Calculus one and several variables 10E Salas solutions manual ch04

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU

JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57

118 SECTION 4.1

CHAPTER 4

SECTION 4.1

1. f is differentiable on (0, 1), continuous on [0, 1]; and f(0) = f(1) = 0.

f ′(c) = 3c2 − 1; 3c2 − 1 = 0 =⇒ c =√

33

(−

√3

3/∈ (0, 1)

)

2. f is differentiable on (−2, 2), continuous on [−2, 2]; and f(−2) = f(2) = 0.

f ′(c) = 4c3 − 4c; 4c(c2 − 1) = 0 =⇒ c = 0, ±1

3. f is differentiable on (0, 2π), continuous on [0, 2π]; and f(0) = f(2π) = 0.

f ′(c) = 2 cos 2c; 2 cos 2c = 0 =⇒ 2c =π

2+ nπ, and c =

π

4+

nπ

2, n = 0, ±1, ±2 . . .

Thus, c =π

4,

3π4,

5π4,

7π4

4. f is differentiable on (0, 8), continuous on [0, 8]; and f(0) = f(8) = 0.

f ′(c) = 23 c

−1/3 − 23 c

−2/3 =23c1/3 − 1c2/3

f ′(c) = 0 =⇒ c = 1.

5. f ′(c) = 2c,f(b) − f(a)

b− a=

4 − 12 − 1

= 3; 2c = 3 =⇒ c = 3/2

6. f ′(c) =3

2√c− 4,

f(b) − f(a)b− a

=−10 − (−1)

4 − 1= −3;

32√c− 4 = −3 =⇒ c = 9/4

7. f ′(c) = 3c2,f(b) − f(a)

b− a=

27 − 13 − 1

= 13; 3c2 = 13 =⇒ c =13

√39

(−1

3

√39 is not in [a, b]

)

8. f ′(c) = 23 c

−1/3,f(b) − f(a)

b− a=

4 − 18 − 1

=37; 2

3 c−1/3 = 3

7 =⇒ c =(14)3

93

9. f ′(c) =−c√1 − c2

,f(b) − f(a)

b− a=

0 − 11 − 0

= −1;−c√1 − c2

= −1 =⇒ c =12

√2

(−12

√2 is not in [a, b])

10. f ′(c) = 3c2 − 3,f(b) − f(a)

b− a=

−2 − 21 − (−1)

= −2; 3c2 − 3 = −2 =⇒ c = ±√

33

11. f is continuous on [−1, 1], differentiable on (−1, 1) and f(−1) = f(1) = 0.

f ′(x) =−x(5 − x2)

(3 + x2)2√

1 − x2, f ′(c) = 0 for c in (−1, 1) implies c = 0.

12. (a) f ′(x) = 23 x

−1/3 =2

3x1/3�= 0 for all x ∈ (−1, 1).

(b) f ′(0) does not exist. Therefore, f is not differentiable on (−1, 1).



SECTION 4.1 119

13. No. By the mean-value theorem there exists at least one number c ∈ (0, 2) such that

f ′(c) =f(2) − f(0)

2 − 0=

32> 1.

14. No, by Rolle’s theorem: f(2) = f(3) = 1 but there is no value c ∈ (2, 3) such that

f ′(c) = 0.

15. By the mean-value theorem there is a number c ∈ (2, 6) such that

f(6) − f(2) = f ′(c)(6 − 2) = f ′(c)4.

Since 1 ≤ f ′(x) ≤ 3 for all x ∈ (2, 6), it follows that

4 ≤ f(6) − f(2) ≤ 12.

16. f(x) = x2 + x + 3, f ′(x) = 2x + 1.

The slope of the line through (−1, 3) and (2, 9) is 2. Setting 2x + 1 = 2, we get x = 12 . The point on

the graph of f where the tangent line is parallel to the line through (−1, 3) and (2, 9) is: (1/2, 15/4).

17. f is everywhere continuous and everywhere differentiable except possibly at x = −1.

f is continuous at x = −1: as you can check,

limx→−1−

f(x) = 0, limx→−1+

f(x) = 0, and f(−1) = 0.

f is differentiable at x = −1 and f ′(−1) = 2: as you can check,

limh→0−

f(−1 + h) − f(−1)h

= 2 and limh→0+

f(−1 + h) − f(−1)h

= 2.

Thus f satisfies the conditions of the mean-value theorem on every closed interval [ a, b ].

f ′(x) =

{2, x ≤ −1

3x2 − 1, x > −1

f(2) − f(−3)2 − (−3)

=6 − (−4)2 − (−3)

= 2.

f ′(c) = 2 with c ∈ (−3, 2) iff c = 1 or −3 < c ≤ −1.

18. f is continuous and differentiable everywhere;

f ′(x) =

{3x2, x ≤ 1

3, x > 1

f(2) − f(−1)2 − (−1)

=6 − 1

3=

53

For c ≤ 1, f ′(c) = 3c2 = 53 =⇒ c = ±

√5

3For c > 1, f ′(c) = 3 �= 5

3



120 SECTION 4.1

19. Let f(x) = Ax2 + Bx + C. Then f ′(x) = 2Ax + B. By the mean-value theorem

f ′(c) =f(b) − f(a)

b− a=

(Ab2 + Bb + C) − (Aa2 + Ba + C)b− a

=A(b2 − a2) + B(b− a)

b− a= A(b + a) + B

Therefore, we have

2Ac + B = A(b + a) + B =⇒ c =a + b

2

20.f(1) − f(−1)

1 − (−1)= 1 and f ′(x) = −1/x2 < 0; f is not continuous at 0.

21.f(1) − f(−1)

1 − (−1)= 0 and f ′(x) is never zero. This result does not violate the mean-value theorem

since f is not differentiable at 0; the theorem does not apply.

22. f(x) =

{2x− 4, x ≥ 1/2

−2x− 2, x < 1/2

f ′(x) =

{2, x > 1/2

−2, x < 1/2

f ′(x) �= 0 for all x �= 12 .

Rolle’s theorem is not violated because f is

not differentiable at x = 12 .

23. Set P (x) = 6x4 − 7x + 1. If there existed three numbers a < b < c at which P (x) = 0, then by Rolle’s

theorem P ′(x) would have to be zero for some x in (a, b) and also for some x in (b, c). This is not the

case: P ′(x) = 24x3 − 7 is zero only at x = (7/24)1/3.

24. Set P (x) = 6x5 + 13x + 1. Note that P (−1) < 0 and P (0) > 0. By the intermediate-value theorem,

the equation P (x) = 0 has at least one real root c. If this equation had another real root d, then

by Rolle’s theorem P ′(x) would have to be zero for some x between c and d. This is not the case:

P ′(x) = 30x4 + 13 is never zero.

25. Set P (x) = x3 + 9x2 + 33x− 8. Note that P (0) < 0 and P (1) > 0. Thus, by the intermediate-value

theorem, there exists some number c between 0 and 1 at which P (x) = 0. If the equation P (x) = 0

had an additional real root, then by Rolle’s theorem there would have to be some real number at

which P ′(x) = 0. This is not the case: P ′(x) = 3x2 + 18x + 33 is never zero since the discriminant

b2 − 4ac = (18)2 − 12(33) < 0.

26. (a) Suppose that f has two zeros, x1, x2 ∈ (a, b). Then, f is differentiable on (x1, x2) and continuous

on [x1, x2]. By Rolle’s theorem, f ′ has a zero in (x1, x2) which contradicts the hypothesis.

(b) If f had three zeros in (a, b), then, by Rolle’s’ theorem, f ′ would have at least two zeros

in (a, b) and f ′′ would have at least one zero in (a, b) which contradicts the hypothesis.



SECTION 4.1 121

27. Let c and d be two consecutive roots of the equation P ′(x) = 0. The equation P (x) = 0 cannot have

two or more roots between c and d for then, by Rolle’s theorem, P ′(x) would have to be zero somewhere

between these two roots and thus between c and d. In this case c and d would no longer be consecutive

roots of P ′(x) = 0.

28. If f(x) = 0 at a1, a2, . . . , an then by Rolle’s theorem, f ′(x) is zero at some number b1 ∈ (a1, a2), at

some number b2 ∈ (a2, a3), . . . , at some number bn−1 ∈ (an−1, an); f ′′(x), in turn, must be zero at

some number c1 ∈ (b1, b2), at some number b2 ∈ (b2, b3), . . . , at some number cn−2 ∈ (bn−2, bn−1).

29. Suppose that f has two fixed points a, b ∈ I, with a < b. Let g(x) = f(x) − x. Then g(a) = f(a) − a =

0 and g(b) = f(b) − b = 0. Since f is differentiable on I, we can conclude that g is differentiable on

(a, b) and continuous on [a, b]. By Rolle’s theorem, there exists a number c ∈ (a, b) such that g′(c) =

f ′(c) − 1 = 0 or f ′(c) = 1. This contradicts the assumption that f ′(x) < 1 on I.

30. Set P (x) = x3 + ax + b. It is obvious that for x sufficiently large, P (x) > 0 and for x sufficiently large

negative, P (x) < 0. Thus, by the intermediate-value theorem, the equation P (x) = 0 has at least one

real root.

If a ≥ 0, then P ′(x) = 3x2 + a is positive, except possibly at 0, where it remains nonnegative. It follows

that P is everywhere increasing and therefore it cannot take on the value 0 more than once.

Suppose now that a < 0. Then − 13

√3 |a| and 1

3

√3 |a| are consecutive roots of the equation P ′(x) = 0

and thus, by Exercise 27, P cannot take on the value zero more than once between these two numbers.

31. (a) f ′(x) = 3x2 − 3 < 0 for all x in (−1, 1). Also, f is differentiable on (−1, 1) and continuous on

[−1, 1]. Thus there cannot be a and b in (−1, 1) such that f(a) = f(b) = 0, or they would contradict

Rolle’s theorem.

(b) When f(x) = 0, b = 3x− x3 = x(3 − x2). When x is in (−1, 1), then |x(3 − x2)| < 2.

Thus |b| < 2.

32. f ′(x) = 3x2 − 3a2 > 0 for all x in(−a, a). Also, f is differentiable on (−a, a) and continuous on [−a, a].

Thus there cannot be b and c in (−a, a) such that f(b) = f(c) = 0, or they would contradict Rolle’s

theorem.

33. For p(x) = xn + ax + b, p′(x) = nxn−1 + a, which has at most one real zero for n even(x = − a

n

1n−1

).

If there were more than two distinct real roots of p(x), then by Rolle’s theorem there would be more

than one zero of p′(x). Thus there are at most two distinct real roots of p(x).

34. For p(x) = xn + ax + b, p′(x) = nxn−1 + a, which has at most two real zeros for n odd. If there were

more than three distinct real roots of p(x), then by Rolle’s theorem there would be more than two

zeros of p′(x). Thus there are at most three distinct real roots of p(x).



122 SECTION 4.1

35. If x1 = x2, then |f(x1) − f(x2)| and |x1 − x2| are both 0 and the inequality holds. If x1 �= x2, then by

the mean-value theoremf(x1) − f(x2)

x1 − x2= f ′(c)

for some number c between x1 and x2. Since |f ′(c)| ≤ 1:∣∣∣∣f(x1) − f(x2)x1 − x2

∣∣∣∣ ≤ 1 and thus |f(x1) − f(x2)| ≤ |x1 − x2|.

36. See the proof of Theorem 4.2.2.

37. Set, for instance, f(x) =

{1, a < x < b

0, x = a, b

38. (a) Let f(x) = cosx. Choose any numbers x and y, (assume x < y). By the mean-value theorem,

there is a number c between x and y such that

f(y) − f(x)y − x

= f ′(c) ⇒ | cos y − cosx||y − x| = | − sin c| ≤ 1 ⇒ | cosx− cos y| ≤ |x− y|

(b) Repeat the in part (a) with f(x) = sinx.

39. (a) By the mean-value theorem, there exists a number c ∈ (a, b) such that f(b) − f(a) = f ′(c)(b− a).

If f ′(x) ≤ M for all x ∈ (a, b), then it follows that

f(b) ≤ f(a) + M(b− a)

(b) If f ′(x) ≥ m for all x ∈ (a, b), then it follows that

f(b) ≥ f(a) + m(b− a)

(c) If |f ′(x)| ≤ K on (a, b), then −K ≤ f ′(x) ≤ K on (a, b) and the result follows from parts (a)

and (b).

40. Assume that g(x) �= 0 for all x ∈ [a, b] and let h(x) =f(x)g(x)

. Then h is defined on [a, b] and h(a) =

h(b) = 0. Therefore, by Rolle’s theorem, there exists a number c ∈ (a, b) such that

h′(c) =g(c)f ′(c) − f(c)g′(c)

g2(c)= 0

Thus g(c)f ′(c) − f(c)g′(c) = 0 which contradicts the given condition f(x)g′(x) − g(x)f ′(x) �= 0 for all

x ∈ I. Thus, g has at least one zero in (a, b).

By reversing the roles of f and g, the same argument can be used to show that g cannot have

two (or more) zeros on (a, b).

41. We show first that the conditions on f and g imply that f and g cannot be simultaneously 0.

Set h(x) = f2(x) + g(x). Then

h′(x) = 2f(x)f ′(x) + 2g(x)g′(x) = 2f(x)[g(x)] + 2g(x)[−f(x)] = 0 =⇒ f2(x) + g2(x) = C constant.



SECTION 4.1 123

If C = 0, then f2(x) = −g2(x) =⇒ f(x) ≡ 0, contradicting the assumptions on f . Therefore,

f2(x) + g2(x) = C, C > 0. If f(α) = 0, then g(α �= 0.

Assume that f(a) = f(b) = 0 and f(x) �= 0 on (a, b). Suppose that g(x) �= 0 on (a, b). We know also

that g(a) �= 0 and g(b) �= 0. Set h(x) = f(x)/g(x). Then h is continuous on the closed interval [a, b]

and differentiable on the open interval (a, b). Therefore, by Rolle’s theorem, there exists at least one

point c ε(a, b) such that h′(c) = 0. But,

h′(x) =g(x)f ′(x) − f(x)g′(x)

g2(x)=

g2(x) + f2(x)g2(x)

=C

g2(x)> 0

for all x ε(a, b), and we have a contradiction. Thus g must have at least one zero in (a, b). Since

g′(x) = −f(x) �= 0 in (a, b), g has exactly one zero in (a, b).

Simply reverse the roles of f and g to show that f has exactly one zero between two consecutive

zeros of g.

42. We prove the result for h > 0. The proof for h < 0 is similar. If f is differentiable on (x, x + h), it is

continuous there and thus, by the hypothesis at x and x + h continuous on [x, x + h]. By the mean-value

theorem, there exists c in (x, x + h) for which

f(x + h) − f(x)x + h− x

= f ′(c).

Multiplying through by (x + h) − x = h, we have

f(x + h) − f(x) = f ′(c)h.

Since c is between x and x + h, c can be written

c = x + θh with 0 < θ < 1.

43. f ′(x0) = limh→0

f(x0 + h) − f(x0)h

= limh→0

f ′(x0 + θh)hh

= limh→0

f ′(x0 + θh)∧

(by the hint)= lim

x→x0f ′(x) = L

∧(by 2.2.6)

44. Suppose that f(a) = f(b) = k, and let g(x) = f(x) − k. Then g is differentiable on (a, b), continuous

on [a, b], and g(a) = g(b) = 0. Therefore, by Rolle’s theorem, there exists at least one number c ∈ (a, b)

such that g′(c) = 0. Since g′(x) = f ′(x), it follows that f ′(c) = 0.

45. Using the hint, F is continuous on [a, b], differentiable on (a, b), and F (a) = F (b). By Exercise 44,

there is a number c in (a, b) such that F ′(c) = 0.

Therefore [f(b) − f(a)]g′(c) − [g(b) − g(a)]f ′(c) = 0 andf(b) − f(a)g(b) − g(a)

=f ′(c)g′(c)

.



124 SECTION 4.1

46. f(x) = 2x3 + 3x2 − 3x− 2 is differentiable on (−2, 1), continuous on [−2, 1], and f(−2) = f(1) = 0.

f ′(x) = 6x2 + 6x− 3

f ′(c) = 0 at c1 ∼= −1.366, c2 ∼= 0.366

47. f(x) = 1 − x3 − cos(π x/2) is differentiable on (0, 1), continuous on [0, 1], and f(0) = f(1) = 0.

f ′(x) = −3x2 +π

2sin(π x/2)

f ′(c) = 0 at c ∼= 0.676

48. f(1) = 0; f satisfies Rolle’s theorem on [0, 1]; f ′(c) = 0 at c ∼= 0.6058.

49. b ∼= 0.5437, c ∼= 0.3045,

f(0.3045) ∼= −0.17490.2 0.4 0.6 0.8

x

y

50. x-intercepts: x = 0, x = 1; f ′(x) = 0 at x ∼= 0.3874

51. 0 c = 0

52. x-intercepts: x = −2, x = 45 x = 2; f ′(x) = 0 at x ∼= −1.1005 and x ∼= 1.5577

53. f(x) = x4 − 7x2 + 2; f ′(x) = 4x3 − 14x

g(x) = 4x3 − 14x− f(3) − f(1)3 − 1

= 4x3 − 14x− 12

g(c) = 0 at c ∼= 2.205



SECTION 4.2 125

54. f(x) = x cosx + 4 sinx; f ′(x) = cosx− x sinx + 4 cosx = 5 cosx− x sinx and

g(x) = 5 cosx− x sinx− f(π/2) − f(−π/2)π

= 5 cosx− x sinx− 8π

g(c) = 0 at c1 ∼= −0.872, c2 ∼= 0.872

55. f(x) = x3 − x2 + x− 1; f ′(x) = 3x2 − 2x + 1; c = 8/3.

1 2 3 4x

20

40

60

y

56. f(x) = x4 − 2x3 − x2 − x + 1; f ′(x) = 4x3 − 6x2 − 2x− 1; c = 1/2, −0.6180, 1.6180.

−2 −1 1 2 3x

5

5

10

y

SECTION 4.2

1. f ′(x) = 3x2 − 3 = 3(x2 − 1

)= 3(x + 1)(x− 1)

f increases on (−∞, −1] and [1, ∞), decreases on [−1, 1]

2. f ′(x) = 3x2 − 6x = 3x(x− 2)

f increases on (−∞, 0] and [2, ∞), decreases on [0, 2]

3. f ′(x) = 1 − 1x2

=x2 − 1x2

=(x + 1)(x− 1)

x2

f increases on (−∞, −1] and [1, ∞), decreases on [−1, 0) and (0, 1 ] (f is not defined at 0)

4. f ′(x) = 3(x− 3)2; f increases on (−∞,∞)



126 SECTION 4.2

5. f ′(x) = 3x2 + 4x3 = x2(3 + 4x)

f increases on[− 3

4 , ∞), decreases on

(−∞, − 3

4

]6. f ′(x) = 3x2 + 6x + 2

f increases on(−∞, −1 − 1

3

√3]

and[−1 + 1

3

√3, ∞

), decreases on

[−1 − 1

3

√3, −1 + 1

3

√3]

7. f ′(x) = 4(x + 1)3

f increases on [−1, ∞), decreases on (−∞, −1]

8. f ′(x) =2(x3 + 1)

x3

f increases on [−∞, −1] and (0,∞), decreases on [−1, 0)

9. f(x) =

⎧⎪⎨⎪⎩

12 − x

, x < 2

1x− 2

, x > 2f ′(x) =

⎧⎪⎨⎪⎩

1(2 − x)2

, x < 2

−1(x− 2)2

, x > 2

f increases on (−∞, 2), decreases on (2, ∞) (f is not defined at 2)

10. f ′(x) =(1 + x2) − x(2x)

(1 + x2)2; f increases on [−1, 1], decreases on (−∞,−1] and [1,∞)

11. f ′(x) = − 4x(x2 − 1)2

f increases on (−∞, −1) and (−1, 0], decreases on [0, 1) and (1, ∞) (f is not defined at ±1)

12. f ′(x) =(x2 + 1)(2x) − x2(2x)

(x2 + 1)2=

2x(x2 + 1)2

f increases on [0, ∞), decreases on (−∞, 0]

13. f(x) =

⎧⎪⎪⎨⎪⎪⎩

x2 − 5, x < −√

5

−(x2 − 5

), −

√5 ≤ x ≤

√5

x2 − 5,√

5 < x

f ′(x) =

⎧⎪⎪⎨⎪⎪⎩

2x, x < −√

5

−2x, −√

5 < x <√

5

2x,√

5 < x

f increases on [−√

5, 0 ] and [√

5, ∞), decreases on (−∞, −√

5 ] and [ 0,√

5 ]

14. f ′(x) = x2(2)(1 + x) + (x + 1)2(2x) = 2x(x + 1)(2x + 1)

f increases on [−1,−1/2] and [0,∞), decreases on (−∞,−1] and [−1/2, 0]

15. f ′(x) =2

(x + 1)2; f increases on (−∞, −1) and (−1, ∞) (f is not defined at −1)

16. f ′(x) = 2x− 32x3

=2(x4 − 16)

x3=

2(x− 2)(x + 2)(x2 + 4)x3

f increases on [−2, 0) and [2,∞), decreases on (−∞,−2] and (0, 2]

17. f ′(x) =x

(2 + x2)2

√2 + x2

1 + x2f increases on [0, ∞), decreases on (−∞, 0 ]



SECTION 4.2 127

18. f(x) =

⎧⎪⎪⎨⎪⎪⎩

x2 − x− 2, x ≤ −1

−x2 + x + 2, −1 < x < 2

x2 − x− 2, x ≥ 2

f ′(x) =

⎧⎪⎪⎨⎪⎪⎩

2x− 1, x ≤ −1

−2x + 1, −1 < x < 2

2x− 1, x ≥ 2

f increases on[−1, 1

2

]and [2, ∞), decreases on (−∞, −1] and

[12 , 2

]19. f ′(x) = 1 + sinx ≥ 0; f increases on [0, 2π]

20. f ′(x) = 1 + cosx ≥ 0; f increases on [0, 2π]

25. f ′(x) = −2 sin 2x− 2 sinx = −2 sinx (2 cosx + 1); f increases on[23π, π

], decreases on

[0, 2

3π]

22. f ′(x) = −2 cosx sinx = −2 sin 2x; f increases on [π/2, π], decreases on [0, π/2]

23. f ′(x) =√

3 + 2 sin 2x; f increases on[0, 2

3π]

and[56π, π

], decreases on

[23π,

56π

]24. f ′(x) = 2 sinx cosx−

√3 cosx = cosx

(2 sinx−

√3)

f increases on [ 13π,12π] and [23π, π], decreases on [0, 1

3π] and [12π,23π]

25.d

dx

(x3

3− x

)= f ′(x) =⇒ f(x) =

x3

3− x + C

f(1) = 2 =⇒ 2 = 13 − 1 + C, so C = 8

3 . Thus, f(x) = 13x

3 − x + 83 .

26.d

dx

(x2 − 5x

)= f ′(x) =⇒ f(x) = x2 − 5x + C

f(2) = 4 =⇒ 4 = 4 − 10 + C, so C = 10. Thus, f(x) = x2 − 5x + 10.

27.d

dx

(x5 + x4 + x3 + x2 + x

)= f ′(x) =⇒ f(x) = x5 + x4 + x3 + x2 + x + C

f(0) = 5 =⇒ 5 = 0 + C, so C = 5. Thus, f(x) = x5 + x4 + x3 + x2 + x + 5.

28.d

dx

(−2x−2

)= f ′(x) =⇒ f(x) = −2x−2 + C

f(1) = 0 =⇒ 0 = −2 + C, so C = 2. Thus, f(x) = −2x−2 + 2, x > 0.

29.d

dx

(34x4/3 − 2

3x3/2

)= f ′(x) =⇒ f(x) =

34x4/3 − 2

3x3/2 + C

f(0) = 1 =⇒ 1 = 0 + C, so C = 1. Thus, f(x) = 34x

4/3 − 23x

3/2 + 1, x ≥ 0.

30.d

dx

(− 1

4x−4 − 25

4 x4/5)

= f ′(x) =⇒ f(x) = − 14x

−4 − 254 x4/5 + C

f(1) = 0 =⇒ 0 = − 14 − 25

2 + C, so C = 132 . Thus, f(x) = − 1

4x−4 − 25

4 x4/5 + 132 , x > 0.

31.d

dx(2x− cosx) = f ′(x) =⇒ f(x) = 2x− cosx + C

f(0) = 3 =⇒ 3 = 0 − 1 + C, so C = 4. Thus, f(x) = 2x− cosx + 4.

32.d

dx(2x2 + sinx) = f ′(x) =⇒ f(x) = 2x2 + sinx + C

f(0) = 1 =⇒ 1 = 0 + C, so C = 1. Thus, f(x) = 2x2 + sinx + 1.



128 SECTION 4.2

33. f ′(x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

1, x < −3

−1, −3 < x < −1

1, −1 < x < 1

−2, 1 < x

f increases on (−∞, −3) and [−1, 1];

decreases on [−3, −1] and [1, ∞)

34. f ′(x) =

⎧⎪⎪⎨⎪⎪⎩

2(x− 1), x < 1

−1, 1 < x < 3

−2, x > 3

f decreases on (−∞, 1), [1, 3) and [3,∞).

35. f ′(x) =

⎧⎪⎪⎨⎪⎪⎩

−2x, x < 1

−2, 1 < x < 3

3, 3 < x

f increases on (−∞, 0] and [3, ∞);

decreases on [0, 1) and [1, 3]

36. f ′(x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

1, x < 0

2(x− 1), 0 < x < 3

−1, 3 < x < 7

2, x > 7

f increases on (−∞, 0], [1, 3], (7, ∞);

decreases on [0, 1] and [3, 7)

37. 38.



