Let S = {t in R mid f(x) = |x - pi|(e^{|x|} - | vedclass

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(A) To check the differentiability of $f(x) = |x - \pi|(e^{|x|} - 1) \sin |x|$ at $x = \pi$ and $x = 0$:$1$. Differentiability at $x = \pi$:$f(\pi) =

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$\phi$ (an empty set)

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(A) To check the differentiability of $f(x) = |x - \pi|(e^{|x|} - 1) \sin |x|$ at $x = \pi$ and $x = 0$:$1$. Differentiability at $x = \pi$:$f(\pi) = 0$.$f'(x) = \frac{d}{dx} [|x - \pi|(e^{|x|} - 1) \sin |x|]$.Since $|x - \pi|$ is differentiable at $x = \pi$ only if the other factors are zero,note that at $x = \pi$,$(e^{|\pi|} - 1) \sin |\pi| = (e^{\pi} - 1) \cdot 0 = 0$.Using the definition of the derivative,$\lim_{h 	o 0} \frac{f(\pi + h) - f(\pi)}{h} = \lim_{h 	o 0} \frac{|h|(e^{|\pi + h|} - 1) \sin |\pi + h|}{h} = \lim_{h 	o 0} \frac{|h|}{h} (e^{\pi} - 1) \sin \pi = 0$.Thus,$f(x)$ is differentiable at $x = \pi$.$2$. Differentiability at $x = 0$:$f(0) = 0$.$\lim_{h 	o 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h 	o 0} \frac{|h - \pi|(e^{|h|} - 1) \sin |h|}{h}$.As $h 	o 0$,$e^{|h|} - 1 \approx |h|$ and $\sin |h| \approx |h|$.So,the limit becomes $\lim_{h 	o 0} \frac{|-\pi| \cdot |h| \cdot |h|}{h} = \lim_{h 	o 0} \pi \cdot \frac{h^2}{h} = \lim_{h 	o 0} \pi h = 0$.Thus,$f(x)$ is differentiable at $x = 0$.Since the function is differentiable everywhere,the set $S$ is an empty set $\phi$.

Let $S = \{t \in R \mid f(x) = |x - \pi|(e^{|x|} - 1) \sin |x| \text{ is not differentiable at } t\}$. Then $S$ is

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