# $f$ strictly increasing does not imply $f'>0$

We know that a function $f: [a,b] \to \mathbb{R}$ continuous on $[a,b]$ and differentiable on $(a,b)$, and if $f'>0 \mbox{ on} (a,b)$ , f is strictly increasing on $[a,b]$. Is there any counterexample that shows the converse fails?

I have been trying to come up with simple examples but they all involve functions that are discontinuous or has derivative $f'=0$ which does not agree with the hypothesis hmmm

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–  Cameron Buie Nov 6 '12 at 6:44

Consider $f(x)=x^3$ on $[-1,1]$. It is strictly increasing, but has zero derivative at $0$.
@jsk Note that you can come up with counterexamples like this by taking a nonnegative continuous function with a $0$ (in this case $x^2/3$) and integrating. –  Alex Becker Jul 10 '12 at 23:49
@AlexBecker You surely meant $3x^2$. –  Pedro Tamaroff Jul 11 '12 at 1:29