# What are the critical values, if any, of  f(x)= sin|x|?

Dec 20, 2015

$\left\{\frac{\pi}{2} + n \pi | n \in \mathbb{Z}\right\} \cup \left\{0\right\}$

#### Explanation:

A critical value $c$ of a function $f \left(x\right)$ is a value where $f ' \left(c\right) = 0$ or $f ' \left(c\right)$ is undefined.

The challenge here is to find $f ' \left(x\right)$ as we do not have a simple rule to handle $| x |$ in differentiation. We could break it into a piecewise function split at $x = 0$, but let's use another trick in this case.

For all $x \in \mathbb{R}$ we have $| x | = \sqrt{{x}^{2}}$. Thus we can rewrite the function as

$f \left(x\right) = \sin \left(\sqrt{{x}^{2}}\right)$

Then, applying the chain rule, we have

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \sin \left(\sqrt{{x}^{2}}\right)$

$= \cos \left(\sqrt{{x}^{2}}\right) \left(\frac{d}{\mathrm{dx}} \sqrt{{x}^{2}}\right)$

$= \cos \left(\sqrt{{x}^{2}}\right) \left(\frac{1}{2 \sqrt{{x}^{2}}}\right) \left(\frac{d}{\mathrm{dx}} {x}^{2}\right)$

$= \cos \left(\sqrt{{x}^{2}}\right) \left(\frac{1}{2 \sqrt{{x}^{2}}}\right) \left(2 x\right)$

$= \frac{x \cos \left(| x |\right)}{|} x |$

Then, $f ' \left(x\right)$ is undefined when $x = 0$, and

$f ' \left(x\right) = 0 \iff \cos \left(| x |\right) = 0$

$\iff x = \frac{\pi}{2} + n \pi$ where $n \in \mathbb{Z}$

Thus the critical values of $\sin | x |$ are

$\left\{\frac{\pi}{2} + n \pi | n \in \mathbb{Z}\right\} \cup \left\{0\right\}$