Math 241 Homework 7 Solutions

Section 4.2

Problem 1. Find the value or values c that satisfy the equationf(b) − f(a)

b − a= f ′(c) in the conclu-

sion of the Mean Value Theorem for functions and intervals:

f(x) = x2 + 2x − 1, [0,1]

Solutionf(b) − f(a)

b − a=

(1 + 2 − 1) − (0 + 0 − 1)

1 − 0= 3

f ′(x) = 2x + 2 so we want to solve the equation 2x + 2 = 3

2x + 2 = 3⇔ 2x = 1⇔ x =1

2

Problem 3. Find the value or values c that satisfy the equationf(b) − f(a)

b − a= f ′(c) in the conclu-

sion of the Mean Value Theorem for functions and intervals:

f(x) = x +1

x, [

1

2,2]

Solutionf(b) − f(a)

b − a=

(1/2 + 2) − (2 + 1/2)

2 − 1/2= 0

f ′(x) = 1 −1

x2=x2 − 1

x2

f ′(x) = 0⇔ x2 − 1 = 0⇔ x = ±1

The solution in the given interval is x = 1 .

Problem 7. Does f(x) =√x(1 − x) on [0,1] satisfy the hypotheses of the Mean Value Theorem?

Solution Continuous forx(1 − x) ≥ 0⇔ 0 ≤ x ≤ 1

and differentiable for x ∈ (0,1) so it does .

1

Problem 9. The function

f(x) =

⎧⎪⎪⎨⎪⎪⎩

x, 0 ≤ x < 1

0, x = 1

is zero at x = 0 and x = 1 and differentiable on (0,1) but its derivative on (0,1) is never zero. Howcan this be? Doesn’t Rolle’s Theorem say the derivative has to be zero somewhere in (0,1)? Givereasons for your answer.

Solution The function is not continuous on [0,1] because it is not continuous at 1.

limx→1−

x = 1 and f(1) = 0

2

Problem 11. (a) Plot the zeros of each polynomial on a line together with the zeros of its firstderivative.

(i) y = x2 − 4

(ii) y = x2 + 8x + 15

(iii) y = x3 − 3x2 + 4 = (x + 1)(x − 2)2

(iv) x3 − 33x2 + 216x = x(x − 9)(x − 24)

(b) Use Rolle’s Theorem to prove that between every two zeros of xn + an−1xn−1 + ... + a1x + a0there lies a zero of

nxn−1 + (n − 1)an−1xn−2 + ... + a1

Solution

(a) (i)y′ = 2x

x2 − 4 = 0⇔ x = ±2 and 2x = −⇔ x = 0

(ii)y′ = 2x + 8 = 2(x + 4)

2(x + 4) = 0⇔ x = −4 and x2 + 8x + 15 = (x + 5)(x + 3) = 0⇔ x = −3,−5

(iii)y′ = 3x2 − 6x = 3x(x − 2)

3x(x − 2) = 0⇔ x = 0,2 and (x + 1)(x − 2)2 = 0⇔ x = −1,2

(iv)y′ = 3x2 − 66x + 216 = 3(x2 − 22 + 72) = 3(x − 18)(x − 4)

3(x − 18)(x − 4) = 0⇔ x = 18,4 and x(x − 9)(x − 24) = 0⇔ x = 0,9,24

The number lines are as follows:

Chapter 4: Answers to Odd-Numbered Exercises A-17

29. Yes 31. (a) 4 (b) 3 (c) 3

33. (a) x2

2 + C (b) x3

3 + C (c) x4

4 + C

35. (a) 1x + C (b) x + 1

x + C (c) 5x - 1x + C

37. (a) - 12 cos 2t + C (b) 2 sin t2 + C

(c) - 12 cos 2t + 2 sin t2 + C

39. ƒ(x) = x2 - x 41. ƒ(x) = 1 + e2x

2

43. s = 4.9t2 + 5t + 10 45. s =1 - cos (pt)

p 47. s = et + 19t + 4 49. s = sin (2t) - 3 51. If T(t) is the temperature of the thermometer at time t, then

T(0) = -19 °C and T(14) = 100 °C. From the Mean Value

Theorem, there exists a 0 6 t0 6 14 such that T(14) - T(0)

14 - 0 = 8.5 °C>sec = T′(t0), the rate at which the temperature was

changing at t = t0 as measured by the rising mercury on the thermometer.

53. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been going that speed at least once during the trip.

57. The conclusion of the Mean Value Theorem yields

1b

- 1a

b - a= - 1

c2 1 c2aa - babb = a - b 1 c = 1ab.

61. ƒ(x) must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that ƒ(x) is zero twice between a and b. Then, by the Mean Value Theorem, ƒ′(x) would have to be zero at least once between the two zeros of ƒ(x), but this can’t be true since we are given that ƒ′(x) ≠ 0 on this interval. Therefore, ƒ(x) is zero once and only once between a and b.

71. 1.09999 … ƒ(0.1) … 1.1

Section 4.3, pp. 228–230 1. (a) 0, 1 (b) Increasing on (-q, 0) and (1, q); decreasing on (0, 1) (c) Local maximum at x = 0; local minimum at x = 1 3. (a) -2, 1 (b) Increasing on (-2, 1) and (1, q); decreasing on (-q, -2) (c) No local maximum; local minimum at x = -2 5. (a) Critical point at x = 1 (b) Decreasing on (-q, 1), increasing on (1, q) (c) Local (and absolute) minimum at x = 1 7. (a) 0, 1 (b) Increasing on (-q, -2) and (1, q); decreasing on (-2, 0)

and (0, 1) (c) Local minimum at x = 1 9. (a) -2, 2 (b) Increasing on (-q, -2) and (2, q); decreasing on (-2, 0)

and (0, 2) (c) Local maximum at x = -2; local minimum at x = 2 11. (a) -2, 0 (b) Increasing on (-q, -2) and (0, q); decreasing on (-2, 0) (c) Local maximum at x = -2; local minimum at x = 0

13. (a) p2 , 2p3 , 4p3

69. Critical point or endpoint

Derivative Extremum Value

x = -45

0 Local max 1225

101>3 ≈ 1.034

x = 0 Undefined Local min 0

71. Critical point or endpoint

Derivative Extremum Value

x = -2 Undefined Local max 0

x = - 22 0 Minimum –2

x = 22 0 Maximum 2

x = 2 Undefined Local min 0

73. Critical point or endpoint

Derivative Extremum Value

x = 1 Undefined Minimum 2

75. Critical point or endpoint

Derivative Extremum Value

x = -1 0 Maximum 5x = 1 Undefined Local min 1x = 3 0 Maximum 5

77. (a) No (b) The derivative is defined and nonzero for x ≠ 2. Also,

ƒ(2) = 0 and ƒ(x) 7 0 for all x ≠ 2. (c) No, because (-q, q) is not a closed interval. (d) The answers are the same as parts (a) and (b), with 2

replaced by a. 79. Yes 81. g assumes a local maximum at -c. 83. (a) Maximum value is 144 at x = 2. (b) The largest volume of the box is 144 cubic units, and it oc-

curs when x = 2.

85. y0

2

2g+ s0

87. Maximum value is 11 at x = 5; minimum value is 5 on the inter-val [-3, 2]; local maximum at (-5, 9).

89. Maximum value is 5 on the interval [3, q); minimum value is -5 on the interval (-q, -2].

Section 4.2, pp. 222–224

1. 1>2 3. 1 5. {A1 - 4p2 ≈ {0.771

7. 13 11 + 272 ≈ 1.22, 13 11 - 272 ≈ -0.549

9. Does not; ƒ is not differentiable at the interior domain point x = 0.

11. Does 13. Does not; ƒ is not differentiable at x = -1. 17. (a) i) x

−2 20

ii) x−5 −4 −3

iii) x−1 0 2

iv) 0 4 9 18 24

x

(b) Let x1 and x2 be two zeros of f(x) = xn + an−1xn−1 + ... + a1x + a0. By Rolle’s theorem wehave that there exists a c between x1 and x2 such that

ncn−1 + (n − 1)an−1cn−2 + ... + a1 = f ′(c) =f(x1) − f(x2)

x1 − x2= 0

3

Problem 17. Show that g(t) =√t +

√1 + t − 4 has exactly one zero in (0,∞)

