If $f\left(x\right)=x^2-2\left|x\right|,g\left(x\right)=\left\{\begin{array}{ll}\min .\left\{f\right(t);-2\le t\le x\}& ,-2\le x<0\\ \max \cdot \left\{f\right(t);0\le t\le x\}& ,\mathrm{ }0\le x\le 3\end{array}\right.$then $g\left(x\right)$ is non differentiable in the interval $(-\mathrm{2,3})$ at a, b then a+b will be

$\because f\left(x\right)=$$\left\{\begin{array}{ll}{x}^2+2x,& x<0\\ x^2-2x,& x\ge 0\end{array}\right.$ and by definition from the graph 

$g\left(x\right)=$$\left\{\begin{array}{cc}{x}^2+2x,& -2\le x<-1\\ -1,& -1\le x<0\\ 0,& 0\le x<2\\ x^2-2x,& 2\le x<3\end{array}\right.$

$\mathrm{g}\text{'}\left(\mathrm{x}\right) = \left\{\begin{array}{ll}2\mathrm{x}+2,& -2\le \mathrm{x}-1\\ 0,& -1\le \mathrm{x}<0\\ 0,& 0\le \mathrm{x}<2\\ 2\mathrm{x}-2.& 2\le \mathrm{x}<3\end{array}\right.$

Clearly g(x) is discontinuous at x= 0 and not differentiable at x= 0, 2