Consider the function $f:R\to R$ defined by $f\left(x\right)=\left\{\begin{array}{ll}\left(2-\sin \left(\frac{1}{x}\right)\right)\left|x\right|,& x\ne 0\\ 0,& x=0\end{array}.\right.$ Then $f$ is:

Given $f\left(x\right)=\left\{\begin{array}{ll}\left(2-\sin \left(\frac{1}{x}\right)\right)\left|x\right|,& x\ne 0\\ 0,& x=0\end{array}\right.$ and we know that

$\left|x\right|=\left\{\begin{array}{ll} x,& x\ge 0\\ -x,& x<0\end{array}\right.$

$\Rightarrow f\left(x\right)=\left\{\begin{array}{ll}-x\left(2-\sin \left(\frac{1}{x}\right)\right),& x<0\\ 0,& x=0\\ x\left(2-\sin \left(\frac{1}{x}\right)\right),& x>0\end{array}\right.$

Now, differentiating using product rule,

$\Rightarrow f^{\text{'}}\left(x\right)=\left\{\begin{array}{ll}-\left(2-\sin \frac{1}{\mathrm{x}}\right)-\mathrm{x}\left(-\cos \frac{1}{\mathrm{x}}·\left(-\frac{1}{{\mathrm{x}}^2}\right)\right),& \mathrm{x}<0\\ \left(2-\sin \frac{1}{\mathrm{x}}\right)+\mathrm{x}\left(-\cos \frac{1}{\mathrm{x}}\left(-\frac{1}{{\mathrm{x}}^2}\right)\right),& \mathrm{x}>0\end{array}\right.$

$\Rightarrow f^{\text{'}}\left(x\right)=\left\{\begin{array}{ll}-2+\sin \frac{1}{x}-\frac{1}{x}\cos \frac{1}{x},& x<0\\ 2-\sin \frac{1}{x}+\frac{1}{x}\cos \frac{1}{x},& x>0\end{array}\right.$

Since $f^{\text{'}}\left(x\right)$ is an oscillating function and hence it is non-monotonic in $(-\infty ,0)\cup (0,\infty ).$