If the function   $f\left(x\right)=\{\begin{array}{ccc}|x-2|& :& x\ge 1\\ \frac{x^2}{5}-\frac{4x}{5}+\frac{8}{5}& :& x<1\end{array}$

$f\left(x\right)=\{\begin{array}{ccc}\frac{x^2-4x+8}{5}& :& x<1\\ -(x-2)& :& 1\le x2(1\le x<2)\\ x-2& :& x\ge 2\end{array}$

$\begin{array}{ccccc}RHL& at& x=1& is& 1\\ LHL& at& x=1& is& 1\end{array}\}f\left(x\right) is continuous at x=1$

$f^1\left(x\right)=\{\begin{array}{ccc}\frac{2x-4}{5}& :& x<1\\ -1& :& 1\le x<2\\ 1& :& x\ge 2\end{array}$

$\begin{array}{ccccc}RHD& at& x=1& is& -1\\ LHD& at& x=1& is& 2/5\end{array}\}$   not differentiable at  $x=1$
                                         not differentiable at $x=2$