Mathematics 242–Principles of Analysis Problem Set 5 – Solutions November 11, 2000

Section 4.1/

1.

a) For f(x) = \frac{1}{x}, for all c ≠ 0,

f′(c) = lim_{x→c} \frac{\frac{1}{x} − \frac{1}{c}}{x − c}

= lim_{x→c} \frac{c − x}{cx(x − c)}

= lim_{x→c} \frac{−1}{cx}

= \frac{−1}{c^{2}}

The derivative function is f′(x) = \frac{−1}{x^{2}}.

b) For g(x) = \frac{1}{x^{2}} , for all c ≠ 0,

g′(c) = lim_{x→c} \frac{\frac{1}{x^{2}} − \frac{1}{c^{2}}}{x − c}

= lim_{x→c} \frac{c^{2} − x^{2}}{c^{2}x^{2}(x − c)}

= lim_{x→c} \frac{−(c + x)}{c^{2}x^{2}}

= \frac{−2c}{c^{4}}

= \frac{2}{c^{3}}

The derivative function is g′(x) = \frac{−2}{x^{3}}.

c) For h(x) = ax^{2} + bx + c,

h′(d) = lim_{x→d} \frac{(ax^{2} + bx + c) − (ad^{2} + bd + c)}{x − d}

= lim_{x→d} \frac{a(x^{2} − d^{2}) + b(x − d)}{(x − d)}

= lim_{x→d} a(x + d) + b

= 2ad + b

The derivative function is h′(x) = 2ax + b. d) For i(x) = \frac{x}{x+1},

i′(c) = lim_{x→c} \frac{\frac{x}{x+1} − \frac{c}{c+1}}{x − c}

= lim_{x→c} \frac{x − c}{(c + 1)(x + 1)(x − c)}

= lim_{x→c} \frac{1}{(c + 1)(x + 1)}

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

The derivative function is i′(x) = \frac{1}{(x+1)^{2}} .

e) For j(x) = \sqrt{x^{2} + 1},

j′(c) = lim_{x→c} \frac{\sqrt{x^{2} + 1} − \sqrt{c^{2} + 1}}{x − c}

= lim_{x→c} \frac{(\sqrt{x^{2} + 1} − \sqrt{c^{2} + 1})(\sqrt{x^{2} + 1} + \sqrt{c^{2} + 1})}{(x − c)(\sqrt{x^{2} + 1} + \sqrt{c^{2} + 1})}

= lim_{x→c} \frac{x^{2} − c^{2}}{(x − c)(\sqrt{x^{2} + 1} + \sqrt{c^{2} + 1})}

= lim_{x→c} \frac{x + c}{(\sqrt{x^{2} + 1} + \sqrt{c^{2} + 1})}

= \frac{2c}{2\sqrt{c^{2} + 1}}

= \frac{c}{\sqrt{c^{2} + 1}}

The derivative function is j′(x) = \frac{x}{\sqrt{x^{2}+1}}.

15. For g(x) = (f(x))^{2},

g′(c) = lim_{x→c} \frac{(f(x))^{2} − (f(c))^{2}}{x − c}

= lim_{x→c} \frac{(f(x) − f(c))(f(x) + f(c))}{(x − c)}

= lim_{x→c} \frac{f(x) − f(c)}{(x − c)} · lim_{x→c}(f(x) + f(c))

= f′(c) · 2f(c)

To get to the last line, we used the fact that f differentiable at c implies f is continuous at c. The derivative function is g′(x) = 2f(x)f′(x).

16. For g(x) = \sqrt{f(x)},

g′(c) = lim_{x→c} \frac{\sqrt{f(x)} − \sqrt{f(c)}}{x − c}

= lim_{x→c} \frac{(\sqrt{f(x)} − \sqrt{f(c)})(\sqrt{f(x)} + \sqrt{f(c)})}{(x − c)(\sqrt{f(x)} + \sqrt{f(c)})}

= lim_{x→c} \frac{f(x) − f(c)}{(x − c)(\sqrt{f(x)} + \sqrt{f(c)})}

= lim_{x→c} \frac{f(x) − f(c)}{(x − c)} · lim_{x→c} \frac{1}{(\sqrt{f(x)} + \sqrt{f(c)})}

= f′(c) · \frac{1}{2\sqrt{f(c)}}

To get to the last line, we used the fact that f differentiable at c implies f is continuous at c. The derivative function is g′(x) = \frac{f′(x)}{2\sqrt{f(x)}}.

Section 4.3/

1. Our hypothesis is that f(x) has a local minimum at c and f′(c) exists. So f(x) ≥ f(c) for all x in some neighborhood of c. Taking x < c, we have f(x) − f(c) ≥ 0, so \frac{f(x)−f(c)}{x−c} ≤ 0. Therefore

lim_{x→c^{−}} \frac{f(x) − f(c)}{x − c} ≤ 0.

On the other hand, taking x > c, we have f(x) − f(c) ≥ 0, so \frac{f(x)−f(c)}{x−c} ≥ 0. Therefore

lim_{x→c^{+}} \frac{f(x) − f(c)}{x − c} ≥ 0.