SECTION 4.2 129

39. 40.

41. 42.

43. 44.

45. Not possible; f is increasing, so f(2) must be greater than f(−1).

46. Not possible; by the intermediate-value theorem, f must have a zero in (3, 5).

47. (a) True. Let x1, x2 ∈ [a, c], x1 < x2. If x1, x2 ∈ [a, b], or if x1, x2 ∈ [b, c], then f(x1) < f(x2).

If x1 ∈ [a, b) and x2 ∈ [b, c], then f(x1) < f(b) ≤ f(x2). Therefore f increases on [a, c]

(b) False. A slight modification of Example 6 is a counterexample. Let g(x) =

⎧⎨⎩

12x + 2, x ≤ 1

x3, x > 1.

48. (a) True. Use the obvious modification of the argument in Exercise 47(a).

(b) False. An example like 47(b): f(x) =

⎧⎨⎩−x + 2, x ≤ 1

−x + 3, x > 1..

49. (a) True. If f ′(c) < 0 at some number c ∈ (a, b), then there exists a number δ such that

f(x) > f(c) > f(z) for x ∈ (c− δ, c) and z ∈ (c, c + δ) (Theorem 4.1.2).

(b) False. f(x) = x3 increases on (−1, 1) and f ′(0) = 0.

50. False. f(x) = x +12π

sin (2x− 1)π is increasing on [0, 4] and f ′(x) = 1 + cos (2x− 1)π x = 0 at

x = 1, x = 2. x = 3.



130 SECTION 4.2

51. Let f(x) = x− sinx. Then f ′(x) = 1 − cosx.

(a) f ′(x) ≥ 0 for all x ∈ (−∞,∞) and f ′(x) = 0 only at x =π

2+ nπ, n = 0, ±1, ±2, . . .

It follows from Theorem 4.2.3 that f is increasing on (−∞,∞).

(b) Since f is increasing on (−∞,∞) and f(0) = 0 − sin 0 = 0, we have:

f(x) > 0 for all x > 0 ⇒ x > sinx on (0,∞);

f(x) < 0 for all x < 0 ⇒ x < sinx on (−∞, 0).

52. A proof is outlined just below the statement of the theorem.

53. f ′(x) = 2 secx(secx tanx) = 2 sec2 x tanx and g′(x) = 2 tanx sec2 x.

Therefore, f ′(x) = g′(x) for all x ∈ I.

54. Evaluating sec2 x− tan2 x = C at x = 0 gives C = 1.

55. Let f and g be functions such that f ′(x) = −g(x) and g′(x) = f(x). Then:

(a) Differentiating f2(x) + g2(x) with respect to x, we have

2f(x)f ′(x) + 2g(x)g′(x) = −2f(x)g(x) + 2g(x)f(x) = 0.

Thus, f2(x) + g2(x) = C (constant).

(b) f(0) = 0 and g(0) = 1 implies C = 1.

(c) The functions f(x) = sin x, g(x) = cos x have these properties.

56. (a) Let h(x) = f(x) − g(x). Then h′(x) = f ′(x) − g′(x) > 0 on (0, c), and h is increasing on (0, c).

Since h(0) = f(0) − g(0) = 0, it follows that h(x) > 0 on (0, c). Thus, f(x) > g(x) on (0, c).

(b) Again let h(x) = f(x) − g(x). Then h is increasing on (−c, 0) which implies that h(x) < 0 on this

interval since h(0) = 0. Therefore, f(x) < g(x) on (−c, 0).

57. Let f(x) = tanx and g(x) = x for x ∈ [0, π/2). Then f(0) = g(0) = 0 and f ′(x) = sec2 x > g′(x) = 1

for x ∈ (0, π/2). Thus, tanx > x for x ∈ (0, π/2) by Exercise 56(a).

58. Let f(x) = cosx−(1 − 1

2 x2)

for x ∈ [0,∞). Then f(0) = 0 and f ′(x) = − sinx + x = x− sinx > 0

for x ∈ (0,∞) by Exercise 51 (b). Thus, f(x) > 0 for x ∈ (0,∞) which implies cosx > 1 − 12 x

2 on

(0,∞).

59. Choose an integer n > 1. Let f(x) = (1 + x)n and g(x) = 1 + nx, x > 0. Then, f(0) = g(0) = 1

and f ′(x) = n(1 + x)n−1 > g′(x) = n since (1 + x)n−1 > 1 for x > 0. The result follows from Exercise

56(a).

60. Let f(x) = sinx−(x− 1

6 x3). Then f(0) = 0 and f ′(x) = cosx−

(1 − 1

2 x2)> 0 by Exercise 58.

Therefore, f(x) > f(0) = 0 for all x ∈ (0,∞) which implies sinx > x− 16 x

3 on (0,∞).

61. 4◦ ∼= 0.06981 radians. By Exercises 51 and 60,

0.6981 − (0.6981)3

6= 0.06975 < sin 4◦ < 0.6981



SECTION 4.2 131

62. (a) Let f(x) = cosx− (1 − 12x2 +

124

x4). Then f(0) = 0 and f ′(x) = − sinx + x− x3

6< 0 by

Exercise 60. Therefore, f(x) < f(0) = 0 on all x ∈ (0,∞), which implies cosx < 1 − 12x2 +

124

x4

on (0,∞).

(b) 6◦ =π

30. Using this for x in 1 − 1

2x2 < cosx < 1 − 1

2x2 +

124

x4,

=⇒ 0.994517 < cos 6◦ < 0.994522.

63. Let f(x) = 3x4 − 10x3 − 4x2 + 10x + 9, x ∈ [−2, 5]. Then f ′(x) = 12x3 − 30x2 − 8x + 10.

f ′(x) = 0 at x ∼= −0.633, 0.5, 2.633

f is decreasing on [−2,−0.633]

and [0.5, 2.633]

f is increasing on [−0.633, 0.5]

and [2.633, 5]

64. Let f(x) = 2x3 − x2 − 13x− 6, x ∈ [−3, 4]. Then f ′(x) = 6x2 − 2x− 13.

f ′(x) = 0 at x ∼= −1.315, 1.648

f is decreasing on [−1.315, 1.648]

f is increasing on [−3,−1.315] and [1.648, 4]

65. Let f(x) = x cosx− 3 sin 2x, x ∈ [0, 6]. Then f ′(x) = cosx− x sinx− 6 cos 2x.

f ′(x) = 0 at x ∼= 0.770, 2.155, 3.798, 5.812

f is decreasing on [0, 0.770], [2.155, 3.798]

and [5.812, 6]

f is increasing on [0.770, 2.155]

and [3.798, 5.812]

66. Let f(x) = x4 + 3x3 − 2x2 + 4x + 4, x ∈ [−5, 3]. Then f ′(x) = 4x3 + 9x2 − 4x + 4.

f ′(x) = 0 at x ∼= −2.747

f is decreasing on [−5,−2.747]

f is increasing on [−2.747, 3]



132 SECTION 4.3

67. (a) f ′(x) = 0 at x = 0, π2 , π, 3π

2 , 2π (b) f ′(x) > 0 on(π, 3π

2

)∪

(3π2 , 2π

)(c) f ′(x) < 0 on

(0, π

2

)∪

(π2 , π

)68. (b) f ′(x) > 0 on (−∞,∞)

69. (a) f ′(x) = 0 at x = 0 (b) f ′(x) > 0 on (0,∞)

(c) f ′(x) < 0 on (−∞, 0)

70. (a) f ′(x) = 0 at x = − 12 ,

85 , 3 (b) f ′(x) > 0 on

(−∞,− 1

2

)∪

(− 1

2 ,85

)∪ (3,∞)

(c) f ′(x) < 0 on(

85 , 3

)71. f = C, constant; f ′(x) ≡ 0

SECTION 4.3

1. f ′(x) = 3x2 + 3 > 0; no critical pts, no local extreme values

2. f ′(x) = 8x3 − 8x = 8x(x2 − 1); critical pts −1, 0, 1

f ′′(x) = 24x2 − 8; f ′′(−1) = f ′′(1) = 16 > 0, f ′′(0) = −8 < 0;

f(0) = 6 local max, f(−1) = 4 local min, f(1) = 4 local min

3. f ′(x) = 1 − 1x2

; critical pts −1, 1

f ′′(x) =2x3

, f ′′(−1) = −2, f ′′(1) = 2 f(−1) = −2 local max, f(1) = 2 local min

4. f ′(x) = 2x +6x3

=2x4 + 6

x3; no critical pts (note: 0 is not in the domain of f),

no local extreme values

5. f ′(x) = 2x− 3x2 = x(2 − 3x); critical pts 0, 23

f ′′(x) = 2 − 6x; f ′′(0) = 2, f ′′( 23 ) = −2

f(0) = 0 local min, f( 23 ) = 4

27 local max

6. f ′(x) = −2(1 − x)(1 + x) + (1 − x)2 = (x− 1)(3x + 1); critical pts − 13 , 1

f ′′(x) = (1 + 3x) + 3(x− 1) = 2(3x− 1); f ′′ (− 13

)= −4, f ′′(1) = 4

f(− 1

3

)= 32

27 local max, f(1) = 0 local min

7. f ′(x) =2

(1 − x)2; no critical pts, no local extreme values

8. f ′(x) =(2 + x)(−3) − (2 − 3x)(1)

(2 + x)2= − 8

(2 + x)2; no critical pts (note: −2 is not

in the domain of f), no local extreme values



SECTION 4.3 133

9. f ′(x) = − 2(2x + 1)x2(x + 1)2

; critical pt −12

f(− 1

2

)= −8 local max

10. f(x) =

⎧⎪⎪⎨⎪⎪⎩

x2 − 16, x < −4

16 − x2, −4 ≤ x < 4

x2 − 16, x ≥ 4

f ′(x) =

⎧⎪⎪⎨⎪⎪⎩

2x, x < −4

−2x, −4 < x < 4

2x, x > 4

critical pts −4, 0, 4; f(−4) = f(4) = 0 local minima, f(0) = 16 local max

11. f ′(x) = x2(5x− 3)(x− 1); critical pts 0, 35 , 1

f

(35

)=

2233

55local max

f(1) = 0 local min

no local extreme at 0

12. f ′(x) = 3(x− 2x + 2

)2 4(x + 2)2

≥ 0; critical pt 2, no local extreme values

13. f ′(x) = (5 − 8x)(x− 1)2; critical pts 58 , 1

f(

58

)= 27

2048 local max


14. f ′(x) = −(1 + x)3 + (1 − x)(3)(1 + x)2 = 2(1 + x)2(1 − 2x); critical pts −1, 12

f(

12

)= 27

16 local max

no local extreme at −1

15. f ′(x) =x(2 + x)(1 + x)2

; critical pts −2, 0

f(−2) = −4 local max

f(0) = 0 local min

16. f ′(x) = (1 − x)1/3 − 13 x(1 − x)−2/3 =

3 − 4x3(1 − x)2/3

; critical pts 34 , 1

f(3/4) = 344/3 local max


17. f ′(x) = 13x(7x + 12)(x + 2)−2/3; critical pts −2, − 12

7 , 0

f(− 12

7

)= 144

49

(27

)1/3 local max

f(0) = 0 local min



134 SECTION 4.3

18. f ′(x) =−1

(x + 1)2+

1(x− 2)2

=3(2x− 1)

(x + 1)2(x− 2)2; critical pt 1

2

f(

12

)= 4

3 local min

19. f(x) =

⎧⎪⎪⎨⎪⎪⎩

2 − 3x, x ≤ −12

x + 4, − 12 < x < 3

3x− 2, 3 ≤ x

f ′(x) =

⎧⎪⎪⎨⎪⎪⎩

−3, x < − 12

1, − 12 < x < 3

3, 3 < x

critical pts − 12 , 3

f(− 1

2

)= 7

2 local min


20. f ′(x) = 73 x

4/3 − 73 x

−2/3 =73x2 − 1x2/3

; critical pts −1, 0, 1

f(−1) = 6 local max,

f(1) = −6 local min


21. f ′(x) = 23x

−4/3(x− 1); critical pt 1

f(1) = 3 local min


22. f ′(x) =(x + 1)3x2 − x3

(x + 1)2=

x2(2x + 3)(x + 1)2

; critical pts − 32 , 0

f(− 3

2

)= 27

4 local min


23. f ′(x) = cosx− sinx; critical pts 14π,

54π

f ′′(x) = − sinx− cosx, f ′′ ( 14π

)= −

√2, f ′′ ( 5

4π)

=√

2

f( 14π) =

√2 local max, f( 5

4π) = −√

2 local min

24. f ′(x) = 1 − 2 sin 2x; critical pts π12 ,

5π12

f( 112π) =

π

12+

√3

2local max

f(512

π) =5π12

−√

32

local min



SECTION 4.3 135

25. f ′(x) = cosx (2 sinx−√

3 ); critical pts 13π,

12π,

23π

f( 13π) = f( 2

3π) = − 34 local mins

f( 12π) = 1 −

√3 local max

26. f ′(x) = 2 sinx cosx; critical pts 12π, π,

32π

f(

12π

)= 1 = f

(32π

)local max

f(π) = 0 local min

27. f ′(x) = cos2 x− sin2 x− 3 cosx + 2 = (2 cosx− 1)(cosx− 1) critical pts 13π,

53π

f(

13π

)= 2

3π − 54

√3 local min

f(

53π

)= 10

3 π + 54

√3 local max

28. f ′(x) = 6 sin2 x cosx− 3 cosx = 3 cosx(2 sin2 x− 1); critical pts 14π,

12π,

34π

f( 14π) = f( 3

4π) = −√

2 local mins

f( 12π) = −1 local max

29. (a) f increases on [−2, 0] and [3,∞); f decreases on (−∞,−2] and [0, 3].

(b) f(−2) and f(3) are local minima; f(0) = 1 is a local maximum.

30. (a) f increases on (−∞,−1] and [0,∞); f decreases on [−1, 0].

(b) f(−1) is a local maximum; f(0) = 1 is a local minimum.

31. Let h(x) = f(x) − g(x). Then h(x) gives the vertical separation between graphs of f and g at x. If

h has a maximum at c, then h′(c) = 0. Since h′(x) = f ′(x) − g′(x), h′(c) = 0 implies f ′(c) = g′(c).

Thus the lines tangent to the graphs of f and g are parallel at x = c.

32. Set g(x) = −f(−x) and apply the proof of the second derivative test already given.

33. Solving f ′(x) = 2ax + b = 0 gives a critical point at x = − b

2a. Since f ′′(x) = 2a,

f has a local maximum at − b

2aif a < 0 and a local minimum at − b

2aif a > 0.

34. Setting f ′(x) = 3ax2 + 2bx + c = 0 and checking the discriminant, we get

(1) 2 local extrema if b2 > 3ac

(2) 1 local extrema if b2 = 3ac

(3) 0 local extrema if b2 < 3ac



136 SECTION 4.3

35. P (x) = x4 − 8x3 + 22x2 − 24x + 4

P ′(x) = 4x3 − 24x2 + 44x− 24

P ′′(x) = 12x2 − 48x + 44

Since P ′(1) = 0, P ′(x) is divisible by x− 1. Division byx− 1 gives

P ′(x) = (x− 1)(4x2 − 20x + 24

)= 4(x− 1)(x− 2)(x− 3).

The critical pts are 1, 2, 3. Since

P ′′(1) > 0, P ′′(2) < 0, P ′′(3) > 0,

P (1) = −5 is a local min, P (2) = −4 is a local max, andP (3) = −5 is a local min.

Since P ′(x) < 0 for x < 0, P decreases on (−∞, 0]. Since P (0) > 0, P does not take on the value 0 on

(−∞, 0].

Since P (0) > 0 and P (1) < 0, P takes on the value 0 at least once on (0, 1). Since P ′(x) < 0 on (0, 1),

P decreases on [0, 1]. It follows that P takes on the value zero only once on [0, 1].

Since P ′(x) > 0 on (1, 2) and P ′(x) < 0 on (2, 3), P increases on [1, 2] and decreases on [2, 3]. Since

P (1), P (2), P (3) are all negative, P cannot take on the value 0 between 1 and 3.

Since P (3) < 0 and P (100) > 0, P takes on the value 0 at least once on (3, 100). Since P ′(x) > 0 on

(3, 100), P increases on [3, 100]. It follows that P takes on the value zero only once on [3, 100].

Since P ′(x) > 0 on (100, ∞), P increases on [100, ∞). Since P (100) > 0, P does not take on the value

0 on [100, ∞).

36. f has a local maximum at x = 0; f has a local minimum at x = −1 and x = 2.

37.

38. Let f(x) = Ax2 + Bx + C. Then f ′(x) = 2Ax + B.

f(−1) = 3 =⇒ A−B + C = 3; f(3) = −1 =⇒ 9A + 3B + C = −1

Since f has a minimum at x = 2, f ′(2) = 4A + B = 0

Solving for A, B, C, we get A = 12 , B = −2, C = 1

2 .



SECTION 4.3 137

39. Let f(x) =ax

x2 + b2. Then f ′(x) =

a(b2 − x2

)(b2 + x2)2

. Now

f ′(0) =a

b2= 1 ⇒ a = b2 and f ′(x) =

b2(b2 − x2

)(b2 + x2)2

f ′(−2) =b2

(b2 − 4

)(b2 + 4)2

= 0 ⇒ b = ±2

Thus, a = 4 and b = ±2.

40. (a) f(x) = xp(1 − x)q, p, q ≥ 2; f ′(x) = xp−1(1 − x)q−1[p− (p + q)x]

f ′(x) = 0 =⇒ x = 0, x = 1, x =p

p + q

(b) p even, p− 1 odd:

f has a local min at x = 0

(c) q even, q − 1 odd:

f has a local min at x = 1

(d) f ′′(

p

p + q

)= − (p + q)

(p

p + q

)p−1 (q

p + q

)q−1

< 0 ⇒ f has a local max at x =p

p + q.

41. If p is a polynomial of degree n, then p′ has degree n− 1. This implies that p′ has at most

n− 1 zeros, and it follows that p has at most n− 1 local extreme values.

42. The function D(x) =√x2 + [f(x)]2 gives the distance from the origin to the point (x, f(x)) on the

graph of f. Since the graph of f does not pass through the origin,

D′(x) =x + f(x)f ′(x)√x2 + [f(x)]2

is defined for all x ∈ dom (f). Suppose that D has a local extreme value at c. Then

D′(c) =c + f(c)f ′(c)√c2 + [f(c)]2

= 0 ⇒ c + f(c)f ′(c) = 0 and f ′(c) = − c

f(c)

Suppose that c �= 0. The slope of the line through (0, 0) and (c, f(c)) is given by m1 =f(c)c

and the

slope of the tangent line to the graph of f at x = c is given by m2 = f ′(c) = − c

f(c). Since m1m2 = −1,

these two lines are perpendicular. If c = 0, then the tangent line to the graph of f is horizontal and

the line through (0, 0) and (0, f(0)) is vertical.

43. If f(x) = x4 − 7x2 − 8x− 3, then f ′(x) = 4x3 − 14x− 8 and f ′′(x) = 12x2 − 14. Since f ′(2) =

−4 < 0 and f ′(3) = 58 > 0, f ′ has at least one zero in (2, 3). Since f ′′(x) > 0 for x ∈ (2, 3), f ′ is

increasing on this interval and so it has exactly one zero. Thus, f has exactly one critical point c in

(2, 3).



138 SECTION 4.3

44. If f(x) = sinx +x2

2− 2x, then f ′(x) = cosx + x− 2 and f ′′(x) = − sinx + 1. Since f ′(2) =

−0.4161 < 0 and f ′(3) = 0.01 > 0, f ′ has at least one zero in (2, 3). Since f ′′(x) > 0 for x ∈ (2, 3),

f ′ is increasing on this interval and so it has exactly one zero. Thus, f has exactly one critical point

c in (2, 3).

45. f(x) =ax2 + b

cx2 + dand f ′(x) =

2(ad− bc)x(cx2 + d)2

; x = 0 is a critical number.

f ′′(x) =2(ad− bc)(cx2 − 4cx + d)

(cx2 + d)3; f ′′(0) =

2(ad− bc)d2

Therefore, ad− bc > 0 implies that f(0) is a local minimum; ad− bc < 0 implies that f(0) is a

local maximum.

46. Let δ be any positive number and consider f on the interval (−δ, δ). Let n be a positive integer

such that

0 <1

π2 + 2nπ

< δ and 0 <1

−π2 + 2nπ

< δ.

Then

f

(1

π2 + 2nπ

)> 0 and f

(1

−π2 + 2nπ

)< 0.

Thus f takes on both positive and negative values in every interval centered at 0 and it follows that

f cannot have a local maximum or minimum at 0.

47. (a)

critical points: x1∼= −0.692, x2

∼= 2.248

local extreme values: f(−0.692) ∼= 29.342, f(2.248) ∼= −8.766

(b) f is increasing on [−3,−0.692], and [2.248, 4]; f is decreasing on [−0.692, 2.248]

48. (a)



SECTION 4.3 139


∼= −1, x3∼= 0.207, x4

∼= 1.096, x5 = 1.544

local extreme values: f(−2.085) ∼= −6.255, f(−1) = 7, f(0.207) ∼= 0.621, f(1.096) ∼= 7.097,

f(1.544) ∼= 4.635

(b) f is increasing on [−2.085,−1], [0.207, 1.096], and [1.544, 4]

f is decreasing on [−4,−2.085], [−1, 0.207], and [1.096, 1.544]

49. (a)


∼= −0.654, x3∼= 0.654, x4

∼= 2.201

local extreme values: f(−2.204) ∼= 2.226, f(−0.654) ∼= −6.634, f(0.654) ∼= 6.634,

f(2.204) ∼= −2.226

(b) f is increasing on [−3.− 2.204], [−0.654, 0.654], and [2.204, 3]

f is decreasing on [−2.204,−0.654], and [0.654, 2.204]

50. f ′(x) = 0 at x = 2, 3; f(2) = 0 is a local minimum.

51. f ′(x) > 0 on(

23 ,∞

); f has no local extrema.

52. f ′(x) = 0 at the multiples of 14π; f(0) = 1 is a local maximum, f(π/4) = 0 is a local minimum,

f(π/2) = 1 is a local maximum, and so on.

53.

critical points of f : x1∼= −1.326, x2 = 0, x3

∼= 1.816

f ′′(−1.326) ∼= −4 < 0 ⇒ f has a local maximum at x = −1.326

f ′′(0) = 4 > 0 ⇒ f has a local minimum at x = 0

f ′′(1.816) ∼= −4 ⇒ f has a local maximum at x = 1.816



140 SECTION 4.4

54.

critical number of f : x1∼= −1.935

f ′′(−1.935) ∼= 14.60 > 0 ⇒ f has a local minimum at x = −1.935

SECTION 4.4

1. f ′(x) = 12 (x + 2)−1/2, x > −2;

f(−2) = 0 endpt and abs min; as x → ∞, f(x) → ∞; so no abs max

2. f ′(x) = 2x− 3; critical pt. 32 ; f

(32

)= − 1

4 local and abs min

3. f ′(x) = 2x− 4, x ∈ (0, 3);

critical pt. 2;

f(0) = 1 endpt and abs max, f(2) = −3 local and abs min, f(3) = −2 endpt max

4. f ′(x) = 4x + 5, x ∈ (−2, 0);

critical pt. − 54 ;

f(−2) = −3 endpt max, f(− 5

4

)= − 33

8 local and abs min, f(0) = −1 endpt and abs max

5. f ′(x) = 2x− 1x2

=2x3 − 1

x2, x �= 0; f ′(x) = 0 at x = 2−1/3

critical pt. 2−1/3 ; f ′′(x) = 2 +2x3

, f ′′(2−1/3

)= 6

f(2−1/3

)= 2−2/3 + 21/3 = 2−2/3 + 2 · 2−2/3 = 3 · 2−2/3 local min

6. f ′(x) = 1 − 2x3

critical pt. 21/3; f(21/3

)= 3(2)−2/3 local min

7. f ′(x) =2x3 − 1

x2, x ∈

(110

, 2)

;

critical pt. 2−1/3;

f(

110

)= 10 1

100 endpt and abs max, f(2−1/3

)= 3 · 2−2/3 local and abs min,

f(2) = 4 12 endpt max



SECTION 4.4 141

8. f ′(x) = 1 − 2x3 , x ∈ (1,

√2)

critical pt. 21/3; f(1) = 2 endpt and abs max

f(21/3

)= 3(2)−2/3 local and abs min, f

(√2)

=√

2 + 12 endpt max

9. f ′(x) = 2x− 3, x ∈ (0, 2);

critical pt. 32 ;

f(0) = 2 endpt and abs max, f(

32

)= − 1

4 local and abs min,

f(2) = 0 endpt max

10. f ′(x) = 2(x− 1)(x− 2)(2x− 3); x ∈ (0, 4)

critical pts. 1, 32 , 2, ; f(0) = 4 endpt max, f(1) = 0 local and abs min,

f(

32

)= 1

16 local max, f(2) = 0 local and abs min, f(4) = 36 endpt and abs max

11. f ′(x) =(2 − x)(2 + x)

(4 + x2)2, x ∈ (−3, 1);

critical pt. −2;

f(−3) = − 313 endpt max, f(−2) = − 1


f(1) = 15 endpt and abs max

12. f ′(x) =2x

(1 + x2)2, x ∈ (−1, 2)

critical pt. 0; f(−1) = 12 endpt max, f(0) = 0 local and abs min,

f(2) = 45 endpt and abs max

13. f ′(x) = 2 (x−√x)

(1 − 1

2√x

), x > 0;

critical pts. 14 , 1 ;

f(0) = 0 endpt and abs min, f(

14

)= 1

16 local max, f(1) = 0 local and abs min;

as x → ∞, f(x) → ∞; so no abs max

14. f ′(x) =2(2 − x2)

(4 − x2)1/2, x ∈ (−2, 2)

critical pts. −√

2,√

2; f(−2) = 0 endpt max, f(−√

2)

= −2 local and abs min,

f(√

2)

= 2 local and abs max, f(2) = 0 endpt min

15. f ′(x) =3(2 − x)2√

3 − x, x < 3

critical pt. 2;

f(2) = 2 local and abs max, f(3) = 0 endpt min;

as x → −∞, f(x) → −∞; so no abs min



142 SECTION 4.4

16. f ′(x) = 12

(x−1/2 + x−3/2

), x > 0

no critical pts; no extreme values.

