Consider the function f: R → R defined by f (x) = \ cases (2 -...

Given f\left(x\right) = \left\{\begin{cases}\left(2 - \sin \left(\dfrac{1}{x}\right)\right) | x | , & x \ne 0 \\ 0, & x = 0\end{cases}\right) and we know that</p><p>\left|x\right| = \left\{$\begin{cases}x, & x \ge 0 \\ - x, & x < 0\end{cases}$\right)

⇒ f\left(x\right) = \left\{\begin{cases}- x\left(2 - \sin \left(\dfrac{1}{x}\right)\right), & x 0\end{cases}\right)</p><p>Now, differentiating using product rule,</p><p>⇒ {f}^{'}\left(x\right) = \left\{$\begin{cases}- \left(2 - sin\dfrac{1}{x}\right) - x\left(- cos\dfrac{1}{x} \cdot \left(- \dfrac{1}{{x}^{2}}\right)\right), & x 0\end{cases}$\right)

⇒ {f}^{'}\left(x\right) = \left\{\begin{cases}- 2 + sin\dfrac{1}{x} - \dfrac{1}{x}cos\dfrac{1}{x}, & x 0\end{cases}\right)</p><p>Since{f}^{'} ( x )$is an oscillating function and hence it is non-monotonic in$( - \infty , 0 ) \cup ( 0, \infty ) .