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?

Suppose $a<b$. We want to prove that $a^3 < b^3$.
Well, $a<b, \ \therefore \exists \delta>0$ such that $b = a+\delta$. So:
$b^3 = (a+ \delta)^3 = a^3 + 3a^2\delta + 3\delta^2a + \delta^3 = a^3 + (3a\delta(a+\delta)+\delta^3)$.
Now we consider five different cases:

*

*$\ a<-\delta$

*$\ a = -\delta$

*$\ -\delta<a<0$

*$\ a=0$

*$\ a > 0$.

In cases $1, 2, 4$ and $5,$ it is fairly easy to see that  $3a\delta(a+\delta)+\delta^3 > 0,\ $ so all that remains is case $3$:
$-\delta<a<0 \implies \exists k$ with $0<k<1,\ $ such that $a = -k\delta.\ $ Then, $\ a\delta(a+\delta) = -k\delta^2(1-k)\delta = -k(1-k)\delta^3$, and using the fact that $k(1-k) \leq \frac{1}{4},\ $ we see that $-k(1-k)\delta^3 \geq -\frac{1}{4}\delta^3,\ $ and so $\ 3a\delta(a+\delta)+\delta^3 \geq (-\frac{3}{4}\delta^3 + \delta^3) = \frac{1}{4}\delta^3 > 0$.
