (Ο/6, 5Ο/6) (d) (0, Ο/6), (5Ο/6, 2Ο)
(e) Ο/6, 5Ο/6
2
β2
0 o
15. (a)
2
4
x
y (b)
2
4
x
y (c)
2
4
x
y
16. (a) p(x) = x3 β x (b) p(x) = x4 β x2
(c) p(x) = x5 β x4 β x3 + x2 (d) p(x) = x5 β x3
17. f β²(x) = 2ax + b; f β²(x) > 0 or f β²(x) < 0 on [0,+β) if f β²(x) = 0 has no positive solution, so thepolynomial is always increasing or always decreasing on [0,+β) provided βb/2a β€ 0.
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Review Exercises, Chapter 5 221
18. f β²(x) = 3ax2 + 2bx+ c; f β²(x) > 0 or f β²(x) < 0 on (ββ,+β) if f β²(x) = 0 has no real solutions sofrom the quadratic formula (2b)2 β 4(3a)c < 0, 4b2 β 12ac < 0, b2 β 3ac < 0. If b2 β 3ac = 0, thenf β²(x) = 0 has only one real solution at, say, x = c so f is always increasing or always decreasingon both (ββ, c] and [c,+β), and hence on (ββ,+β) because f is continuous everywhere. Thusf is always increasing or decreasing if b2 β 3ac β€ 0.
19. The maximum increase in yseems to occur near x = β1, y = 1/4.
β2 β1 1 2
0.25
0.5
x
y
20. y =ax
1 + ax+k
yβ² =ax ln a
(1 + ax+k)2
yβ²β² = βax(ln a)2(ax+k β 1)(1 + ax+k)3
yβ²β² = 0 when x = βk and yβ²β² changes sign there
22. (a) False; an example is y =x3
3β x
2
2on [β2, 2]; x = 0 is a relative maximum and x = 1 is a
relative minimum, but y = 0 is not the largest value of y on the interval, nor is y = β16the
smallest.(b) true(c) False; for example y = x3 on (β1, 1) which has a critical number but no relative extrema
24. (a) f β²(x) = 3x2 + 6xβ 9 = 3(x+ 3)(xβ 1), f β²(x) = 0 when x = β3, 1 (stationary points).(b) f β²(x) = 4x(x2 β 3), f β²(x) = 0 when x = 0, Β±
β3 (stationary points).
25. (a) f β²(x) = (2β x2)/(x2 + 2)2, f β²(x) = 0 when x = Β±β2 (stationary points).
(b) f β²(x) = 8x/(x2 + 1)2, f β²(x) = 0 when x = 0 (stationary point).
26. (a) f β²(x) =4(x+ 1)3x2/3 , f β²(x) = 0 when x = β1 (stationary point), f β²(x) does not exist when
x = 0.
(b) f β²(x) =4(xβ 3/2)3x2/3 , f β²(x) = 0 when x = 3/2 (stationary point), f β²(x) does not exist when
x = 0.
27. (a) f β²(x) =7(xβ 7)(xβ 1)
3x2/3 ; critical numbers at x = 0, 1, 7;
neither at x = 0, relative maximum at x = 1, relative minimum at x = 7 (First DerivativeTest)
(b) f β²(x) = 2 cosx(1 + 2 sinx); critical numbers at x = Ο/2, 3Ο/2, 7Ο/6, 11Ο/6;relative maximum at x = Ο/2, 3Ο/2, relative minimum at x = 7Ο/6, 11Ο/6
(c) f β²(x) = 3β 3βxβ 12
; critical numbers at x = 5; relative maximum at x = 5
28. (a) f β²(x) =xβ 918x3/2 , f
β²β²(x) =27β x36x5/2 ; critical number at x = 9;
f β²β²(9) > 0, relative minimum at x = 9
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222 Chapter 5
(b) f β²(x) = 2x3 β 4x2 , f β²β²(x) = 2
x3 + 8x3 ;
critical number at x = 41/3, f β²β²(41/3) > 0, relative minimum at x = 41/3
(c) f β²(x) = sinx(2 cosx+ 1), f β²β²(x) = 2 cos2 xβ 2 sin2 x+ cosx; critical numbers atx = 2Ο/3, Ο, 4Ο/3; f β²β²(2Ο/3) < 0, relative maximum at x = 2Ο/3; f β²β²(Ο) > 0, relativeminimum at x = Ο; f β²β²(4Ο/3) < 0, relative maximum at x = 4Ο/3
29. limxβββ
f(x) = +β, limxβ+β
f(x) = +β
f β²(x) = x(4x2 β 9x+ 6), f β²β²(x) = 6(2xβ 1)(xβ 1)relative minimum at x = 0,points of inflection when x = 1/2, 1,no asymptotes
y
x1
2
3
4
1 2
(0,1))(
(1,2)12 ,23
16
30. limxβββ
f(x) = ββ, limxβ+β
f(x) = +β
