MATH 1A: CALCULUS HOMEWORK #8

JOHN VOIGHT

§4.2: The Mean Value Theorem

Problem 2. We compute that f(0) = 5 = 2^{3} − 3(2^{2}) + 2(2) + 5 = f(2), so the starting and ending values are the same. The function f is continuous on [0, 2] (in fact, on all real numbers) because it is a polynomial. It is differentiable on (0, 2) also because it is a polynomial: to check this, we see that

f′(x) = 3x^{2} − 6x + 2

which has domain all real numbers.

We want to find all values c such that f′(c) = 0:

f′(c) = 3c^{3} − 6c + 2 = 0;

by the quadratic formula, we get

c = \frac{6 ± \sqrt{36 − 24}}{6} = 1 ± \frac{\sqrt{3}}{3}.

In terms of Rolle’s theorem, for the conclusion we consider only values in the open interval (0, 2), and both of these values lie in this interval (since \sqrt{3}/3 = 1/\sqrt{3} < 1).

Problem 4. We see that

f(−6) = 0 = f(0);

the function f is continuous on the interval [−6, 0] since it is the product of a root function and a polynomial; and that

f′(x) = (x(x + 6)^{1/2})′ = \frac{1}{2}x(x + 6)^{−1/2} + (x + 6)^{1/2} = \frac{x}{2\sqrt{x + 6}} + \sqrt{x + 6}

which has domain x > −6 so f is differentiable on (−6, 0). Therefore f satisfies the conditions of Rolle’s theorem, and

f′(x) = \frac{x + 2(x + 6)}{2\sqrt{x + 6}} = 0
3x + 12 = 0

so c = −4, which is indeed in (−6, 0).

Problem 6. We compute that f′(x) = −2(x − 1)^{−3} = −2/(x − 1)^{3}, which is never zero. This does not contradict Rolle’s theorem because the function f is discontinuous at x = 1, so Rolle’s theorem does not apply to the function f on the interval [0, 2].

Problem 8. We see that f(1) = 5 and f(7) = 2, so we look for values c such that

f′(c) = \frac{f(7) − f(1)}{7 − 1} = \frac{2 − 5}{6} = −\frac{1}{2}.

Looking at the graph, we see that the slope is about −1/2 and we have the values

c ≈ 1.2, 2.8, 4.7, 5.8.

Date: March 16, 2004.

§4.2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 19, 21(a), 24, 30, 31, 32; §4.3: 2, 6, 8, 12, 14, 20, 22, 32, 34, 38, 42; Updated March 17, 2004.

Problem 10(a). By plotting points, we have:

From the graph, we estimate that the x coordinates are −1.1, 1.1.

Problem 10(b). On the interval [−2, 2], we compute

\frac{f(b) − f(a)}{b − a} = \frac{4 − (−4)}{2 − (−2)} = 2.

So we want to solve

f′(x) = 3x^{2} − 2 = 2

which has the roots x^{2} = 4/3 or x = ±2/\sqrt{3} = ±(2\sqrt{3})/3 ≈ ±1.15. These compare well with the value estimated in (a).

Problem 12. The function f(x) = x^{3} + x − 1 is continuous on [0, 2] and differentiable on (0, 2) because it is a polynomial, therefore it satisfies the hypotheses of the Mean Value Theorem.

We compute that

\frac{f(2) − f(0)}{2 − 0} = \frac{9 + 1}{2} = 5

and

f′(x) = 3x^{2} + 1 = 5
x^{2} = 4/3
x = ±2/\sqrt{3} = ±(2\sqrt{3})/3.

In the interval (0, 2), we have only c = (2\sqrt{3})/3 satisfying the conditions of the theorem.

Problem 14. The function f(x) = x/(x + 2) is continuous on [1, 4] because this is a rational function on its domain (x ≠ −2). It is differentiable on (1, 4) because

f′(x) = \frac{(x + 2) − x}{(x + 2)^{2}} = \frac{2}{(x + 2)^{2}}

is also a rational function with domain x ≠ −2. Therefore it satisfies the conditions of the Mean Value Theorem.

We compute that

\frac{f(4) − f(1)}{4 − 1} = \frac{2/3 − 1/3}{3} = \frac{1}{9}

and

f′(x) = \frac{2}{(x + 2)^{2}} = \frac{1}{9}
18 = (x + 2)^{2}
±3\sqrt{2} = x + 2
x = −2 ± 3\sqrt{2}

In the interval (1, 4), we take only c = −2 + 3\sqrt{2} ≈ 2.2 satisfying the conditions of the theorem.

