I am not sure if this is true, but intuitively it seems that if a function is strictly increasing and it is also continuous...it is differentiable.
It may be because there are no bumps like in the absolute value.
I am not sure if this is true, but intuitively it seems that if a function is strictly increasing and it is also continuous...it is differentiable.
It may be because there are no bumps like in the absolute value.
Not necessarily. Counterexample: $$ f(x)=\begin{cases} x & \text{if }x<0,\\ 2x & \text{if }x\ge 0.\end{cases} $$ Is continuous, strictly increasing but not differentiable at $x=0$.
No, as Julián Aguirre's answer shows. However, what is true is:
Lebesgue's Theorem: Every monotone function $f$ (increasing or decreasing) defined on an open interval $(a,b)$ is differentiable almost everywhere on $(a,b).$
Take a look at Billingsley' Probability and Measure, example 31.1