17. f ′(x) = − 13 (x− 1)−2/3, x �= 1;

critical pt. 1;

no local extremes;as x → ∞, f(x) → −∞as x → −∞, f(x) → ∞

}no abs extremes

18. f ′(x) =83

3x− 1(4x− 1)2/3(2x− 1)1/3

;

critical pts. 14 ,

13 ,

12 ; no extreme value at 1

4

f(

13

)= 1

3 local max, f(

12

)= 0 local min

19. f ′(x) = sinx(2 cosx +

√3

), x ∈ (0, π);

critical pt.56π ;

f(0) = −√

3 endpt and abs min, f(

56π

)=

74

local and abs max, f(π) =√

3 endpt min

20. f ′(x) = − csc2 x + 1, x ∈ (0, 2π/3) ;

critical pt. 12π (note: 0 is not in the domain)

No extreme value at 12π, f

(23π

)= 1

3

(2π −

√3)

endpt and abs max

21. f ′(x) = −3 sinx(2 cos2 x + 1

)< 0, x ∈ (0, π); no critical pts.

f(0) = 5 endpt and abs max, f(π) = −5 endpt and abs min

22. f ′(x) = 2 cos 2x− 1, x ∈ (0, π) :

critical pts. 16π,

56π ; f(0) = 0 endpt min, f

(16π

)= 1

2

√3 − 1

6π local and abs max,

f(

56π

)= − 1

2

√3 − 5

6π local and abs min, f(π) = −π endpt max

23. f ′(x) = sec2 x− 1 ≥ 0, x ∈(− 1

3π,12π

); critical pt. 0 ;

f(− 1

3π)

= 13π −

√3 endpt and abs min, no abs max

24. f ′(x) = 2 sinx cosx(2 sin2 x− 1), x ∈(0, 2

3 π);

critical pts. 14π,

12π; f(0) = 0 endpt and abs max, f

(14π

)= − 1


f(

12π

)= 0 local and abs max, f

(23π

)= − 3

16 endpt min



SECTION 4.4 143

25.

f ′(x) =

⎧⎪⎪⎨⎪⎪⎩

−2, 0 < x < 1

1, 1 < x < 4

−1, 4 < x < 7

critical pts. 1, 4;

f(0) = 0 endpt max, f(1) = −2 local and abs min,

f(4) = 1 local and absolute max, f(7) = −2 endpt and abs min

26. f ′(x) =

⎧⎪⎪⎨⎪⎪⎩

1, −8 < x < −3

2x + 1, −3 < x ≤ 2

5, 2 < x < 5

critical pts. −3, − 12 ;

f(−8) = 1 endpt min, f(−3) = 6 local max

f(− 1

2

)= − 1

4 local and abs min

27.

f ′(x) =

⎧⎪⎪⎨⎪⎪⎩

2x, −2 < x < −1

2 − 2x, −1 < x < 3

1, 3 < x < 6

critical pts. −1, 1, 3

f(−2) = 5 endpt max, f(−1) = 2 local and abs min,

f(1) = 6 local and abs max, f(3) = 2 local and abs min

28. f ′(x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

−2x− 2, −2 < x < 0

−1, 0 < x < 2

1, 2 < x < 3

(x− 2)2, 3 < x < 4

critical pts. −1, 0, 2, 3 ;

f(−2) = 2 endpt min, f(−1) = 3 local and abs max,

f(0) not an extreme value, f(2) = 0 local and abs min,

f(3) = 13 local min, f(4) = 8

3 endpt max



144 SECTION 4.4

29.

f ′(x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

−1, −3 < x < −1

1, −1 < x < 0

2x− 4, 0 < x < 3

2, 3 ≤ x < 4

critical pts. −1, 0, 2

f(−3) = 2 endpt and abs max, f(−1) = 0 local min,

f(0) = 2 local and abs max, f(2) = −2 local and abs min

30. f ′(x) =

⎧⎪⎪⎨⎪⎪⎩

−2x, 0 < x < 1

−2, 1 < x < 2

−x, 2 < x < 3

Note: 1 is not in the domain of f .

critical pts. 2;

f(0) = 0 endpt and abs max, f(2) = −2 local max,

f(3) = − 92 endpt and abs min

31. 32.

33. Not possible: f(1) = f(3) = 0 implies f ′(c) = 0 for some c ∈ (1, 3) (Rolle’s theorem).

34. f(x) = x− 12π

sin 2πx

35. Let p(x) = x3 + ax2 + bx + c. Then p′(x) = 3x2 + 2ax + b is a quadratic with discriminant Δ = 4a2 −12b = 4(a2 − 3b). If a2 ≤ 3b, then Δ ≤ 0. This implies that p′(x) does not change sign on (−∞,∞).

On the other hand, if a2 − 3b > 0, then Δ > 0 and p′ has two real zeros, c1 and c2, from which it

follows that p has extreme values at c1 and c2. Therefore, if p has no extreme values, then we must

have a2 − 3b ≤ 0.



SECTION 4.4 145

36. f(x) = (1 + x)r − (1 + rx), x ≥ −1.

f ′(x) = r[(1 + x)r−1 − 1

]; f ′(x) = 0 =⇒ x = 0

f ′′(x) = r(r − 1)(1 + x)r−2; f ′′(0) = r(r − 1) > 0 =⇒ f has a local minimum at x = 0

Since 0 is the only critical number of f , the local minimum must be an absolute minimum.

37. By contradiction. If f is continuous at c, then, by the first-derivative test (4.3.4), f(c) is not a local

maximum.

38. Since f(c) ≥ f(x) for all x in some open interval around c, and likewise f(c) ≤ f(x) for all x in some

open interval around c, it follows that f must be constant on some open interval containing c.

39. If f is not differentiable on (a, b), then f has a critical point at each point c in (a, b) where f ′(c) does

not exist. If f is differentiable on (a, b), then by the mean-value theorem there exists c in (a, b) where

f ′(c) = [f(b) − f(a)]/(b− a) = 0. This means c is a critical point of f .

40. We give a proof by contradiction. Suppose for no c in (c1, c2) is f(c) a local minimum. By Theorem

2.6.2, f has a minimum on [c1, c2] and therefore, this minimum must occur at c1 or c2. Suppose that

f(c1) is an endpoint minimum. Then for some δ1 > 0,

(*) f(x) ≥ f(c), x ∈ [c1, c1 + δ1).

Since f(c1) is a local maximum, there exists δ2 > 0 such that

(**) f(x) ≤ f(c1), x ∈ (c1 − δ2, c1 + δ2).

Set δ = min [δ1, δ2]. From (*) and (**), it follows that

f(x) = f(c1), x ∈ (c1, c1 + δ).

This means that f has a local minimum on (c1, c2). The argument at c2 is similar.

41. Let f(x) =

⎧⎨⎩1, if x is a rational number

0, if x is an irrational number

42. f(x) = sinx and f(x) = cosx each have an infinite number of local maxima and local minima occur-

ring at distinct points. However, the local maximum values are all the same, 1, and the local minimum

values are all the same, −1. The function f(x) = 12x + sinx has an infinite number of local maxima

and local minima, and the local maximum values and the local minimum values are all different.

43. Let M be a positive number. Then

P (x) −M ≥ anxn −

(|an−1|xn−1 + · · · + |a1|x + |a0| + M

)for x > 0

≥ anxn − xn−1 (|an−1| + · · · + |a1| + |a0| + M) for x > 1

≥ xn−1 [anx− (|an−1| + · · · + |a1| + |a0| + M)]

It now follows that

P (x) −M ≥ 0 for x ≥ K =|an−1| + · · · + |a1| + |a0| + M

an

1/n

.



146 SECTION 4.4

44. Let R be a rectangle with its diagonals having length c, and let x be the length of one of its sides.

Then the length of the other side is y =√c2 − x2 and the area of R is given by

A(x) = x√c2 − x2

Now

A′(x) =√c2 − x2 − x2

√c2 − x2

=c2 − 2x2

√c2 − x2

,

and

A′(x) = 0 =⇒ x =√

22

c

It is easy to verify that A has a maximum at x =√

22

c. Since y =√

22

c when x =√

22

c, it follows that

the rectangle of maximum area is a square.

45. f(0) = f(1) = 0 and f(x) > 0 on (0, 1).

f ′(x) = −qxp(1 − x)q−1 + pxp−1(1 − x)q; f ′(x) = 0 implies x =p

p + q. The absolute maximum value

of f is f(p/(p + q)) =(

p

p + q

)p

·(

q

p + q

)q

46. Let S = x3 + y3 where x + y = 16. Then S(x) = x3 + (16 − x)3 and

S′(x) = 3x2 − 3(16 − x)2 = 96(x− 8); S ′(x) = 0 =⇒ x = 8 =⇒ y = 8;

S′′(x) = 96; S ′′(8) = 96 > 0 =⇒ S has a local minimum at x = 8.

It now follows that S(8) is the absolute minimum of S.

47. Setting R′(θ) =v2 cos 2θ

16= 0, gives θ =

π

4. Since R′′

(π

4

)= −v2

8< 0, θ =

π

4is a maximum.

48. Cut the wire into two pieces, one of length x and the other of length L− x. Suppose that the wire of

length x is used to form the equilateral triangle, and the other piece is used to form the square. Then

the area of the triangle is√

3x2/36, and the area of the square is (L− x)2/16. Now, let

S(x) =√

336

x2 +116

(L− x)2

Then

S ′(x) =√

318

x− 18

(L− x)

=4√

3 + 972

x− 18L

Setting S′(x) = 0 we find that

x =9

4√

3 + 9L. ∼= 0.5650L

Now,

S(0) =116

L2 = 0.0625L2 (absolute maximum)



SECTION 4.4 147

S

(9

4√

3 + 9L

)∼= 0.0390L2 (absolute minimum)

S(L) =√

336

L2 = 0.0481L2

To maximize the sum of the areas, use the wire to form a square; to minimize the sum, use x ∼= 0.5650L

to form the triangle and the remainder to form the square.

49. critical pts: x1 = −1.452, x2 = 0.760

f(−1.452) local maximum

f(0.727) local minimum

f(3) absolute maximum

f(−2.5) absolute minimum

50. critical pts: x1 = −2.179, x2 = 1, x3 = 1.158

f(−3) endpoint maximum

f(1) local maximum

f(−2.158), f(1.158) local minima


f(−2.179) absolute minimum

51. critical points: x1 = −1.683, x2 = −0.284,

x3 = 0.645, x4 = 1.760

f(−1.683), f(0.645) local maxima

f(−0.284), f(1.760) local minima

f(π) absolute maximum

f(−π) absolute minimum

52. critical pts: x1 = −0.667, x2 = 0

f(−0.667) local maximum

f(0) local minimum


f(−3) absolute minimum



148 SECTION 4.5

53. Yes; M = f(2) = 1; m = f(1) = f(3) = 0

1 2 3x

0.5

1

y

54. No; f is discontinuous at x = 3; M = f(3) = 72 ; m = f(0) = − 19

4

55. Yes; M = f(6) = 2 +√

3; m = f(1) = 32

1 4 6x

1

2

3

y

SECTION 4.5

1. Set P = xy and y = 40 − x. We want to maximize

P (x) = x(40 − x), 0 ≤ x ≤ 40.

P ′(x) = 40 − 2x, P ′(x) = 0 =⇒ x = 20.

Since P increases on (0, 20] and decreases on [20, 40), the abs max of P occurs when x = 20. Then,

y = 20 and xy = 400.

The maximal value of xy is 400.

2. Set A = xy and 2x + 2y = 24 or y = 12 − x. We want to maximize

A(x) = x(12 − x), 0 ≤ x ≤ 12.

A′(x) = 12 − 2x, P ′(x) = 0 =⇒ x = 6.

Since A increases on [0, 6] and decreases on [6, 12], the abs max of A occurs when x = 6. Then, y = 6.

The dimensions of the rectangle having perimeter 24 and maximum area are: 6 × 6.

3.Minimize P

P = x + 2y, 200 = xy, y = 200/x

P (x) = x +400x

, x > 0.



SECTION 4.5 149

P ′(x) = 1 − 400x2

, P ′(x) = 0 =⇒ x = 20.

Since P decreases on (0, 20 ] and increases on [20, ∞), the abs min of P occurs when x = 20.

To minimize the fencing, make the garden 20 ft (parallel to barn) by 10 ft.

4. Maximize A

A = 2xy, y = 4 − x2

A(x) = 2x(4 − x2) = 8x− 2x3, 0 ≤ x ≤ 2.

A′(x) = 8 − 6x2, P ′(x) = 0 =⇒ x =2√3.

Since A increases on [0, 2/√

3] and decreases on [2/√

3, 2], the abs max of A occurs when x = 2/√

3.

The maximal area is 329

√3.

5. Maximize A

A = xy, x2 + y2 = 82, y =√

64 − x2

A(x) = x√

64 − x2, 0 ≤ x ≤ 8.

A′(x) =√

64 − x2 + x

( −x√64 − x2

)=

64 − 2x2

√64 − x2

, A′(x) = 0 =⇒ x = 4√

2 .

Since A increases on (0, 4√

2 ] and decreases on [4√

2, 8), the abs max of A occurs when x = 4√

2 .

Then, y = 4√

2 and xy = 32.

The maximal area is 32.

6. Maximize P = 2x + 2y; xy = A (constant) =⇒ y =A

x

P (x) = 2x +2Ax

, x > 0; P ′(x) = 2 − 2Ax2

;

P ′(x) = 0 =⇒ x =√A =⇒ y =

√A

The rectangle having a given area A and minimum perimeter is a square of side length√A.

7. Minimize P = 3x + 2y

A = xy = 15, 000, y =15, 000

x



150 SECTION 4.5

P (x) = 3x +30, 000

x, 0 < x < ∞.

P ′(x) = 3 − 30, 000x2

=3(x2 − 10, 000)

x2; P ′(x) = 0 =⇒ x = 100.

Since P decreases on (0, 100] and increases on [100,∞), the abs min of P occurs when x = 100. When

x = 100, y = 150; at least 600 feet of fencing is needed.

8. Minimize C = 300y + 400x

A = 5000 = xy ⇒ y =5000x

; p

C(x) =1, 500, 000

x+ 400x, x > 0;

C ′(x) = − 1, 500, 000x2

+ 400;

C ′(x) = 0 =⇒ x ∼= 61.24.

C ′′(x) =3, 000, 000

x3; C ′′(61.24) > 0.

C has an abs min at x = 61.24. The dimensions that will minimize the cost are: x = 61.24, y = 81.65.

9. Maximize L

To account for the semi-circular portion admitting less

light per square foot, we multiply its area by 1/3.

L = 2xy +13

(πx2

2

),

2x + 2y + πx = p, y = 12 (p− 2x− πx)

L = 2x(p− 2x− πx

2

)+

16πx2

L(x) = px−(

2 +56π

)x2, 0 ≤ x ≤ p

2 + π.

L′(x) = p−(

4 +53π

)x; L′(x) = 0 =⇒ x =

3p12 + 5π

.

Since L′′(x) < 0 for all x in the domain of L, the local max at x = 3p/(12 + 5π) is the abs max.

For the window that admits the most light, take the radius of the semicircle as3p

12 + 5πft.

10. Maximize A

A = xy, x + 2y = 800

A(y) = (800 − 2y)y = 800y − 2y2, 0 ≤ y ≤ 400.

A′(y) = 800 − 4y, A′(y) = 0 =⇒ y = 200.

Since A increases on [0, 200] and decreases on [200, 400], the abs max of A occurs when y = 200.

The dimensions of the field of maximum area are: 200 × 400.



SECTION 4.5 151

11. Maximize A

A = xy,34

=y

4 − x(similar triangles)

y = 34 (4 − x)

A(x) =3x4

(4 − x), 0 ≤ x ≤ 4.

A′(x) = 3 − 3x2, A′(x) = 0 =⇒ x = 2.

Since A increases on (0, 2] and decreases on [2, 4), the abs max of A occurs when x = 2.

To maximize the area of the rectangle, take P as the point(2, 3

2

).

12. The equation of the third side is: y = mx + (1 −m).

The base of the triangle is: b =m− 1m

.

The two lines intersect when 3x = mx + (1 −m);

=⇒ x =1 −m

3 −m=⇒ h =

3(1 −m)3 −m

.

We want to minimize

A(m) =12m− 1m

3(1 −m)3 −m

= − 32

(1 −m)2

3m−m2, m < 0.

A′(m) = − 32

(m + 3)(m− 1)(3m−m2)2

, A′(m) = 0 =⇒ m = −3.

The area of the triangle is a minimum when the slope of the line is −3.

13. Minimize A

A = 12 (x-intercept) (y-intercept)

Equation of line: y − 5 = m(x− 2)

x-intercept: 2 − 5m

y-intercept: 5 − 2m

A =12

(2 − 5

m

)(5 − 2m) = 10 − 2m− 25

2m

A(m) = 10 − 2m− 252m

, m < 0.

A′(m) = −2 +25

2m2, A′(m) = 0 =⇒ m = −5

2.

Since A′′(m) = −25/m3 > 0 for m < 0, the local min at m = −5/2 is the abs min.

The triangle of minimal area is formed by the line of slope −5/2.



152 SECTION 4.5

14. Since limm→0−

A(m) = +∞, no minimum exists.

15. Maximize V

V = 2x2h, 2(2x2 + xh + 2xh

)= 100, h =

50 − 2x2

3x

V = 2x2

(50 − 2x2

3x

)

V (x) = 1003 x− 4

3x3, 0 ≤ x ≤ 5.

V ′(x) = 1003 − 4x2, V ′(x) = 0 =⇒ x = 5

3

√3 .

Since V ′′(x) = −8x < 0 on (0, 5), the local max at x = 53

√3 is the abs max.

The base of the box of greatest volume measures 53

√3 in. by 10

3

√3 in.

16. With no top, we have 2x2 + 2xh + 4xh = 100, or h =50 − x2

3x.

Maximize V (x) = 2x2

(50 − x2

3x

)=

2x3

(50 − x2), 0 ≤ x ≤ 5√

2.

V ′(x) =1003

− 2x2, V ′(x) = 0 =⇒ x =53

√6.

Since V ′′(x) = −4x < 0 on (0, 5√

2), the local max at x = 53

√6 is the abs max.

The base of the box of greatest volume measures 53

√6 in. by 10

3

√6 in.

17. Maximize A

A = 12hy

2x + y = 12 =⇒ y = 12 − 2x

Pythagorean Theorem:

h2 +(y

2

)2

= x2 =⇒ h =

√x2 −

(y

2

)2

Thus, h =√x2 − (6 − x)2 =

√12x− 36.

A(x) = (6 − x)√

12x− 36, 3 ≤ x ≤ 6.

A′(x) = −√

12x− 36 + (6 − x)(

6√12x− 36

)=

72 − 18x√12x− 36

,

A′(x) = 0 =⇒ x = 4.

Since A increases on (3, 4] and decreases on [4, 6), the abs max of A occurs at x = 4.

The triangle of maximal area is equilateral with side of length 4.

18. It is sufficient to minimize the square of the distance:

S = (x− 0)2 + (y − 6)2 = 8y + (y − 6)2 since x2 = 8y where y ≥ 0.

S ′(y) = 2y − 4, S′(y) = 0 =⇒ y = 2.

The points on the parabola that are closest to (0, 6) are: (4, 2) and (−4, 2).



SECTION 4.5 153

19. Minimize d

d =√

(y2 − 0)2 + (y − 3)2

The square-root function is increasing;

d is minimal when D = d2 is minimal.

D(y) = y4 + (y − 3)2, y real.

D′(y) = 4y3 + 2(y − 3) = (y − 1)(4y2 + 4y + 6

), D′(y) = 0 at y = 1.

Since D′′(y) = 12y2 + 2 > 0, the local min at y = 1 is the abs min.

The point (1, 1) is the point on the parabola closest to (0, 3).

20. f(x) = Ax−1/2 + Bx1/2, f(9) = 6 =⇒ 13 A + 3B = 6.

f ′(x) =−A

2x3/2+

B

2x1/2; f ′(9) = 0 =⇒ −A

54+

B

6= 0.

Solving the two equations gives: A = 9, B = 1.

21. The figure shows a rectangle inscribed in the ellipse 16x2 + 9y2 = 144. The area of the rectangle is:

A = (2x)(2y) = 4xy. Solving the equation of the ellipse for y, we get y = 43

√9 − x2.

Maximize A = 163 x

√9 − x2, 0 ≤ x ≤ 3.

A′(x) =163

[√9 − x2 − x2

√9 − x2

]=

163

(9 − 2x2

√9 − x2

)

A′(x) = 0 =⇒ x = 3/√

2.

Since A(0) = A(3) = 0, we can conclude that A(3/√

2)

is the absolute maximum value of A.

At x = 3/√

2, y = 4/√

2 and A = 24; the maximum possible

area for a rectangle inscribed in the ellipse is 24 sq. units.

x

y

22. Simply repeat the solution of Exercise 21, replacing 9 by a2 and 16 by b2. The result is:

x = a/√

2, y = b/√

2; the maximum possible area of a rectangle inscribed in the ellipse is A =

4(a/√

2)(b/√

2) = 2ab square units.

23. Maximize A

A = xy +√

34

x2, 30 = 3x + 2y, y =30 − 3x

2

A(x) = 15x− 32x2 +

√3

4x2, 0 ≤ x ≤ 10.

A′(x) = 15 − 3x +√

32

x, A′(x) = 0 =⇒ x =30

6 −√

3=

1011

(6 +√

3).



154 SECTION 4.5

Since A′′(x) = −3 +√

32

< 0 on (0, 10), the local max at x =1011

(6 +

√3

)is the abs max.

The pentagon of greatest area is composed of an equilateral triangle with side1011

(6 +

√3)∼= 7.03

in. and rectangle with height1511

(5 −

√3

)∼= 4.46 in.

24. To maximize the area, use the cross-section that is wider at the top.

A(h) = 4h + h√

16 − h2, 0 ≤ h ≤ 4;

A′(h) = 4 +16 − 2h2

√16 − h2

;

A′(h) = 0 =⇒ h = 2√

3.

The depth of the gutter that has maximum carrying capacity is: 2√

3 inches.

25. Maximize V

V = x(8 − 2x)(15 − 2x)

x ≥ 0

8 − 2x ≥ 0

15 − 2x ≥ 0

⎫⎪⎪⎬⎪⎪⎭ =⇒ 0 ≤ x ≤ 4

V (x) = 120x− 46x2 + 4x3, 0 ≤ x ≤ 4.

V ′(x) = 120 − 92x + 12x2 = 4(3x− 5)(x− 6), V ′(x) = 0 at x = 53 .

Since V increases on(0, 5

3

)and decreases on [ 53 , 4), the abs max of V occurs when x = 5

3 .

The box of maximal volume is made by cutting out squares 5/3 inches on a side.

26. Minimize P

P = 2x + 2y;

(x− 4)(y − 6) = 81; y =81

x− 4+ 6.

P (x) = 2x + 2(

81x− 4

+ 6), x > 4.

P ′(x) = 2 − 162(x− 4)2

, P ′(x) = 0 =⇒ x = 13.

Since P ′′(x) =324

(x− 4)3> 0 when x > 4, x = 13 is the abs min.

The most economical page has dimensions: width 13 cm, length 15 cm.



SECTION 4.5 155

27. MinimizeAP + BP + CP = S

length AP =√

9 + y2

length BP = 6 − y

length CP =√

9 + y2

S(y) = 6 − y + 2√

9 + y2, 0 ≤ y ≤ 6.

S′(y) = −1 +2y√

9 + y2, S′(y) = 0 =⇒ y =

√3.

Since

S(0) = 12, S(√

3)

= 6 + 3√

3 ∼= 11.2, and S(6) = 6√

5 ∼= 13.4,

the abs min of S occurs when y =√

3.

To minimize the sum of the distances, take P as the point(0,√

3).

28. Refer to Exercise 27. Here we want to minimize

S(y) = 3 − y + 2√

36 + y2, 0 ≤ y ≤ 3.

S′(y) = −1 +2y√

36 + y2, S′(y) = 0 =⇒ y =

√12 > 3.

Thus, the minimum must occur at one of the endpoints: S(0) = 15, S(3) = 2√

45 < S(0).

To minimize the sum of the distances, take P = (0, 3).

29. Minimize L

L2 = y2 + (x + 1)2.

By similar trianglesy

x + 1=

8x, y =

8x

(x + 1).

L2 =[(

8x

)(x + 1)

]2

+ (x + 1)2 = (x + 1)2(

64x2

+ 1)

Since L is minimal when L2 is minimal, we consider the function

f(x) = (x + 1)2(

64x2

+ 1), x > 0.

f ′(x) = 2(x + 1)(

64x2

+ 1)

+ (x + 1)2(−128

x3

)

=2(x + 1)

x3

[x3 − 64

], f ′(x) = 0 =⇒ x = 4.

