Continuously differentiable function $f'(0)>0$. Can we claim that $f$ is locally strictly increasing around $0$?

Question

Continuously differentiable function $f'(0)>0$. Can we claim that $f(x)$ is locally strictly increasing in an interval $(-\epsilon,\epsilon)$ around $x=0$?

-> Proof by contradiction,

Suppose that $f$ is not strictly increasing around $0$. Then, for any $\epsilon>0$, there exists $0<\delta<\epsilon$ such that $f'(\delta)\not >0$. Then $f'$ is not continuous. $f$ is not continuously differentiable. Contradiction!?

Is this plausible?

Answer 1

If $f$ is continuously differentiable, then that means that $f'(x)$ is continuous. Since it is strictly positive at $0$, there is, by the definition of continuity, a small region around $0$ where $f'$ is strictly positive. Then by the mean value theorem it has to be increasing on that region.