How to prove $x^3$ is strictly increasing I am trying to use $f(x)=x^3$ as a counterexample to the following statement. 
If $f(x)$ is strictly increasing over $[a,b]$ then for any $x\in (a,b), f'(x)>0$. 
But how can I show that $f(x)=x^3$ is strictly increasing?

Here is an intuitive proof that could be made rigorous using Lebesgue measure. 
Start with $x>0$. Place a cube with edge length $x$ so that it has three faces on the coordinate planes in three-dimensional space and two corners at $(0,0,0)$ and at $(x,x,x)$. Then $f(x)=x^3$ is the volume of the cube. 
Strict monotonicity of $f$ is a consequence of the volume being strictly bigger for a strictly bigger cube. To incorporate negative values of $x$, argue with the fact that $f$ is odd.
