Let f(x) = x |sin x|,x in R. Then, | vedclass

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(B) Given $f(x) = x |\sin x|$.At $x = n\pi$ (where $n \in Z$),we check the differentiability using the limit definition:$f'(n\pi) = \lim_{x 	o n\pi}

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$f$ is differentiable for all $x$,except at $x = n\pi, n = \pm 1, \pm 2, \pm 3, \dots$

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(B) Given $f(x) = x |\sin x|$.At $x = n\pi$ (where $n \in Z$),we check the differentiability using the limit definition:$f'(n\pi) = \lim_{x 	o n\pi} \frac{f(x) - f(n\pi)}{x - n\pi} = \lim_{x 	o n\pi} \frac{x |\sin x| - 0}{x - n\pi} = \lim_{x 	o n\pi} \frac{x |\sin x|}{x - n\pi}$.Let $x = n\pi + h$,where $h 	o 0$.Then $f'(n\pi) = \lim_{h 	o 0} \frac{(n\pi + h) |\sin(n\pi + h)|}{h} = \lim_{h 	o 0} \frac{(n\pi + h) |(-1)^n \sin h|}{h} = \lim_{h 	o 0} (n\pi + h) \frac{|\sin h|}{|h|} \cdot |h| \cdot \frac{(-1)^n}{h}$.Since $\frac{|\sin h|}{h} 	o 1$ as $h 	o 0$,the limit becomes $\lim_{h 	o 0} (n\pi + h) \cdot 1 \cdot \frac{|h|}{h} \cdot (-1)^n$.For $n = 0$,$f'(0) = \lim_{h 	o 0} h \cdot \frac{|h|}{h} = \lim_{h 	o 0} |h| = 0$. Thus,$f(x)$ is differentiable at $x = 0$.For $n 
eq 0$,the limit $\lim_{h 	o 0} n\pi \cdot (-1)^n \cdot \frac{|h|}{h}$ does not exist because the left-hand limit is $-n\pi(-1)^n$ and the right-hand limit is $n\pi(-1)^n$.Therefore,$f(x)$ is differentiable for all $x$ except at $x = n\pi$ for $n = \pm 1, \pm 2, \pm 3, \dots$.

Let $f(x) = x |\sin x|$,$x \in R$. Then,

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