Since f decreases on (0, 4] and increases on [4,∞), the abs min of f occurs when x = 4.

The shortest ladder is 5√

5 ft long.



156 SECTION 4.5

30. Maximize L = x + y.

By similar triangles,y

6=

x√x2 − 64

L(x) = x +6x√

x2 − 64, x > 8

L′(x) = 1 − 384(x2 − 64)3/2

L′(x) = 0 =⇒ x =√

64 + (384)2/3 ∼= 10.81

L(10.81) ∼= 19.73; the longest ladder is approximately 19.7 ft.

31. Maximize A

(We use feet rather than inches to reduce arithmetic.)

A = (L− 1)(W − 4

3

)LW = 27 =⇒ W =

27L

A = (L− 1)(

27L

− 43

)=

853

− 27L

− 43L

A(L) =853

− 27L

− 43L, 1 ≤ L ≤ 81

4.

A′(L) =27L2

− 43, A′(L) = 0 =⇒ L =

92.

Since A′(L) = −54/L3 < 0 for 1 < L < 814 , the max at L = 9

2 is the abs max.

The banner has length 9/2 ft = 54 in. and height 6 ft = 72 in.

32. Assume V =13πr2h = 1, or h =

3πr2

.

Minimize surface area S = πr√r2 + h2 = πr

√r2 +

9π2r4

=√π2r6 + 9

r.

dS

dr=

2π2r6 − 9r2√π2r6 + 9

= 0 =⇒ r =(

92π2

)1/6

=31/3

21/6π1/3

=⇒ h =3

π

(9

2π2

)1/3=

31/321/3

π1/3= r

√2.

33. Find the extreme values of A

A = πr2 + x2

2πr + 4x = 28 =⇒ x = 7 − 12πr.

A(r) = πr2 +(

7 − 12πr

)2

, 0 ≤ r ≤ 14π.



SECTION 4.5 157

Note: the endpoints of the domain correspond to the instances when the string is not cut: r = 0when no circle is formed, r = 14/π when no square is formed.

A′(r) = 2πr − π

(7 − 1

2πr

), A′(r) = 0 =⇒ r =

144 + π

.

Since A′′(r) = 2π + π2/2 > 0 on (0, 14/π) , the abs min of A occurs when r = 14/(4 + π) and

the abs max of A occurs at one of the endpts: A(0) = 49, A(14/π) = 196/π > 49.

(a) To maximize the sum of the two areas, use all of the string to form the circle.

(b) To minimize the sum of the two areas, use 2πr = 28π/(4 + π) ∼= 12.32 inches of string for the

circle.

34. Maximize V = x2h given that x2 + 4xh = 12 =⇒ h =12 − x2

4x.

V (x) = x2

(12 − x2

4x

)= 3x− 1

4 x3, 0 < x ≤

√12.

V ′(x) = 3 − 34 x

2, V ′(x) = 0 =⇒ x = 2. Since V increases on (0, 2] and decreases on [2,√

12),

V has an abs max at x = 2; the maximum volume is V (2) = 4 cu ft.

35. Maximize V

V = πr2h

By similar triangles85

=h

5 − ror h =

85(5 − r).

V (r) =8π5r2(5 − r), 0 ≤ r ≤ 5.

V ′(r) =8π5

(10r − 3r2

), V ′(r) = 0 =⇒ r = 10/3.

Since V increases on (0, 10/3] and decreases on [10/3, 5), the abs max of V occurs when r = 10/3.

The cylinder with maximal volume has radius 10/3 and height 8/3.

36. Maximize A = 2πrh =16π5

r(5 − r), 0 ≤ r ≤ 5,(h = 8

5 (5 − r) from Exercise 33).

A′(r) =16π5

(5 − 2r), A′(r) = 0 =⇒ r = 52

The curved surface is a maximum when r = 52 , h = 4.

37. Minimize C

In dollars,

C = cost base + cost top + cost sides

= .35(x2

)+ .15

(x2

)+ .20(4xy)

= 12x

2 + 45xy

Volume = x2y = 1250 y =1250x2



158 SECTION 4.5

C(x) =12x2 +

1000x

, x > 0.

C ′(x) = x− 1000x2

, C ′(x) = 0 =⇒ x = 10.

Since C ′′(x) = 1 + 2000/x3 > 0 for x > 0, the local min of C at x = 10 is the abs min.

The least expensive box is 12.5 ft tall with a square base 10 ft on a side.

38. Maximize A = xy. By similar triangles

y

b− x=

h

b, so

A(x) =h

bx(b− x), 0 ≤ x ≤ b.

A′(x) =h

b(b− 2x), A′(x) = 0 ⇒ x =

b

2.

Since A is increasing on [0, b/2] and decreasing on [b/2, b], A has an abs max at x = b/2

A(b/2) = 14 hb = 1

2 area of triangle ABC.

39.Minimize A

A = 12 (h)(2x) = hx

Triangles ADC and ABE are similar:

AD

DC=

AB

BEor

h

x=

AB

r.

Pythagorean Theorem:

r2 + (AB)2 = (h− r)2.

Thus

r2 +(hr

x

)2

= (h− r)2.

Solving this equation for h we find that

h =2x2r

x2 − r2.

A(x) =2x3r

x2 − r2, x > r.

A′(x) =

(x2 − r2

) (6x2r

)− 2x3r(2x)

(x2 − r2)2=

2x2r(x2 − 3r2

)(x2 − r2)2

,

A′(x) = 0 =⇒ x = r√

3.



SECTION 4.5 159

Since A decreases on(r, r

√3

]and increases on

[r√

3, ∞), the local min at x = r

√3 is the abs

min of A. When x = r√

3, we get h = 3r so that FC = 2r√

3 and AF = FC =√h2 + x2 =

2r√

3. The triangle of least area is equilateral with side of length 2r√

3.

40. Maximize A(x) = 12 (r + x)2

√r2 − x2

= (r + x)√r2 − x2, 0 ≤ x ≤ r.

A′(x) =r2 − rx− 2x2

√r2 − x2

;

A′(x) = 0 =⇒ x =r

2.

Since A increases on [0, r/2] and decreases on [r/2, r], A has an abs max at x = r/2;

A(r/2) =3√

34

r2.

41. Maximize V

V = πr2h

By the Pythagorean Theorem,

(2r)2 + h2 = (2R)2

so

h = 2√R2 − r2.

V (r) = 2πr2√R2 − r2, 0 ≤ r ≤ R.

V ′(r) = 2π[2r√R2 − r2 − r3

√R2 − r2

]=

2πr(2R2 − 3r2

)√R2 − r2

V ′(r) = 0 =⇒ r = 13R

√6.


3R√

6]

and decreases on[13R

√6, R

), the local max at r = 1

3R√

6 is

the abs max.

The cylinder of maximal volume has base radius 13R

√6 and height 2

3R√

3.

42. Maximize A = 2πrh = 4πr√R2 − r2, 0 ≤ r ≤ R,

(h = 2

√R2 − r2 from Exercise 39

).

A′(r) =4π(R2 − 2r2)√

R2 − r2, A′(r) = 0 =⇒ r =

R√2.

The curved surface is a maximum when r =R√2, h = R

√2.



160 SECTION 4.5

43. Maximize V

V = 13πr

2h

Pythagorean Theorem

Case 1 : (h−R)2 + r2 = R2

Case 2 : (R− h)2 + r2 = R2

Case 1 : h ≥ R Case 2 : h ≤ R In both cases

r2 = R2 − (R− h)2 = 2hR− h2.

V (h) = 13π

(2h2R− h3

), 0 ≤ h ≤ 2R.

V ′(h) = 13π

(4hR− 3h2

), V ′(h) = 0 at h =

4R3

.


3R]

and decreases on[43R, 2R

), the local max at h = 4

3R is the abs

max.

The cone of maximal volume has height 43R and radius 2

3R√

2.

44. Maximize V = 13 πr

2h, where r2 + h2 = a2.

V (h) = 13 π(a2 − h2)h, 0 ≤ h ≤ a,

V ′(h) = 13 π(a2 − 3h2), V ′(h) = 0 ⇒ h =

a√3.

Maximum volume V(a/

√3)

= 227 πa

3√

3.

45. Minimize C

In units of $10, 000,

C = cost of cableunderground

+ cost of cableunder water

= 3(4 − x) + 5√x2 + 1.

Clearly, the cost is unnecessarily high if

x > 4 or x < 0.

C(x) = 12 − 3x + 5√x2 + 1, 0 ≤ x ≤ 4.

C ′(x) = −3 +5x√x2 + 1

, C ′(x) = 0 =⇒ x = 3/4.



SECTION 4.5 161

Since the domain of C is closed, the abs min can be identified by evaluating C at each critical point:

C(0) = 17, C(

34

)= 16, C(4) = 5

√17 ∼= 20.6.

The minimum cost is $160, 000.

46. Maximize α− θ.

Since the tangent function is an increasing function

on [0, π/2), it suffices to maximize tan(α− θ).

tan(α− θ) =tanα− tan θ

1 + tanα tan θ; tanα =

16x, tan θ =

9x.

Thus, we maximize T (x) =

16x

− 9x

1 +16x

· 9x

=7x

x2 + 144, x > 0.

T ′(x) =7(144 − x2)(x2 + 144)2

, T ′(x) = 0 =⇒ x = 12.

Stand 12 ft from the wall for the most favorable view.

47. P ′(θ) =−mW (m cos θ − sin θ)

(m sin θ + cos θ)2; P is minimized when tan θ = m.

48. R(θ) =2v2

g cos2 αcos θ sin(θ − α), 0 < θ < 1

2 π

R′(θ) =2v2

g cos2 α[− sin θ sin(θ − α) + cos θ cos(θ − α)]

=2v2

g cos2 αcos(2θ − α)

R′(θ) = 0 =⇒ 2θ − α = 12 π =⇒ θ = 1

4 π + 12 α.

49. Minimize I =a

x2+

b

(s− x)2.

I ′(x) = −2ax3

+2b

(s− x)3, I ′(x) = 0 =⇒ x =

a13 s

a13 + b

13

.

50. Minimize D = (y − y1)2 + (x− x1)2, where y = − 1B

(Ax + C).

D′ = 2(− 1B

(Ax + C) − y1

) (−A

B

)+ 2(x− x1) = 0

=⇒ x =B2x1 −AC −ABy1

A2 + B2and thus y =

A2y1 −BC −ABx1

A2 + B2.

Thus d =√D =

√(A2y1 −BC −ABx1

A2 + B2− y1

)2

+(B2x1 −AC −ABy1

A2 + B2− x1

)2

=|Ax1 + By1 + C|√

A2 + B2.



162 SECTION 4.5

51. The slope of the line through (a, b) and (x, f(x)) isf(x) − b

x− a.

Let D(x) = [x− a]2 + [b− f(x)]2. Then D′(x) = 0

=⇒ 2[x− a] − 2[b− f(x)]f ′(x) = 0

=⇒ f ′(x) =x− a

b− f(x).

52. Let P = (x1, x21) and Q = (x2, x

22). The slope of the line PQ is

x21 − x2

2

x1 − x2= x1 + x2. This is

perpendicular to the slope of the tangent to the parabola through point P, which is 2x1. Thus

x1 + x2 = − 12x1

=⇒ x2 = −x1 −1

2x1. Now minimize:

D = (x21 − x2

2)2 + (x1 − x2)2 = (x2

1 − x22)

2(1 + (x1 + x2)2)

=(

2x1 +1

2x1

)2 (1 +

14x2

1

)

Thus D′ =(

2x1 +1

2x1

)2 (− 1

2x31

)+ 2

(2x1 +

12x1

) (2 − 1

2x21

) (1 +

14x2

1

)= 0

=⇒(

2x1 +1

2x1

) (1

2x31

)= 2

(2 − 1

2x21

) (1 +

14x2

1

)

=⇒ x1 = ±√

22

, and y1 =12.

53. Set F (x) = 6x4 − 16x3 + 9x2. For integral values of x, F (x) = f(x).

F ′(x) = 24x3 − 48x2 + 18x = 6x(4x2 − 8x + 3) = 6x(2x− 1)(2x− 3)

F ′(x) = 0 at x = 0, 1/2, 3/2.

F ′(x) < 0 on (−∞, 0) ∪ (1/2, 3/2); F ′(x) > 0 on (0, 1/2) ∪ (3/2,∞). F has local minima at x = 0

and x = 3/2; F has a local maximum at x = 1/2. F (0) = f(0) = 0, F (1) = f(1) = −1, F (2) =

f(2) = 4; n = 1 minimizes f(n).

54. Let x be the number of passengers and R the revenue in dollars.

R(x) =

{37x, 16 ≤ x ≤ 35[

37 − 12 (x− 35)

]x, 35 < x ≤ 48;

R′(x) =

{37, 16 < x < 35

1092 − x, 35 < x < 48.

The critical number is 35. From R(16) = 592, R(35) = 1295, and R(48) = 1464 we conclude that

the revenue is maximized by taking a full load of 48 passengers.



SECTION 4.5 163

55. Let x be the number of customers and P the net profit in dollars. Then 0 ≤ x ≤ 250 and

P (x) =

{12x, 0 ≤ x ≤ 50

[12 − 0.06(x− 50)]x, 50 < x ≤ 250;

P ′(x) =

{12, 0 ≤ x ≤ 50

15 − 0.12x, 50 < x ≤ 250.

The critical points are: x = 50, x = 125. From P (0) = 0, P (50) = 600, P (125) = 937.50, and

P (250) = 0, we conclude that the net profit is maximized by servicing 125 customers.

56. Maximize P (x) = 2cy + cx, where c is the price of low-grade steel and y =(

40 − 5x10 − x

).

P (x) = 2c(40 − 5x10 − x

) + cx

P ′(x) = 2c[(10 − x)(−5) − (40 − 5x)(−1)

(10 − x)2

]+ c,

P ′(x) = 0 =⇒ x = 10 −√

20 or about 512

tons.

57. y = mx− 1400

(m2 + 1)x2. When y = 0, x =800mm2 + 1

.

Differentiating x with respect to m, x′ =800 − 800m2

(m2 + 1)2= 0

=⇒ m = 1.

58. When x = 300, y = 300m− 2252

(m2 + 1).

Differentiating y with respect to m, y′ = 300 − 225m = 0

=⇒ m =43.

59. Driving at ν mph, the trip takes300ν

hours and uses(

1 +1

400ν2

)300ν

gallons of fuel.

Thus, the expenses are:

E(ν) = 2.60(

1 +1

400ν2

)300ν

+ 20(

300ν

)= 1.95ν +

6780ν

, 35 ≤ x ≤ 70.

Differentiating, we get

E′(ν) = 1.95 − 6780ν2

, and E ′(ν) = 0 at ν ∼= 59

E is decreasing on [35, 59] and increasing on [59, 70]; the minimal expenses occur

when the truck is driven at 59 mph.

60. Let ν be the speed of the boat measured in kilometers per hour. A 100 kilometer trip will take

100/ν hours.



164 SECTION 4.5

Fixed costs: F (ν) = 2500(

100ν

)dollars.

Fuel costs: G(ν) = kν2; 400 = k(10)2 implies k = 4. Therefore, G(ν) = 4ν2

(100ν

)dollars.

Total cost: C(ν) = 2500(

100ν

)+ 4ν2

(100ν

)= 100

(2500ν

+ 4ν), 0 < ν < ∞. Differentiating, we

get

C ′(ν) = 100(−2500

ν2+ 4

)= 100

(4ν2 − 2500

ν2

)and C ′(ν) = 0 at ν = 25.

C is decreasing on (0, 25] and increasing on [25,∞); the speed that minimizes the expenses is 25

kilometers per hour. If we replace the 100 kilometers by M kilometers, the total cost will be

C(ν) = M

(2500ν

+ 4ν)

and we will get exactly the same result. The minimizing speed is independent of the length of the trip.

61. Minimize SA = 2πrh + 2πr2, where 2r ≤ h < 6 and πr2h = 16π (hence h =16r2

).

Thus SA =32πr

+ 2πr2. Differentiating, SA′ = −32πr2

+ 4πr = 0

=⇒ r3 = 8, so r = 2 feet and h = 4 feet. Thus no minimum exists.

62. We want to maximize the ratioincomecost

=200, 000n

1, 000, 000n + 100, 000(1 + 2 + · · · + n− 1) + 5, 000, 000

=2n

10n + 12 (n− 1)n + 50

=4n

n2 + 19n + 100.

Let f(x) =4x

x2 + 19x + 100, x > 0

Then f ′(x) =(x2 + 19x + 100)4 − 4x(2x + 19)

(x2 + 19x + 100)2=

4(100 − x2)(x2 + 19x + 100)2

= 0

=⇒ x = 10.

Since f ′(x) > 0 for x < 0, f ′(x) < 0 for x > 10, f has an absolute maximum at x = 10.

A ten story building provides the greatest return on investment.

63. Swimming at 2 miles per hour, Maggie will reach point B in 1 hour; walking along the shore (a distance

of π miles), she will reach point B in π/5 ∼= 0.63 hours. Suppose that she swims to a point C and then

walks to B. Let θ be the central angle (measured in radians) determined by the points B and C. See

the figure. By the law of cosines, the square of the distance from A to C is given by

d2 = 12 + 12 − 2(1)(1) cos (π − θ) = 2 + 2 cos θ.

Therefore the distance d from A to C is d =√

2 + 2 cos θ. The distance from C to B is θ.



SECTION 4.5 165

Now, the total length of time to swim to C and then walk to B is

T (θ) = 2√

2 + 2 cos θ + 5θ

Maggie wants to minimize T .

T ′(θ) =−2 sin θ

2√

2 + 2 cos θ+ 5.

Setting T ′(θ) = 0, we get

−2 sin θ

2√

2 + 2 cos θ+ 5 = 0 A θ B

C

which reduces to 2 cos2 θ + 25 cos θ + 23 = 0

and factors into (cos θ + 1)(2 cos θ + 23) = 0.

This implies θ = π. Maggie should walk the entire distance!

64. We modify the solution of Exercise 63, replacing the walking rate of 2 miles per hour by the rowing

rate of 3 miles per hour.

The total length of time to row to C and then walk to B is

T (θ) = 3√

2 + 2 cos θ + 5θ

The derivative of T is

T ′(θ) =−3 sin θ

2√

2 + 2 cos θ+ 5.

Setting T ′(θ) = 0, we get

−3 sin θ

2√

2 + 2 cos θ+ 5 = 0,

which reduces to

9 cos2 θ + 50 cos θ + 41 = (cos θ + 1)(9 cos θ + 41) = 0.

This implies θ = π. Again, Maggie should walk the entire distance!

65. (b) The point on the graph of f that is closest to P is:(1 +

√2, 2 +

√2)

(c) lPQ : y −(2 +

√2)

=1 −

√2

3 −√

2

(x−

[1 +

√2])

(d) and (e) lPQ = lN .

66. The point on the graph of f(x) = x− x3 that is closest to (1, 8) is (approximately) (−2.1474, 7.7548).

67. D(x) =√x2 + (7 − 3x)2; the point on the line that is closest to the origin is:

(2110 ,

710

)



166 SECTION 4.6

68. The point on the graph of f that minimizes the distance to the point (4, 3) is (approximately)

(1.3918, 2.0629).

PROJECT 4.5

1. Distance over water:√

36 + x2.

Distance over land: 12 − x.

Total energy: E(x) = W√

36 + x2 + L(12 − x), 0 ≤ x ≤ 12.

2. W = 1.5L, so E(x) = 1.5L√

36 + x2 + L(12 − x), for 0 ≤ x ≤ 12.

E′(x) =1.5Lx√36 + x2

− L = 0 =⇒ x =12√

5� 5.36.

E′(x) < 0 on (0,12√

5) and E ′(x) > 0 on (

12√5, 12), so E has an absolute minimum at

12√5.

3. (a) W = kL, k > 1, so E(x) = kL√

36 + x2 + L(12 − x), for 0 ≤ x ≤ 12.

E′(x) =kLx√36 + x2

− L = 0 =⇒ x =6√

k2 − 1.

E′(x) < 0 on (0,6√

k2 − 1) and E ′(x) > 0 on (

6√k2 − 1

, 12), so E has an absolute minimum

at6√

k2 − 1.

(b) As k increases, x decreases. As k → 1+, x increases.

(c) x = 12 =⇒ k =√

52

� 1.12.

(d) No

SECTION 4.6

1. (a) f is increasing on [a, b], [d, n]; f is decreasing on [b, d], [n, p].

(b) The graph of f is concave up on (c, k), (l,m);

The graph of f is concave down on (a, c), (k, l), (m, p).

The x-coordinates of the points of inflection are: x = c, x = k, x = l, x = m.

2. (a) g is increasing on [a, b], [c, e], [m,n]; g is decreasing on [b, c], [e,m].

(b) The graph of g is concave up on (a, b), (b, d);

The graph of g is concave down on (d,m), (m,n).

The x-coordinate of the point of inflection is: x = d.

3. (i) f ′, (ii) f, (iii) f ′′.

4. (a) x = −2 local max, no local minima.

(b) points of inflection at x = 0, 2.



SECTION 4.6 167

(c)

−2 2x

y

5. f ′(x) = −x−2, f ′′(x) = 2x−3;

concave down on (−∞, 0), concave up on (0, ∞); no pts of inflection

6. f ′(x) = 1 − x−2, f ′′(x) = 2x−3;


7. f ′(x) = 3x2 − 3, f ′′(x) = 6x;

concave down on (−∞, 0), concave up on (0, ∞); pt of inflection (0, 2)

8. f ′(x) = 4x− 5, f ′′(x) = 4; concave up on (−∞,∞)

9. f ′(x) = x3 − x, f ′′(x) = 3x2 − 1;

concave up on(−∞, − 1

3

√3

)and

(13

√3, ∞

), concave down on

(− 1

3

√3, 1

3

√3

);

pts of inflection(− 1

3

√3, − 5

36

)and

(13

√3, − 5

36

)10. f ′(x) = 3x2 − 4x3, f ′′(x) = 6x− 12x2 = 6x(1 − 2x);

concave down on (−∞, 0), and(

12 , ∞

), concave up on

(0, 1

2

);

pts of inflection (0, 0),(

12 ,

116

)

11. f ′(x) = − x2 + 1(x2 − 1)2

, f ′′(x) =2x

(x2 + 3

)(x2 − 1)3

;

concave down on (−∞, −1) and (0, 1), concave up on (−1, 0) and on (1, ∞);

pts of inflection (0, 0)

12. f ′(x) =−4

(x− 2)2, f ′′(x) =

8(x− 2)3

;


13. f ′(x) = 4x3 − 4x, f ′′(x) = 12x2 − 4;

concave up on(−∞, − 1

3

√3

)and

(13

√3, ∞

), concave down on

(− 1

3

√3, 1

3

√3

);

pts of inflection(− 1

3

√3, 4

9

)and

(13

√3, 4

9

)



168 SECTION 4.6

14. f ′(x) =6(1 − x2)(x2 + 1)2

, f ′′(x) =12x(x2 − 3)(x2 + 1)3

;

concave down on(−∞,−

√3)

and(0,

√3), concave up on

(−√

3, 0)

and(√

3, ∞);

pts of inflection(−√

3,− 32

√3), (0, 0),

(√3, 3

2

√3)

15. f ′(x) =−1

√x (1 +

√x )2

, f ′′(x) =1 + 3

√x

2x√x (1 +

√x )3

;

concave up on (0, ∞); no pts of inflection

16. f ′(x) = 15 (x− 3)−4/5, f ′′(x) = − 4

25 (x− 3)−9/5;

concave up on (−∞, 3), concave down on (3,∞); pt of inflection (3, 0)

17. f ′(x) = 53 (x + 2)2/3, f ′′(x) = 10

9 (x + 2)−1/3;

concave down on (−∞, −2), concave up on (−2, ∞); pt of inflection (−2, 0)

18. f ′(x) =4 − 2x2

(4 − x2)1/2, f ′′(x) =

2x(x2 − 6)(4 − x2)3/2

Note: dom (f) = [−2, 2]

concave up on (−2, 0), concave down on (0, 2); pt of inflection (0, 0)

19. f ′(x) = 2 sinx cosx = sin 2x, f ′′(x) = 2 cos 2x;

concave up on(0, 1

4π)

and(

34π, π

), concave down on

(14π,

34π

);

pts of inflection(

14π,

12

)and

(34π,

12

)20. f ′(x) = −4 cosx sinx− 2x, f ′′(x) = −4(cos2 x− sin2 x) − 2 = −4 cos 2x− 2;

concave down on(0, 1

3π)

and(

23π, π

), concave up on

(13π,

23π

);

pts of inflection(

13π,

9 − 2π2

18

)and

(23π,

9 − 8π2

18

)

21. f ′(x) = 2x + 2 cos 2x, f ′′(x) = 2 − 4 sin 2x;

concave up on(0, 1

12π)

and on(

512π, π

), concave down on

(112π,

512π

);

pts of inflection(

112

π,72 + π2

144

)and

(512

π,72 + 25π2

144

)

22. f ′(x) = 4 sin3 x cosx, f ′′(x) = 4 sin2 x[3 cos2 x− sin2 x];

concave up on(0, 1

3π)

and(

23π, π

), concave down on

(13π,

23π

);

pts of inflection(

13π,

916

)and

(23π,

916

)23. points of inflection: (±3.94822, 10.39228)

24. f ′′(x) = 0 at x ∼= ±0.94, ±2.57, ±3.71, ±5.35

25. points of inflection: (−3, 0), (−2.11652, 2, 39953), (−0.28349,−18.43523)

26. f ′′(x) > 0 for all x ∈ dom f



SECTION 4.6 169

27. f(x) = x3 − 9x

(a) f ′(x) = 3x2 − 9 = 3(x2 − 3)

f ′(x) ≥ 0 ⇒ x ≤ −√

3 or x ≥√

3;

f ′(x) ≤ 0 ⇒ −√

3 ≤ x ≤√

3.

