Due: December 1st, 2017

Math 142A Assignment 7

Answers to these problems are now available here. The $\LaTeX$ file used to produce them is here.

1. For each of the following functions, compute the derivative where it exists, and state the points at which it does not exist.
1. $\frac{x^2-1}{x^2+2x+1}$
2. $\frac{x^2-2x+1}{1-x^2}$
3. $\sin\left(\cos\left(2x^2\right)\right)$
4. $\frac{x-4}{\sqrt{|x|+1}}$
2. Let $a > 0$, and suppose that $f : \mathbb{R}\to\mathbb{R}$ is defined by $f(x) = a^x$. Show that if $f$ is differentiable at $0$, then $f$ is differentiable everywhere, and $f'(x) = a^xf'(0)$.
3. Let the function $f$ be defined as follows: \begin{align}f : \mathbb{R} &\longrightarrow \mathbb{R} \\ x &\longmapsto \begin{cases}x^2\sin\left(\frac1{x}\right)&\text{ if } x \neq 0\\0 & \text{ if } x = 0\end{cases}.\end{align} Show that $f$ is differentiable everywhere, but its derivative is discontinuous at $0$. (Hint: a lot of this problem is very similar to a problem on assignment 5.)

4. Suppose that $a < b$, and $g : [a, b] \to \mathbb{R}$ is continuous on its domain, and invertible. Note that, as a consequence, $g([a,b])$ is a closed interval.
1. Suppose that $g$ is differentiable at $x_0 \in (a, b)$, and that $g'(x_0) \neq 0$. Show that $g^{-1}$ is differentiable at $g(x_0)$, and find its derivative there.
2. Suppose that $g$ is differentiable at $x_0 \in (a, b)$, and that $g'(x_0) = 0$. Show that $g^{-1}$ is not differentiable at $g(x_0)$.
1. Suppose that $a < b$, and that $f : [a, b] \to \mathbb{R}$ is strictly increasing and differentiable at $x_0 \in (a, b)$. Show that $f'(x_0) \geq 0$.
2. Give an example of a strictly increasing differentiable function $f$ such that $f'(x) = 0$ for some $x$ in its domain.
5. Bonus:

Recall the definition of uniform convergence and of pointwise convergence from homework 5:
The supremum norm of a function $f : X \to \mathbb{R}$ is defined to be the quantity $$\|f\|_\infty = \sup\{|f(x)| : x \in X\}$$ if the supremum exists, and infinity otherwise. A sequence of functions $(f_n)$, with $f_n : X \to \mathbb{R}$, is said to converge uniformly to a function $f : X \to \mathbb{R}$ if for every $\epsilon > 0$ there is $N \in \mathbb{N}$ such that for all $n > N$, $$\|f_n - f\|_\infty < \epsilon.$$ The sequence is said to converge to $f$ pointwise if for every $x \in X$, the sequence of real numbers $(f_n(x))$ converges to $f(x)$.
One of the results from that assignment was that if a sequence of continuous functions converges uniformly, then its limit is a continuous function. In this problem we will demonstrate a function which is continuous everywhere, but differentiable nowhere.

For each $n \in \mathbb{N}$, let $f_n$ be the function defined as follows: \begin{align}f_n : \mathbb{R} &\longrightarrow \mathbb{R} \\ x &\longmapsto \sum_{j=0}^n \left(\frac{3}{4}\right)^{\color{red} j}\sin\left(9^{\color{red} j}\pi x\right).\end{align}

1. Show that the sequence $(f_n)$ converges pointwise to some function $f$.
2. Show that the convergence is uniform. (It follows, then, that $f$ must be continuous.)
3. Show that for any $x_0 \in \mathbb{R}$, there are points $y, z \in \mathbb{R}$ so that $|x_0 - y|, |x_0-z| \leq \frac1{9^n}$, but $|f(y) - f(z)| \geq \left(\frac34\right)^n$.
4. Conclude that $f$ is not differentiable at $x_0$.