find the absolute max and min values for f(x) = x^3 - 12x on the interval [0, 3]

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.

Discover how to find the absolute maximum and minimum values of the function f(x) = x^3 - 12x on the interval [0, 3]. This step-by-step solution includes finding critical points and evaluating the function at key points, suitable for advanced calculus students.

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