The problem asks to find the absolute maximum value of the function $f(x) = x^3 - 3x^2 + 12$ on the closed interval $[-2, 4]$.

Solution:

1. Find the critical points: To find critical points, we first compute the derivative of $f(x)$: $f'(x) = 3x^2 - 6x$

   Set $f'(x) = 0$ to find the critical points: $3x^2 - 6x = 0$ Factorize: $3x(x - 2) = 0$ Thus, $x = 0$ and $x = 2$ are the critical points.

2. Evaluate $f(x)$ at the critical points and endpoints of the interval: The interval is $[-2, 4]$, so we evaluate $f(x)$ at $x = -2, 0, 2,$ and $4$:

   • At $x = -2$: $f(-2) = (-2)^3 - 3(-2)^2 + 12 = -8 - 12 + 12 = -8$
   • At $x = 0$: $f(0) = (0)^3 - 3(0)^2 + 12 = 12$
   • At $x = 2$: $f(2) = (2)^3 - 3(2)^2 + 12 = 8 - 12 + 12 = 8$
   • At $x = 4$: $f(4) = (4)^3 - 3(4)^2 + 12 = 64 - 48 + 12 = 28$

3. Determine the maximum value: Comparing the function values: $f(-2) = -8, \, f(0) = 12, \, f(2) = 8, \, f(4) = 28$ The maximum value is 28, which occurs at $x = 4$.

Final Answer:

The absolute maximum value of $f(x)$ on $[-2, 4]$ is 28, occurring at $x = 4$.

Would you like further clarification, or should I explain any step in more detail?

Here are 5 follow-up questions related to this topic:

1. What is the process for finding the absolute minimum of the same function on the interval?
2. Why is it important to evaluate both critical points and endpoints for absolute extrema on closed intervals?
3. What are the differences between local and absolute extrema?
4. How does the shape of the graph of $f(x)$ relate to its extrema?
5. How can this process be applied to functions with more complex derivatives?

Tip: Always check endpoints when finding extrema on closed intervals, as they may contain the maximum or minimum value.