Question

The function $f\left(x\right)=\left|\sin x\right|,\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ is not differentiable at x = ____________.

Answer 1

We know that, $\left|x\right|$ is not differentiable at x = 0.

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

$\sin x=0$

$\Rightarrow x=n\mathrm{\pi }$, n ∈ Z

$\Rightarrow x=...,-2\mathrm{\pi },-\mathrm{\pi },0,\mathrm{\pi },2\mathrm{\pi },...$

Now, the only value of x lying in given interval $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ at which the function $f\left(x\right)=\left|\sin x\right|$ is not differentiable is 0.

Thus, the function $f\left(x\right)=\left|\sin x\right|,x\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ is not differentiable at x = 0.


The function $f\left(x\right)=\left|\sin x\right|,\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ is not differentiable at x = ___0___.