$$\text { If } f(x)=\left\{\begin{array}{ll} x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0 & , x=0 \end{array}\right. \text {, then }$$

<p>Given the function:</p> <p>$ f(x)=\left\{\begin{array}{ll} x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{array}\right. $</p> <p>we need to find its second derivative at specific points.</p> <p>First, let’s compute the first derivative $ f^{\prime}(x) $:</p> <p>$ f^{\prime}(x) = 3x^2 \sin \left( \frac{1}{x} \right) - x \cos \left( \frac{1}{x} \right) $</p> <p>Next, the second derivative $ f^{\prime \prime}(x) $ is:</p> <p>$ f^{\prime \prime}(x) = 6x \sin \left(\frac{1}{x}\right) - 3x \cos \left(\frac{1}{x}\right) - \cos \left(\frac{1}{x}\right) - \frac{1}{x} \sin \left(\frac{1}{x}\right) $</p> <p>Therefore, evaluating the second derivative at $ x = \frac{2}{\pi} $:</p> <p>$ f^{\prime \prime} \left(\frac{2}{\pi}\right) = 6 \left( \frac{2}{\pi} \right) \sin \left(\frac{\pi}{2}\right) - 3 \left( \frac{2}{\pi} \right) \cos \left(\frac{\pi}{2}\right) - \cos \left( \frac{\pi}{2} \right) - \frac{\pi}{2} \sin \left( \frac{\pi}{2} \right) $</p> <p>Since $\sin(\frac{\pi}{2}) = 1$ and $\cos(\frac{\pi}{2}) = 0$, this simplifies to:</p> <p>$ f^{\prime \prime} \left( \frac{2}{\pi} \right) = \frac{12}{\pi} - \frac{\pi}{2} = \frac{24 - \pi^2}{2\pi} $</p> <p>Finally, note that $ f^{\prime}(0) $ is not defined, as it involves terms like $\frac{1}{x}$ when $ x = 0 $.</p>