Since we assume lim_{x→c} \frac{f(x)−f(c)}{x−c} exists, the two one-sided limits must exist and be equal. Therefore lim_{x→c} \frac{f(x)−f(c)}{x−c} = 0.

5.

a) f(x) = x^{33} − 33x. We have f′(x) = 33x^{32} − 33 = which is zero only when x = ±1. x = −1 is a local maximum and x = 1 is a local minimum since f′(x) > 0 for x < −1, f′(x) < 0 for −1 < x < 1 and f′(x) > 0 for x > 1.

b) f′(x) = 34(x^{33} − 1) is zero only at x = 1, which is a local minimum.

c) f′(x) = 35x^{34} − 4480x^{27} = 35x^{27}(x^{7} − 128) which is zero at x = 0 and x = 2. x = 0 is a local maximum, x = 2 is a local minimum.

d) f′(x) = 36x^{35} − 576x^{31} = 36x^{31}(x − 2)(x + 2)(x^{2} + 4). x = −2 is a local minimum, x = 0 is a local maximum, x = 2 is a local minimum.

6. Here the idea is to compute f′(x), find critical points, then compare the values of f at the critical points and the endpoints of the interval.

a) f(x) = x^{4} − 4x^{3} + 1 on [−1, 4]. f′(x) = 4x^{3} − 12x^{2} = 4x^{2}(x − 3). The derivative is zero at x = 0, x = 3. f(−1) = −2, f(0) = 1, f(3) = −26, f(4) = 1. So the maximum on the interval is M = 1, attained at x = 0 (local max), and also at x = 4 (endpoint). The minimum is m = −26, attained at x = 3.

12.

a) The MVT does apply since f(x) = \sqrt{x} is continuous on [0, b] and differentiable on (0, b) (not at x = 0, but that is not necessary). We have f′(c) = \frac{f(b)−f(0)}{b−0} = 1, when c = \frac{b}{4}.

b) The MVT does not apply since g(x) = [x] (greatest integer function) is not continuous on [1, 2] (it has a jump at the right endpoint).

c) The MVT does not apply since h(x) = 2x + |x − 2| is not differentiable at x = 2. This is true because lim_{x→2^{−}} \frac{h(x)−h(2)}{x−2} = 2 − 1 = 1 but lim_{x→2^{+}} \frac{h(x)−h(2)}{x−2} = 2 + 1 = 3.

17. If f(x) has another zero x ∈ [a, b], then apply Rolle’s Theorem to f on the interval [a, x]. We get that f′(c) = 0 for some c between a and x. But this contradicts the hypothesis that f′(x) ≠ 0 for all x ∈ (a, b). Therefore f can have only one zero.

20.

a) Since cos(π/x) is undefined at 0, we must use the definition to compute f′(0):

f′(0) = lim_{x→0} \frac{x + x^{2} cos(π/x)}{x}

= lim_{x→0} 1 + x cos(π/x)

= 1,

since cos(π/x) is bounded as x → 0. b)

f (\frac{1}{2n}) = \frac{1}{2n} + \frac{1}{(2n)^{2}} cos(2nπ) = \frac{1}{2n} + \frac{1}{(2n)^{2}}
f (\frac{1}{2n − 1}) = \frac{1}{2n − 1} + \frac{1}{(2n − 1)^{2}} cos((2n − 1)π) = \frac{1}{2n − 1} − \frac{1}{(2n − 1)^{2}}
f (\frac{1}{2n}) − f (\frac{1}{2n − 1}) = \frac{1}{2n} + \frac{1}{(2n)^{2}} − \frac{1}{2n − 1} + \frac{1}{(2n − 1)^{2}}

= \frac{2n(2n − 1)^{2} + (2n − 1)^{2} − (2n)^{2}(2n − 1) + (2n)^{2}}{(2n)^{2}(2n − 1)^{2}}

= \frac{8n^{3} − 8n^{2} + 2n + 4n^{2} − 4n + 1 − 8n^{3} + 4n^{2} + 4n^{2}}{(2n)^{2}(2n − 1)^{2}}

= \frac{4n^{2} − 2n + 1}{(2n)^{2}(2n − 1)^{2}}

= \frac{3n^{2} + (n − 1)^{2}}{(2n)^{2}(2n − 1)^{2}}

> 0

This showsf (\frac{1}{2n}) > f (\frac{1}{2n−1}). This problem shows the surprising result that it is not sufficient to have f′(c) > 0 to

conclude that f is increasing on an interval containing c. In this case, there is no interval containing 0 on which f(x) is increasing, even though f′(0) = 1. (!)

