$f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{1 + x,{\rm{ if }}x \le 2}\\{5 - x,{\rm{ if }}x > 2}\end{array}} \right.$at $x = 2$ }

We have $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{1 + x,{\rm{ if }}x \le 2}\\{5 - x,{\rm{ if }}x > 2}\end{array}} \right.$at $x = 2$

For differentiability at $x = 2$,

$L{f^\prime }(2) = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{f(x) - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{(1 + x) - (1 + 2)}}{{x - 2}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{(1 + 2 - h) - 3}}{{2 - h - 2}} = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{ - h}} = 1$
$R{f^\prime }(2) = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{f(x) - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{(5 - x) - 3}}{{x - 2}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{5 - (2 + h) - 3}}{{2 + h - 2}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{5 - 2 - h - 3}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{ + h}}$
$= - 1$

$L{f^\prime }(2) \ne R{f^\prime }(2)$

Therefore we can say that $f(x)$ is not differentiable at $x = 2$.