Question

$f\left(x\right)=\{\begin{array}{cc}1+x,& ifx\le 2\\ 5-x,& ifx>2\end{array}$

Answer 1

We have $f\left(x\right)=\{\begin{array}{cc}1+x,& ifx\le 2\\ 5-x& ifx>2\end{array}$

For differentiability at x=2
$L{f}^{\prime }\left(2\right)=\underset{x\to 2^{-}}{lim}\frac{f\left(x\right)-f\left(2\right)}{x-2}=\underset{x\to 2^{-}}{lim}\frac{(1+x)-(1+2)}{x-2}=\underset{h\to 0}{lim}\frac{(1+2-h)-3}{2-h-2}=\underset{h\to 0}{lim}\frac{-h}{-h}=1$

$R{f}^{\prime }\left(2\right)=\underset{x\to 2^{+}}{lim}\frac{f\left(x\right)-f\left(2\right)}{x-2}=\underset{x\to 2^{+}}{lim}\frac{(5-x)-3}{2+h-2}=\underset{h\to 0}{lim}\frac{5-(2+h)-3}{2+h-2}=\underset{h\to 0}{lim}\frac{5-2-h-3}{h}\underset{h\to 0}{lim}\frac{-h}{+h}=-1\because L{f}^{\prime }\left(x\right)\ne R{f}^{\prime }\left(2\right)$

So, f(x) is not differentiable at x=2.