If f(x) = begin{cases} x^3 sin left(frac{1}{x | vedclass

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(B) First,we find $f^{\prime}(0)$ using the definition of the derivative: $f^{\prime}(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac

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$f^{\prime \prime}\left(\frac{2}{\pi}\right) = \frac{24-\pi^2}{2 \pi}$

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(B) First,we find $f^{\prime}(0)$ using the definition of the derivative: $f^{\prime}(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{h^3 \sin(1/h) - 0}{h} = \lim_{h 	o 0} h^2 \sin(1/h) = 0$. Next,for $x 
eq 0$,$f^{\prime}(x) = 3x^2 \sin(1/x) - x \cos(1/x)$. Now,we find $f^{\prime \prime}(0)$ using the definition: $f^{\prime \prime}(0) = \lim_{h 	o 0} \frac{f^{\prime}(h) - f^{\prime}(0)}{h} = \lim_{h 	o 0} \frac{3h^2 \sin(1/h) - h \cos(1/h) - 0}{h} = \lim_{h 	o 0} (3h \sin(1/h) - \cos(1/h))$. Since $\lim_{h 	o 0} 3h \sin(1/h) = 0$ but $\lim_{h 	o 0} \cos(1/h)$ does not exist,$f^{\prime \prime}(0)$ does not exist. For $x 
eq 0$,$f^{\prime \prime}(x) = \frac{d}{dx} [3x^2 \sin(1/x) - x \cos(1/x)] = 6x \sin(1/x) - 3 \cos(1/x) - (\cos(1/x) + x(-\sin(1/x))(-1/x^2)) = 6x \sin(1/x) - 4 \cos(1/x) - \frac{1}{x} \sin(1/x)$. Evaluating at $x = \frac{2}{\pi}$: $f^{\prime \prime}\left(\frac{2}{\pi}ight) = 6(\frac{2}{\pi}) \sin(\frac{\pi}{2}) - 4 \cos(\frac{\pi}{2}) - \frac{\pi}{2} \sin(\frac{\pi}{2}) = \frac{12}{\pi}(1) - 4(0) - \frac{\pi}{2}(1) = \frac{12}{\pi} - \frac{\pi}{2} = \frac{24-\pi^2}{2 \pi}$.

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

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