# Recitation 8

Problem 1 Consider the function $f: [0, \infty)\to\mathbb{R}$ defined by $f(x) = x^a\sin(1/x)$ for $x >0$ and $f(0) = 0$. (a) For which values of the parameter $a$ the function $f$ is differentiable at $x=0$? (b) For which values of $a$ is the derivative $f'(x)$ a continuous function?

Proof sketch (a) The function is differentiable at 0 if and only if $\lim_{x\to 0^+}\frac{f(x) - f(0)}{x - 0} = \lim_{x\to 0^+}x^{a-1}\sin(1/x)$ exists. Note that $\lim_{x\to 0^+}x^{a-1}\sin(1/x) = 0$ when $a > 1$. When $a = 1$, we already know that $\lim_{x\to 0^+}\sin(1/x)$ does not exist. By problem 3 in homework 4, when $a < 1$, because $\lim_{x\to 0^+}x^{a-1}=\infty$, $\lim_{x\to 0^+}x^{a-1}\sin(1/x)$ does not exist. Therefore $f$ is differentiable at 0 if and only if $a>1$, moreover, the derivative at 0 is 0.

(b) For $x > 0$, $f'(x) = ax^{a-1}\sin(1/x) + x^a\cos(1/x)(-1/x^2) = ax^{a-1}\sin(1/x) - x^{a-2}\cos(1/x)$ and $$\lim_{x\to 0^+}f'(x) = a\lim_{x\to 0^+}x^{a-1}\sin(1/x) - \lim_{x\to 0^+}x^{a-2}\cos(1/x) = - \lim_{x\to 0^+}x^{a-2}\cos(1/x).$$ An argument similar to the one for part (a) shows that $f'(x)$ is continuous if and only if $a > 2$.

Problem 2 Give an example of a differentiable function $f\colon [0,1] \to \mathbb{R}$ such that $x=0$ is not a local minimum or a local maximum.

Proof By Problem 1, $f(x) = x^2\sin(1/x)$ is differentiable on $[0,1]$. For every $\epsilon > 0$, there exists $n\in\mathbb{N}$ (sufficiently large) such that both $x_1 := 1/[(2n + 1/2)\pi]$ and $x_2 := 1/[(2n - 1/2)\pi]$ are less than $\epsilon$. Notice that $f(x_1) = x_1^2 > 0$, $f(x_2) = -x_2^2 < 0$ and $f(0) = 0$. Therefore 0 is not a local maximum or a local minimum point.

Problem 3 Find the maximum and minimum values of the function $f(x) = x^5 - 5x^4 + 5x^3 + 1$ on the interval $[-1,2]$.

Solution Note that $f'(x) = 5x^4 - 20x^3 + 15x^2 = 5x^2(x-1)(x-3)$ has roots $0, 1, 3$ (we may ignore 3). We compute the value of the function at the end points of the interval and those roots: $f(-1) = -10, f(0) = 1, f(1) = 2, f(2) = -7$. Therefore the maximum value is 2 and the minimum value is -10.

Problem 4 Use Lagrange’s mean value theorem to prove the following inequality for $0 < a \le b$, $$\frac{b-a}{b} \le \ln \frac{b}{a} \le \frac{b-a}{a}.$$

Proof If $a = b$, it is clear. Hereafter we assume $a < b$. Rewrite the inequality as $$\frac{1}{b} \le \frac{\ln b - \ln a}{b - a} \le \frac{1}{a}.$$ This inspires us to consider the function $f(x) = \ln x$. By the mean value theorem, there is $a < c < b$ such that $f'(c) = \frac{\ln b - \ln a}{b - a}$. Notice that $f'(c) = 1/c$ and $1/b < 1/c < 1/a$.

Problem 5 Prove that the equation $x^n + px + q = 0$ can have at most two distinct (real) solutions if $n$ is even, and it can have at most three distinct (real) solutions if $n$ is odd.

Proof sketch By Rolle’s theorem, if there are $k$ distinct solutions to $f(x) = 0$, then there are at least $k-1$ distinct solutions to $f'(x) = 0$ (why?). Now notice that the derivative $f'(x) = x^{n-1}+p$ has 1 solution when $n$ is even, and it has 2 solutions when $n$ is odd.

Problem 6 Suppose $f\colon \mathbb{R} \to \mathbb{R}$ satisfies $f(n) = 2$ for every $n\in\mathbb{N}$ and suppose that $f$ is differentiable and $\lim_{x\to\infty} f'(x) = 0$. Show that $\lim_{x\to\infty} f(x) = 2$.

Proof Take a sequence $\{a_n\}_{n=1}^\infty$ such that $\lim_{n\to\infty} a_n = \infty$. By Heine’s theorem, it suffices to show that $\lim_{n\to\infty} f(a_n) = 2$. For every $a_n$, by the mean value theorem, there is $[a_n]\le b_n \le a_n$ such that $f(a_n) - f([a_n]) = (a_n - [a_n])f'(b_n)$. Because $[a_n]$ is a natural number, $f([a_n]) = 2$. Thus $|f(a_n)-2| = |a_n - [a_n]||f'(b_n)| \le |f'(b_n)|$. Since $b_n \ge [a_n] > a_n - 1$, by Heine’s theorem, $\lim_{n\to\infty}f'(b_n) = 0$. By the sandwich theorem, $\lim_{n\to\infty}|f(a_n)-2| = 0$ and so $\lim_{n\to\infty}f(a_n) = 2$.

Problem 7 Prove that for every $x\in\mathbb{R}$ and $n\in\mathbb{N}$ we have $$1 + x +\frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots + \frac{1}{(2n)!}x^{2n} > 0.$$

Proof Denote the left hand side of the inequality by $f(x)$. Notice that $\lim_{x\to -\infty}f(x) = \lim_{x\to \infty}f(x) = \infty$. The function has minimum points. Let $x_0$ be a minimum point. We know that $f'(x_0) = 0$. However, $$f'(x) = 1 + x +\frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots + \frac{1}{(2n-1)!}x^{2n-1} = f(x) - \frac{1}{(2n)!}x^{2n}.$$ This implies that $f(x_0) = f'(x_0) + \frac{1}{(2n)!}x_0^{2n} > 0$. In other words, $f(x) \ge f(x_0) > 0$.