Question

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

• A. $f''(0) = 1$
• B. $f''\left(\frac{2}{\pi}\right) = \frac{24 - \pi^2}{2\pi}$
• C. $f''\left(\frac{2}{\pi}\right) = \frac{12 - \pi^2}{2\pi}$
• D. $f''(0) = 0$

Answer 1

Answer: (B)

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

Answer 2

The given function is:

$f(x) = x^3 \sin\left(\frac{1}{x}\right) - x \cos\left(\frac{1}{x}\right).$

The second derivative of $f(x)$ is computed as:

$f''(x) = 6x \sin\left(\frac{1}{x}\right) - 3 \cos\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) + \sin\left(\frac{1}{x}\right) \left(-\cos\left(\frac{1}{x}\right)\right).$

Substitute $x = \frac{2}{\pi}$:

$f''\left(\frac{2}{\pi}\right) = 6\left(\frac{2}{\pi}\right) \sin\left(\frac{\pi}{2}\right) - 3 \cos\left(\frac{\pi}{2}\right) - \cos\left(\frac{\pi}{2}\right).$

Simplify:

$f''\left(\frac{2}{\pi}\right) = \frac{12}{\pi} - \frac{\pi^2}{2\pi} = \frac{24 - \pi^2}{2\pi}.$

Thus, the final value is:

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