f(x) = x3(xβ 2)2, f β²(x) = x2(5xβ 6)(xβ 2),f β²β²(x) = 4x(5x2 β 12x+ 6)
critical numbers at x = 0,8Β± 2
β31
5
relative maximum at x =8β 2
β31
5β β0.63
relative minimum at x =8 + 2
β31
5β 3.83
points of inflection at x = 0,6Β±β66
5β 0,β0.42, 2.82
no asymptotes
y
x
β200
β100
β1 1 2 3 4
(0,0)(β0.42, 0.16)
(2.82, β165.00)(3.83, β261.31)
(β0.63, 0.27)
31. limxβΒ±β
f(x) doesnβt exist
f β²(x) = 2x sec2(x2 + 1),f β²β²(x) = 2 sec2(x2 + 1)
[1 + 4x2 tan(x2 + 1)
]critical number at x = 0; relative minimum at x = 0point of inflection when 1 + 4x2 tan(x2 + 1) = 0
vertical asymptotes at x = Β±βΟ(n+ 1
2 )β 1, n = 0, 1, 2, . . .
y
x
β4
β2
2
4
β2 β1 1 2
32. limxβββ
f(x) = ββ, limxβ+β
f(x) = +β
f β²(x) = 1 + sinx, f β²β²(x) = cosxcritical numbers at x = 2nΟ + Ο/2, n = 0,Β±1,Β±2, . . .,no extrema because f β² β₯ 0 and by Exercise 59 of Section 5.1,f is increasing on (ββ,+β)inflections points at x = nΟ + Ο/2, n = 0,Β±1,Β±2, . . .no asymptotes
y
x
β6
β4β2
2
4
-c c
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Review Exercises, Chapter 5 223
33. f β²(x) = 2x(x+ 5)
(x2 + 2x+ 5)2, f β²β²(x) = β22x
3 + 15x2 β 25(x2 + 2x+ 5)3
critical numbers at x = β5, 0;relative maximum at x = β5,relative minimum at x = 0points of inflection at x = β7.26,β1.44, 1.20horizontal asymptote y = 1 as xβ Β±β
y
x0.2
0.4
0.6
0.8
1
β20 β10 10 20
34. f β²(x) = 33x2 β 25x4 , f β²β²(x) = β63x
2 β 50x5
critical numbers at x = Β±5β3/3;
relative maximum at x = β5β3/3,
relative minimum at x = +5β3/3
inflection points at x = Β±5β2/3
horizontal asymptote of y = 0 as xβ Β±β,vertical asymptote x = 0
y
x
β5
5
β4
35. limxβββ
f(x) = +β, limxβ+β
f(x) = ββ
f β²(x) ={x
β2x if{x β€ 0x > 0
critical number at x = 0, no extremainflection point at x = 0 (f changes concavity)no asymptotes
x
y
β2
1
2
β2 1
36. f β²(x) =5β 3x
3(1 + x)1/3(3β x)2/3 ,
f β²β²(x) =β32
9(1 + x)4/3(3β x)5/3
critical number at x = 5/3;relative maximum at x = 5/3cusp at x = β1;point of inflection at x = 3oblique asymptote y = βx as xβ Β±β
y
x
β3
β1
2
4
β4 β2 2
37. f β²(x) = 3x2 + 5; no relative extrema because there are no critical numbers.
38. f β²(x) = 4x(x2 β 1); critical numbers x = 0, 1,β1f β²β²(x) = 12x2 β 4; f β²β²(0) < 0, f β²β²(1) > 0, f β²β²(β1) > 0relative minimum of 6 at x = 1,β1, relative maximum of 7 at x = 0
39. f β²(x) = 45xβ1/5; critical number x = 0; relative minimum of 0 at x = 0 (first derivative test)
40. f β²(x) = 2 + 23xβ1/3; critical numbers x = 0,β1/27
relative minimum of 0 at x = 0, relative maximum of 1/27 at x = β1/27
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224 Chapter 5
41. f β²(x) = 2x/(x2 + 1)2; critical number x = 0; relative minimum of 0 at x = 0
42. f β²(x) = 2/(x+ 2)2; no critical numbers (x = β2 is not in the domain of f) no relative extrema
43. f β²(x) = 2x/(1 + x2); critical point at x = 0; relative minimum of 0 at x = 0 (first derivative test)
44. f β²(x) = x(2 + x)ex; critical points at x = 0,β2; relative minimum of 0 at x = 0 and relativemaximum of 4/e2 at x = β2 (first derivative test)
45. (a) 40
β40
β5 5
(b) f β²(x) = x2 β 1400
, f β²β²(x) = 2x
critical points at x = Β± 120;
relative maximum at x = β 120,
relative minimum at x =120
(c) The finer details can be seen whengraphing over a much smaller x-window.
