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

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(B) $(i)$ Check for differentiability at $x=0$: $LHD = f'(0^-) = \lim_{h 	o 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h 	o 0^+} \frac{(-h)^2 |\cos(-\pi/

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differentiable at $x=0$ but not differentiable at $x=2$

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(B) $(i)$ Check for differentiability at $x=0$: $LHD = f'(0^-) = \lim_{h 	o 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h 	o 0^+} \frac{(-h)^2 |\cos(-\pi/h)| - 0}{-h} = \lim_{h 	o 0^+} -h |\cos(\pi/h)| = 0$. $RHD = f'(0^+) = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{h^2 |\cos(\pi/h)| - 0}{h} = \lim_{h 	o 0^+} h |\cos(\pi/h)| = 0$. Since $LHD = RHD = 0$,$f(x)$ is differentiable at $x=0$. $(ii)$ Check for differentiability at $x=2$: $f(2) = 2^2 |\cos(\pi/2)| = 0$. $RHD = f'(2^+) = \lim_{h 	o 0^+} \frac{f(2+h) - f(2)}{h} = \lim_{h 	o 0^+} \frac{(2+h)^2 |\cos(\pi/(2+h))|}{h}$. Since $\cos(\pi/(2+h)) > 0$ for small $h > 0$,$|\cos(\pi/(2+h))| = \cos(\pi/(2+h)) = \sin(\pi/2 - \pi/(2+h)) = \sin(\pi h / (2(2+h)))$. $RHD = \lim_{h 	o 0^+} \frac{(2+h)^2 \sin(\pi h / (2(2+h)))}{h} = 4 \cdot \frac{\pi}{4} = \pi$. $LHD = f'(2^-) = \lim_{h 	o 0^+} \frac{f(2-h) - f(2)}{-h} = \lim_{h 	o 0^+} \frac{(2-h)^2 |\cos(\pi/(2-h))|}{-h}$. Since $\cos(\pi/(2-h))  0$,$|\cos(\pi/(2-h))| = -\cos(\pi/(2-h)) = -\sin(\pi/2 - \pi/(2-h)) = -\sin(-\pi h / (2(2-h))) = \sin(\pi h / (2(2-h)))$. $LHD = \lim_{h 	o 0^+} \frac{(2-h)^2 \sin(\pi h / (2(2-h)))}{-h} = 4 \cdot (-\pi/4) = -\pi$. Since $LHD 
eq RHD$,$f(x)$ is not differentiable at $x=2$.

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

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