# The function

Question:

The function $f(x)=\cos ^{-1}(\cos x), x \in(-2 \pi, 2 \pi)$ is not differentiable at $x=$________________

Solution:

$f(x)=\cos ^{-1}(\cos x)= \begin{cases}x+2 \pi & -2 \pi \leq x \leq-\pi \\ -x, & -\pi \leq x \leq 0 \\ x, & 0 \leq x \leq \pi \\ 2 \pi-x, & \pi \leq x \leq 2 \pi\end{cases}$

Let us check the differentiability of the function at $x=-\pi, x=0$ and $x=\pi$.

At $x=-\pi$

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

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

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

$\Rightarrow L f^{\prime}(-\pi)=1$

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

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

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

$\Rightarrow R f^{\prime}(-\pi)=-1$

$\therefore L f^{\prime}(-\pi) \neq R f^{\prime}(-\pi)$

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

At $x=0$

$L f^{\prime}(0)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$

$\Rightarrow L f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{-x-0}{x}$

$\Rightarrow L f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{-x}{x}$

$\Rightarrow L f^{\prime}(0)=-1$

$R f^{\prime}(0)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$

$\Rightarrow R f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{x-0}{x-0}$

$\Rightarrow R f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{x}{x}$

$\Rightarrow R f^{\prime}(0)=1$

$\therefore L f^{\prime}(0) \neq R f^{\prime}(0)$

So, the function $f(x)$ is not differentiable at $x=0$.

At $x=\pi$,

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

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

$\Rightarrow L f^{\prime}(\pi)=1$

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

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

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

$\Rightarrow R f^{\prime}(\pi)=-1$

$\therefore L f^{\prime}(\pi) \neq R f^{\prime}(\pi)$

So, the function $f(x)$ is not differentiable at $x=\pi$.

Thus, the function $f(x)=\cos ^{-1}(\cos x), x \in(-2 \pi, 2 \pi)$ is not differentiable at $x=-\pi, x=0$ and $x=\pi$.

The function $f(x)=\cos ^{-1}(\cos x), x \in(-2 \pi, 2 \pi)$ is not differentiable at $x=$ $-\pi, 0, \pi$