Q:

$f$ 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

A:

Consider $f(x)=x^3$ on $[-1,1]$. It is strictly increasing, but has zero derivative at $0$.