Problem 16. We see that

f(2) − f(0) = 3 + 1 = 4

and

f′(x) = \frac{(x − 1) − (x + 1)}{(x − 1)^{2}} = \frac{−2}{(x − 1)^{2}}

so we look for c such that 4 = f′(c)(2) or f′(c) = 2, i.e.

\frac{−2}{(c − 1)^{2}} = 2
−1 = (c − 1)^{2}

This equation has no solution, since the square of a real number is never negative. Therefore no such c satisfying the Mean Value Theorem exists.

This does not contradict the Mean Value Theorem because the function f is discontinuous at x = 1, so the Mean Value Theorem does not apply to the function f on the interval [0, 2].

Problem 18. First, we show that 2x − 1 − sin x = 0 has at least one real root. Let f(x) = 2x − 1 − sin x. Then f(0) = −1 and f(π) = 2π − 1 > 0. The function f is continuous on the interval [0, π] (in fact, all real numbers) because it is the difference of a polynomial and a trigonometric function. Therefore by the Intermediate Value Theorem, it has a root.

Suppose the equation has two real roots. The function f is continuous and differentiable for all real numbers (since f′(x) = 2 − cos x arises from a trigonometric function). Therefore if f(a) = f(b) = 0, f satisfies the conditions of Rolle’s theorem on the interval [a, b]; but

f′(x) = 2 − cos x = 0

has no solution, since cos x ≤ 1. This is a contradiction, so f has at most one real root. Hence f has exactly one real root.

Problem 19. Suppose that f has more than one real root in the interval [−2, 2]. The function f is continuous and differentiable on this interval (it is a polynomial), so by Rolle’s theorem, f′(x) = 0 somewhere in this interval. But

f′(x) = 3x^{2} − 15 = 3(x^{2} − 5) = 0

has only the roots x = ±\sqrt{5} ≈ 2.23, which do not lie in the interval (−2, 2). This is a contradiction, so f has at most one real root.

Problem 21(a). Suppose that the cubic polynomial f has at least 4 roots. A polynomial is continuous and differentiable for all real numbers, therefore by Rolle’s theorem, the derivative f′(x) must take on the value 0 at least 3 times, in each of the three consecutive intervals with endpoints among the 4 roots. But since f has degree 3, we know that f′ has degree 2, which by the quadratic formula can have at most 2 roots. This is a contradiction, therefore f has at most 3 real roots.

Problem 24. If we can apply the Mean Value Theorem (which we must be able to, see Example 5), we will conclude that there exists a c in (a, b) such that

f′(c)(b − a) = f(b) − f(a).

This looks like the inequality

18 ≤ f(8) − f(2) ≤ 30

if we take b = 8 and a = 2.

Therefore, let a = 2 and b = 8, i.e. look at f on the interval [2, 8]. By the Mean Value Theorem, there is a c in (2, 8) such that

6f′(c) = f(8) − f(2)

But now we are given that

3 ≤ f′(c) ≤ 5,

which multiplying by 6 gives

18 ≤ 6f′(c) ≤ 30

which by the previous equality is just

18 ≤ f(8) − f(2) ≤ 30

as desired.

Problem 30. Let g(x) = cx. Then g′(x) = c, so f′(x) = g′(x) for all real numbers. Therefore by Corollary 7, f − g is a constant, which we call d. That is,

f(x) = g(x) + d = cx + d.

Problem 31. We see indeed that f′(x) = −1/x^{2} and that in both cases x > 0 and x < 0, g′(x) = −1/x^{2}.

We can conclude from Corollary 7 that f − g is constant on the interval (0, ∞), and separately that f − g is also constant on the interval (−∞, 0), but these constants may very well be different, since neither f nor g is defined for x = 0.

Problem 32. Let

f(x) = 2 sin^{−1} x

and

g(x) = cos^{−1}(1 − 2x^{2}).

for x ≥ 0. By trig formulas, we have

f′(x) = \frac{2}{\sqrt{1 − x^{2}}} .

By the chain rule, with u = 1 − 2x^{2} we compute that

g′(x) = −\frac{1}{\sqrt{1 − u^{2}}} u′ = −\frac{1}{\sqrt{1 − (1 − 2x^{2})^{2}}} (−4x) = \frac{4x}{\sqrt{1 − (1 − 4x^{2} + 4x^{4})}}

= \frac{4x}{\sqrt{4x^{2} − 4x^{4}}} = \frac{4x}{\sqrt{4x^{2}(1 − x^{2})}} = \frac{4x}{2x\sqrt{1 − x^{2}}} = \frac{2}{\sqrt{1 − x^{2}}} .