Thus, f is increasing on (−∞,−√

3] ∪ [√

3,∞)

and decreasing on [−√

3,√

3].(b) f(−

√3) ∼= 10.39 is a local maximum;

f(√

3) ∼= −10.39 is a local minimum.

(c) f ′′(x) = 6x;

The graph of f is concave up on (0,∞) and concave down on (−∞, 0).

(d) point of inflection: (0, 0)

28. f(x) = 3x4 + 4x3 + 1

(a) f ′(x) = 12x3 + 12x2 = 12x2(x + 1)

f ′(x) ≥ 0 ⇒ x ≥ −1;

f ′(x) ≤ 0 ⇒ x ≤ −1.

Thus, f is increasing on [−1,∞)

and decreasing on (−∞,−1].

(b) f(−1) = 0 is a local minimum;

no local maxima.

(c) f ′′(x) = 36x2 + 24x = 36x(x + 2

3

);

The graph of f is concave up on(−∞,− 2

3

)and (0,∞); and concave down on

(− 2

3 , 0).

(d) points of inflection:(− 2

3 ,1127

), (0, 1)

29. f(x) =2x

x2 + 1

(a) f ′(x) = − 2(x + 1)(x− 1)(x2 + 1)2

f ′(x) ≥ 0 ⇒ −1 ≤ x ≤ 1;

f ′(x) ≤ 0 ⇒ x ≤ −1 or x ≥ 1.

Thus, f is increasing on [−1, 1];

and decreasing on (−∞,−1] ∪ [1,∞).

(b) f(−1) = −1 is a local minimum;

f(1) = 1 is a local maximum.

(c) f ′′(x) =4x(x +

√3)(x−

√3)

(x2 + 1)3

f ′′(x) > 0 ⇒ x ≤ −√

3 or x ≥√

3;

f ′′(x) < 0 ⇒ −√

3 < x <√

3.

The graph of f is concave up on (−√

3 , 0) ∪ (√

3,∞) and concave down

on (−∞,−√

3) ∪ (0,√

3).

(d) points of inflection: (−√

3,−√

3/2), (0, 0), (√

3,√

3/2)



170 SECTION 4.6

30. f(x) = x1/3(x− 6)2/3

(a) f ′(x) =x− 2

x2/3(x− 6)1/3

f ′(x) ≥ 0 ⇒ x ≤ 2, x �= 0, or x > 6;

f ′(x) ≤ 0 ⇒ 2 ≤ x < 6.

Thus, f is increasing on (−∞, 2] ∪ [6,∞)

and decreasing on [2, 6].

(b) f(2) = 2(4)1/3 is a local maximum;

f(6) = 0 is a local minimum.

(c) f ′′(x) =−8

x5/3(x− 6)4/3;

The graph of f is concave down on (0,∞) and concave down up (−∞, 0).


31. f(x) = x + sinx, x ∈ [−π, π]

(a) f ′(x) = 1 + cosx

f ′(x) > 0 on (−π, π)

Thus, f is increasing on [−π, π].

(b) No local extrema

(c) f ′′(x) = − sinx

f ′′(x) > 0 for x ∈ (−π, 0);

f ′′(x) < 0 for x ∈ (0, π).

The graph of f is concave up on (−π, 0) and concave down on (0, π).


32. f(x) = sinx + cosx, x ∈ [0, 2π]

(a) f ′(x) = cosx− sinx

f ′(x) ≥ 0 ⇒ 0 ≤ x ≤ 14π or 5

4π ≤ x ≤ 2π

f ′(x) ≤ 0 ⇒ 14π ≤ x ≤ 5

4π

Thus, f is increasing on[0, 1

4π]∪

[54π, 2π

];

f is decreasing on[14π,

54π

].

(b) f(π/4) =√

2 is a local maximum;

f(5π/4) = −√

2 is a local minimum.

(c) f ′′(x) = − sinx− cosx

f ′′(x) > 0 ⇒ 34π < x < 7

4π;

f ′′(x) < 0 ⇒ 0 < x < 34π or 7

4π < x < 2π.

The graph of f is concave up on(

34π,

74π

)and concave down on

(0, 3

4π)∪

(74π, 2π

).

(d) points of inflection:(

34π, 0

),

(74π, 0

)



SECTION 4.6 171

33. f(x) =

{x3, x < 1

3x− 2, x ≥ 1.

(a) f ′(x) =

{3x2, x < 1

3, x ≥ 1;

f ′(x) > 0 on (−∞, 0) ∪ (0,∞)

Thus, f is increasing on (−∞,∞).

(b) No local extrema

(c) f ′′(x) =

{6x, x < 1

0, x ≥ 1;

f ′′(x) > 0 for x ∈ (0, 1); f ′′(x) < 0 for x ∈ (−∞, 0).

Thus, the graph of f is concave up on (0, 1) and concave down on (−∞, 0).

The graph of f is a straight line for x ≥ 1.


34. f(x) =

{2x + 4, x ≤ −1

3 − x2, x > −1.

(a) f ′(x) =

{2, x ≤ −1

−2x, x > −1;

f ′(x) ≥ 0 on (−∞, 0];

f ′(x) ≤ 0 on [0,∞).

Thus, f is increasing on (−∞, 0] and decreasing on [0,∞).

(b) f(0) = 3 is a local maximum.

(c) f ′′(x) =

{0, x < −1

−2, x > −1;

f ′′(x) < 0 for x ∈ (−1,∞).

Thus, the graph of f is concave down on (−1,∞).

The graph of f is a straight line for x ≤ −1.

(d) There are no points of inflection.

35. 36.



172 SECTION 4.6

37. 38.

39. Since f ′′(x) = 6x− 2(a + b + c), set d = 13 (a + b + c). Note that f ′′(d) = 0 and that f is concave

down on (−∞, d) and concave up on (d, ∞); (d, f(d)) is a point of inflection.

40. f ′(x) = 2cx− 2x−3, f ′′(x) = 2c + 6x−4. To have a point of inflection at 1 we need

f ′′(1) = 0 =⇒ 2c + 6 = 0 =⇒ c = −3

41. Since (−1, 1) lies on the graph, 1 = −a + b.

Since f ′′(x) exists for all x and there is a pt of inflection at x = 13 , we must have f ′′ ( 1

3

)= 0.

Therefore

0 = 2a + 2b.

Solving these two equations, we find a = − 12 and b = 1

2 .

Verification: the function

f(x) = −12x3 +

12x2

has second derivative f ′′(x) = −3x + 1. This does change sign at x = 13 .

42. f(x) = Ax1/2 + Bx−1/2; f(1) = 4 =⇒ A + B = 4.

f ′(x) = 12 Ax−1/2 − 1

2 Bx−3/2, f ′′(x) = − 14 Ax−3/2 + 3

4 Bx−5/2.

To have a point of inflection at (1, 4), we need f ′′(1) = 0 =⇒ − 14 A + 3

4 B = 0.

Solving the two equations gives A = 3, B = 1.

43. First, we require that(

16π, 5

)lie on the curve:

5 = 12A + B.

Next we require thatd2y

dx2= −4A cos 2x− 9B sin 3x be zero (and change sign) at x = 1

6π :

0 = −2A− 9B.

Solving these two equations, we find A = 18, B = −4.

Verification: the function

f(x) = 18 cos 2x− 4 sin 3x

has second derivative f ′′(x) = −72 cos 2x + 36 sin 3x. This does change sign at x = 16π.



SECTION 4.6 173

44. f(x) = Ax2 + Bx + C; f ′(x) = 2Ax + B; f ′′(x) = 2A.

(a) Concave up =⇒ f ′′(x) > 0 =⇒ A > 0; to decrease between A and B we need

f ′(x) < 0, for x between A and B =⇒ B ≤ −2A2.

(b) Concave down =⇒ f ′′(x) < 0 =⇒ A < 0; to have f ′(x) > 0 for x between A and B we

need 2A2 + B ≥ 0 and 2AB + B ≥ 0, that is, B ≥ −2A2 and B(2A + 1) ≥ 0. If A > − 12 , then

we need B ≥ 0 (and automatically B ≥ −2A2). If A ≤ − 12 , then we need B ≤ 0 and B ≥ −2A2.

The conditions are: − 12 < A < 0, B ≥ 0, or A ≤ − 1

2 , −2A2 ≤ B ≤ 0.

45. Let f ′(x) = 3x2 − 6x + 3. Then we must have f(x) = x3 − 3x2 + 3x + c for some constant c. Note

that f ′′(x) = 6x− 6 and f ′′(1) = 0. Since (1,−2) is a point of inflection of the graph of f, (1,−2)

must lie on the graph. Therefore,

13 − 3(1)2 + 3(1) + c = −2 which implies c = −3

and so f(x) = x3 − 3x2 + 3x− 3.

46. f ′(x) = cosx and f ′′(x) = − sinx = −f(x).

Thus f is concave down when f ′′(x) < 0 =⇒ f(x) > 0.

Similarly, f is concave up when f ′′(x) > 0 =⇒ f(x) < 0.

g′(x) = − sinx and g′′(x) = − cosx = −g(x), so g(x) has the same property.

47. (a) p′′(x) = 6x + 2a is negative for x < −a/3, and positive for x > −a/3. Therefore, the graph of p

has a point of inflection at x = −a/3.

(b) p′(x) = 3x2 + 2ax + b. The discriminant of this quadratic is 4a2 − 12b = 4(a2 − 3b). Thus, p′ has

two real zeros iff a2 > 3b.

(c) If a2 ≤ 3b, then b ≥ a2/3 and p′(x) = 3x2 + 2ax + b ≥ 3x2 + 2ax + a2/3 = 3(x + 13a)

2 ≥ 0 for

all x; p is increasing in this case.

48. It is sufficient to show that the x-coordinate of the point of inflection is the x-coordinate of the mid-

point of the line segment connecting the local extrema. It is easy to show that the x-coordinate of the

point of inflection is x0 = − 13 a. Now suppose that p has local extrema at x1 and x2, x1 �= x2. Then

p′(x1) = p′(x2) = 0 ⇒ 3x21 + 2ax1 + b− (3x2

2 + 2ax2 + b) = 0 ⇒ x1 + x2 = − 23 a.

Thus,x1 + x2

2= − 1

3 a = x0.

49. (a)(b) No. If f ′′(x) < 0 and f ′(x) < 0 for all x, then

f(x) < f ′(0)x + f(0) on (0,∞), which implies that

f(x) → −∞ as x → ∞.



174 SECTION 4.6

50. Let f(x) = anxn + an−1x

n−1 + · · · + a2x2 + a1x + a0.

Then f ′′(x) = n(n− 1)anxn−2 + · · · + 2a2 is a degree n− 2 polynomial which can have at most

n− 2 roots. Hence f has at most n− 2 points of inflection.

51.

(a) concave up on (−4,−0.913) ∪ (0.913, 4)

concave down on (−0.913, 0.913)

(b) pts of inflection at x = −0.913, 0.913

52.

(a) up: (−2π,−3.64), (−1.077, 1.077), (3.64, 2π)

down: (−3.64,−1.077) ∪ (1.077, 3.64)

(b) pts of inflection at x = ±3.64, ±1.077

53.

(a) concave up on:

(−π,−1.996) ∪ (−.0345, 2.550)

concave down on:

(−1.996,−0.345) ∪ (2.550, π)

(b) pts of inflection at:

x = −1.996, −0.345, 2, 550

54.

(a) concave up on (−5, 5)

(b) no pts of inflection

55. (a) f ′′(x) = 0 at x = 0.68824, 2.27492, 4.00827, 5.59494

(b) f ′′(x) > 0 on (0.68824, 2.27492) ∪ (4.00827, 5.59494)

(c) f ′′(x) < 0 on (0, 0.68824) ∪ (2.27492, 4.00827) ∪ (5.59494, 2π)

56. (a) f ′′(x) �= 0 for all x (b) f ′′(x) > 0 on (−∞,−1) ∪ (1,∞)

(c) f ′′(x) < 0 on (−1, 1)

57. (a) f ′′(x) = 0 at x = ±1, 0, ±0.32654, ±0.71523

(b) f ′′(x) > 0 on (−0.71523,−0.32654) ∪ (0, 0.32654) ∪ (0.71523, 1) ∪ (1,∞)

(c) f ′′(x) < 0 on (−∞,−1) ∪ (−1,−0.71523) ∪ (−0.32654, 0) ∪ (0, 32654, 0.71523)



SECTION 4.7 175

58. (a) f ′′(x) = 0 at x = 0 (b) f ′′(x) > 0 on (−4, 0)

(c) f ′′(x) < 0 on (0, 4)

SECTION 4.7

1. (a) ∞ (b) −∞ (c) ∞ (d) 1

(e) 0 (f) x = −1, x = 1 (g) y = 0, y = 1

2. (a) d (b) c (c) x = a, x = b

(d) y = d (e) p (f) q

3. vertical: x = 13 ; horizontal: y = 1

3 4. vertical: x = −2; horizontal: none

5. vertical: x = 2 ; horizontal: none 6. vertical: none; horizontal: y = 0

7. vertical: x = ±3 ; horizontal: y = 0 8. vertical: x = 16; horizontal: y = 0

9. vertical: x = − 43 ; horizontal: y = 4

9 10. vertical: x = 13 ; horizontal: y = 4

9

11. vertical: x = 52 ; horizontal: y = 0 12. vertical: x = 1

2 ; horizontal: y = − 18

13. vertical: none ; horizontal: y = ± 32 14. vertical: x = 8; horizontal: y = 0

15. vertical: x = 1 ; horizontal: y = 0 16. vertical: x = ± 1; horizontal: y = ± 2

17. vertical: none ; horizontal: y = 0 18. vertical: none; horizontal: y = 0

19. vertical: x =(2n + 1

2

)π ; horizontal: none 20. vertical: x = 2nπ; horizontal: none

21. f ′(x) = 43 (x + 3)1/3; neither 22. f ′(x) = 2

5 x−3/5; cusp

23. f ′(x) = − 45 (2 − x)−1/5; cusp 24. neither; f(−1) is not defined

25. f ′(x) = 65x

−2/5(1 − x3/5

); tangent 26. f ′(x) = 7

5 (x− 5)2/5; neither

27. f(−2) undefined; neither 28. f ′(x) = 37 (2 − x)−4/7; tangent

29. f ′(x) =

⎧⎨⎩

12 (x− 1)−1/2, x > 1

− 12 (1 − x)−1/2, x < 1;

cusp

30. f ′(x) = (4x− 3)(x− 1)−2/3; tangent



176 SECTION 4.7

31. f ′(x) =

⎧⎨⎩

13 (x + 8)−23, x > −8

− 13 (x + 8)−2/3, x < −8;

cusp

32. f is not defined for x > 2; neither 33. f not continuous at 0; neither

34. f is not continuous at 0; neither

35. 36.

37. 38.

39. f(x) = x− 3x1/3

(a) f ′(x) = 1 − 1x2/3

f is increasing on (−∞,−1] ∪ [1,∞)

f is decreasing on [−1, 1]

(b) f ′′(x) = 23 x

−5/3

concave up on (0,∞); concave down on (−∞, 0)

vertical tangent at (0, 0)

40. f(x) = x2/3 − x1/3

(a) f ′(x) = 23 x

−1/3 − 13 x

−2/3 =2x1/3 − 1

3x2/3

f is increasing on[18 ,∞

)f is decreasing on

(−∞, 1

8

](b) f ′′(x) = − 2

9 x−4/3 + 2

9 x−5/3 =

2(1 − x1/3)9x5/3

concave up on (0, 1)

concave down on (−∞, 0) ∪ (1,∞)




SECTION 4.7 177

41. f(x) = 35 x

5/3 − 3x2/3

(a) f ′(x) = x2/3 − 2x−1/3 =x− 2x1/3

f is increasing on (−∞, 0] ∪ [2,∞)

f is decreasing on [0, 2]

(b) f ′′(x) = 23 x

−1/3 + 23 x

−4/3 =2x + 23x4/3

concave up on (−1, 0) ∪ (0,∞); concave down on

(−∞,−1)

vertical cusp at (0, 0)

42. f(x) =√|x|

=

{x1/2, x ≥ 0

(−x)1/2, x < 0.

(a) f ′(x) =

{12 x

−1/2, x > 0

− 12 (−x)−1/2, x < 0;

f is increasing on [0,∞)

f is decreasing on (−∞, 0]

(b) f ′′(x) =

{− 1

4 x−3/2, x > 0

− 14 (−x)−3/2, x < 0;

concave down on (−∞, 0) ∪ (0,∞)


43.

no asymptotes


44.

vertical asymptotes: x = 1, x = −1

horizontal asymptote: y = 1

no vertical tangents or cusps



178 SECTION 4.7

45.

horizontal asymptotes: y = −1, y = 1


46.

horizontal asymptote: y = 4


47. (a) p odd; (b) p even.

48. [r(x) − (ax + b)] =Q(x)q(x)

→ 0 as x → ±∞, since deg [Q(x)] < deg [q(x)].

49.

vertical asymptote: x = 0

oblique asymptote: y = x

50.

vertical asymptote: x = −1

oblique asymptote: y = 2x + 1

51.


oblique asymptote: y = x

52.


oblique asymptote: y = −3x + 1

53. oblique asymptote: y = 3x− 4 54. oblique asymptote: y = 5x− 3



SECTION 4.8 179

55. y = 1 is a horizontal asymptote.

f(x) =√x2 + 2x− x =

(√x2 + 2x− x

)·√x2 + 2x + x√x2 + 2x + x

=2x√

x2 + 2x + x→ 1.

56. y = − 12 is a horizontal asymptote.

f(x) =√x4 − x2 − x2 =

(√x4 − x2 − x2

)·√x4 − x2 + x2

√x4 − x2 + x2

=−x2

√x4 − x2 + x2

→ − 12 .

SECTION 4.8

[Rough sketches; not scale drawings]

1. f(x) = (x− 2)2

f ′(x) = 2(x− 2)

f ′′(x) = 2

2. f(x) = 1 − (x− 2)2

f ′(x) = −2(x− 2)

f ′′(x) = −2

3. f(x) = x3 − 2x2 + x + 1

f ′(x) = (3x− 1)(x− 1)

f ′′(x) = 6x− 4



180 SECTION 4.8

4. f(x) = x3 − 9x2 + 24x− 7

f ′(x) = 3x2 − 18x + 24

f ′′(x) = 6x− 18

5. f(x) = x3 + 6x2, x ∈ [−4, 4]

f ′(x) = 3x(x + 4)

f ′′(x) = 6x + 12

6. f(x) = x4 − 8x2, x ∈ [0,∞)

f ′(x) = 4x3 − 16x

f ′′(x) = 12x2 − 16

7. f(x) = 23x

3 − 12x

2 − 10x− 1

f ′(x) = (2x− 5)(x + 2)

f ′′(x) = 4x− 1



SECTION 4.8 181

8. f(x) = x(x2 + 4)2 = x5 + 8x3 + 16x

f ′(x) = 5x4 + 24x2 + 16

f ′′(x) = 20x3 + 48x = 4x(5x2 + 12)

9. f(x) = x2 + 2x−1

f ′(x) = 2x− 2x−2 = 2(x3 − 1

)/x2

f ′′(x) = 2 + 4x−3

asymptote: x = 0

10. f(x) = x− x−1,

f ′(x) = 1 + x−2

f ′′(x) = −x−3

asymptotes: x = 0, y = x



182 SECTION 4.8

11. f(x) = (x− 4)/x2

f ′(x) = (8 − x)/x3

f ′′(x) = (2x− 24)/x4

asymptotes: x = 0, y = 0

12. f(x) =x + 2x3

=1x2

+2x3

f ′(x) = − 2x3

− 6x4

=−2x− 6

x4

f ′′(x) =6x4

+24x5

=6(x + 4)

x5


13. f(x) = 2x1/2 − x, x ∈ [0, 4]

f ′(x) = x−1/2(1 − x1/2

)f ′′(x) = − 1

2x−3/2

14. f(x) = 14 x−√

x, x ∈ [0, 9]

f ′(x) = 14 − 1

2 x−1/2

f ′′(x) = 14 x

−3/2



SECTION 4.8 183

15. f(x) = 2 + (x + 1)6/5

f ′(x) = 65 (x + 1)1/5

f ′′(x) = 625 (x + 1)−4/5

16. f(x) = 2 + (x + 1)7/5

f ′(x) = 75 (x + 1)2/5

f ′′(x) = 1425 (x + 1)−3/5

17. f(x) = 3x5 + 5x3

f ′(x) = 15x2(x2 + 1

)f ′′(x) = 30x

(2x2 + 1

)

18. f(x) = 3x4 + 4x3

f ′(x) = 12x3 + 12x2 = 12x2(x + 1)

f ′′(x) = 36x2 + 24x = 12x(3x + 2)



184 SECTION 4.8

19. f(x) = 1 + (x− 2)5/3

f ′(x) = 53 (x− 2)2/3

f ′′(x) = 109 (x− 2)−1/3

20. f(x) = 1 + (x− 2)4/3

f ′(x) = 43 (x− 2)1/3

f ′′(x) = 49 (x− 2)−2/3

21. f ′(x) =8x

(x2 + 4)2

f ′′(x) = 488(4 − 3x2)(x2 + 4)3

f ′(x) < 0 on (−∞, 0)

f ′(x) > 0 on (0, ∞)

f ′′(x) < 0 on(−∞, −2/

√3)∪

(2/√

3, ∞);

f ′′(x) > 0 on(−2/

√3, 2/

√3);

asymptote: y = 1

22. f(x) =2x2

x + 1

f ′(x) =2x(x + 2)(x + 1)2

f ′′(x) =4

(x + 1)3

asymptote: y = 2x− 2



SECTION 4.8 185

23. f(x) =x

(x + 3)2

f ′(x) =3 − x

(x + 3)3

f ′′(x) =2x− 12(x + 3)4

asymptotes: x = −3, y = 0

24. f(x) =x

x2 + 1

f ′(x) =1 − x2

(x2 + 1)2

f ′′(x) =2x(x2 − 3)(x2 + 1)3

asymptote: y = 0

25. f(x) =x2

x2 − 4

f ′(x) =−8x

(x2 − 4)2

f ′′(x) =8(3x2 + 4)(x2 − 4)3

asymptotes: x = −2, x = 2, y = 1



186 SECTION 4.8

26. f(x) =1

x3 − x

f ′(x) =1 − 3x2

(x3 − x)2

increasing on(−1/

√3, 0

)∪

(0, 1/

√3);

decreasing otherwise

f ′′(x) =2

(6x4 − 3x2 + 1

)(x3 − x)3

concave up on (−1, 0) ∪ (1,∞)

concave down on (−∞, 0) ∪ (0, 1)

asymptotes: x = −1, x = 0, x = 1, y = 0

27. f(x) = x(1 − x)1/2

f ′(x) = 12 (1 − x)−1/2(2 − 3x)

f ′′(x) = 14 (1 − x)−3/2(3x− 4)

28. f(x) = (x− 1)4 − 2(x− 1)2

f ′(x) = 4(x− 1)3 − 4(x− 1)

f ′′(x) = 12(x− 1)2 − 4

29. f(x) = x + sin 2x, x ∈ [0, π]

f ′(x) = 1 + 2 cos 2x

f ′′(x) = −4 sin 2x



SECTION 4.8 187

30. f(x) = cos3 x + 6 cosx, x ∈ [0, π]

f ′(x) = −3 sinx(cos2 x + 2)

f ′′(x) = −9 cos3 x

31. f(x) = cos4 x, x ∈ [0, π]

f ′(x) = −4 cos3 x sinx

f ′′(x) = 4 cos2 x(3 sin2 x− cos2 x

)

32. f(x) =√

3x− cos 2x, x ∈ [0, π]

f ′(x) =√

3 + 2 sin 2x

f ′′(x) = 4 cos 2x

33. f(x) = 2 sin3 x + 3 sinx, x ∈ [0, π]

f ′(x) = 3 cosx(2 sin2 x + 1

)f ′′(x) = 9 sinx

(1 − 2 sin2 x

)



188 SECTION 4.8

34. f(x) = sin4 x, x ∈ [0, π]

f ′(x) = 4 sin3 x cosx

f ′′(x) = 4 sin2 x(3 cos2 x− sin2 x

)

35. f(x) = [(x + 1) − 1]3 + 1

f ′(x) = 3x2

f ′′(x) = 6x

36. f(x) = x3(x + 5)2

f ′(x) = 5x2(x + 3)(x + 5)

f ′′(x) = 10x(2x2 + 12x + 15)

37. f(x) = x2(5 − x)3

f ′(x) = 5x(2 − x)(5 − x)2

f ′′(x) = 10(5 − x)(2x2 − 8x + 5

)



SECTION 4.8 189

38. f(x) =

{4 − 2x + x2, 0 < x < 2

4 + 2x− x2, x ≤ 0, x ≥ 2

f ′(x) =

{−2 + 2x, 0 < x < 2

2 − 2x, x < 0, x > 2

f ′′(x) =

{2, 0 < x < 2

−2, x < 0, x > 2

39. f(x) =

{4 − x2, |x| > 1

x2 + 2, −1 ≤ x ≤ 1

f ′(x) =

{−2x, |x| > 1

2x, −1 < x < 1

f ′′(x) =

{−2, |x| > 1

2, −1 < x < 1

40. f(x) = x− x1/3

f ′(x) = 1 − 13 x

−2/3

f ′′(x) = 29 x

−5/3




190 SECTION 4.8

41. f(x) = x(x− 1)1/5

f ′(x) = 15 (x− 1)−4/5(6x− 5)

f ′′(x) = 225 (x− 1)−9/5(3x− 5)


42. f(x) = x2(x− 7)1/3

f ′(x) =7x(x− 6)

3(x− 7)2/3

f ′′(x) =14(2x2 − 24x + 63)

9(x− 7)5/3


43. f(x) = x2 − 6x1/3

f ′(x) = 2x−2/3(x5/3 − 1

)f ′′(x) =

23x−5/3

(3x5/3 + 2

)




SECTION 4.8 191

44. f(x) =2x√x2 + 1

f ′(x) =2

(x2 + 1)3/2

f ′′(x) =−6x

(x2 + 1)5/2

asymptotes: y = 2, y = −2

45. f(x) =(

x

x− 2

)1/2

; x ≤ 0, x > 2

f ′(x) = −(

x

x− 2

)−1/2

(x− 2)−2

f ′′(x) = (2x− 1)(

x

x− 2

)−3/2

(x− 2)−4


46. f(x) =(

x

x + 4

)1/2

; x < −4, x ≥ 0

f ′(x) = 2(

x

x + 4

)−1/2

(x + 4)−2

f ′′(x) =−4(x + 1)

(x + 4)5/2x3/2

asymptotes: x = −4, y = 1



192 SECTION 4.8

47. f(x) = x2(x2 − 2

)−1/2, |x| >

√2

f ′(x) = x(x2 − 4

) (x2 − 2

)−3/2

f ′′(x) = 2(x2 + 4

) (x2 − 2

)−5/2

asymptotes: x = −√

2, x =√

2

48. f(x) = 3 cos 4x, x ∈ [0, π]

f ′(x) = −12 sin 4x

f ′′(x) = −48 cos 4x

49. f(x) = 2 sin 3x, x ∈ [0, π]

f ′(x) = 6 cos 3x

f ′′(x) = −18 sin 3x

50. f(x) = 3 + 2 cotx + csc2 x, x ∈ (0, π/2)

f ′(x) = −2 csc2 x (1 + cotx)

f ′′(x) = 2 csc2 x(3 cot2 x + 2 cotx + 1

)

asymptote: x = 0



SECTION 4.8 193

51. f(x) = 2 tanx− sec2 x, x ∈ (0, π/2)

= −(1 − tanx)2

f ′(x) = 2 sec2 x (1 − tanx)

f ′′(x) = −2 sec2 x(3 tan2 x− 2 tanx + 1

)

asymptote: x = 12π

52. f(x) = 2 cosx + sin2 x

f ′(x) = 2 sinx (cosx− 1)

f ′′(x) = 2(2 cos2 x− cosx− 1)

53. f(x) =sinx

1 − sinx, x ∈ (−π, π)

f ′(x) =cosx

(1 − sinx)2

f ′′(x) =1 − sinx + cos2 x

(1 − sinx)3

asymptote: x = 12π



194 SECTION 4.8

54. f(x) =1

1 − cosx, x ∈ (−π, π)

f ′(x) =− sinx

(1 − cosx)2

f ′′(x) =2 − cosx− cos2 x

(1 − cosx)3

asymptote: x = 0

55. (a) f increases on (−∞,−1] ∪ (0, 1] ∪ [3,∞);

f decreases on [−1, 0) ∪ [1, 3]; critical points: x = −1, 0, 1, 3.

