Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f(x)=\{\begin{array}{cc}-1,& -2\le x<0\\ x^2-1,& 0\le x\le 2\end{array}$ $\text{ Then, }f(x\mid )=\{\begin{array}{cc}-1,& -2\le |x|<0\\ |x{|}^2-1,& 0\le |x|\le 2\end{array}$ $\Rightarrow f(|x|)=x^2-1,-2\le x\le 2$ $\Rightarrow g(x)=\{\begin{array}{rl}1+x^2-1,& -2\le x<0\\ \left(x^2-1\right)+\left|x^2-1\right|,& 0\le x\le 2\end{array}$ $=\{\begin{array}{cc}{x}^2,& -2\le x<0\\ 0,& 0\le x<1\\ 2\left(x^2-1\right),& 1\le x\le 2\end{array}$ $g^{\prime }\left(0^{-}\right)=0,g^{\prime }\left(0^{\ast }\right)=0,g^{\prime }\left(1^{-}\right)=0,g^{\prime }\left(1^{+}\right)=4$ $\Rightarrow g(x)$ is non-differentiable at $x=1$ $\Rightarrow g(x)$ is not differentiable at one point.