Does $f'(x_0)>0$ imply $f$ is increasing in $(x_0-\delta, x_0+\delta)$?

Question

Let $f$ be a function differentiable in $x_0 \in \mathbb{R}$ (and its neighborhood) and $f'(x_0)>0$. Does this imply that there exists a $\delta>0$ for which $f$ is increasing in $(x_0-\delta, x_0+\delta)$?

My intuition is that since the condition is $f'(x_0)>0$ and not $f'(x_0)\ge0$, there is a neighborhood in which the derivative stays positive, and the answer to the question would be “yes”. Is this correct? If so, how could I write it down more formally?

Answer 1

Consider $f(x)=x^2\sin(\frac{1}{x})+a x$ when $x \neq 0$ and $f(0)=0$. Then, it is easy to prove that $f$ is differentiable at $0$ and $f'(0)=a$.

But when $x \neq 0$ you have $$f'(x)=2x \sin(\frac{1}{x})+\cos(\frac{1}{x})+a$$

It is easy to pick $a$ such that $f'(0)=a>0$ but $f'(x)$ is not positive in any interval $(- \delta, \delta)$.

Your argument would be true if you know that $f'$ is continuous at $x_0$.