To find the absolute maximum and minimum values of the function $f(x) = x^3 - 12x$ on the interval $[0, 3]$, follow these steps:

1. Find the derivative: $f'(x) = 3x^2 - 12$

2. Set the derivative equal to zero to find critical points: $3x^2 - 12 = 0 \implies x^2 = 4 \implies x = \pm 2$ Only $x = 2$ is in the interval $[0, 3]$.

3. Evaluate $f(x)$ at the critical point and the endpoints of the interval:

   • At $x = 0$: $f(0) = 0^3 - 12(0) = 0$
   • At $x = 2$: $f(2) = 2^3 - 12(2) = 8 - 24 = -16$
   • At $x = 3$: $f(3) = 3^3 - 12(3) = 27 - 36 = -9$

4. Compare the values:

   • $f(0) = 0$
   • $f(2) = -16$
   • $f(3) = -9$

5. Determine absolute max and min:

   • The absolute maximum value is $0$ at $x = 0$.
   • The absolute minimum value is $-16$ at $x = 2$.

Thus, the absolute maximum value is $0$ and the absolute minimum value is $-16$.

If you have any questions or need further details, feel free to ask! Here are some related questions for further exploration:

1. How does the behavior of $f(x)$ change outside the interval $[0, 3]$?
2. What would happen if the interval were $[-3, 3]$?
3. Can you find the second derivative and discuss the concavity of $f(x)$?
4. How do the critical points relate to the graph of $f(x)$?
5. What are the implications of the function's behavior at endpoints?

Tip: When finding extrema, always check the endpoints in addition to critical points.