Solutiong(1) = 1 +

√2 − 4 < 0 and g(5) =

√5 +

√6 − 4 > 0

Thus by the Intermediate Value Theorem we have that there is at least on zero on (0,∞). Assumethere are two zeroes. Then by Rolle’s Theorem there is a c in (0,∞) such that g′(c) = 0

g′(t) =1√t+

1√

1 + t

=

√1 + t +

√t

√t + t2

⋅

√1 + t −

√t

√1 + t −

√t

=1

√t + t2(

√1 + t −

√t)

f ′(c) = 0⇔1

√c + c2(

√1 + c −

√c)

= 0⇔ 1 = 0

which is impossible thus there cannot be two solutions.

Problem 27. Find all possible functions with the given derivative:

(i) y′ = x

(ii) y′ = x2

(iii) y′ = x3

Solution

(i)x2

2has derivative x so the answer is

x2

2+C

(ii)x3

3has derivative x2 so the answer is

x3

3+C

(iii)x4

4has derivative x3 so the answer is

x4

4+C

4

Problem 35. Find the function with the given derivative whose graph passes through the point P :

r′(θ) = 8 − csc2 θ, (π

4,0)

Solution 8θ + cot θ has derivative 8− csc2 θ so we have the general form is 8θ + cot θ +C. We alsowant it to pass through P so we have

0 = 8π

4+ cot

π

4+C⇔ 0 = 2π + 1 +C

⇔ C = −1 − 2π

⇒ r(θ) = 8θ + cot θ − 1 − 2π

Problem 47. Classical accounts tell us that a 170-oar trireme (ancient Greek or Roman warship)once covered 184 sea miles in 24 hours. Explain why at some point during this feat, the trireme’sspeed exceeded 7.5 knots (sea miles per hour).

Solution The average speed is184

24≈ 7.7

Thus by the Mean Value Theorem there has to be a point where the boat is going 7.7 knots whichexceeds 7.5 knots.

5

Problem 54. (i) Construct a polynomial f(x) that has zeros at x = −2,−1,0,1, and 2

(ii) Graph f and its derivative f ′ together. How is what you see related to Rolle’s Theorem?

(iii) Do g(x) = sinx and its derivative g′ illustrate the same phenomenon?

Solution

(i) We have that f(x) looks like x(x + 2)(x + 1)(x − 1)(x − 2)

(ii) Graphing this we have

Which shows that the derivative is zero (i.e. intercepts the x-axis) between every two zerosof f

(iii) Yes

Problem 55. Assume that f is continuous on [a, b] and differentiable on (a, b). Also assume thatf(a) and f(b) have opposite signs and that f ′ ≠ 0 between a and b. Show that f(x) = 0 exactly oncebetween a and b.

Solution By the Intermediate Value Theorem, there must be at least one zero between a andb. If there were more than one, then by Rolle’s Theorem, there would have to be a point betweenthem such that the derivative f ′ is zero. However, we are given that the derivative is never zerobetween a and b and thus there can only be exactly one solution.

6

Section 4.3

Problem 1. Answer the following questions about the function whose derivative is given by f ′(x) =x(x − 1):

(i) What are the critical points of f?

(ii) On what intervals is f increasing or decreasing?

(iii) At what points, if any, does f assume local maximum and minimum values?