0.0001
β0.0001
β0.1 0.1
46. (a) 200
β200
β5 5
(b) critical points at x = Β±β2, 3
2 , 2;
relative maximum at x = ββ2,
relative minimum at x =β2,
relative maximum at x = 32 ,
relative minimum at x = 2
(c) 10
β4
β2.2 3.5
β2.909
β2.9121.3 1.6
47. (a) 6
β6
β5 5
(b) Divide y = x2 + 1 into y = x3 β 8to get the asymptote ax+ b = x
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Review Exercises, Chapter 5 225
48. cosx β (sin y)dydx
= 2dy
dx;dy
dx= 0 when cosx = 0. Use the first derivative test:
dy
dx=
cosx2 + sin y
and 2+ sin y > 0, so critical points when cosx = 0, relative maxima when x = 2nΟ+ Ο/2, relativeminima when x = 2nΟ β Ο/2, n = 0,Β±1,Β±2, . . .
49. f(x) =(2xβ 1)(x2 + xβ 7)(2xβ 1)(3x2 + xβ 1) =
x2 + xβ 73x2 + xβ 1 , x = 1/2
horizontal asymptote: y = 1/3,vertical asymptotes: x = (β1Β±
β13)/6
y
x
β5
5
β4 2 4
50. (a) f(x) =(xβ 2)(x2 + x+ 1)(x2 β 2)(xβ 2)(x2 β 2)2(x2 + 1)
=x2 + x+ 1
(x2 β 2)(x2 + 1)
(b)
β1 1
5
x
y
x = Β€x = βΒ€
53. (a) If f has an absolute extremum at a point of (a, b) then it must, by Theorem 5.4.3, be at acritical point of f ; since f is differentiable on (a, b) the critical point is a stationary point.
(b) It could occur at a critical point which is not a stationary point: for example, f(x) = |x| on[β1, 1] has an absolute minimum at x = 0 but is not differentiable there.
54. (a) f β²(x) = β1/x2 = 0, no critical points; by inspection M = β1/2 at x = β2; m = β1 atx = β1
(b) f β²(x) = 3x2 β 4x3 = 0 at x = 0, 3/4; f(β1) = β2, f(0) = 0, f(3/4) = 27/256,f(3/2) = β27/16, so m = β2 at x = β1, M = 27/256 at x = 3/4
(c) f β²(x) = 1β sec2 x, f β²(x) = 0 for x in (βΟ/4, Ο/4) when x = 0; f(βΟ/4) = 1β Ο/4, f(0) = 0,f(Ο/4) = Ο/4β 1 so the maximum value is 1β Ο/4 at x = βΟ/4 and the minimum value isΟ/4β 1 at x = Ο/4.
(d) critical point at x = 2; m = β3 at x = 3, M = 0 at x = 2
55. (a) f β²(x) = 2xβ 3; critical point x = 3/2. Minimum value f(3/2) = β13/4, no maximum.(b) No maximum or minimum because lim
xβ+βf(x) = +β and lim
xβββf(x) = ββ.
(c) limxβ0+
f(x) = limxβ+β
f(x) = +β and f β²(x) =ex(xβ 2)x3 , stationary point at x = 2; by
Theorem 5.4.4 f(x) has an absolute minimum at x = 2, and m = e2/4.(d) f β²(x) = (1 + lnx)xx, critical point at x = 1/e; lim
xβ0+f(x) = lim
xβ0+ex ln x = 1,
limxβ+β
f(x) = +β; no absolute maximum, absolute minimum m = eβ1/e at x = 1/e
56. (a) f β²(x) = 10x3(x β 2), critical points at x = 0, 2; limxβ3β
f(x) = 88, so f(x) has no maximum;
m = β9 at x = 2(b) lim
xβ2βf(x) = +β so no maximum; f β²(x) = 1/(xβ2)2, so f β²(x) is never zero, thus no minimum
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226 Chapter 5
(c) f β²(x) = 23β x2
(x2 + 3)2, critical point at x =
β3. Since lim
xβ0+f(x) = 0, f(x) has no minimum,
and M =β3/3 at x =
β3.
