Question:
Let $f: \mathrm{R} \rightarrow \mathrm{R}$ be differentiable at $\mathrm{c} \in \mathrm{R}$ and $f(c)=0$. If $\mathrm{g}(\mathrm{x})=|f(\mathrm{x})|$, then at $\mathrm{x}=\mathrm{c}, \mathrm{g}$ is :
Correct Option: 1
Solution:
$g^{\prime}(c)=\lim _{h \rightarrow 0} \frac{|f(c+h)|-|f(c)|}{h}$
$=\lim _{h \rightarrow 0} \frac{|f(c+h)|}{h}=\lim _{h \rightarrow 0} \frac{|f(c+h)-f(c)|}{h}$
$=\lim _{h \rightarrow 0}\left|\frac{f(c+h)-f(c)}{h}\right| \frac{|h|}{h}$
$=\lim _{h \rightarrow 0}\left|f^{\prime}(c)\right| \frac{|h|}{h}=0$, if $f^{\prime}(c)=0$
i.e., $\mathrm{g}(\mathrm{x})$ is differentiable at $\mathrm{x}=\mathrm{c}$, if $\mathrm{f}^{\prime}(\mathrm{c})=0$