The set of points where f(x) = |sin x| is not differentiable, is ____________.
Let $g(x)=|x|= \begin{cases}x, & x \geq 0 \\ -x, & x<0\end{cases}$
Now,
$L g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{g(0-h)-g(0)}{-h}$
$\Rightarrow L g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{-(-h)-0}{-h}$
$\Rightarrow L g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{h}{-h}$
$\Rightarrow L g^{\prime}(0)=-1$
And
$R g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{g(0+h)-g(0)}{h}$
$\Rightarrow R g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{h-0}{h}$
$\Rightarrow R g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{h}{h}$
$\Rightarrow R g^{\prime}(0)=1$
$\therefore L g^{\prime}(0) \neq R g^{\prime}(0)$
So,
Therefore, $f(x)=|\sin x|$ is not differentiable when $\sin x=0$.
$\sin x=0$
$\Rightarrow x=n \pi, n \in \mathrm{Z}$
Thus, the set of points where $f(x)=|\sin x|$ is not differentiable is $\{x=n \pi: n \in \mathrm{Z}\}$.
The set of points where $f(x)=|\sin x|$ is not differentiable, is $\{x=n \pi: n \in \mathrm{Z}\}$
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