A function that is differentiable at exactly two points

Question

As homework we were given 2 questions:

1. Find a function that is continuous at exactly one point and not differentiable there.

2. Find a function that is differentiable at exactly two points.

The answer to the first one was somewhat simple - $ f(x) = x*D(x) $, wherein $ D(x) $ stands for Dirichlet function.

I've been thinking about the second question for a while, and someone suggested me the following: $ g(x) = x^2*D(x) $

$ f(x) = (x-1)^2 * g(x) $

Claiming that $f(x)$ is only differentiable in $ x = 0, x = 1 $. Why is that?

Answer 1

It is simplier to say that$$g(x)=\begin{cases}x^2(x-1)^2&\text{ if }x\in\mathbb Q\\0&\text{ otherwise.}\end{cases}$$And, yes, this example works. Outside $\{0,1\}$, $g$ is discontinuous, and therefore non-differentiable. And it follows from the definition of $g'$ that $g'(0)=g'(1)=0$.

Answer 2

Clearly, $f$ is not differentiable away from $0$ and $1$: the $x^2$ and $(x-1)^2$ factors don't fix the horribleness of the Dirichlet function there: a rational $y$ near some $x$ is sent near to $x^2(x-1)^2$, while an irrational $y$ near $x$ is sent to $0$, and those differ, so $f$ isn't even continuous away from $0$ and $1$.

So, why is it differentiable at $0$ and $1$: well, let's just check the definition.

$$\lim_{x\to 0}\frac{f(x)-f(0)}{x} = \lim_{x\to 0}\frac{x^2(x-1)^2D(x)}{x} = \lim_{x\to 0}x(x-1)^2D(x),$$ and $|x(x-1)^2D(x)| \leq |x|(x-1)^2 \to 0$ as $x \to 0$, so the limit exists, and is zero, at $0$. Similarly, $$\lim_{x\to 1}\frac{f(x)-f(1)}{x-1} = \lim_{x\to 1}x^2(x-1)D(x),$$ and $|x^2(x-1)D(x)| \leq x^2|x-1| \to 0$ as $x \to 1$.