We need the piecewise function

to be differentiable on all of $\mathbb R$.

Since each part is differentiable on its own interval, we only need to check the junction points: $x=2 \quad \text{and} \quad x=-2.$

1. Rewrite the outer part without modulus

For $x\ge 2$, $|x|=x$, so $f(x)=\frac1x.$ For $x\le -2$, $|x|=-x$, so $f(x)=\frac1{-x}=-\frac1x.$

Thus,

• for $x\ge 2$: $f(x)=\dfrac1x$
• for $x\le -2$: $f(x)=-\dfrac1x$
• for $|x|<2$: $f(x)=ax^2+2b$

2. Continuity at $x=2$

For differentiability, continuity is necessary.

From the middle piece, $f(2^-)=a(2)^2+2b=4a+2b.$ From the outer piece, $f(2)=\frac12.$ So continuity at $x=2$ gives $4a+2b=\frac12. \qquad (1)$

3. Equality of derivatives at $x=2$

Derivative from the middle piece: $\frac{d}{dx}(ax^2+2b)=2ax,$ so at $x=2$, $f'_-(2)=4a.$

Derivative from the right-side outer piece $\dfrac1x$: $\frac{d}{dx}\left(\frac1x\right)=-\frac1{x^2},$ so at $x=2$, $f'_+(2)=-\frac14.$

Differentiability at $x=2$ gives $4a=-\frac14 \implies a=-\frac1{16}. \qquad (2)$

4. Use continuity equation to find $b$

Substitute $a=-\dfrac1{16}$ into (1): $4\left(-\frac1{16}\right)+2b=\frac12.$ That is, $-\frac14+2b=\frac12,$ so $2b=\frac12+\frac14=\frac34,$ $b=\frac38.$

5. Check at $x=-2$

Now verify differentiability at $x=-2$.

Continuity

From middle piece, $f((-2)^+)=a(-2)^2+2b=4a+2b=\frac12.$ From outer piece for $x\le -2$, $f(x)=-\dfrac1x$, hence $f(-2)= -\frac1{-2}=\frac12.$ So continuity holds.

Derivatives

Middle piece derivative at $x=-2$: $f'_+(-2)=2a(-2)=-4a= -4\left(-\frac1{16}\right)=\frac14.$

Outer piece derivative: $\frac{d}{dx}\left(-\frac1x\right)=\frac1{x^2},$ so at $x=-2$, $f'_-(-2)=\frac1{(-2)^2}=\frac14.$ Hence differentiability at $x=-2$ also holds.

6. Compute $48(a+b)$

$a+b=-\frac1{16}+\frac38=-\frac2{32}+\frac{12}{32}=\frac{10}{32}=\frac5{16}.$ Therefore, $48(a+b)=48\cdot \frac5{16}=3\cdot 5=15.$

Final Answer

$\boxed{15}$

The derived answer matches the stored correct answer.