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

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

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!?
• i think your proof will work, but maybe you could mention that this $\delta$ exists because of the mean value theorem. – Nathanael Skrepek Nov 26 '19 at 9:39
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.