If f(x) = begin{cases} x^2 left| cos frac{pi} | vedclass

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(A) To check the differentiability of $f(x)$ at $x = 2$, we first note that for $x$ in a neighborhood of $2$, $\cos(\frac{\pi}{x})$ is positive becaus

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Differentiable

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(A) To check the differentiability of $f(x)$ at $x = 2$, we first note that for $x$ in a neighborhood of $2$, $\cos(\frac{\pi}{x})$ is positive because $\frac{\pi}{x}$ is near $\frac{\pi}{2}$. Specifically, for $x$ near $2$, $\frac{\pi}{x}$ is in the interval $(0, \pi)$, where $\cos(\frac{\pi}{x})$ is positive. Thus, for $x$ in a small neighborhood of $2$, $f(x) = x^2 \cos(\frac{\pi}{x})$. Now, we find the derivative $f'(x)$ using the product rule: $f'(x) = \frac{d}{dx} [x^2 \cos(\frac{\pi}{x})] = 2x \cos(\frac{\pi}{x}) + x^2 [-\sin(\frac{\pi}{x})] \cdot (-\frac{\pi}{x^2}) = 2x \cos(\frac{\pi}{x}) + \pi \sin(\frac{\pi}{x})$. Evaluating at $x = 2$: $f'(2) = 2(2) \cos(\frac{\pi}{2}) + \pi \sin(\frac{\pi}{2}) = 4(0) + \pi(1) = \pi$. Since the derivative exists at $x = 2$, the function is differentiable at $x = 2$.

If $f(x) = \begin{cases} x^2 \left| \cos \frac{\pi}{x} \right|, & x \neq 0 \\ 0, & x = 0 \end{cases}$, then at $x = 2$, $f(x)$ is

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