Let f(x) = begin{cases} 0, & x

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(C) Given $f(x) = \begin{cases} 0, & x < 0 \ x^2, & x \ge 0 \end{cases}$.First,we check the continuity at $x = 0$:$\lim_{x 	o 0^-} f(x) = \lim_{h

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$f'$ is continuous but not differentiable

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(C) Given $f(x) = \begin{cases} 0, & x First,we check the continuity at $x = 0$:$\lim_{x 	o 0^-} f(x) = \lim_{h 	o 0} f(0 - h) = 0$.$\lim_{x 	o 0^+} f(x) = \lim_{h 	o 0} f(0 + h) = \lim_{h 	o 0} (0 + h)^2 = 0$.Since $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = 0$,$f(x)$ is continuous at $x = 0$.Next,we check the differentiability at $x = 0$:$Lf'(0) = \lim_{h 	o 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h 	o 0} \frac{0 - 0}{-h} = 0$.$Rf'(0) = \lim_{h 	o 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h 	o 0} \frac{h^2 - 0}{h} = \lim_{h 	o 0} h = 0$.Since $Lf'(0) = Rf'(0) = 0$,$f(x)$ is differentiable at $x = 0$ and $f'(0) = 0$.The derivative is $f'(x) = \begin{cases} 0, & x Now,check the continuity of $f'(x)$ at $x = 0$:$\lim_{x 	o 0^-} f'(x) = 0$ and $\lim_{x 	o 0^+} f'(x) = \lim_{x 	o 0^+} 2x = 0$.Since $\lim_{x 	o 0^-} f'(x) = \lim_{x 	o 0^+} f'(x) = f'(0) = 0$,$f'(x)$ is continuous at $x = 0$.Finally,check the differentiability of $f'(x)$ at $x = 0$:$Lf''(0) = \lim_{h 	o 0} \frac{f'(0 - h) - f'(0)}{-h} = \lim_{h 	o 0} \frac{0 - 0}{-h} = 0$.$Rf''(0) = \lim_{h 	o 0} \frac{f'(0 + h) - f'(0)}{h} = \lim_{h 	o 0} \frac{2(0 + h) - 0}{h} = \lim_{h 	o 0} \frac{2h}{h} = 2$.Since $Lf''(0) 
eq Rf''(0)$,$f'(x)$ is not differentiable at $x = 0$.

Let $f(x) = \begin{cases} 0, & x < 0 \\ x^2, & x \ge 0 \end{cases}$,then for all values of $x$

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