Let $f(x)=\left\{\begin{array}{cl}(x-1)^{2} \sin \frac{1}{x-1}-|x|, & ext { if } x
eq 1 \ -1 &

Question

Let $f(x)=\left\{\begin{array}{cl}(x-1)^{2} \sin \frac{1}{x-1}-|x|, & 	ext { if } x 
eq 1 \ -1 & , 	ext { if } x=1\end{array}ight.$ be a real valued function. Then the real number where $f(x)$ is not differentiable is

Professor PC Thomas Classes · Answer

Since $(x-1)^{2} \sin \left(\frac{1}{x-1}ight)$\ is differentiable at $x=0$ and $|x|$ is not differentiable there $f(x)$ is not differentiable at $x=0$\ Now as regards differentiability at $x=1$,\ we have $f^{\prime}\left(1^{+}ight)=\lim _{h ightarrow 0^{+}}\left[\frac{f(1+h)-f(1)}{h}ight]$\ $=\lim _{h ightarrow 0^{+}} \frac{h^{2} \sin \frac{1}{h}-(1+h)+1}{h}$\ $=\lim _{h ightarrow 0^{+}} \frac{h^{2} \sin \left(\frac{1}{h}ight)-h}{h}=\lim _{h ightarrow 0^{+}} h \sin \frac{1}{h}-1$\ $=0-1=-1$\ Similarly $f^{\prime}\left(1^{-}ight)=-1$.\ $	herefore \quad f$ is differentiable at $x=1$\ $	herefore \quad f$ is differentiable except at $x=0$.\