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 _______.

Continuity at $$x = 2$$

For $$f(x)$$ to be differentiable, it must first be continuous.

• $$\lim_{x \to 2^+} f(x) = \frac{1}{2}$$

• $$\lim_{x \to 2^-} f(x) = a(2)^2 + 2b = 4a + 2b$$

• Equation 1: $$4a + 2b = \frac{1}{2} \implies 8a + 4b = 1$$

 Differentiability at $$x = 2$$

• For $$x > 2$$, $$f(x) = x^{-1}$$, so $$f'(x) = -x^{-2}$$. At $$x=2$$, $$f'(2^+) = -\frac{1}{4}$$.

• For $$x < 2$$, $$f(x) = ax^2 + 2b$$, so $$f'(x) = 2ax$$. At $$x=2$$, $$f'(2^-) = 4a$$.

• Equation 2: $$4a = -\frac{1}{4} \implies a = -\frac{1}{16}$$

Step 3: Solve for $$b$$ and final value

Substitute $$a$$ into Equation 1:

$$8(-\frac{1}{16}) + 4b = 1 \implies -\frac{1}{2} + 4b = 1 \implies 4b = \frac{3}{2} \implies b = \frac{3}{8}$$

• $$a + b = -\frac{1}{16} + \frac{6}{16} = \frac{5}{16}$$

• $$48(a + b) = 48 \times \frac{5}{16} = 3 \times 5 = \mathbf{15}$$