Solution

(i)f ′(x) = 0⇔ x = 0,1

Thus the critical points occur at x = 0 and x = 1

(ii) Plotting the critical points we get

10

Thus f is increasing on (−∞,0), (1,∞) and decreasing on (0,1)

(iii) There is a local max at x = 0 and a local min at x = 1

7

Problem 7. Answer the following questions about the function whose derivative is given by f ′(x) =x−1/3(x + 2):

(i) What are the critical points of f?

(ii) On what intervals is f increasing or decreasing?

(iii) At what points, if any, does f assume local maximum and minimum values?

Solution

(i)

f ′(x) =x + 2

x1/3

f ′ is undefined at x = 0 and zero at x = −2 so the critical points are x = 0 and x = −2

(ii) Testing the intervals we have

-2 0

Thus f is increasing on (−∞,−2), (0,∞) and decreasing on (−2,0)

(iii) There is a local min at x = 0 and a local max at x = −2

8

Problem 9. Let g(t) = −t2 − 3t + 3

(i) Find the intervals on which the function is increasing and decreasing.

(ii) Then identify the function’s local extreme values, if any, saying where they are taken on.

(iii) Which, if any, of the extreme value are absolute?

(iv) Support your findings with a graphing calculator or computer grapher.

Solution

(i)

g′(t) = 0⇔ −2t − 3 = 0⇔ t = −3

2

Testing the intervals we get

-3/2

Thus g is increasing on (−∞,−3/2) and decreasing on (−3/2,∞)

(ii)

f (−3

2) = −(−

3

2)

2

− 3(−3

2) + 3

= −9

4+

18

4+

12

4

=21

4

There is a local max of 21/4 at x = −3/2

(iii) Since the function has only one local max, 21/4 at x = −3/2 is an absolute max.

(iv) The graph looks like

9

Problem 15. Let f(r) = 3r3 + 16r

(i) Find the intervals on which the function is increasing and decreasing.

(ii) Then identify the function’s local extreme values, if any, saying where they are taken on.

(iii) Which, if any, of the extreme value are absolute?

(iv) Support your findings with a graphing calculator or computer grapher.

Solution

(i)f ′(r) = 9r2 + 16 > 0

Thus f is increasing on (−∞,∞)

(ii) There are no local extrema

(iii) There are no absolute extrema

(iv) The graph looks like

10

Problem 19. Let H(t) =3

2t4 − t6

(i) Find the intervals on which the function is increasing and decreasing.

(ii) Then identify the function’s local extreme values, if any, saying where they are taken on.

(iii) Which, if any, of the extreme value are absolute?

(iv) Support your findings with a graphing calculator or computer grapher.

Solution

(i)H ′

(t) = 6t3 − 6t5 = 6t3(1 − t2) = 6t3(1 − t)(1 + t)

H ′(t) = 0⇔ t = ±1,0 Plotting these critical points and testing the intervals we have

-1 0 1

Thus H is increasing on (−∞,−1), (0,1) and decreasing on (−1,0), (1,∞)

(ii)

H(−1) =1

2, H(0) = 0, H(1) =

1

2

Thus there are local maxes of 1/2 at x = −1 and x = 1 and a local min of 0 at x = 0

(iii) Since the graph points downward at its ends there is no local minimum and

the two local maxes are absolute

(iv) The graph looks like

11

Problem 35. Let h(x) =x3

3− 2x2 + 4x, 0 ≤ x <∞

(i) Identify the function’s local extreme values in the given domain, and say where they areassumed.

(ii) Which, if any, of the extreme value are absolute?

(iii) Support your findings with a graphing calculator or computer grapher.

Solution

(i)h′(x) = x2 − 4x + 4 = (x − 2)2 and h′(x) = 0⇔ x = 2

Since h′(x) > 0 we have that h is increasing on [0,∞) Thus there is a local minimum of 0 at x = 0

(ii) Since the local minimum is attained at the endpoint and the function is always increasing,

the local minimum is an absolute minimum

(iii) The graph is

12