(d) f β²(x) =x(7xβ 12)3(xβ 2)2/3 , critical points at x = 12/7, 2; m = f(12/7) =
14449
(β27
)1/3
β β1.9356
at x = 12/7, M = 9 at x = 3
57. (a) (x2 β 1)2 can never be less than zero because it is thesquare of x2 β 1; the minimum value is 0 for x = Β±1,no maximum because lim
xβ+βf(x) = +β.
10
0β2 2
(b) f β²(x) = (1β x2)/(x2 +1)2; critical point x = 1. Maxi-mum value f(1) = 1/2, minimum value 0 because f(x)is never less than zero on [0,+β) and f(0) = 0.
0.5
00 20
(c) f β²(x) = 2 secx tanx β sec2 x = (2 sinx β 1)/ cos2 x,f β²(x) = 0 for x in (0, Ο/4) when x = Ο/6; f(0) = 2,f(Ο/6) =
β3, f(Ο/4) = 2
β2β1 so the maximum value
is 2 at x = 0 and the minimum value isβ3 at x = Ο/6.
2
1.50 3
(d) f β²(x) = 1/2 + 2x/(x2 + 1),
f β²(x) = 0 on [β4, 0] for x = β2Β±β3
if x = β2ββ3,β2 +
β3 then
f(x) = β1ββ3/2 + ln 4 + ln(2 +
β3) β 0.84,
β1 +β3/2 + ln 4 + ln(2β
β3) β β0.06,
absolute maximum at x = β2ββ3,
absolute minimum at x = β2 +β3
1
β0.5
β4 0
58. Let f(x) = sinβ1 x β x for 0 β€ x β€ 1. f(0) = 0 and f β²(x) = 1β1β x2
β 1 = 1ββ1β x2
β1β x2
. Note
thatβ1β x2 β€ 1, so f β²(x) β₯ 0. Thus we know that f is increasing. Since f(0) = 0, it follows that
f(x) β₯ 0 for 0 β€ x β€ 1.
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Review Exercises, Chapter 5 227
59. (a) 2.1
β0.5
β10 10
(b) minimum: (β2.111985, β0.355116)maximum: (0.372591, 2.012931)
.
60. Let k be the amount of light admitted per unit area of clear glass. The total amount of lightadmitted by the entire window is
T = k Β· (area of clear glass) + 12k Β· (area of blue glass) = 2krh+ 1
4Οkr2.
But P = 2h+ 2r + Οr which gives 2h = P β 2r β Οr so
T = kr(P β 2r β Οr) + 14Οkr2 = k
[Pr β
(2 + Ο β Ο
4
)r2]
= k[Pr β 8 + 3Ο
4r2]for 0 < r <
P
2 + Ο,
dT
dr= k
(P β 8 + 3Ο
2r
),dT
dr= 0 when r =
2P8 + 3Ο
.
This is the only critical point and d2T/dr2 < 0 there so the most light is admitted whenr = 2P/(8 + 3Ο) ft.
61. If one corner of the rectangle is at (x, y) with x > 0, y > 0, then A = 4xy, y = 3β1β (x/4)2,
A = 12xβ1β (x/4)2 = 3x
β16β x2,
dA
dx= 6
8β x2β16β x2
, critical point at x = 2β2. Since A = 0
when x = 0, 4 and A > 0 otherwise, there is an absolute maximum A = 24 at x = 2β2.
62. (a) y x
β2
β1.5
β1
β0.5
0.2 0.6 1
(b) The distance between the boat and the origin isβx2 + y2, where y = (x10/3 β 1)/(2x2/3).
The minimum distance is 0.8247 mi when x = 0.6598 mi. The boat gets swept downstream.
