A point where the derivative is $\infty$

Question

Let $f:I \to \mathbb{R}$, where $I$ is an interval and $a \in I$ such that $f'(a)= \infty$. Prove that there is a neighborhood $(a-\epsilon, a+\epsilon)$ such that $f$ is convex on $(a-\epsilon, a)$ and $f$ is concave on $(a, a+\epsilon)$.

I tried to use the definition of the derivative, but I only got that there is a left-neighborhood of $a$ where $f(x)< f(a)$ and a right-neighborhood of $a$ where $f(x)>f(a)$.

I suppose that the statement is false, but I couldn't construct an elementary counter example. Is there any?

Answer 1

Hint

The result is wrong. Look at

$$f(x)=\begin{cases} -\sqrt{-x} + x \sin \left(\frac{1}{x}\right) & x <0\\ 0 & x=0\\ \sqrt{x} + x \sin \left(\frac{1}{x}\right)& x>0 \end{cases}$$

And prove that around zero the derivative is not monotonic on each side of zero.

Answer 2

This is false. If $f$ is continuously differentiable on $\Bbb R\setminus\{a\}$ and $\lim_{x\to a}f'(x)=\infty$ then it follows from the Mean Value Theorem that $f'(a)=\infty$, but $f'$ certainly need not be monotone on $(a-\epsilon,a)$.