The function $f\left( x \right)=\max { \left\{ \left( 1-x \right) ,\left( 1+x \right) ,2 \right\} } ,x\in \left( -\infty ,\infty \right)$ is

$f\left( x \right) =\begin{cases} \begin{matrix} 1-x, & x\le -1 \end{matrix} \\ \begin{matrix} 2, & -1<x\le 1 \end{matrix} \\ \begin{matrix} 1+x, & x>1 \end{matrix} \end{cases}$

$\displaystyle \lim _{ x\rightarrow -{ 1 }^{ - } }{ f\left( x \right) } =\displaystyle \lim _{ x\rightarrow -{ 1 }^{ - } }{ \left( 1-x \right) } =2=\displaystyle \lim _{ x\rightarrow -{ 1 }^{ + } }{ f\left( x \right) }$

and, $\displaystyle \lim _{ x\rightarrow 1 }{ f\left( x \right) } =2,$ so $f$ is continuous at all points.

$\displaystyle f'\left( { -1 }^{ - } \right) =\displaystyle \lim _{ h\rightarrow { o }^{ - } }{ \frac { f\left( -1-h \right) -f\left( -1 \right) }{ -h } } =\displaystyle \lim _{ h\rightarrow { o }^{ - } }{ \frac { 1+1+h-2 }{ -h } } =-1$

$f'\left( { -1 }^{ + } \right) =0.$

Similarly, $f'\left( { 1 }^{ - } \right) =0$ and $f'\left( 1+ \right) =1,$

so $f$ differentiable everywhere except at $x=-1,1.$