If the function f(x) = begin{cases} frac{1}{| | vedclass

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(A) Given $f(x) = \begin{cases} \frac{1}{x} & , x \geq 2 \ ax^2 + 2b & , -2 < x < 2 \ -\frac{1}{x} & , x \leq -2 \end{cases}$.For $f(x)$ to be diffe

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$15$

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(A) Given $f(x) = \begin{cases} \frac{1}{x} & , x \geq 2 \ ax^2 + 2b & , -2 For $f(x)$ to be differentiable on $\mathbb{R}$,it must be continuous and differentiable at $x = 2$ and $x = -2$.At $x = 2$,continuity implies $\lim_{x 	o 2^+} f(x) = \lim_{x 	o 2^-} f(x) = f(2)$.$\frac{1}{2} = a(2)^2 + 2b \Rightarrow 4a + 2b = \frac{1}{2} \Rightarrow 8a + 4b = 1$.For differentiability at $x = 2$,$f'(2^+) = f'(2^-)$.$f'(x) = -\frac{1}{x^2}$ for $x > 2$ and $f'(x) = 2ax$ for $-2 $-\frac{1}{2^2} = 2a(2) \Rightarrow -\frac{1}{4} = 4a \Rightarrow a = -\frac{1}{16}$.Substituting $a = -\frac{1}{16}$ into $8a + 4b = 1$:$8(-\frac{1}{16}) + 4b = 1 \Rightarrow -\frac{1}{2} + 4b = 1 \Rightarrow 4b = \frac{3}{2} \Rightarrow b = \frac{3}{8}$.Now,calculate $48(a+b)$:$48(-\frac{1}{16} + \frac{3}{8}) = 48(\frac{-1+6}{16}) = 48(\frac{5}{16}) = 3 	imes 5 = 15$.

If the function $f(x) = \begin{cases} \frac{1}{|x|} & , |x| \geq 2 \\ ax^2 + 2b & , |x| < 2 \end{cases}$ is differentiable on $\mathbb{R}$,then $48(a+b)$ is equal to . . . . . . .

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