Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f\left(x\right)=\left\{\begin{array}{cc}{e}^x,& x\le 0\\ 1-x,& 0<x\le 1\\ x-1,& x>1\end{array}\right.$

$RHD: f^{\text{'}}\left(0\right)=\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0}{\lim }-\left[\frac{h-1}{h}\right]=-1$

$LHD: f^{\text{'}}\left(0\right)=\underset{h\to 0}{\lim }\frac{f\left(0-h\right)-f\left(0\right)}{-h}=\underset{h\to 0}{\lim }\frac{e^{-h}-1}{-h}=1$

So, it is not differentiable at $x=0$.

Similarly, it is not differentiable at $x=1$ but it is continuous at $x=0$ and $1$.