63. V = x(12β 2x)2 for 0 β€ x β€ 6;dV/dx = 12(xβ 2)(xβ 6), dV/dx = 0when x = 2 for 0 < x < 6. If x = 0, 2, 6then V = 0, 128, 0 so the volume is largestwhen x = 2 in.
x
xx
x
x
x
x
x
12
12
12 β 2x
12 β 2x
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228 Chapter 5
65. x = β2.11491, 0.25410, 1.86081 66. x = 2.3561945
67. At the point of intersection, x3 = 0.5xβ 1,x3 β 0.5x+ 1 = 0. Let f(x) = x3 β 0.5x+ 1. Bygraphing y = x3 and y = 0.5x β 1 it is evident thatthere is only one point of intersection and it occursin the interval [β2,β1]; note that f(β2) < 0 andf(β1) > 0. f β²(x) = 3x2 β 0.5 so
xn+1 = xn βx3n β 0.5x+ 13x2
n β 0.5;
x1 = β1, x2 = β1.2,x3 = β1.166492147, . . . ,x5 = x6 = β1.165373043
2
β2
β2 2
68. Solve Οβ 0.0167 sinΟ = 2Ο(90)/365 to get Ο = 1.565978 sor = 150Γ 106(1β 0.0167 cosΟ) = 149.988Γ 106 km.
69. Solve Οβ 0.0934 sinΟ = 2Ο(1)/1.88 to get Ο = 3.325078 sor = 228Γ 106(1β 0.0934 cosΟ) = 248.938Γ 106 km.
70. Yes; by the Mean-Value Theorem there is a point c in (a, b) such that f β²(c) =f(b)β f(a)bβ a = 0.
71. (a) yes; f β²(0) = 0
(b) no, f is not differentiable on (β1, 1)(c) yes, f β²(
βΟ/2) = 0
72. (a) no, f is not differentiable on (β2, 2)
(b) yes,f(3)β f(2)3β 2 = β1 = f β²(1 +
β2)
(c) limxβ1β
f(x) = 2, limxβ1+
f(x) = 2 so f is continuous on [0, 2]; limxβ1β
f β²(x) = limxβ1β
β2x = β2 and
limxβ1+
f β²(x) = limxβ1+
(β2/x2) = β2, so f is differentiable on (0, 2);
andf(2)β f(0)2β 0 = β1 = f β²(
β2).
73. f(x) = x6β 2x2+x satisfies f(0) = f(1) = 0, so by Rolleβs Theorem f β²(c) = 0 for some c in (0, 1).
74. If f β²(x) = gβ²(x), then f(x) = g(x) + k. Let x = 1,f(1) = g(1) + k = (1)3 β 4(1) + 6 + k = 3 + k = 2, so k = β1. f(x) = x3 β 4x+ 5.
75. (a) If a = k, a constant, then v = kt+ b where b is constant;so the velocity changes sign at t = βb/k.
v
t
bβ b / k
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Review Exercises, Chapter 5 229
(b) Consider the equation s = 5β t3/6, v = βt2/2, a = βt.Then for t > 0, a is decreasing and av > 0, so theparticle is speeding up.
vt
76. s(t) = t/(2t2 + 8), v(t) = (4β t2)/2(t2 + 4)2, a(t) = t(t2 β 12)/(t2 + 4)3
(a) 0.20
β0.05
0
0.15
β0.10
0
0.25
00 20
20
20
s (t) v(t) a (t)
(b) v changes sign at t = 2
(c) s = 1/8, v = 0, a = β1/32
(d) a changes sign at t = 2β3, so the particle is speeding up for 2 < t < 2
β3, and it is slowing
down for 0 < t < 2 and 2β3 < t
(e) v(0) = 1/5, limtβ+β
v(t) = 0, v(t) has one t-intercept at t =β5 and v(t) has one critical
point at t =β15. Consequently the maximum velocity occurs when t = 0 and the minimum
velocity occurs when t =β15.
77. (a) v = β2 t(t4 + 2t2 β 1)(t4 + 1)2
, a = 23t8 + 10t6 β 12t4 β 6t2 + 1
(t4 + 1)3
(b) s
t
1
2
v
t
β0.2
0.2
2
a
t1
2
(c) It is farthest from the origin at approximately t = 0.64 (when v = 0) and s = 1.2(d) Find t so that the velocity v = ds/dt > 0. The particle is moving in the positive direction
for 0 β€ t β€ 0.64 s.(e) It is speeding up when a, v > 0 or a, v < 0, so for 0 β€ t < 0.36 and 0.64 < t < 1.1, otherwise
it is slowing down.(f) Find the maximum value of |v| to obtain: maximum speed = 1.05 m/s when t = 1.10 s.
78. No; speeding up means the velocity and acceleration have the same sign, i.e. av > 0; the velocity isincreasing when the acceleration is positive, i.e. a > 0. These are not the same thing. An exampleis s = tβ t2 at t = 1, where v = β1 and a = β2, so av > 0 but a < 0.