Since f′(x) = g′(x), i.e. f′(x) − g′(x) = 0, by Corollary 7, we see that f(x) − g(x) is constant, or f(x) = g(x) + c. Now

f(0) = 2 sin^{−1}(0) = 0

and

g(0) = cos^{−1}(1) = 0 = 0

so in fact c = 0, i.e.

2 sin^{−1} x = cos^{−1}(1 − 2x^{2}).

§4.3: How derivatives affect the shape of a graph

Problem 2(a). f is increasing on (1, 3.9) and (5, 6.5), approximately. (Note they ask for open intervals!)

Problem 2(b). f is decreasing on (0, 1), (3.9, 5), (6.5, 9), approximately.

Problem 2(c). f is concave upward on (0, 3) and (8, 9).

Problem 2(d). f is concave downward on (3, 8).

Problem 2(e). f has only the inflection point x = 3. At x = 5, it stays from concave down to concave down; at x = 8, it is not continuous.

Problem 6(a). f is increasing if f′ > 0 (including appropriate endpoints); therefore f is increasing on [0, 1] and [3, 5].

f is decreasing if f′ < 0 (including appropriate endpoints); therefore f is decreasing on [1, 3] and [5, 6].

Problem 6(b). f has a local maximum or minimum only if f′(x) = 0, i.e. x = 1, 3, 5. At x = 1 and x = 5, f′ goes from + to − so it is a local maximum; at x = 3 it goes from − to + so it is a local minimum.

Problem 8(a). f is increasing if f′ > 0, including appropriate endpoints; therefore f is increasing on [2, 4] and [6, 9]. (By default, if there is no dot at x = 0 or x = 9, we assume it is defined there, i.e. it is a solid dot.)

Problem 8(b). At x = 2 and x = 6, f′ goes from − to +, so f has a local minimum there; at x = 4, f′ goes from + to −, so f has a local maximum there.

Problem 8(c). f is concave upward if f′′ > 0 and concave downward if f′′ < 0. Looking at the graph of f′, we see that the slope of f′ is positive on (1, 3), (5, 7), and (8, 9) so the function is concave upward on these intervals, and f′ has negative slope on (0, 1), (3, 5), and (7, 8), so f is concave downward there.

Problem 8(d). The inflection points are where f′′(x) = 0 and changes sign; these are the points x = 1, 3, 5, 7, 8.

Problem 12(a). We compute that

f′(x) = −6x + 3x^{2} = 3x(x − 2).

Therefore f′(x) > 0 and f is increasing when 3x > 0 and x−2 > 0, i.e. x > 2, or when 3x < 0 and x−2 < 0, i.e. x < 0. Similarly, f′(x) < 0 and f is decreasing when 3x > 0 and x − 2 < 0, i.e. 0 < x < 2, or 3x < 0 and x − 2 > 0, which cannot happen.

f is increasing on the intervals (−∞, 0] and [2, ∞) and f is decreasing on the interval [0, 2].

Problem 12(b). We see that f′(x) = 0 for x = 0, 2. Here,

f′′(x) = −6 + 6x = 6x − 6

so f′′(0) = −6 < 0 and x = 0 is a local maximum, and f′′(2) = 6 > 0 so x = 2 is a local minimum.

Problem 12(c). We have f′′(x) = 6x − 6 = 0 for x = 1. Since from (b) f′′(0) < 0 and f′′(2) > 0, we see that on the interval (−∞, 1), f is concave downward, and on the interval (1, ∞), f is concave upward. f has an inflection point at x = 1.

Problem 14(a). We compute that

f′(x) = \frac{(x^{2} + 3)(2x) − x^{2}(2x)}{(x^{2} + 3)^{2}} = \frac{6x}{(x^{2} + 3)^{2}} .

Since (x^{2} + 3)^{2} > 0 for all x, we see that f′(x) > 0 and f is increasing for x > 0 and f′(x) < 0 and f is decreasing for x < 0. Therefore f is decreasing on the interval (−∞, 0] and is increasing on the interval [0, ∞).

Problem 14(b). Since f changes from decreasing to increasing at x = 0, we see that it is a local minimum.

