Question

The function f(x) = cos^–1(cos x), x ∈ (–2π, 2π) is not differentiable at x = ____________.

Answer 1

$f\left(x\right)={\cos }^{-1}\left(\cos x\right)=\left\{\begin{array}{ll}x+2\mathrm{\pi }& -2\mathrm{\pi }\le x\le -\mathrm{\pi }\\ -x,& -\mathrm{\pi }\le x\le 0\\ x,& 0\le x\le \mathrm{\pi }\\ 2\mathrm{\pi }-x,& \mathrm{\pi }\le x\le 2\mathrm{\pi }\end{array}\right.$

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

At $x=-\mathrm{\pi }$,

$$\begin{gathered} Lf\text{'}\left(-\mathrm{\pi }\right)=\underset{x\to -{\mathrm{\pi }}^{-}}{\lim }\frac{f\left(x\right)-f\left(-\mathrm{\pi }\right)}{x-\left(-\mathrm{\pi }\right)} \\ \Rightarrow Lf\text{'}\left(-\mathrm{\pi }\right)=\underset{x\to -\mathrm{\pi }}{\lim }\frac{x+2\mathrm{\pi }-\mathrm{\pi }}{x+\mathrm{\pi }} \\ \Rightarrow Lf\text{'}\left(-\mathrm{\pi }\right)=\underset{x\to -\mathrm{\pi }}{\lim }\frac{x+\mathrm{\pi }}{x+\mathrm{\pi }} \\ \Rightarrow Lf\text{'}\left(-\mathrm{\pi }\right)=1 \end{gathered}$$

$$\begin{gathered} Rf\text{'}\left(-\mathrm{\pi }\right)=\underset{x\to -{\mathrm{\pi }}^{+}}{\lim }\frac{f\left(x\right)-f\left(-\mathrm{\pi }\right)}{x-\left(-\mathrm{\pi }\right)} \\ \Rightarrow Rf\text{'}\left(-\mathrm{\pi }\right)=\underset{x\to -\mathrm{\pi }}{\lim }\frac{-x-\mathrm{\pi }}{x+\mathrm{\pi }} \\ \Rightarrow Rf\text{'}\left(-\mathrm{\pi }\right)=\underset{x\to -\mathrm{\pi }}{\lim }\frac{-\left(x+\mathrm{\pi }\right)}{x+\mathrm{\pi }} \\ \Rightarrow Rf\text{'}\left(-\mathrm{\pi }\right)=-1 \end{gathered}$$

$\therefore Lf\text{'}\left(-\mathrm{\pi }\right)\ne Rf\text{'}\left(-\mathrm{\pi }\right)$

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

At $x=0$,

$$\begin{gathered} Lf\text{'}\left(0\right)=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ \Rightarrow Lf\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{-x-0}{x} \\ \Rightarrow Lf\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{-x}{x} \\ \Rightarrow Lf\text{'}\left(0\right)=-1 \end{gathered}$$

$$\begin{gathered} Rf\text{'}\left(0\right)=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ \Rightarrow Rf\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{x-0}{x-0} \\ \Rightarrow Rf\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{\mathit{x}}{\mathit{x}} \\ \Rightarrow Rf\text{'}\left(0\right)=1 \end{gathered}$$

$\therefore Lf\text{'}\left(0\right)\ne Rf\text{'}\left(0\right)$

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

At $x=\mathrm{\pi }$,

$$\begin{gathered} Lf\text{'}\left(\mathrm{\pi }\right)=\underset{x\to {\mathrm{\pi }}^{-}}{\lim }\frac{f\left(x\right)-f\left(\mathrm{\pi }\right)}{x-\mathrm{\pi }} \\ \Rightarrow Lf\text{'}\left(\mathrm{\pi }\right)=\underset{x\to -\mathrm{\pi }}{\lim }\frac{x-\mathrm{\pi }}{x-\mathrm{\pi }} \\ \Rightarrow Lf\text{'}\left(\mathrm{\pi }\right)=1 \end{gathered}$$

$$\begin{gathered} Rf\text{'}\left(\mathrm{\pi }\right)=\underset{x\to {\mathrm{\pi }}^{+}}{\lim }\frac{f\left(x\right)-f\left(\mathrm{\pi }\right)}{x-\mathrm{\pi }} \\ \Rightarrow Rf\text{'}\left(\mathrm{\pi }\right)=\underset{x\to \mathrm{\pi }}{\lim }\frac{2\mathrm{\pi }-x-\mathrm{\pi }}{x-\mathrm{\pi }} \\ \Rightarrow Rf\text{'}\left(\mathrm{\pi }\right)=\underset{x\to \mathrm{\pi }}{\lim }\frac{-\left(x-\mathrm{\pi }\right)}{x-\mathrm{\pi }} \\ \Rightarrow Rf\text{'}\left(\mathrm{\pi }\right)=-1 \end{gathered}$$

$\therefore Lf\text{'}\left(\mathrm{\pi }\right)\ne Rf\text{'}\left(\mathrm{\pi }\right)$

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

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


The function f(x) = cos^–1(cos x), x ∈ (–2$\mathrm{\pi }$, 2$\mathrm{\pi }$) is not differentiable at x = $\underline{-\mathrm{\pi },0,\mathrm{\pi }}$.