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Question Number 95638 by Ar Brandon last updated on 26/May/20

$$\mathrm{a}\mathrm{\setminus }\mathrm{Show}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}\mathrm{is}\mathrm{derivable}\mathrm{at}\mathrm{all}\mathrm{points}{\mathrm{x}}_0>0$$ $$\mathrm{and}\mathrm{that}{\mathrm{f}}^{\prime }\left({\mathrm{x}}_0\right)=\frac{1}{{2\mathrm{x}}_0}$$ $$\mathrm{b}\mathrm{\setminus }\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}(\mathrm{continuous}\mathrm{at}{\mathrm{x}}_0=0)$$ $$\mathrm{is}\mathrm{not}\mathrm{derivable}\mathrm{at}{\mathrm{x}}_0=0$$

Answered by john santu last updated on 26/May/20

$$\left(\mathrm{b}\right)\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}\mathrm{continuous}\mathrm{at}{\mathrm{x}}_0=0$$ $$\left(\mathrm{i}\right)\underset{x\to 0^{-}}{\lim }\sqrt{\mathrm{x}}=\underset{x\to 0^{+}}{\lim }\sqrt{\mathrm{x}}=0$$ $$\left(\mathrm{ii}\right)\mathrm{f}\left(0\right)=\sqrt{0}=0$$ $$\mathrm{since}\underset{x\to 0}{\lim }\sqrt{\mathrm{x}}=\mathrm{f}\left(0\right),\mathrm{hence}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}$$ $$\mathrm{continuous}\mathrm{at}{\mathrm{x}}_0=0$$

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