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?