Question Number 155083 by Tawa11 last updated on 25/Sep/21
Answered by physicstutes last updated on 25/Sep/21
$$\left(\mathrm{a}\right)\:{f}\left({x}\right)=\:\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\:,\:{x}\:\neq\mathrm{0} \\ $$$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{differentiable}\:\mathrm{on}\:−\mathrm{1}<{x}<\mathrm{1} \\ $$$$\left(\mathrm{b}\right)\:{g}\left({x}\right)=\mid{x}\mid\:\mathrm{is}\:\mathrm{not}\:\mathrm{differentiable}\:\mathrm{on} \\ $$$$−\mathrm{1}<{x}<\mathrm{1} \\ $$
Commented by Tawa11 last updated on 25/Sep/21
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$