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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?

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    $\begingroup$ i think your proof will work, but maybe you could mention that this $\delta$ exists because of the mean value theorem. $\endgroup$
    – Nathanael Skrepek
    Nov 26, 2019 at 9:39

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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.

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