The set of points where

Question:

The set of points where f(x) = |sin x| is not differentiable, is ____________.

Solution:

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, x">|x|x is not differentiable at x = 0.

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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