(b) concave up on (−∞,−3) ∪ (2,∞)

concave down on (−3, 0) ∪ (0, 2).

(c) The graph does not necessarily have

a horizontal asymptote.

56. (a)



SECTION 4.8 195

(b) (c)

(d) F is discontinuous at 0; limx→0

sin(1/x) does not exist

G is continuous at 0: |x sin(1/x)| = |x| | sin(1/x)| ≤ |x| → 0,

so that limx→0

G(x) = 0 = G(0).

H is continuous at 0:∣∣x2 sin(1/x)

∣∣ = |x|2 | sin(1/x)| ≤ |x|2 → 0,

so that limx→0

H(x) = 0 = H(0).

(e) F is not differentiable at 0 since it is not continuous at 0.

G is not differentiable at 0: limh→0

G(h) −G(0)h

= limh→0

sin(1/h) does not exist.

H is differentiable at 0: H ′(0) = limh→0

H(h) −H(0)h

= limh→0

h sin(1/h) = 0.

57. f(x) − x2 =x3 − x1/3

x− x2 =

x3 − x1/3 − x3

x= − 1

x2/3→ 0 as x → ±∞.

The figure shows the graphs of f and g(x) = x2 for x > 0. Since f and g are even functions,

the graphs are symmetric about the y-axis.

x

y

58. (a)

−a ax

b

y

(b)b

a

√x2 − a2 =

b

ax

√1 − a2

x2→ b

ax as x → ∞. Therefore,

b

a

√x2 − a2 − b

ax → 0 as x → ∞.

(c) If x < 0, then the second quadrant arc is given by y = − b

ax

√1 − a2

x2→ − b

ax as x → −∞.



196 SECTION 4.9

SECTION 4.9

1. x(5) = −6; v(t) = 3 − 2t so v(5) = −7 and speed = 7; a(t) = −2 so a(5) = −2.

2. x(3) = −12; v(t) = 5 − 3t2 so v(3) = −22 and speed = 22; a(t) = −6t so a(3) = −18.

3. x(1) = 6; v(t) = −18/(t + 2)2 so v(1) = −2 and speed = 2,

a(t) = 36/(t + 2)3 so a(1) = 4/3.

4. x(3) = 1; v(t) =6

(t + 3)2so v(3) = 1/6 = speed; a(t) = −12/(t + 3)3 so a(3) = −1/18.

5. x(1) = 0, v(t) = 4t3 + 18t2 + 6t− 10 so v(1) = 18 and speed = 18,

a(t) = 12t2 + 36t + 6 so a(1) = 54.

6. x(2) = −20; v(t) = 4t3 − 18t so v(2) = −4 and speed = 4,

a(t) = 12t2 − 18 so a(2) = 30.

7. x(t) = 5t + 1, x′(t) = v(t) = 5, x′′(t) = a(t) = 0.

v(t) �= 0, a(t) = 0 for all t.

8. x(t) = 4t2 − t + 3, x′(t) = v(t) = 8t− 1, x′′(t) = a(t) = 8.

v(t) = 0 at t = 1/8; a(t) �= 0 for all t.

9. x(t) = t3 − 6t2 + 9t− 1, x′(t) = v(t) = 3t2 − 12t + 9, x′′(t) = a(t) = 6t− 12.

v(t) = 3(t− 1)(t− 3), v(t) = 0 at t = 1, 3; a(t) = 0 at t = 2.

10. x(t) = t4 − 4t3 + 4t2 + 2, x′(t) = v(t) = 4t3 − 12t2 + 8t, x′′(t) = a(t) = 12t2 − 24t + 8.

v(t) = 4t(t− 1)(t− 2), v(t) = 0 at t = 0, 1, 2; a(t) = 0 at t = 1 +√

3/3.

11. A 12. C 13. A 14. C 15. A and B

16. B 17. A 18. A 19. A and C 20. B

21. The object is moving right when v(t) > 0. Here,

v(t) = 4t3 − 36t2 + 56t = 4t(t− 2)(t− 7); v(t) > 0 when 0 < t < 2 and 7 < t.

22. The object is moving left when v(t) < 0. Here,

v(t) = 3t2 − 24t + 21 = 3(t− 7)(t− 1); v(t) < 0 when 1 < t < 7.

23. The object is speeding up when v(t) and a(t) have the same sign.

v(t) = 5t3(4 − t) sign of v(t) :

a(t) = 20t2(3 − t) sign of a(t) :

Thus, 0 < t < 3 and 4 < t.



SECTION 4.9 197

24. The object is slowing down when v(t) and a(t) have opposite sign.

v(t) = 12t− 4t3 = 4t(3 − t2) sign of v(t) :

a(t) = 12(1 − t) sign of a(t) :

Thus, 1 < t <√

3.

25. The object is moving left and slowing down when v(t) < 0 and a(t) > 0.

v(t) = 3(t− 5)(t + 1) sign of v(t) :

a(t) = 6(t− 2) sign of a(t) :

Thus, 2 < t < 5.

26. The object is moving right and slowing down when v(t) > 0 and a(t) < 0.

v(t) = 3(t− 5)(t + 1) sign of v(t) :

a(t) = 6(t− 2) sign of a(t) :

This never happens.

27. The object is moving right and speeding up when v(t) > 0 and a(t) > 0.

v(t) = 4t(t− 2)(t− 4) sign of v(t) :

a(t) = 4(3t2 − 12t + 8) sign of a(t) :

Thus, 0 < t < 2 − 23

√3 and 4 < t.

28. The object is moving left and speeding up when v(t) < 0 and a(t) < 0.

v(t) = 4t(t− 2)(t− 4) sign of v(t) :

a(t) = 4(3t2 − 12t + 8) sign of a(t) :

Thus, 2 < t < 2 + 23

√3.

29. x(t) = (t + 1)2(t− 9)3, x′(t) = v(t) = 3(t + 1)2(t− 9)2 + 2(t + 1)(t− 9)3 = 5(t + 1)(t− 9)2(t− 3).

The object changes direction at time t = 3.

30. x(t) = t(t− 8)3, x′(t) = v(t) = (t− 8)3 + 3t(t− 8)2 = 4(t− 8)2(t− 2).


31. x(t) = (t3 − 12t)4, x/(t) = v(t) = 4(t3 − 12t)3(3t2 − 12) = 12t3(t + 2√

3)3(t− 2√

3)3(t + 2)(t− 2).

The object changes direction at times t = 2, 2√

3.

32. x(t) = (t2 − 8t + 15)3, x′(t) = v(t) = 3(t2 − 8t + 15)2(2t− 8) = 6(t2 − 8t + 15)2(t− 4).


Exercises 33–38. The object is moving right with increasing speed when both v(t) and a(t) are positive.

33. x(t) = sin 3t, x′(t) = v(t) = 3 cos 3t, x′′(t) = a(t) = −9 sin 3t.

v(t) > 0 on (0, π/6) ∪ (π/2, 5π/6) ∪ (7π/6, 3π/2) ∪ (11π/6, 2π).

a(t) > 0 on (π/3, 2π/3) ∪ (π, 4π/3) ∪ (5π/3, 2π).

v(t) and a(t) are positive on (π/2, 2π/3) ∪ (7π/6, 4π/3) ∪ (11π/6, 2π).



198 SECTION 4.9

34. x(t) = cos 2t, x′(t) = v(t) = −2 sin 2t, x′′(t) = a(t) = −4 cos 2t.

v(t) > 0 on (π/2, π) ∪ (3π/2, 2π); a(t) > 0 on (π/4, 3π/4) ∪ (5π/4, 7π/4).

v(t) and a(t) are positive on (π/2, 3π/4) ∪ (3π/2, 7π/4).

35. x(t) = sin t− cos t, x′(t) = v(t) = cos t + sin t, x′′(t) = a(t) = − sin t + cos t.

v(t) > 0 on (0, 3π/4) ∪ (7π/4, 2π); a(t) > 0 on (0, π/4) ∪ (5π/4, 2π).

v(t) and a(t) are positive on (0, π/4) ∪ (7π/4, 2π).

36. x(t) = sin t + cos t, x′(t) = v(t) = cos t− sin t, x′′(t) = a(t) = − sin t− cos t.

v(t) > 0 on (0, π/4) ∪ (5π/4, 2π); a(t) > 0 on (3π/4, 7π/4).

v(t) and a(t) are positive on (5π/4, 7π/4).

37. x(t) = t + 2 cos t, x′(t) = v(t) = 1 − 2 sin t, x′′(t) = a(t) = −2 cos t.

v(t) > 0 on (0, π/6) ∪ (5π/6, 2π); a(t) > 0 on (π/2, 3π/2).

v(t) and a(t) are positive on (5π/6, 3π/2).

38. x(t) = t−√

2 sin t, x′(t) = v(t) = 1 −√

2 cos t, x′′(t) = a(t) =√

2 sin t.

v(t) > 0 on (π/4, 7π/4); a(t) > 0 on (0, π).

v(t) and a(t) are positive on (π/4, π).

39. Since v0 = 0 the equation of motion is

y(t) = −16t2 + y0.

We want to find y0 so that y(6) = 0. From

0 = −16(6)2 + y0

we get y0 = 576 feet.

40. The equation of motion is: y(t) = −4.9t2 + y0. Therefore, the velocity is given by v(t) = −9.8t.

Since the object hits the ground at 98 m/sec., we have −9.8t = −98, and t = 10.

Therefore, y(10) = 0 = −4.9(10)2 + y0 and y0 = 490 meters.

41. The object’s height and velocity at time t are given by

y(t) = − 12gt2 + v0t and v(t) = −gt + v0

Since the object’s velocity at its maximum height is 0, it takes v0/g seconds to reach

maximum height, and

y(v0/g) = − 12 g(v0/g)2 + v0(v0/g) = v2

0/2g or v20/19.6 (meters)



SECTION 4.9 199

42. Since y0 = 0, we have y(t) = −16t2 + v0t = t(−16t + v0). Now,

y(8) = 0 =⇒ v0 = (16)8 = 128 =⇒ the initial velocity was 128 ft/sec.

43. At time t, the object’s height is y(t) = − 12 gt

2 + v0t + y0, and its velocity is v(t) = −gt + v0. Suppose

that y(t1) = y(t2), t1 �= t2. Then

− 12 gt

21 + v0t1 + y0 = − 1

2 gt22 + v0t2 + y0

12 g(t

22 − t21) = v0(t2 − t1)

gt2 + gt1 = 2v0

From this equation, we get −(−gt1 + v0) = −gt2 + v0 and so |v(t1)| = |v(t2)|.

44. Since y0 = 0, we have y(t) = −4.9t2 + v0t = t(v0 − 4.9t) The object hits the ground at t = v0/4.9 sec.,

that is, the object is in the air for v0/4.9 sec. At its maximum height, the velocity of the object is 0.

Since v(t) = −9.8t + v0, we have −9.8t + v0 = 0 and t = v0/9.8 = 12 (v0/4.9). The result follows from

this.

45. In the equation

y(t) = −16t2 + v0t + y0

we take v0 = −80 and y0 = 224. The ball first strikes the ground when

−16t2 − 80t + 224 = 0;

that is, at t = 2. Since

v(t) = y′(t) = −32t− 80,

we have v(2) = −144 so that the speed of the ball the first time it strikes theground is 144 ft/sec. Thus, the speed of the ball the third time it strikes the

ground is14

[14(144)

]= 9 ft/sec.

46. Since y0 = 0, we have y(t) = −16t2 + v0t.

y(2) = 64 =⇒ −16(2)2 + 2v0 = 64 =⇒ v0 = 64 and y(t) = −16t2 + 64t

Now, at the maximum height, v(t) = −32t + 64 = 0 =⇒ t = 2. We already know

the height at t = 2, namely 64 ft.

47. The equation is y(t) = −16t2 + 32t. (Here y0 = 0 and v0 = 32.)

(a) We solve y(t) = 0 to find that the stone strikes the ground at t = 2 seconds.

(b) The stone attains its maximum height when v(t) = 0. Solving

v(t) = −32t + 32 = 0, we get t = 1 and, thus, the maximum height is y(1) = 16 feet.



200 SECTION 4.9

(c) We want to choose v0 in

y(t) = −16t2 + v0t

so that y(t0) = 36 when v(t0) = 0 for some time t0.

From v(t) = −32t + v0 = 0 we get t0 = v0/32 so that

−16( v0

32

)2

+ v0

( v0

32

)= 36, or

v02

64= 36.

Thus, v0 = 48 ft/sec.

48. (a) Measuring height from the water surface, we have y(t) = −16t2 + y0, since v0(0) = 0.

If the stone hits the water 3 seconds later, then y(3) = −16(3)2 + y0 = 0. so y0 = 144.

(b) It takes y0/1080 seconds for the sound of the splash to reach the man so the stone hits the

at time t = 3 − y0/1080. Thus,

y(t) = −16(3 − y0

1080

)2

+ y0 = 0 =⇒ y0∼= 132.47 ft.

49. For all three parts of the problem the basic equation is

y(t) = −16t2 + v0t + y0

with

(∗) y(t0) = 100 and y(t0 + 2) = 16

for some time t0 > 0.

We are asked to find y0 for a given value of v0.

From (∗) we get

16 − 100 = y(t0 + 2) − y(t0)

= [−16(t0 + 2)2 + v0(t0 + 2) + y0] − [−16t02 + v0t0 + y0]

= −64t0 − 64 + 2v0

so that

t0 = 132 (v0 + 10).

Substituting this result in the basic equation and noting that y(t0) = 100, we have

−16(v0 + 10

32

)2

+ v0

(v0 + 10

32

)+ y0 = 100

and therefore

(∗∗) y0 = 100 − v02

64+

2516

.

We use (∗∗) to find the answer to each part of the problem.

(a) v0 = 0 so y0 = 162516 ft (b) v0 = −5 so y0 = 6475

64 ft (c) v0 = 10 so y0 = 100 ft



SECTION 4.9 201

50. Let v0 > 0 be the initial velocity. The equation of motion prior to the impact is: y(t) = −16t2 − v0t + 4.

The ball hits the ground at time t =

√v20 + 256 − v0

32with velocity v =

√v20 + 256. The equation of

motion following the impact is: y(t) = −16t2 +

√v20 + 2562

t. It reaches its maximum height at time

T =

√v20 + 25664

. Now, y(T ) = 4 =⇒ v0 = 16√

3.

51. Let y0 > 0 be the initial height. The equation of motion becomes:

0 = −16(8)2 + 5(8) + y0, so y0 = 984 ft.

52. Using 0 = −16t2 − 5t + 984, yields t =12316

or about 7.7 sec.

53. Let f1(t) and f2(t) be the positions of the horses at time t. Consider f(t) = f1(t) − f2(t). Let T be

the time the horses finish the race. Then f(0) = f(T ) = 0. By the mean-value theorem there is a c in

(0, T ) such that f ′(c) = 0. Hence f ′1(t) = f ′

2(t), so the horses had the same speed at time c.

54. Apply the argument given in Exercise 53 to the velocity functions v1(t), v2(t).

55. The driver must have exceeded the speed limit at some time during the trip. Let s(t) denote the car’s

position at time t, with s(0) = 0 and s(5/3) = 120. Then, by the mean-value theorem, there exists at

least one number (time) c such that

v(c) = s′(c) =s(5/3) − s(0)

53 − 0

=120

53

=3605

= 72 mi./hr.

56. Let f(t) be the function that gives the car’s velocity after t hours. Then f(0) = 30 and f(14) = 60. f

is differentiable on (1,14) and continuous on [1,

14], so by the mean-value theorem there is a c in (1,

14)

such that f ′(c) =60 − 300 − t1

= 120. i.e. The acceleration at time c was 120 mph.

57. If the speed s(t) of the car is less than 60 mi/hr= 1 mi/min, then the distance traveled in 20 minutes

is less than 20 miles. Therefore, the car must have gone at least 1 mi/min at some time t < 20. Let

t1 be the first instant the car’s speed is 1 mi/min. Then the speed s(t) is less than 1 mi/min on the

interval [0, t1) and the distance r traveled in t1 minutes is less than t1 miles. Now, by the mean-value

theorem, there is a time c ∈ [t1, 20] such that

s′(c) =20 − r

20 − t1>

20 − t120 − t1

= 1 (= 60 mi/hr).

58. Let s(t) denote the distance that the car has traveled in t seconds since applying the brakes, 0 ≤ t ≤ 6.

Then s(0) = 0 and s(6) = 280. Assume that s is differentiable on (0, 6) and continuous on [0, 6]. Then,

by the mean-value theorem, there exists a time c ∈ (0, 6) such that

s′(c) = v(c) =s(6) − s(0)

6 − 0=

2806

∼= 46.67 ft/sec

Now v(0) ≥ v(c) = 46.7 ft/sec. Thus, the driver must have been exceeding the speed limit (44 ft/sec)

at the instant he applied his brakes.



202 SECTION 4.9

59. y(t) = A sin(ωt + φ0), y′(t) = v(t) = ωA cos(ωt + φ0), y′′(t) = a(t) = −ω2A sin(ωt + φ0)

(a) y′′(t) + ω2y(t) = −ω2A sin(ωt + φ0) + ω2A sin(ωt + φ0) = 0.

(b) v′(t) = a(t) = −ω2A sin(ωt + φ0) = 0 when ωt + φ0 = nπ; t =nπ − φ0

ω.

The extreme values of v occur at these times. Now

y

(nπ − φ0

ω

)= A sin

(ωnπ − φ0

ω+ φ0

)= A sin(nπ) = 0.

The bob attains maximum speed at the equilibrium position.

(c) a′(t) − ω3A cos(ωt + φ0) = 0 when ωt + φ0 = (2n− 1)π/2; t =(2n− 1)π/2 − φ0

ω.

The extreme values of a occur at these times. Now

a

((2n− 1)π/2 − φ0

ω

)= −ω2A sin

((2n− 1)π/2 − φ0

ω+ φ0

)= ±ω2A.

The bob attains these values at

y

((2n− 1)π/2 − φ0

ω

)= A sin

(ω

(2n− 1)π/2 − φ0

ω+ φ0

)= A sin (2n− 1)π/2 = ±A.

60. (a) x(t) = t3 − 7t2 + 10t + 5, v(t) = 3t2 − 14t + 10, 0 ≤ t ≤ 5

(b) The object is moving to the right when 0 < t < 0.88 and when 3.79 < t < 5.

The object is moving to the left when 0.88 < t < 3.79

(c) The object stops at times t ∼= 0.88 and t ∼= 3.79.

The maximum speed is v ∼= 6.33 at t ∼= 2.33.

(d) a(t) = 6t− 14

The object is speeding up when v(t) and a(t) have

the same sign: 0.88 < t < 2.33 and 3.79 < t < 5.

The object is slowing down when v(t) and a(t) have

opposite sign: 0 < t < 0.88 and 2.33 < t < 3.79.

PROJECT 4.9a

1. length of arc = rθ, speed =d

dt[rθ] = r

dθ

dt= rω

2. v = rω so KE = 12mr2ω2.

3. We know that dθ/dt = ω and, at time t = 0, θ = θ0. Therefore θ = ωt + θ0. It follows that

x(t) = r cos (ωt + θ0) and y(t) = r sin (ωt + θ0).

x(t) = r cos(ωt + θ0), y(t) = r sin(ωt + θ0)

v(t) = x′(t) = − rω sin(ωt + θ0) = −ω y(t)

a(t) = − rω2 cos(ωt + θ0) = −ω2 x(t)

v(t) = y′(t) = rω cos(ωt + θ0) = ω x(t)

a(t) = − rω2 sin(ωt + θ0) = −ω2 y(t)



SECTION 4.10 203

4. For the sector A =12r2θ,

dA

dt=

12r2 dθ

dt=

12r2ω is constant.

For triangle T

A = 12 (2r sin 1

2θ)(r cos 12θ)

= 12r

2(2 sin 12θ cos 1

2θ) = 12r

2 sin θ,

dA

dt=

12r2 cos θ

dθ

dt=

12r2ω cos θ varies with θ.

For segment S

A = 12r

2θ − 12r

2 sin θ = 12r

2(θ − sin θ),

dA

dt=

12r2

(dθ

dt− cos

dθ

dt

)=

12r2ω(1 − cos θ) varies with θ.

5. From Exercise 4,dAT

dt= 1

2r2ω cos θ and

dAS

dt= 1

2r2ω − 1

2r2ω cos θ

Now,12r2ω cos θ =

12r2ω − 1

2r2ω cos θ =⇒ cos θ =

12

=⇒ θ =π

3.

PROJECT 4.9B

1. d

dt

[mgy + 1

2 mv2]

= mgdy

dt+

12m

d

dt(v2)

= mgv +12m

[2v

dv

dt

]

= mgv + mvdv

dt

= mgv + mv(−g) (since dv/dt = a = −g)

= mgv −mgv = 0

2. By Problem 1, mgy + 12 mv2 = C (constant). Since v = 0 at height y = y0, we have C = mgy0.

Thus,

mgy0 = mgy + 12 mv2 and |v| =

√2g(y0 − y)

3. y(t) = − 12 gt

2 + y0 =⇒ gt =√

2g(y0 − y)

v(t) = y′(t) = −gt Therefore, |v(t)| =√

2g(y0 − y).

SECTION 4.10

1. x + 2y = 2,dx

dt+ 2

dy

dt= 0

(a) Ifdx

dt= 4, then

dy

dt= −2 units/sec. (b) If

dy

dt= −2, then

dx

dt= 4 units/sec.



204 SECTION 4.10

2. x2 + y2 = 25, 2xdx

dt+ 2y

dy

dt= 0 and

dx

dt= − y

x

dy

dt.

At the point (3, 4),dy

dt= −2. Therefore,

dx

dt=

83; the x-coordinate is

increasing at the rate of 8/3 units per second.

3. y2 = 4(x + 2), 2ydy

dt= 4

dx

dtand

dx

dt= 1

2 ydy

dt

At the point (7, 6),dy

dt= 3. Therefore

dx

dt= 1

2 · 6 · 3 = 9 units/sec.

4. We are givendx

dt= 2. Also 4y = (x + 2)2 so 4

dy

dt= 2(x + 2)

dx

dtor

dy

dt= 1

2 (x + 2)dx

dt.

At x = 2,dy

dt= 4. The distance from a point on the parabola to the point (−2, 0) is given by

S =√

(x + 2)2 + y2 =√

4y + y2 since (x + 2)2 = 4y. Now

dS

dt=

12

(4y + y2)−1/2(4 + 2y)dy

dt=

2 + y√4y + y2

dy

dt.