Problem 14(c). We compute that

f′′(x) = \frac{(x^{2} + 3)^{2}(6) − (6x)(2(x^{2} + 3)(2x))}{(x^{2} + 3)^{4}}

= \frac{6(x^{2} + 3)^{2} − 24x^{2}(x^{2} + 3)}{(x^{2} + 3)^{4}}

= \frac{(x^{2} + 3)(6(x^{2} + 3) − 24x^{2})}{(x^{2} + 3)^{4}}

= \frac{−18x^{2} + 18}{(x^{2} + 3)^{3}} = \frac{−18(x^{2} − 1)}{(x^{2} + 3)^{3}}.

Therefore f′′(x) = 0 for x = −1, 1. We saw that f′′(0) > 0, so f is concave upward on that interval. On the interval (−∞, −1) we see e.g. f′′(−2) = −18(3)/7^{3} < 0 so f is concave downward, and similarly f′′(2) > 0 so f is also concave downward there.

In sum, f is concave upward on (−1, 1) and f is concave downward on (−∞, −1) and (1, ∞). f has inflection points at x = ±1.

Problem 20(a). The domain of the function, since ln x is defined only for x > 0, is x > 0. We have

f′(x) = x(1/x) + ln x = 1 + ln x

so f′(x) < 0 and f is decreasing for 1 + ln x < 0, i.e. ln x < −1 or x < e^{−1} = 1/e, and f′(x) > 0 and f is increasing for x > 1/e.

In sum, f is decreasing on (0, 1/e] and is increasing on [1/e, ∞).

Problem 20(b). Now

f′′(x) = 1/x

so f′′(1/e) = e > 0, so x = 1/e is a local minimum.

Problem 20(c). Since f′′(x) = 1/x is never zero, and f is defined only for x > 0, we conclude that f is always concave upward, and has no inflection point.

Problem 22(a). For f(x) = x/(x^{2} + 4), we have

f′(x) = \frac{(x^{2} + 4) − x(2x)}{(x^{2} + 4)^{2}} = \frac{−x^{2} + 4}{(x^{2} + 4)^{2}} .

This is zero only when −x^{2} + 4 = 0, i.e. x = ±2.

Since f′(−3) = (−9 + 4)/(9 + 4)^{2} < 0 and f′(−1) = (−1 + 4)/(1 + 4)^{2} > 0, by the First Derivative Test, x = −2 is a local minimum. Since f′(1) = (−1 + 4)/(1 + 4)^{2} > 0 and f′(3) = (−9 + 4)/(9 + 4)^{2} < 0, x = 2 is a local maximum.

For the Second Derivative Test, we compute that

f′′(x) = \frac{(x^{2} + 4)^{2}(−2x) − (−x^{2} + 4)(2(x^{2} + 4)(2x))}{(x^{2} + 4)^{4}}

= \frac{−2x(x^{2} + 4) − 4x(−x^{2} + 4)(x^{2} + 4)}{(x^{2} + 4)^{4}}

so f′′(−2) = (4(8)−0)/(4+4)^{4} > 0, and again x = −2 is a local minimum, and f′′(2) = (−4(8)−0)/(4+4)^{4} < 0, so x = 2 is a local maximum.

Problem 32(a). f is increasing when f′ > 0, i.e. on [1, 6] and [8, 9], and decreasing when f′ < 0, i.e. on [0, 1] and [6, 8].

Problem 32(b). x has a local maximum or minimum possibly only when f′ = 0, when x = 1, 6, 8. We see that f′′(1) > 0 (the slope of the graph of f′ at x = 1 is positive), so x = 1 is a local minimum, and similarly f′′(6) < 0 so x = 6 is a local maximum and f′′(8) > 0 so x = 8 is a local minimum.

Problem 32(c). f is concave upward where f′′ > 0, on the intervals (0, 2), (3, 5), and (7, 9). f is concave downward where f′′ < 0, on (2, 3) and (5, 7).

Problem 32(d). The points of inflection are where f′′ = 0 and f′′ changes sign. These are the values x = 2, 3, 5, 7.

Problem 32(e). We use the data from (a)–(d): first, we locate the local maximum at x = 1 and local minima at x = 6 and x = 8; then we note it is concave upward on [0, 2), (3, 5), and (7, 9] and concave downward on (2, 3) and (5, 7). Since f is increasing on [1, 6] and [8, 9] and decreasing on [0, 1] and [6, 8], starting at the origin we can draw the following graph:

Problem 34(a). We have f′(x) = 3 − 3x^{2} = 3(1 − x^{2}) = 3(1 − x)(1 + x), so f′ > 0 and f is increasing when 1 − x > 0 and 1 + x > 0, i.e. when x < 1 and x > −1 or on the interval (−1, 1), or when 1 − x < 0 and 1 + x < 0, i.e. x > 1 and x < −1 which cannot happen.

