Can a continuous function on [0,1] be constructed which is differentiable exactly at two points in [0,1]? We see that we can find a function nowhere differentiable or finitely not differentiable.
But I want to understand, can a continuous function on $[0,1]$ be constructed which is differentiable exactly at two points in $[0,1]$?
How can I construct such a function which is finitely differentiable?

Take a function that is continuous everywhere and differentiable nowhere, call it $g$.
Take $2$ points, $a$ and $b$.  Then $f(x) = (x-a)(x-b)g(x)$ is continuous everywhere, differentiable only at $a$ and $b$.  

You can use continuity of $g$ and the definition of the derivative to verify very quickly that $f$ is differentiable at $a$ and $b$.   To see that $f$ is differentiable nowhere else, you could note that if $f$ were differentiable at $c\not\in \{a,b\}$, then so would be $g(x) =\dfrac{f(x)}{(x-a)(x-b)}$ by the quotient rule.
