A function not differentiable exactly two points of $[0,1]$. construction of such a function is possible?

Question

Can a continuous function on $[0,1]$ be constructed which is not differentiable exactly at two points on $[0,1]$ ?

Answer 1

Consider: $$f(x)=\left|x-\frac{1}{2}\right|+\left|x-\frac{1}{3}\right|.$$

Answer 2

Sure. Presuming you allow one-sided limits at interval endpoints in your limit of differentiability (otherwise any otherwise differentiable function is automatically an example because it fails to be differentiable at $0, 1$), you can take as an example the function $f$ defined by $0$ on $[0, \frac{1}{3}]$, $3x - 1$ on $[\frac{1}{3}, \frac{2}{3}]$, and $1$ on $[\frac{2}{3}, 1]$. It is nondifferentiable precisely at $\frac{1}{3}, \frac{2}{3}$.

Answer 3

Let me give another, potentially more common example: the upper-half circle of radius $1/2$ with center at $1/2$, i.e. $$f(x)=\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}$$