Question

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

Answer 1

Let $g\left(x\right)=\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}\right.$

Now,

$Lg\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{g\left(0-h\right)-g\left(0\right)}{-h}$

$\Rightarrow Lg\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{-\left(-h\right)-0}{-h}$

$\Rightarrow Lg\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{h}{-h}$

$\Rightarrow Lg\text{'}\left(0\right)=-1$

And

$Rg\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{g\left(0+h\right)-g\left(0\right)}{h}$

$\Rightarrow Rg\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{h-0}{h}$

$\Rightarrow Rg\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{h}{h}$

$\Rightarrow Rg\text{'}\left(0\right)=1$

$\therefore Lg\text{'}\left(0\right)\ne Rg\text{'}\left(0\right)$

So, $\left|x\right|$ is not differentiable at x = 0.

Therefore, $f\left(x\right)=\left|\sin x\right|$ is not differentiable when $\sin x=0$.

$\sin x=0$

$\Rightarrow x=n\mathrm{\pi },n\in \mathrm{Z}$

Thus, the set of points where $f\left(x\right)=\left|\sin x\right|$ is not differentiable is $\left\{x=n\mathrm{\pi }:n\in \mathrm{Z}\right\}$.


The set of points where f(x) = |sin x| is not differentiable, is $\underline{\left\{x=n\mathrm{\pi }:n\in \mathrm{Z}\right\}}$.