To find the critical points, we first differentiate the function $f(x) = x^3 - 12x$ with respect to $x$: $$f'(x) = 3x^2 - 12$$
Set $f'(x) = 0$ to find the critical points: $$3x^2 - 12 = 0$$ $$3(x^2 - 4) = 0$$ $$x^2 = 4 \implies x = 2, -2$$ Since the interval is $[0, 3]$, we only consider $x = 2$ as the critical point within the domain.
We evaluate $f(x)$ at the critical point $x = 2$ and the endpoints $x = 0$ and $x = 3$: $$f(0) = (0)^3 - 12(0) = 0$$ $$f(2) = (2)^3 - 12(2) = 8 - 24 = -16$$ $$f(3) = (3)^3 - 12(3) = 27 - 36 = -9$$
Comparing the values $0, -16,$ and $-9$, the minimum value is $-16$.
Final Answer: -16
AI generated content. Review strictly for academic accuracy.