We see then that f is decreasing on (−∞, −1] and [1, ∞), and increasing on [−1, 1].

Problem 34(b). We have f′(x) = 0 for x = −1, 1. Since f′′(x) = −6x, f′′(−1) = 6 so x = −1 is a local minimum and f′′(1) = −6 so x = 1 is a local maximum.

Problem 34(c). We have f′′(x) = −6x = 0 for x = 0, the only inflection point. For x < 0 f′′(x) > 0 so f is concave upward; for x > 0, f′′(x) < 0 so f is concave downward.

Problem 34(d). We graph using the data in (a)–(d).

Problem 38(a). We have h′(x) = 3(x^{2} − 1)(2x) = 6x(x^{2} − 1)^{2}. Since (x^{2} − 1)^{2} > 0, we see that h′ > 0 and h is increasing for x > 0 and h′ < 0 and h is decreasing for x < 0, i.e. h is increasing on [0, ∞) and decreasing on (−∞, 0].

Problem 38(b). The solutions to h′(x) = 6x(x^{2} − 1)^{2} = 0 are x = 0, −1, 1. We have

h′′(x) = 6(x^{2} − 1)^{2} + 6x(2)(x^{2} − 1)(2x) = 6(x^{2} − 1)^{2} + 24x^{2}(x^{2} − 1)

and h′′(0) = 6 > 0 so x = 0 is a local minimum. However, h′′(−1) = 0 which is inconclusive. So we test that h′(−1/2) = −3(1/4 − 1)^{2} < 0 and h′(−3/2) = −9(9/4 − 1)^{2} < 0, so x = −1 is neither a local minimum nor a local maximum. Similary, we see that h′(1/2) = 3(1/4 − 1) > 0 and h′(3/2) = 9(9/4 − 1)^{2} > 0, so x = 1 is neither a local minimum nor a local maximum.

Problem 38(c). We factor

h′′(x) = (x^{2} − 1)(6(x^{2} − 1) − 24x^{2}) = (x^{2} − 1)(30x^{2} − 6) = 6(x^{2} − 1)(5x^{2} − 1) = 0.

This gives the roots x = ±1, ±1/\sqrt{5}. We test the values

h′′(−2) = 6(4 − 1)(12 + 1) > 0

and

h′′(−2/3) = 6(4/9 − 1)(5(4/9) − 1)) < 0

and

h′′(0) = 6(−1)(1) > 0

so x = −1, −1/\sqrt{5} are both inflection points. The same calculation shows that x = 1, 1/\sqrt{5} are also inflection points.

The function is concave upward on (−∞, −1), (−1/\sqrt{5}, 1/\sqrt{5}) and (1, ∞) and is concave downward on (−1, −1/\sqrt{5}) and (1/\sqrt{5}, 1). The inflection points are at x = ±1, ±1/\sqrt{5}.

Problem 38(d). Using the data from (a)–(c), we have the following graph:

Problem 42(a). We have

f′(x) = \frac{4x^{3}}{x^{4} + 27}

so since x^{4} + 27 > 0 always, we have f′ > 0 and f is increasing for x > 0 and f′ < 0 and f is decreasing for x < 0.

Problem 42(b). We have f′(x) = 0 only for x = 0. We see f′(−1) = −4/28 < 0 and f′(1) = 4/28 > 0 so x = 0 is a local minimum.

Problem 42(c). We compute

f′′(x) = \frac{(x^{4} + 27)(12x^{2}) − (4x^{3})(4x^{3})}{(x^{4} + 27)^{2}}

= \frac{−4x^{6} + 324x^{2}}{(x^{4} + 27)^{2}}

This has

f′′(x) = −4x^{6} + 324x^{2} = −4x^{2}(x^{4} − 81) = −4x^{2}(x^{2} − 9)(x^{2} + 9)
= −4x^{2}(x − 3)(x + 3)(x^{2} + 9)

1A: HOMEWORK #3 9

so we must consider the possible values x = 0, 3, −3. We see that

f′′(−4) = −4(16)(−7)(−1)(16 + 9) < 0
f′′(−1) = −4(1)(−4)(2)(16 + 9) > 0
f′′(1) = −4(1)(−2)(4)(1 + 9) > 0
f′′(4) = −4(16)(1)(7)(16 + 9) < 0

So x = −3, 3 are inflection points and x = 0 is not. The function f is concave downward on (−∞, −3) and (3, ∞) and concave upward on the interval (−3, 3). Problem 42(d). Using (a)–(d), we have: