The number of points

Question:

The number of points in $[-\pi, \pi]$ where $f(x)=\sin ^{-1}(\sin x)$ is not differentiable is._____________

Solution:

$f(x)=\sin ^{-1}(\sin x)= \begin{cases}-x-\pi, & -\pi \leq x \leq-\frac{\pi}{2} \\ x, & -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \\ \pi-x, & \frac{\pi}{2} \leq x \leq \pi\end{cases}$

Let us check the differentiability of the function at $x=-\frac{\pi}{2}$ and $x=\frac{\pi}{2}$.

At $x=-\frac{\pi}{2}$

$L f^{\prime}\left(-\frac{\pi}{2}\right)=\lim _{x \rightarrow-\frac{\pi}{2}^{-}} \frac{f(x)-f\left(-\frac{\pi}{2}\right)}{x-\left(-\frac{\pi}{2}\right)}$

$\Rightarrow L f^{\prime}\left(-\frac{\pi}{2}\right)=\lim _{x \rightarrow-\frac{\pi}{2}} \frac{-x-\pi-\left(-\frac{\pi}{2}\right)}{x+\frac{\pi}{2}}$

$\Rightarrow L f^{\prime}\left(-\frac{\pi}{2}\right)=\lim _{x \rightarrow-\frac{\pi}{2}} \frac{-\left(x+\frac{\pi}{2}\right)}{x+\frac{\pi}{2}}$

$\Rightarrow L f^{\prime}\left(-\frac{\pi}{2}\right)=-1$

$R f^{\prime}\left(-\frac{\pi}{2}\right)=\lim _{x \rightarrow-\frac{\pi}{2}^{+}} \frac{f(x)-f\left(-\frac{\pi}{2}\right)}{x-\left(-\frac{\pi}{2}\right)}$

$\Rightarrow R f^{\prime}\left(-\frac{\pi}{2}\right)=\lim _{x \rightarrow-\frac{\pi}{2}} \frac{x-\left(-\frac{\pi}{2}\right)}{x+\frac{\pi}{2}}$

$\Rightarrow R f^{\prime}\left(-\frac{\pi}{2}\right)=\lim _{x \rightarrow-\frac{\pi}{2}} \frac{x+\frac{\pi}{2}}{x+\frac{\pi}{2}}$

$\Rightarrow R f^{\prime}\left(-\frac{\pi}{2}\right)=1$

$\therefore L f^{\prime}\left(-\frac{\pi}{2}\right) \neq R f^{\prime}\left(-\frac{\pi}{2}\right)$

So, the function $f(x)$ is not differentiable at $x=-\frac{\pi}{2}$.

At $x=\frac{\pi}{2}$

$L f^{\prime}\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}^{-}} \frac{f(x)-f\left(\frac{\pi}{2}\right)}{x-\frac{\pi}{2}}$

$\Rightarrow L f^{\prime}\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{x-\frac{\pi}{2}}{x-\frac{\pi}{2}}$

$\Rightarrow L f^{\prime}\left(\frac{\pi}{2}\right)=1$

$R f^{\prime}\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}^{+}} \frac{f(x)-f\left(\frac{\pi}{2}\right)}{x-\frac{\pi}{2}}$

$\Rightarrow R f^{\prime}\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\pi-x-\frac{\pi}{2}}{x-\frac{\pi}{2}}$

$\Rightarrow R f^{\prime}\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\left(x-\frac{\pi}{2}\right)}{x-\frac{\pi}{2}}$

$\Rightarrow R f^{\prime}\left(\frac{\pi}{2}\right)=-1$

$\therefore L f^{\prime}\left(\frac{\pi}{2}\right) \neq R f^{\prime}\left(\frac{\pi}{2}\right)$

Thus, the function $f(x)=\sin ^{-1}(\sin x), x \in[-\pi, \pi]$ is not differentiable at $x=-\frac{\pi}{2}$ and $x=\frac{\pi}{2}$.

The number of points in $[-\pi, \pi]$ where $f(x)=\sin ^{-1}(\sin x)$ is not differentiable is 2

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