Question Number 25495 by rita1608 last updated on 11/Dec/17
Commented by prakash jain last updated on 11/Dec/17
$$\mid{x}\mid=\begin{cases}{−{x}}&{{x}<\mathrm{0}}\\{{x}}&{{x}\geqslant\mathrm{0}}\end{cases} \\ $$$${f}\left({x}\right)=\begin{cases}{−{x}^{\mathrm{2}} }&{{x}<\mathrm{0}}\\{{x}^{\mathrm{2}} }&{{x}\geqslant\mathrm{0}}\end{cases} \\ $$$$\mathrm{clearly}\:\mathrm{it}\:\mathrm{differential}\:\mathrm{at}\:\mathrm{every}\:\mathrm{pt}\:\mathrm{other} \\ $$$$\mathrm{than}\:\mathrm{0}\:\left(\mathrm{polynomial}\:\mathrm{function}\right) \\ $$$$\mathrm{You}\:\mathrm{need}\:\mathrm{to}\:\mathrm{check}\:\mathrm{differentiability} \\ $$$$\mathrm{at}\:{x}=\mathrm{0}\:\mathrm{by}\:\mathrm{taking}\:\mathrm{LHD}\:\mathrm{and}\:\mathrm{RHD}. \\ $$