Therefore, at the point (2, 4),dS

dt=

6√32

4 = 3√

2.

5. Let s =√x2 + y2 denote the distance to the origin at time t. Since x = 4 cos t and y = 2 sin t,

we have

s(t) =√

16 cos2 t + 4 sin2 t =√

12 cos2 t + 4

ds

dt=

12

(12 cos2 t + 4)−1/2(−24 cos t sin t)

=−12 cos t sin t√

12 cos2 t + 4

At t = π/4,ds

dt=

−12 cos(π/4) sin(π/4)√12 cos2(π/4) + 4

= − 35

√10.

6. y = x√x = x3/2;

dy

dt=

32x1/2 dx

dt.

Nowdx

dt=

dy

dt= z �= 0, =⇒ 3

2 x1/2 = 1 =⇒ x = 4

9 and y = 827 .

Both coordinates are changing at the same rate at the point (4/9, 8/27).

7. Finddx

dtand

dS

dtwhen V = 27m3

given thatdV

dt= −2m3/min.

(∗) V = x3, S = 6x2

Differentiation of equations (∗) gives

dV

dt= 3x2 dx

dtand

dS

dt= 12x

dx

dt.



SECTION 4.10 205

When V = 27, x = 3. Substituting x = 3 and dV/dt = −2, we get

−2 = 27dx

dtso that

dx

dt= −2/27 and

dS

dt= 12(3)

(−227

)= −8/3.

The rate of change of an edge is −2/27 m/min; the rate of change of the surface area is

−8/3 m2/min.

8. Finddr

dtand

dS

dtwhen r = 10 ft

given thatdV

dt= 8 ft3/min.

(∗) V = 43πr

3, S = 4πr2

Differentiation of equations (∗) with respect to t gives

dV

dt= 4πr2 dr

dtand

dS

dt= 8πr

dr

dt.

Substituting r = 10 and dV/dt = 8, we get

8 = 4π(10)2dr

dtso that

dr

dt=

150π

anddS

dt= 8π(10)

150π

=85.

The radius is increasing1

50πft/min; the surface area is increasing

85

ft2/min.

9. The area of an equilateral triangle of side x is given by

A =12x

(x√

32

)=

√3

4x2. Thus

dA

dt=

√3

2xdx

dt.

When x = α,dx

dt= k, and

dA

dt=

√3

2 αkcm2/min.

10. We have 2x + 2y = 24, or x + y = 12. Thus, A = xy = x(12 − x) = 12x− x2.

When A = 32, x = 4 or x = 8, and it follows thatdA

dt= (12 − 2x)

dx

dt= ± 4

dx

dt.

11. FinddA

dtwhen l = 6 in.

given thatdl

dt= −2 in./sec.

By the Pythagorean theorem

l2 + w2 = 100.

Also, A = lw. Thus, A = l√

100 − l2. Differentiation with respect to t gives

dA

dt= l

( −l√100 − l2

)dl

dt+

√100 − l2

dl

dt.

Substituting l = 6 and dl/dt = −2, we get

dA

dt= 6

(−68

)(−2) + (8)(−2) = −7.

The area is decreasing at the rate of 7 in.2/sec.



206 SECTION 4.10

12. x2 + 62 = y2; 2xdx

dt= 2y

dy

dt

If y = 30 ft anddx

dt= 8 ft/min, then

dy

dt=

x

y

dy

dt=

√(30)2 − 36

308 =

165

√6 ft/min

13. Comparedy

dtto

dx

dt= −13 mph

given that z = 16 anddz

dt= −17

when x = y.

By the Pythagorean theorem x2 + y2 = z2. Thus,

2xdx

dt+ 2y

dy

dt= 2z

dz

dt.

Since x = y when z = 16, we have x = y = 8√

2 and

2(8√

2 )(−13) + 2(8√

2 )dy

dt= 2(16)(−17).

Solving for dy/dt, we get

−13√

2 +√

2dy

dt= −34 or

dy

dt=

1√2(13

√2 − 34) ∼= −11.

Thus, boat A wins the race.

14. V =43πr3, so

dV

dt= 4πr2 dr

dt= 4πr2(−1

5). Thus at r = 12 we have

dV

dt= −576

5π cm3/min.

15. We want to finddA

dtwhen

dx

dt= 2 and x = 12.

A =12x√

169 − x2, sodA

dt=

[12

√169 − x2 − x2

2√

169 − x2

]dx

dt

Whendx

dt= 2 and x = 12,

dA

dt= −119

5ft2/sec.

16. x2 + y2 = (13)2; 2xdx

dt+ 2y

dy

dt= 0 and

dy

dt= − x

2y

sincedx

dt= 0.5. When x = 5, y = 12 and

dy

dt= − 5

24; the top of the ladder is

dropping 5/24 ft/sec.

17. We want to find dV/dt when V = 1000 ft3 and P = 5 lb/in.2 given that dP/dt = −0.05 lb/in.2/hr.

Differentiating PV = C with respect to t, we get

PdV

dt+ V

dP

dt= 0 so that 5

dV

dt+ 1000(−0.05) = 0. Thus,

dV

dt= 10.

The volume increases at the rate of 10 ft3/hr.



SECTION 4.10 207

18. PV 1.4 = C; V 1.4 dP

dt+ (1.4)PV 0.4 dV

dt= 0 and

dP

dt= − 1.4P

V

dV

dt.

With V = 10, P = 50 anddV

dt= −1, we have

dP

dt= − (1.4)50

10(−1) = 7

The pressure is increasing 7 lb/in2/sec.

19. Findds

dtwhen x = 3 ft (and s = 4 ft)

given thatdx

dt= 400 ft/min.

By similar trianglesL

x + s=

6s.

Substitution of x = 3 and s = 4 gives usL

7=

64

so that the lamp post is

L = 10.5 ft tall. Rewriting10.5x + s

=6s

as s =43x

and differentiating with respect to t, we find thatds

dt=

43dx

dt=

16003

.

The shadow lengthens at the rate of 1600/3 ft/min.

If the tip of his shadow is at the point z, then

z = x + s anddz

dt=

dx

dt+

ds

dt= 400 +

16003

=2800

3.

The tip of his shadow is moving at the rate of 2800/3 ft./min.

20. y(t) = 64 − 16t2;dy

dt= −32t

By similar triangles,y

x=

6420 + x

.

Thus, y =64x

20 + x.

Now,dy

dt=

(20 + x)(64) − 64x(20 + x)2

dx

dt=

1280(20 + x)2

=⇒ dx

dt=

(20 + x)2

1280dy

dt.

At t = 1, y = 48, x = 60, anddy

dt= −32. Therefore,

dx

dt=

(80)2

1280(−32) = −160 ft/sec.

21. Let W (t) = 150(1 + 1

4000 r)−2

. We want to find dW/dt when r = 400 given that

dr/dt = 10 mi/sec. Differentiating with respect to t, we get

dW

dt= −300

(1 +

14000

r

)−3 (1

4000

)dr

dt

Now set r = 400 and dr/dt = 10. Then

dW

dt= − 300

(1 +

4004000

)−3 104000

∼= −0.5634 lbs/sec



208 SECTION 4.10

22. M =m√

1 − v2/c2;

dM

dt= m

(− 1

2

) (1 − v2/c2

)−3/2 (−2v/c2

) dv

dt=

mv

c2 (1 − v2/c2)3/2dv

dt.

If v =c

2and

dv

dt=

c

100, then

dM

dt=

m(c/2)

c2 (1 − c2/4c2)3/2c

100=

√3

225m.

23. Finddh

dtwhen h = 3 in.

given thatdV

dt= −1

2cu in./min.

By similar triangles

r = 13h.

Thus V = 13πr

2h = 127πh

3. Differentiating with respect to t , we get

dV

dt=

19πh2 dh

dt.

When h = 3,

−12

=19π(9)

dh

dtand

dh

dt= − 1

2π.

The water level is dropping at the rate of 1/2π inches per minute.

24. V =13πr2h and

r

h=

46

(similar triangles)

so V =427

πh3. ThusdV

dt=

49πh2 dh

dt,

so atdh

dt= 0.5 and h = 2,

dV

dt=

8π9

cubic ft per sec.

25.dV

dt= 4πr2 dr

dtand

dSA

dt= 8πr

dr

dt. Thus when

dSA

dt= 4 and

dr

dt= 0.1

we get r =5π

anddV

dt=

10π

cm3/min.

26. V = πrh2 − 13πh

3;dV

dt= 2πrh

dh

dt− πh2 dh

dt, and

dh

dt=

2π(14h− h2)

since r = 7 anddV

dt= 2.

(a) When h = 7/2,dh

dt=

8147π

in./sec. (b) When h = 7,dh

dt=

249π

in./sec.

27. Finddθ

dtwhen x = 4 ft

given thatdx

dt= 2 in./min.

(∗) tanθ

2=

3x

Differentiation of (∗) with respect to t gives

12

sec2 θ

2dθ

dt= − 3

x2

dx

dtor

dθ

dt= − 6

x2cos2

θ

2dx

dt.



SECTION 4.10 209

Note that dx/dt = 2 in./min=1/6 ft/min. When x = 4, we have cos θ/2 = 4/5 and thus

dθ

dt= − 6

16

(45

)2 (16

)= − 1

25.

The vertex angle decreases at the rate of 0.04 rad/min.

28. tanα =60x

; sec2 αdα

dt= − 60

x2

dx

dt

Now, sec2 α =(y

x

)2

sodα

dt= − 60

y2

dx

dt.

With y = 100, anddx

dt= −10, we have

dα

dt= − 60

(100)2(−10) = 0.06 rad/min.

29. Finddx

dtwhen x = 1 mi

given thatdθ

dt= 2π rad/min.

(∗) tan θ =x

1/2= 2x

Differentiation of (∗) with respect to t gives

sec2 θdθ

dt= 2

dx

dt.

When x = 1, we get sec θ =√

5 and thusdx

dt= 5π.

The light is traveling at 5π mi/min.

30. (a) We have tan θ = x, so sec2 θdθ

dt=

dx

dt. Switching to radians,

dθ

dt= 4π.

Thus at θ =π

4,

dx

dt= 8π mi/min.

(b) At θ = 0,dx

dt= 4π mi/min.

31. We have tan θ =x

40, so sec2 θ

dθ

dt=

140

dx

dt, and

dx

dt= 4.

At t = 15, x = 60 and sec θ =√

520040

, sodθ

dt=

265

∼= 0.031 rad/sec.

32. We have V = hπr2, so V = 500π cm3. Thus h =500r2

, anddh

dt=

−1000r3

dr

dt.

At r = 5 anddr

dt=

12, we have

dh

dt= −4 cm/sec.

33. We have sin θ =45

so tan θ =43

=x

h. Thus x =

43h. V = 12

(3 + 2x + 3

2h

)= 36h + 16h2.

ThusdV

dt= (36 + 32h)

dh

dt, so at

dV

dt= 10 and h = 2,

dh

dt=

110

ft/min.



210 SECTION 4.10

34. tanα =y

x; sec2 α

dα

dt=

xdy

dt− y

dx

dtx2

dα

dt=

xdy

dt− y

dx

dtx2

· x2

x2 + y2

=xdy

dt− y

dx

dtx2 + y2

Nowdx

dt= −30 mph = −44 ft/sec and

dy

dt= −22.5 mph = −33 ft/sec.

At x = 300, y = 400, we have

dα

dt=

300(−33) − 400(−44)(300)2 + (400)2

= 0.0308 the angle is increasing 0.0308 rad/sec.

35. Finddθ

dtwhen y = 4 ft

given thatdx

dt= 3 ft/sec.

tan θ =16x, x2 + (16)2 = (16 + y)2

Differentiating tan θ = 16/x with respect to t, we obtain

sec2 θdθ

dt=

−16x2

dx

dtand thus

dθ

dt=

−16x2

cos2 θdx

dt.

From x2 + (16)2 = (16 + y)2 we conclude that x = 12, when y = 4. Thus

cos θ =x

16 + y=

1220

=35

anddθ

dt=

−16(12)2

(35

)2

(3) =−325

.

The angle decreases at the rate of 0.12 rad/sec.

36. tan(α/2) =3x

; sec2(α/2) 12

dα

dt= − 3

x2

dx

dt,

anddα

dt= − 6

y2

dx

dt

Initially, y = 5 so x = 4. 8 seconds later x = 12 so y =√

(12)2 + (3)2 =√

153. Therefore,

dα

dt= − 6

y2

dx

dt= − 6

153(1) = − 6

153; the angle is decreasing − 6/153 rad/sec.

37. Finddθ

dtwhen t = 6 min.

tan θ =100t

500 + 75t=

4t20 + 3t

Differentiation with respect to t gives

sec2 θdθ

dt=

(20 + 3t)4 − 4t(3)(20 + 3t)2

=80

(20 + 3t)2.

When t = 6

tan θ = 2438 = 12

19 and sec2 θ = 1 +(

1219

)2 = 505361



SECTION 4.10 211

so thatdθ

dt=

80(20 + 3t)2

· 1sec2 θ

=80

(38)2· 361

505=

4101

.

The angle increases at the rate of 4/101 rad/min.

38. tanα =x

H; sec2 α

dα

dt=

1H

dx

dt

dα

dt=

1H

dx

dtcos2 α =

H

H2 + x2

dx

dt

We have H = 2 mi. anddx

dt= 400 mph. After 2 seconds, x = 400

(2

3600

)=

29

miles.

dα

dt=

222 + (2/9)2

(400) =200(81)

82rad/hr =

9164

rad/sec.

39. Let x be the distance from third base to the player. Then the distance from home plate to the player

is given by: y =√

(90)2 + x2. Differentiation with respect to t gives

dy

dt=

x√(90)2 + x2

dx

dt

We are given that dx/dt = −15. Therefore,

dy

dt= − 15x√

(90)2 + x2and

dy

dt

∣∣∣∣x=10

=−15√

82∼= 1.66

The distance between home plate to the player is decreasing 1.66 ft./sec.

40. The plane is flying at an altitude 6 miles. If y denotes the horizontal distance between the radar station

and the plane, then

tan θ =6y

=⇒ sec2 θdθ

dt=

−6y

dy

dt

Therefore,

dy

dt=

y2 sec2 θ

−6dθ

dt

At the instant the plane is 12 miles from the station, y =√

108, θ = 16π and the speed is∣∣∣∣dydt

∣∣∣∣ =∣∣∣∣108−6

43

π

360

∣∣∣∣ =4π60

mi/sec. ∼= 754 mi/hr.

41. Let x be the position of the runner on the track, let y be the distance between the runner and the

spectator, and let θ be the central angle indicated in the figure. The runner’s speed is 5 meters per

second so

50dθ

dt= 5 and

dθ

dt=

110

.

By the law of cosines: Sθ

R

y2 = (50)2 + (200)2 − 2(50)(200) cos θ.

Differentiating with respect to t, we get

2ydy

dt= 20, 000 sin θ dθ

dt



212 SECTION 4.11

dy

dt=

10, 000 sin θ

y110

dy

dt=

1, 000y

sin θ

Now,dy

dt

∣∣∣∣y=200

= 5

√(200)2 − (25)2

200= 5

√39375200

= 4.96.

The runner is approaching the spectator at the rate of 4.96 meters per second.

42.dx

dt=

234

43.dx

dt= 4 44.

dy

dt= − 1

2√

3

SECTION 4.11

1. ΔV = (x + h)3 − x3

= (x3 + 3x2h + 3xh2 + h3) − x3

= 3x2h + 3xh2 + h3,

dV = 3x2h,

ΔV − dV = 3xh2 + h3 (see figure)

2. The area of the ring can be thought of as the increase of the area of a disk as the radius increases from

r to r + h.

A = πr2; so dA = 2πr dr = 2πrh

The exact area is: π(r + h)2 − πr2 = π(r2 + 2rh + h2) − πr2 = 2πrh + πh2.

3. f(x) = x1/3, x = 1000, h = 2, f ′(x) = 13x

−2/3

3√

1002 = f(x + h) ∼= f(x) + hf ′(x) = 3√

1000 + 2[13 (1000

)−2/3] = 10 + 1150

∼= 10.0067

4. f(x) =1√x, x = 25, h = −0.5, f ′(x) =

−12x3/2

1√24.5

= f(x + h) ∼= f(x) + hf ′(x) =15

+ (−12)

−12(5)3

=101500

= 0.202

5. f(x) = x1/4, x = 16, h = −0.5, f ′(x) = 14x

−3/4

(15.5)1/4 = f(x + h) ∼= f(x) + hf ′(x) = (16)1/4 + (−0.5)[14 (16)−3/4

]= 1 63

64∼= 1.9844

6. f(x) = x2/3, x = 27, h = −1, f ′(x) = 23x

−1/3

(26)2/3 = f(x + h) ∼= f(x) + hf ′(x) = (27)2/3 + (−1)[23 (27)−1/3

]=

799

∼= 8.778



SECTION 4.11 213

7. f(x) = x3/5, x = 32, h = 1, f ′(x) = 35x

−2/5

(33)3/5 = f(x + h) ∼= f(x) + hf ′(x) = (32)3/5 + (1)[35 (32)−2/5

]= 8.15

8. f(x) = x−1/5, x = 32, h = 1, f ′(x) = − 15x

−6/5

(33)−1/5 = f(x + h) ∼= f(x) + hf ′(x) = (32)−1/5 − (1)[15 (32)−6/5

]=

159320

∼= 0.497

9. f(x) = sinx, x =π

4, h =

π

180, f ′(x) = cosx

sin 46◦ = f(x + h) ∼= f(x) + hf ′(x) = sinπ

4+

π

180cos

π

4=

√2

2

(1 +

π

180

)∼= 0.719

10. f(x) = cosx, x =π

3, h =

π

90, f ′(x) = − sinx

cos 62◦ = f(x + h) ∼= f(x) + hf ′(x) = cosπ

3+

π

90

(− sin

π

3

)=

12−

( π

90

) (√3

2

)∼= 0.470

11. f(x) = tanx, x =π

6, h =

−π

90, f ′(x) = sec2 x

tan 28◦ = f(x + h) ∼= f(x) + hf ′(x) = tanπ

6+

(−π

90

) (43

)=

√3

3− 2π

135∼= 0.531

12. f(x) = sinx, x =π

4, h = − π

90, f ′(x) = cosx

sin 43◦ = f(x + h) ∼= f(x) + hf ′(x) = sinπ

4+

(− π

180

)cos

π

4=

√2

2−

( π

180

) (√2

2

)∼= 0.682

13. f(2.8) ∼= f(3) + (−0.2)f ′(3) = 2 + (−0.2)(2) = 1.6

14. f(5.4) ∼= f(5) + (0.4)f ′(5) = 1 + (0.4)(3) = 2.2

15. V (r) = πr2h; volume = V (r + t) − V (r) ∼= tV ′(r) = 2πrht

16. Error in diameter = 0.3 =⇒ error in radius = 0.15.

(a) dS = 8πrh = 8π(8)(0.15) ∼= 9.6π cm2

(b) dV = 4πr2h = 4π(8)2(0.15) ∼= 38.4 cm3

17. V (x) = x3, V ′(x) = 3x2, ΔV ∼= dV = V ′(10)h = 300h

|dV | ≤ 3 =⇒ |300h| ≤ 3 =⇒ |h| ≤ 0.01, error ≤ 0.01 feet

18. (a) Let f(x) =√x. Then f ′(x) =

12√x

and√x + 1 −

√x ∼= (1)f ′(x) =

12√x.

Now,1

2√x< 0.01 =

1100

=⇒√x > 50 =⇒ x > 2500.

(b) Let f(x) = x1/4. Then f ′(x) = 14 x

−3/4 and 4√x + 1 − 4

√x ∼= (1)f ′(x) = 1

4 x−3/4.

Now, 14 x

−3/4 < 0.002 =2

1000=⇒ x3/4 > 125 =⇒ x > 625.



214 SECTION 4.11

19. V (r) = 23 πr

3 and dr = 0.01.

V (r + 0.01) − V (r) ∼= V ′(r)(0.01) = 2πr2(0.01)

= 2π(600)2(0.01) (50 ft = 600 in)

= 22619.5 in3 or 98 gallons (approx.)

20. V (r) = 43 πr

3;dV

dr= 4πr2.

Now, V (r + h) − V (r) = 8 × (10)6 ∼= dV

drh = 4πr2h. Therefore,

h =8 × (10)6

4π(4000)2∼= 0.0398 (miles) ∼= 210 (feet)

21. P = 2π√

L

gimplies P 2 = 4π2 L

g

Differentiating with respect to t, we have

2PdP

dt=

4π2

g· dLdt

=P 2

L· dLdt

sinceP 2

L=

4π2

g.

ThusdP

P=

12· dLL

22.dP

P= −15 sec/hour = − 15

3600. By Exercise 21,

12dL

L= − 15

3600and dL = − 30

3600L = − 1

120L

With L = 90, we have dL = −90/120 = − 0.75; the pendulum should be shortened

0.75 cm to 89.25 cm.

23. L = 3.26 ft, P = 2 sec, and dL = 0.01 ft

dP

P=

12· dLL

dP =12· dLL

· P =12· 0.013.26

· 2 dP ∼= 0.00307 sec

24. Each edge increases by 0.1%; take h = 0.001x.

S = 6x2, dS = 12xh, anddS

S=

12x(0.001x)6x2

= 0.002 = 0.2%.

A = x3, dA = 3x2h, anddA

A=

3x2(0.001x)x3

= 0.003 = 0.3%.

25. A(x) =14πx2, dA =

12πxh,

dA

A= 2

h

x

dA

A≤ 0.01 ⇐⇒ 2

h

x≤ 0.01 ⇐⇒ h

x≤ 0.005 within

12%



SECTION 4.11 215

26. (a) Let y = xn. Then dy = nxn−1h.

To getdy

y=

nxn−1h

xn< 0.01, we need

h

x<

0.01n

, that is, within 1n%.

(b) Let y = x1/n. Then dy = 1nx

(1−n)/nh.

To getdy

y=

1nx

(1−n)/nh

x1/n< 0.01, we need

h

x< (0.01)n, that is, within n%.

27. (a) and (b) 28. limh→0

g(h) = limh→0

g(h)h

· h =(

limh→0

g(h)h

) (limh→0

h

)= 0

29. limh→0

g1(h) + g2(h)h

= limh→0

g1(h)h

+ limh→0

g2(h)h

= 0 + 0 = 0

limh→0

g1(h)g2(h)h

= limh→0

hg1(h)g2(h)

h2=

(limh→0

h

) (limh→0

g1(h)h

) (limh→0

g2(h)h

)= (0)(0)(0) = 0

30. (a) g(h) = f(x + h) − f(x) −mh.

(b) limh→0

g(h)h

= limh→0

[f(x + h) − f(x)

h−m

]= f ′(x) −m = 0 iff m = f ′(x).

31. Suppose that f is differentiable at x. Then there is a number m such that

limh→0

f(x + h) − f(x)h

= m or limh→0

f(x + h) − f(x)h

−m = 0.

Let g(h) = f(x + h) − f(x) −mh. Then

g(h)h

=f(x + h) − f(x)

h−m and lim

h→0

g(h)h

= limh→0

f(x + h) − f(x)h

−m = 0.

Now suppose that f(x + h) − f(x) −mh = g(h) where g(h) = o(h). Then

limh→0

(f(x + h) − f(x)

h−m

)= lim

h→0

g(h)h

= 0.

Therefore

limh→0

f(x + h) − f(x)h

= m

and f is differentiable at x. m = f ′(x).



216 SECTION 4.11

PROJECT 4.11

1. C(x) = 2000 + 50x− x2

20.

Marginal cost: C ′(x) = 50 − x

10; C ′(20) = 48

Exact cost of the 21st component: C(21) − C(20) = 3027.95 − 2980 = 47.95.

2. Profit function: P (x) = R(x) − C(x) = 650x− 5x2 − (12, 000 + 30x) = 620x− 5x2 − 12000.

Breakeven points: P (x) = 0, implies x2 − 124x + 2400 = (x− 24)(x− 100) = 0; x = 24, x = 100

units.

Maximum profit: P ′(x) = 620 − 10x; P ′(x) = 0 at x = 62 units.

3. (a) Profit function: P (x) = R(x) − C(x) = 20x− x2

50− (4x + 1400) = 16x− x2

50− 1400.

Break-even points: 16x− x2

50− 1400 = 0 or x2 − 800x + 70, 000 = 0

Thus x = 100, or x = 700 units.

(b) P ′(x) = 16 − x

25; P ′(x) = 0 at x = 400 units.

(c)

4. (a) (b)

Break-even points at x = 81.11 and x = 631.19 Maximum profit at x = 336.11 units

5. (a) (b) P (x) =10x

1 + 0.25x2− 4 − 0.75x

Breakeven points at x ∼= 0.46, and x ∼= 4.53 Maximum profit at 175 units

6. A(x) =C(x)x

, A′(x) =xC ′(x) − C(x)

x2.

A′(x) = 0 implies xC ′(x) − C(x) = 0 or C ′(x) =C(x)x

= A(x). The critical points of A are

the points where C ′(x) = C(x)/x.

A′′(x) =x2[xC ′′(x)] − [xC ′(x) − C(x)](2x)

x4

At the points where xC ′(x) − C(x) = 0, A′′(x) =C ′′(x)x

. If C ′′(x) > 0 at such a point, then A(x)

is a minimum.



SECTION 4.12 217

7. B(x) =R(x)x

, B′(x) =xP ′(x) − P (x)

x2.

