Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given$g\left(x\right)=\left|f\left(x\right)\right|$and$f\left(c\right)=0$

$g^{\text{'}}\left(c^{+}\right)=\underset{x\to c^{+}}{\lim }\frac{\left|f\left(x\right)\right|-f\left(c\right)}{x-c}=\underset{x\to c^{+}}{\lim }\left|\frac{f\left(x\right)-f\left(c\right)}{x-c}\right|, \mathrm{as} \mathrm{x}>\mathrm{c}$

$\Rightarrow g^{\text{'}}\left(c^{+}\right)=\left|f^{\text{'}}\left(c\right)\right|$

$g\left(c^{-}\right)=\underset{x\to c^{-}}{\lim }\frac{\left|f\left(x\right)\right|-f\left(c\right)}{x-c}=-\underset{x\to c^{-}}{\lim }\left|\frac{f\left(x\right)-f\left(c\right)}{x-c}\right|, \mathrm{as} \mathrm{x}<\mathrm{c}$

$\Rightarrow g^{\text{'}}\left(c^{-}\right)=-\left|f^{\text{'}}\left(c\right)\right|$

For $g\left(x\right)$ to be differentiable at $x=c$, $f^{\text{'}}\left(c\right)$ must be $0$. Else, it is non-differentiable.