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Question Number 186445 by myint last updated on 04/Feb/23

$$\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{function}\mathrm{y}=\mid \mathrm{x}-5\mid \mathrm{has}\mathrm{no}\mathrm{derivative}\mathrm{at}\mathrm{x}=5.$$

Answered by ARUNG_Brandon_MBU last updated on 04/Feb/23

$$f{}^{\prime }\left(5\right)=\underset{x\to 5}{\lim }\frac{f\left(x\right)-f\left(5\right)}{x-5}=\underset{x\to 5}{\lim }\frac{\mid x-5\mid }{x-5}$$$$\underset{x\to 5^{>}}{\lim }\frac{\mid x-5\mid }{x-5}=\underset{x\to 5^{>}}{\lim }\frac{x-5}{x-5}=1$$$$\underset{x\to 5^{<}}{\lim }\frac{\mid x-5\mid }{x-5}=\underset{x\to 5^{<}}{\lim }\frac{5-x}{x-5}=-1$$$$\underset{x\to 5^{>}}{\lim }\frac{\mid x-5\mid }{x-5}\ne \underset{x\to 5^{<}}{\lim }\frac{\mid x-5\mid }{x-5}$$$$\mathrm{hence}f{}^{\prime }\left(5\right)\mathrm{does}\mathrm{not}\mathrm{exist}.$$

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