Answer
$a = \frac{1}{3}$
$b = -\frac{4}{3}$
Work Step by Step
For any function to be differentiable at all points, it must be continous. This means that, when $x = 2$, both $ax^{3}$ and $x^{2} + b$ must yield the same value for $f(x)$. As such, we can equate both expressions when $x = 2$: $$ax^{3} = x^{2} + b$$ $$a(2)^{3} = (2)^{2} + b$$ $$ 8a = 4 + b$$ Furthermore, since we've established that $ax^{3} = x^{2} + b$ is differentiable, we can proceed to find its first derivative using the Addition and Powers Rules: $$ax^{3} (\frac{d}{dx})= (x^{2} + b)(\frac{d}{dx})$$ $$3ax^{3-1} = 2x^{2-1} + 0$$ $$3ax^{2} = 2x$$ since $a$ and $b$ are considered to be constants and are treated as such. When $x=2$, we have: $$3a(2)^{2} = 2(2)$$ $$3a(4) = 4$$ $$3a = 1$$ $$a = \frac{1}{3}$$ and solving for $b$ using the first equation we constructed, we get: $$8a = 4+b$$ $$8(\frac{1}{3}) = \frac{12}{3} + b$$ $$\frac{8}{3} - \frac{12}{3} = b$$ $$-\frac{4}{3} = b$$
