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Question Number 155083 by Tawa11 last updated on 25/Sep/21

Answered by physicstutes last updated on 25/Sep/21

$$\left(\mathrm{a}\right)f\left(x\right)=\frac{2}{x^2},x\ne 0$$$$f\left(x\right)\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{on}-1<x<1$$$$\left(\mathrm{b}\right)g\left(x\right)=\mid x\mid \mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{on}$$$$-1<x<1$$

Commented by Tawa11 last updated on 25/Sep/21

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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