Question

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 \\_\\_\\_\\_\\_\\_\\_\\_\\_.

Answer 1

Continuity at $x = \pm 2$: For $f(x)$ to be continuous at $x = 2$, we need:
$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^-} f(x) = f(2)$
Since $f(x) = \frac{1}{x}$ for $|x| \geq 2$, the limit as $x \to 2$ from the right is $\frac{1}{2}$. For $f(x) = ax^2 + 2b$ on $|x| < 2$, setting $f(2) = \frac{1}{2}$:
$a \times 4 + 2b = \frac{1}{2} \Rightarrow 4a + 2b = \frac{1}{2}$
Similarly, for $x = -2$, we get the same equation, ensuring continuity:
$4a + 2b = \frac{1}{2}$
Differentiability at $x = \pm 2$: For differentiability at $x = 2$, calculate $f'(x)$ for both cases:
$f'(x) = -\frac{1}{x^2} \quad \text{for } |x| \geq 2$
Using $f(x) = ax^2 + 2b$ for $|x| < 2$:
$f'(2) = -\frac{1}{4} = 2a \Rightarrow a = -\frac{1}{8}$
Substitute $a = -\frac{1}{8}$ into $4a + 2b = \frac{1}{2}$ to solve for $b$:
$b = \frac{3}{8}$
Calculate $48(a + b)$: Substitute $a = -\frac{1}{8}$ and $b = \frac{3}{8}$:
$48(a + b) = 48 \left(-\frac{1}{8} + \frac{3}{8}\right) = 48 \times \frac{1}{4} = 15$