$f:[a,0]\rightarrow \mathbb{R}; a>0$ continuous and monotone increasing implies differentiability?

Question

I had to prove an inequality of certain integrals using this continuous and monotone increasing function $f:[a,0]\rightarrow \mathbb{R}$, $a>0$, with $f(0)=0$ and the $g$ is the inverse functions of the $f$. I managed to solve and to prove the inequality but at a certain step I assumed that $f$ differentiable. Can I assume differentiabilty, and if so, how can I prove it?

Answer 1

Consider $f:[0,a]\rightarrow\mathbb{R}$ given by $$f(x)=2x-\frac{1}{2}a+|x-\frac{1}{2}a|=\begin{cases}x \text{ if }x\leq\frac{1}{2}a\\ 3x-a \text{ if }x\geq\frac{1}{2}a\end{cases}.$$ Note that $f$ is continuous, and monotone increasing but not differentiable.