Let $\mathrm{f}=\mathbf{R} \rightarrow \mathbf{R}$ be differentiable at $\mathrm{c} \in \mathbf{R}$ and $\mathrm{f}(\mathrm{c})=0$. If $g(x)=|f(x)|$, then at $x=c, g$ is :
Correct Option: , 3
$g^{\prime}(c)=\lim _{x \rightarrow c} \frac{g(x) \quad g(c)}{x \quad c}$
$\Rightarrow g^{\prime}(c)=\lim _{x \rightarrow c} \frac{|f(x)|-|f(c)|}{x-c}$
Since, $f(c)=0$
Then, $g^{\prime}(c)=\lim _{x \rightarrow c} \frac{|f(x)|}{x-c}$
$\Rightarrow g^{\prime}(c)=\lim _{x \rightarrow c} \frac{f(x)}{x-c} ;$ if $f(x)>0$
$\Rightarrow g^{\prime}(c)=f^{\prime}(c)=-f^{\prime}(c)$
$\Rightarrow 2 f^{\prime}(c)=0 \Rightarrow f^{\prime}(c)=0$
Hence, $g(x)$ is differentiable if $f^{\prime}(c)=0$
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