If f(x) = begin{cases} ax^2 + b; & x le 0 x^ | vedclass

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(C) For $f(x)$ to possess a derivative at $x = 0$,it must be continuous at $x = 0$. For continuity at $x = 0$: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^

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$a \in R, b = 0$

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(C) For $f(x)$ to possess a derivative at $x = 0$,it must be continuous at $x = 0$. For continuity at $x = 0$: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$ $\lim_{x 	o 0^-} (ax^2 + b) = b$ $\lim_{x 	o 0^+} (x^2) = 0$ $f(0) = b$ Thus,$b = 0$. Now,for differentiability at $x = 0$,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$: $LHD = \lim_{h 	o 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^-} \frac{a(h)^2 + b - b}{h} = \lim_{h 	o 0^-} (ah) = 0$ $RHD = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{(h)^2 - 0}{h} = \lim_{h 	o 0^+} (h) = 0$ Since $LHD = RHD = 0$ for any value of $a$,the function is differentiable at $x = 0$ for all $a \in R$ provided $b = 0$.

If $f(x) = \begin{cases} ax^2 + b; & x \le 0 \\ x^2; & x > 0 \end{cases}$ possesses a derivative at $x = 0$,then:

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