B′(x) = 0 implies xP ′(x) − P (x) = 0 or P ′(x) =P (x)x

= B(x). The critical points of B are

the points where P ′(x) = P (x)/x.

B′′(x) =x2[xP ′′(x)] − [xP ′(x) − P (x)](2x)

x4

At the points where xP ′(x) − P (x) = 0, B′′(x) =P ′′(x)x

. If P ′′(x) < 0 at such a point, then B(x)

is a maximum.

SECTION 4.12

1. (a) xn+1 =12xn + 12

(1xn

)(b) x4

∼= 4.89898

2. (a) xn+1 =2x3

n − 13x2

n − 4(b) x4

∼= 1.86081

3. (a) xn+1 =23xn +

253

(1xn

)2

(b) x4∼= 2.92402

4. (a) xn+1 =45xn + 6

(1xn

)4

(b) x4∼= 1.97435

5. (a) xn+1 =xn sinxn + cosxn

sinxn + 1(b) x4

∼= 0.73909

6. (a) xn+1 =xn cosxn − sinxn − x2

n

cosxn − 2xn(b) x4

∼= 0.87673

7. (a) xn+1 =6 + xn

2√xn + 3 − 1

(b) x4∼= 2.30278

8. (a) xn+1 =xn sec2 xn − tan xn

1 + sec2 xn(b) x4

∼= 2.30278

9. Let f(x) = x1/3. Then f ′(x) = 13 x

−2/3. The Newton-Raphson method applied to

this function gives:

xn+1 = xn − f(xn)f ′(xn)

= xn − x1/3n

13 x

−2/3n

= −2xn.

Choose any x1 �= 0. Then x2 = −2x1, x3 = −2x2 = 4x1, · · · ,xn = −2xn−1 = (−1)n−12n x1, · · · .

10. xn+1 = xn for all n.

11. (a) f is a continuous function, and f(1) = −2 < 0, f(2) = 3 > 0. Thus, f has a root in (1, 2).

(b) f ′(x) = 6x2 − 6x and f ′(1) = 0. Therefore, x1 = 1 will fail to generate values that

will approach the root in (1, 2).

(c) xn+1 = xn − 2x3n − 3x2

n − 16x2

n − 6xn;

x1 = 2, x2 = 1.75, x3 = 1.68254, x4 = 1.67768, f(x4) ∼= 0.00020.



218 SECTION 4.12

12. (a) Let f(x) = x4 − 2x2 − 1716 . Then f ′(x) = 4x3 − 4x. The Newton-Raphson method applied to this

function gives:

xn+1 = xn − x4n − 2x2

n − 1716

4x3n − 4xn

If x1 = 12 , then x2 = − 1

2 , x3 = 12 , · · ·xn = (−1)n−1 1

2 , · · · .

(b) x1 = 2, x2 = 1.71094, x3 = 1.58569, x4 = 1.56165; f(x4) = 0.00748

13. (a) f(x) = x2 − a; f ′(x) = 2x. Substituting into (4.12.1), we have

xn+1 = xn − x2n − a

2xn=

x2n + a

2xn=

12

(xn +

a

xn

), n ≥ 1

(b) Let a = 5, x1 = 2, and xn+1 =12

(xn +

a

xn

), n ≥ 1.

Then x2 = 2.25, x3 = 2.23611, x4 = 2.23607, and f(x4) ∼= 0.000009045.

14. (a) Let f(x) = xk − a. Then f ′(x) = kxk−1. The Newton-Raphson method applied to this function

gives:

xn+1 = xn − xkn − a

kxk−1n

= xn − 1kxn +

1k

a

xk−1n

=1k

[(k − 1)xn +

a

xk−1n

]

(b) Let a = 23, k = 3 and x1 = 3. Then

x1 = 3, x2 = 2.85185, x3 = 2.84389, x4 = 2.84382; f(x4) = −0.00114

15. (a) Let f(x) = 1x − a. Then f ′(x) = − 1

x2 . The Newton-Raphson method applied to

this function gives:

xn+1 = xn −1xn

− a

− 1x2n

= xn + xn − ax2n = 2xn − ax2

n

(b) Let a = 2.7153, and x1 = 0.3. Then

x2 = 0.35562, x3 = 0.36785, x4 = x5 = 0.36828,

Thus 12.7153 � 0.36828.

16. (a) f(x) = x4 − 7x2 − 8x− 3, f ′(x) = 4x3 − 14x− 8, f ′′(x) = 12x2 − 14.

f ′(2) = −4 < 0 and f ′(3) = 58 > 0; f ′ has a zero in (2, 3).

f ′′(x) = 12x2 − 14 > 0 on (2, 3). Therefore, f ′ has exactly one zero in this interval.

(b) xn+1 = xn − 4x3n − 14xn − 812x2

n − 14=

4x3n + 4

6x2n − 7

; x3∼= 2.1091. Since f ′′(x3) > 0, f has a local mini-

mum at c.

17. (a) f(x) = sin x + 12 x

2 − 2x, f ′(x) = cos x + x− 2, f ′′(x) = − sin x + 1.

f ′(2) = cos 2 ∼= −0.4161 < 0 and f ′(3) = cos 3 + 1 ∼= 0.0100 > 0; f ′ has a zero in (2, 3).

f ′′(x) = − sin x + 1 > 0 on (2, 3). Therefore, f ′ has exactly one zero in this interval.

(b) xn+1 = xn − cos xn + xn − 21 − sin xn

=2 − xn sin xn − cos xn

1 − sin xn; x3

∼= 2.9883. Since f ′′(x3) > 0,

f has a local minimum at c.



REVIEW EXERCISES 219

18. (a) xn+1 = xn − sin xn

cos xn= xn − tan xn, x1 = 3; x4

∼= 3.14159

(b) x4∼= 6.28319

19. (a) xn+1 = xn − xn + tan xn

1 + sec2 xn, x1 = 2π/3; r1 ∼= 2.029

(b) x1 = 5π/3; r2 ∼= 4.913

REVIEW EXERCISES

1. f is differentiable on (−1, 1) and continuous on [−1, 1]; f(1) = f(−1) = 0.

f ′(x) = 3x2 − 1; 3c2 − 1 = 0 =⇒ c = ±√

33

.

2. f is differentiable on (0, 2π) and continuous on [0, 2π];f(0) = f(2π) = 0.

f ′(x) = cosx− sinx; cos c− sin c = 0 =⇒ c = 14π,

54π.

3. f is differentiable on (−2, 3) and continuous on [−2, 3].

f ′(c) = 3c2 − 2 =f(3) − f(−2)

5= 5 =⇒ 3c2 = 7 =⇒ c = ±

√73

4. f is differentiable on (2, 5) and continuous on [2, 5].

f ′(c) =1

2√c− 1

=f(5) − f(2)

3=

13

=⇒ c =134.


f ′(c) = − 2(c− 1)2

=f(4) − f(2)

2= −2

3=⇒ c = 1 +

√3.


f ′(c) =3

4c1/4=

f(16) − f(0)16

=12

=⇒ c =8116

.

7. f ′(x) =1 + x2/3

3x2/3�= 0 for all x ∈ (−1, 1). f ′(0) does not exist. Therefore f is not differentiable on

(−1, 1).

8. f ′(x) =−3

(x− 2)2< 0 for all x ∈ (1, 4);

f(4) − f(1)3

=96> 0. f ′(2) does not exist. Therefore f is

not differentiable on (1, 4).

9. No. Reason: If such a function did exist, then, by the mean-value theorem, there is a number c ∈ (1, 4)

such that f ′(c) = − 43 < −1.

10. (a) f ′(x) = 3x2 − 3 = 3(x2 − 1) < 0 on (−1, 1). Therefore, by Rolle’s theorem, f can have at most

one zero in [−1, 1].

(b) Since f decreases on [−1, 1], we must have f(−1) = −1 + 3 + k ≥ 0 and f(1) = 1 − 3 + k ≤ 0.

These conditions imply that k ∈ [−2, 2].



220 REVIEW EXERCISES

11. f ′(x) = 6x(x + 1).

f increases on (−∞,−1] ∪ [0,∞) and decreases on [−1, 0].

The critical points are x = −1, 0; f(−1) = 2 is a local max and f(0) = 1 is a local min.

12. f ′(x) = 4(x− 1)(x2 + x + 1).

f increases on [1,∞) and decreases on (−∞, 1]. The critical point is x = 1; f(1) = 0 is a local min.

13. f ′ = (x + 2)(x− 1)2(5x + 4).

f increases on (−∞,−2] ∪ [− 45 ,∞) and decreases on [−2,− 4

5 ].

The critical points are x = −2, 1, − 45 . f(−2) = 0 is a local max and f(− 4

5 ) ∼= −8.981 is a local min.

14. f ′(x) = 1 − 8x3

=x3 − 8x3

.

f increases on (−∞, 0) ∪ [2,∞) and decreases on (0, 2]. x = 2 is the only critical point; 0 is not a

critical point. f(2) = 3 is a local min.

15. f ′(x) =1 − x2

(1 + x2)2.

f increases on [−1, 1] and decreases on (−∞,−1] ∪ [1,∞). The critical points are x = −1, 1. f(−1) =

− 12 is a local min and f(1) = 1

2 is a local max.

16. f ′(x) = sinx + cosx.

f increases on [0, 3π4 ] ∪ [ 7π4 , 2π] and decreases on [ 3π4 , 7π

4 ]. The critical points are x = 3π4 , 7π

4 . f( 3π4 ) =

√2 is a local max and f( 7π

4 ) = −√

2 is a local min.

17. f ′(x) = 3x2 + 4x + 1 = (3x + 1)(x + 1); critical points: x = − 13 , −1.

f(−2) = −1 endpoint and abs. min; f(−1) = 1 local max; f(− 13 ) = 23

27 local min; f(1) = 5 endpoint

and abs. max.




18. f ′(x) = 4x3 − 16x = 4x(x− 2)(x + 2); critical points: x = 0, 2.

f(−1) = −5 endpoint min; f(0) = 2 local max; f(2) = −14 local and abs. min; f(3) = 11 endpoint

and abs. max.

19. f ′(x) = 2x− 8x3

=2(x4 − 4)

x3; critical point: x = 41/4 =

√2.

f(1) = 5 endpoint max; f(√

2) = 4 local and abs. min; f(4) = 654 endpoint and abs. max.

20. f ′(x) = cosx(1 − 2 sinx); critical points: x = π6 ,

π2 ,

5π6 , 3π

2 .

f(0) = 1 endpoint min; f(π6 ) = 5/4 local and abs. max; f(π2 ) = 1 local min; f( 5π6 ) = 5

4 local and abs.

max; f( 3π2 ) = −1 local and abs. min; f(2π) = 1 endpoint max.

21. f ′(x) =−x

2√

1 − x+√

1 − x =2 − 3x

2√

1 − x; critical point: x = 2

3 .

f(1) = 0 endpoint min; f( 23 ) = 2

√3

9 local and abs. max.

22. f ′(x) =(x− 2)2x− x2

(x− 2)2=

x(x− 4)(x− 2)2

; critical point: x = 4;

f(4) = 8 local and abs. min.

23. f(x) =3x(x− 3)

(x− 4)(x + 3). Vertical asymptotes: x = −3 and x = 4; horizontal asymptote: y = 3.

24. f(x) =(x− 2)(x + 2)(x− 2)(x− 3)

=x + 2x− 3

, x �= 2. Vertical asymptote: x = 3; horizontal asymptote: y = 1.

25. f(x) = x− x

(x− 1)(x2 + x + 1). Vertical asymptote: x = 1; oblique asymptote: y = x.

26. f ′(x) =3

5(x− 1)2/5, vertical tangent.

27. f ′(x) = x2/5 − 2x−3/5 =x− 2x3/5

, vertical cusp.

28. f ′(x) = 2x−2/3 + 4x1/3 =2 + 4xx2/3

, vertical tangent.




29. f(x) = 6 + 4x3 − 3x4, domain: (−∞,∞)

f ′(x) = 12x2(1 − x)

critical pts. x = 0, 1

f ′′(x) = 12x(2 − 3x)

f ′(x) > 0 on (−∞, 1),

f ′(x) < 0 on (1,∞)

f ′′(x) > 0 on (0, 2/3)

f ′′(x) < 0 on (−∞, 0) ∪ (2/3,∞) −1 1x

2

4

6

y

30. f(x) = 3x5 − 5x3 + 1, domain: (−∞, ∞)

f ′(x) = 15x2(x− 1)(x + 1)

critical pts. x = −1, 0, 1

f ′′(x) = 30x(2x2 − 1

)f ′(x) > 0 on (−∞,−1) ∪ (1,∞)

f ′(x) < 0 on (−1, 1)

f ′′(x) > 0 on(−

√2

2 , 0)∪

(√2

2 ,∞)

f ′′(x) < 0 on(−∞,−

√2

2

)∪

(0,

√2

2

) −1 1x

−1

1

3

5

y

31. f(x) =2x

x2 + 4, domain: (−∞, ∞)

symmetric with respect to the origin.

f ′(x) =2

(4 − x2

)(x2 + 4)2

critical pts. x = 2, x = −2

f ′′(x) =4x

(x2 − 12

)(x2 + 4)3

f ′(x) > 0 on (−2, 2),

f ′(x) < 0 on (−∞, 2) ∪ (2, ∞)

f ′′(x) > 0 on(−√

12, 0)∪

(√12, ∞

)f ′′(x) < 0 on

(−∞, −

√12

)∪

(0,

√12

)−2 2

x

0.5

−0.5

y




32. f(x) = x2/3(x− 10), domain: (−∞, ∞)

f ′(x) =5(x− 4)3x1/3

critical pts. x = 4, x = 0

f ′′(x) =10(x + 2)

9x4/3

f ′(x) > 0 on (−∞, 0) ∪ (4, ∞),

f ′(x) < 0 on (0, 4)

f ′′(x) > 0 on (−2, 0) ∪ (0, ∞)

f ′′(x) < 0 on (−∞, −2)

−2 4 10x

−10

10

y

33. f(x) = x√

4 − x, domain: (−∞, 4]

f ′(x) =8 − 3x

2√

4 − x

f ′′(x) =3x− 16

4(4 − x)3/2

f ′(x) > 0 on(−∞, 8

3

],

f ′(x) < 0 on[83 , 4

)f ′′(x) < 0 on (−∞, 4)

−2 2 4x

−2

2

y

34. f(x) = x4 − 2x2 + 3, domain (−∞,∞);

symmetric with respect to the y-axis

f ′(x) = 4x3 − 4x = 4x(x− 1)(x + 1)

critical pts. x = −1, 0, 1

f ′′(x) = 12x2 − 4 = 12(x− 1/√

3)(x + 1/√

3)

f ′(x) > 0 on [−1, 0] ∪ [1,∞)

f ′(x) < 0 on (−∞,−1] ∪ [0, 1]

f ′′(x) > 0 on (∞,−1/√

3) ∪ (1/√

3,∞)

f ′′(x) < 0 on [−1/√

3, 1/√

3) -2 -1 1 2x

1

2

3

y

35. f(x) = sinx +√

3 cosx, domain [0, 2π]

f ′(x) = cosx−√

3 sinx

critical pts. x = 16 π, x = 7

6 π

f ′′(x) = − sinx−√

3 cosx

f ′(x) > 0 on [0, π/6) ∪ (7π/6, 2π]

f ′(x) < 0 on (π/6, 7π/6)

f ′′(x) > 0 on (2π/3, 5π/3)

f ′′(x) < 0 on [0, π/3) ∪ (5π/3, 2π]

π 2 πx

−2

2

y




36. f(x) = sin2 x− cosx, x ∈ [0, 2π]

f ′(x) = 2 sinx cosx + sinx

critical pts. x = 23π, π, 4

3π

f ′′(x) = 4 cos2 x + cosx− 2

f ′(x) > 0 on(0, 2

3π)∪

(π, 4

3π),

f ′(x) < 0 on(

23π, π

)∪

(43π, 2π

)1 2 3 4 5 6

x

-1

1

y

37.

−1 2x

y

38. Let r be the radius of the sphere and x the side length of the cube. Then the sum of the surface areas

equals constant implies

4πr2 + 6x2 = C =⇒ x =

√C − 4πr2

6.

The sum of the volumes is:

V =43πr3 +

(C − 4πr2

6

)3/2

, 0 ≤ r ≤√

C

4π.

Now

V ′ = 4πr2 +32

(C − 4πr2

6

)1/2 (−4

3πr

)

= 4πr2 − 2πr(C − 4πr2

6

)1/2

Set V ′(r) = 0:

2r =(C − 4πr2

6

)1/2

=⇒ 4r2 =C − 4πr2

6=⇒ r =

√C

24 + 4π, and x = 2

√C

24 + 4π.

Without loss of generality, set C = 1. Then

V (0) = (1/6)3/2 ∼= 0.07, V (1/√

24 + 4π) ∼= 0.055, V (1/√

4π) ∼= 0.09.

Thus, the sum of the volumes is a minimum when the side length of the cube equals the diameter of

the sphere. The sum of the volumes is a maximum when the side length of the cube is 0.

39. Let x be the side length of the square base. Then, since the volume is 27, the height y = 27/x2. The

conditions: x2 ≤ 18, 0 ≤ y ≤ 4 imply 3√

32 ≤ x ≤ 3

√2. Thus, the surface area S of the box is




given by

S(x) = 2x2 +108x

,3√

32

≤ x ≤ 3√

2

S′(x) = 4x− 108x2

; S ′(x) = 0 =⇒ 4x3 = 27 =⇒ x = 3

Now,

S(

3√

32

)= 27

2 + 24√

3 ∼= 55.07,

S(3) = 54

S(3√

2) = 36 + 18√

2 ∼= 61.46

(a) The minimal surface area is 54 sq. ft. at x = 3.

(b) The maximal surface area is 61.46 sq. ft. (approx.) at x = 3√

2.

40. An equation for the line with x-intercept a and y-intercept b is:x

a+

y

b= 1

Since the line passes through (1, 2),1a

+2b

= 1 =⇒ b =2a

a− 1

The area of the right triangle OAB is: S = 12ab =

a2

a− 1Now,

S ′(a) =a2 − 2a(a− 1)2

and S ′(a) = 0 at a = 0, a = 2

The triangle OAB will have minimum area when A = (2, 0) and B = (0, 4).

41. Introduce a rectangular coordinate system so that the rectangle is in the first quadrant and the base

and left side lie on the coordinate axes. Let x denote the length of the base, and y the height. Let

P = 2x + 2y be the given perimeter. If the rectangle is rotated about the y-axis, then the volume V

is given by:

V (x) =πP

2x2 − πx3, 0 ≤ x ≤ P/2.

Now,

V ′(x) = πPx− 3πx2 and V ′(x) = 0 =⇒ x = 0 or x = 13P.

At x = 0, V = 0; at x = P/3, y = P/6, V = πP 3/54; at x = P/2, V = 0. The dimensions that

generate the cylinder of maximum volume are x = P/3, y = P/6.

42. Let x be the width of the page and y the height. Then xy = 80. The area available for print is:

A = (x− 2)(y − 2.5) = (x− 2)(

80x

− 2.5)

= 85 − 2.5x− 160x

, 0 < x < ∞.

A′(x) = −2.5 +160x2

; A′(x) = 0 : 2.5x2 = 160, x2 = 64, x = 8




Since A′′(x) < 0, A has an absolute maximum at x = 8. The dimensions of the page that will maximize

the printed area are: width 8, height 10.

43. x′(t) = 1 − 2 sin t, x′′(t) = −2 cos t.

The object is slowing down when x′(t)x′′(t) < 0 =⇒ t ∈ [0, 16π) ∪ ( 1

2 π,56 π) ∪ ( 3

2 π, 2π].

44. x(t) = (4t− 1)(t− 1)2, t ≥ 0; x′(t) = v(t) = 2(4t− 1)(t− 1) + 4(t− 1)2 = 6(2t2 − 3t + 1)

= 6(2t− 1)(t− 1)

(a) The object is moving to the right: v(t) > 0 when 0 ≤ t < 12 and when t > 1; moving to the

left: v(t) < 0 when 12 < t < 1; the object changes direction at t = 1

2 and t = 1.

(b) Speed when moving left: ν = |v| = 6(3t− 2t2 − 1) on[12 , 1

]. dν/dt = 6(3 − 4t); dν/dt = 0 at

t = 3/4. The maximum speed when moving left is ν(3/4) = 3/4 units per unit time.

45. (a) x(t) =√t + 1

x′(t) = v(t) =1

2√t + 1

x′′(t) = a(t) = − 14(t + 1)3/2

= −2 [v(t)]3,

x′′′(t) = a′(t) =3

8(t + 1)5/2

(b) Position: x(17) = x(15 + 2) ∼= x(15) + 2x′(15) = 4.25;

Velocity: v(17) = v(15 + 2) ∼= v(15) + 2v′(15) =15128

∼= 0.1172;

Acceleration: a(17) = a(15 + 2) ∼= a(15) + 2a′(15) ∼= 0.0032.

46. The height of the rocket at time t is given by: y(t) = 128 t− 16 t2.

(a) At maximum height, v(t) = y′(t) = 128 − 32t = 0 =⇒ t = 4. The rocket reaches its

maximum height at 4 seconds; the maximum height is: y(4) = 256 feet.

(b) y(t) = 0 =⇒ 16t(8 − t) = 0 =⇒ t = 0, t = 8; the rocket hits the ground at t = 8

seconds; v(8) = −128 ft/sec.

47. The height at time t is: y(t) = −16t2 + 8t + y0. y(10) = 0 = −1600 + 80 + y0 =⇒ y0 = 1520 ft.

48. The height at time t is: y(t) = −16t2 + v0t. y(1) = 24 = −16 + v0 =⇒ v0 = 40. Therefore,

y(t) = −16t2 + 40t. The ball achieves maximum height at the instant v(t) = −32t + 40 = 0 =⇒t = 5

4 . The maximum height is y(5/4) = −16(5/4)2 + 40(5/4) = 25.

49. Let x be the position of the boy and let y be the length of his shadow. Thendx

dt= 168 and

x + y

12=

y

5=⇒ 7y = 5x and y =

57x

Differentiating implicitly with respect to t, we get

7dy

dt= 5

dx

dt=⇒ dy

dt=

(57

)(168) = 120

The shadow is lengthening at the rate of 120 ft/min.




50. V = 13πr

2h, V constant

0 =d

dtV =

d

dt

(13πr2h

)=

2πrh3

dr

dt+

πr2

3dh

dt

Solving fordh

dtwe get:

dh

dt= −2h

r· drdt

.

At the instant r = 4 and h = 15,dh

dt= −30

4(0.3) = −2.25; the height is decreasing at the rate of

2.25 in/min.

51. Let x be the position of the locomotive at time t and let y be the position of the car. By the law of

cosines the distance between the locomotive and the car is given by

s2 = x2 + y2 − 2xy cos 60◦ = x2 + y2 − xy.


2sds

dt= 2x

dx

dt+ 2y

dy

dt− x

dy

dt− y

dx

dt.

Setting dx/dt = 60, dy/dt = −30 and converting to feet, gives

2sds

dt= 2x(60)(5280) − 2y(30)(5280) + x(30)(5280) − y(60)(5280) = 5280(150x− 120y).

When x = y = 500, s = 500 and

2(500)ds

dt= 5280(30)(500) =⇒ ds

dt= 15(5280).

The distance between the locomotive and the car is increasing at the rate of 15 miles per hour.

52. Let r be the radius of the circle. Then dr/dt = 5. The square has side length r√

2 and area A = 2r2.


dA

dt= 4r

dr

dt= 20r; and

dA

dt

∣∣∣∣r=10

= 200.

53. The cross-section of the water at time t is an isosceles triangle with base x and height y where x and

y are related by12x

y= 1

2 (similar triangles). Thus x = y.

The volume of water in the trough at time t is

V = 12( 12xy) = 6y2.

Differentiating with respect to t givesdV

dt= 12y

dy

dt=⇒ dy

dt=

112y

dV

dt.

Since dV/dt = −3,dy

dt= − 1

4yand

dy

dt

∣∣∣∣t=1.5

= −16. The water level is falling at the rate of

1/6 feet per minute.

54. f(3.8) ∼= f(4) + f ′(4)(−0.2) = 2 + 2(−0.2) = 1.6.




55. f(x) =√x + 1/

√x, f ′(x) = 1

2x−1/2 − 1

2x−3/2

f(4.2) ∼= f(4) + f ′(4)(0.2) = 52 + 3

16 (0.2) = 2.5375.

56. f(x) = x1/4, f ′(x) = 14x

−3/4

f(83) ∼= f(81) + f ′(81)(2) = 3 + 154

∼= 3.01852.

57. f(x) = tan x, f ′(x) = sec2 x

f(43◦) = f( 14π − 1

90π) ∼= f(π/4) + f ′(π/4)(−π/90) = 1 − 2(π/90) = 1 − (π/45) ∼= 0.9302.

58. V = 43πr

3, dV = 4πr2 dr. Calculate dV at r = 10 ft. and dr = 0.05 in. ∼= 0.00417 ft.

dV |r=10, dr=0.00417 = 4π (10)2 (0.00417) ∼= 5.240 cu. ft. = 9055.025 cu. in.;

= 9055.025/231 ∼= 39.20 gal.

59. f(x) = x3 − 10

(a) xn+1 = xn − x3n − 103x2

n

=2x3

n + 103x2

n

(b) x4∼= 2.15443; f(2.15443) ∼= −0.00007

60. f(x) = x sin x− cos x

(a) xn+1 =x2n cos xn + xn sin xn + cos xn

2 sin xn + xn cos xn

(b) x4∼= 0.86057; f(0.86057) ∼= 0.00049

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Calculus one and several variables 10E Salas solutions manual ch04