Section 5.1

9.

a) Isolate the discontinuity by choosing a suitable partition. For example, let δ < 1, and let P be the partition P = {0, 1 − δ, 1 + δ, 2}. Then on the first subinterval we have minimum value m_{1} = 1 and maximum value M_{1} = 1. The second interval has m_{2} = 1 and M_{2} = 2 = f(1). The third interval has m_{3} = 1 = M_{3}. Therefore the upper sum is

U_{P} (f) = M_{1}(1 − δ) + M_{2}(2δ) + M_{3}(1 − δ) = 2 + 2δ

and the lower sum is

L_{P} (f) = m_{1}(1 − δ) + m_{2}(2δ) + m_{3}(1 − δ) = 2

The difference U_{P} (f) − L_{P} (f) = 2δ can be made less than ϵ by taking δ < ϵ/2. By Theorem 5.1.5 in the text, f is integrable. The value of the integral is the greatest lower bound of the upper sums, which equals the least upper bound of the lower sums, namely ∫_{0}^{2} f(x) dx = 2. b,c) The idea is similar – take partitions isolating the discontinuities. For example, in c,

[2x] = \{ \begin{matrix}0 & if 0 ≤ x < 1/2 \\ 1 & if 1/2 ≤ x < 1 \\ 2 & if 1 ≤ x < 3/2 \\ 3 & if 3/2 ≤ x < 2 \\ 4 & if x = 2 \end{matrix}

Let δ < 1/4 and

P = {0, 1/2 − δ, 1/2 + δ, 1 − δ, 1 + δ, 3/2 − δ, 3/2 + δ, 2 − δ, 2}

Then

U_{P} (f) = 0 · (1/2 − δ) + 1 · (2δ) + 1 · (1/2 − 2δ)2 · (2δ)
+ 2 · (1/2 − 2δ) + 3 · (2δ) + 3 · (1/2 − 2δ) + 4δ
= 3 + 4δ

The lower sum L_{P} (f) = 3. Hence, letting δ → 0, the value of the integral is 3.

10. Let P_{n} be the regular partition of [0, 2] into n equal subintervals – x_{i} = \frac{2i}{n}, and Δx = \frac{2}{n}. Then since f(x) = 2x + 1 is increasing on the whole interval we have

L_{P} (f) = ∑_{i=1}^{n} (2\frac{2(i − 1)}{n} + 1) \frac{2}{n}

and

U_{P} (f) = ∑_{i=1}^{n} (2\frac{2i}{n} + 1) \frac{2}{n}

When we subtract, all but the last term in U_{P} (f) cancel with all but the first term in L_{P} (f) and the difference is

U_{P} (f) − L_{P} (f) = \frac{2}{n}(4 + 1 − 1) = \frac{8}{n}.

This can be made < ϵ by chosing n > \frac{8}{ϵ}. Hence Theorem 5.1.5 in the text shows that f(x) is integrable over [0, 2].

To find the value of the integral, we can take the limit of the upper sums as n → ∞. (Since the difference between the upper and lower sums for the regular partitions goes to 0 an n → ∞, this shows that lim_{n→∞} U_{P_{n}}(f) equals the greatest lower bound of the upper sums, which equals the integral.) The summation formula 1+2+⋯+n = ∑_{i=1}^{n} i = \frac{n(n+1)}{2} will be used at a key step: We have

U_{P_{n}}(f) = ∑_{i=1}^{n} (2\frac{2i}{n} + 1) \frac{2}{n}

= \frac{8}{n^{2}} ∑_{i=1}^{n} i + \frac{2}{n} ∑_{i=1}^{n} 1

= \frac{8}{n^{2}} \frac{n(n + 1)}{2} + 2

= 6 + \frac{4}{n}

As n → ∞, this goes to ∫_{0}^{2} 2x + 1 dx = 6.

11. Similar to 10, using the summation formula for ∑_{i=1}^{n} i^{3} given in the problem to compute the integral.

∫_{0}^{1} x^{3} dx = lim_{n→∞} ∑_{i=1}^{n} (\frac{i}{n})^{3} \frac{1}{n}

= lim_{n→∞} \frac{1}{n^{4}} ∑_{i=1}^{n} i^{3}

= lim_{n→∞} \frac{1}{n^{4}} \frac{n^{2}(n + 1)^{2}}{4}

= lim_{n→∞} \frac{1}{4} + \frac{1}{2n} + \frac{1}{4n^{2}}

= \frac{1}{4}

12. The idea for all three parts here is similar to what we did in problem 9. For instance, in part a, if you let δ < 1/2, and

P = {0, 1 − δ, 1 + δ, 2 − δ, 2 + δ, 3 − δ, 3 + δ, 4 − δ, 4 + δ, 5 − δ}.

Then U_{P} (f) = 9δ, and L_{P} (f) = 0. By picking δ < ϵ/9, the difference can be made less than any ϵ > 0. Hence f is integrable by Theorem 5.1.5 (and the value of the integral is 0, since all lower sums for this function equal 0).

13. Our hypothesis is that ∫_{a}^{b} f(x) dx exists and equals some number I > 0. Since the integral is the least upper bound of the lower sums for the function, for every ϵ > 0, there exists some partition P such that L_{P} (f) > I − ϵ. If we take 0 < ϵ < I, this shows that L_{P} (f) > 0 for some P. But then some term m_{i}Δx_{i} in the lower sum must be strictly positive. As a result, m_{i} > 0 for at least one of subintervals from P. Since m_{i} = glb{f(x) : x ∈ [x_{i−1}, x_{i}]}, this shows that f(x) > m_{i} > 0 for all x ∈ J = [x_{i−1}, x_{i}], the interval from the partition.