ext { If } f(x)=\left\{\begin{array}{ll} x^3 \sin \left(\frac{1}{x} ight), & x
eq 0 \ 0 & , x=0 \end{array} ight. ext {, then }

Question

ext { If } f(x)=\left\{\begin{array}{ll} x^3 \sin \left(\frac{1}{x}ight), & x 
eq 0 \ 0 & , x=0 \end{array}ight. 	ext {, then }

Accepted Answer

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

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

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