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To find the absolute maximum and minimum of \( f(x) = (x - 2)^3 + 4 \) on the interval \([-7, 7]\), follow these steps:

1. **Differentiate \( f(x) \):**
   To find the critical points, we need the derivative \( f'(x) \).
   \[
   f(x) = (x - 2)^3 + 4
   \]
   Using the chain rule, the derivative is:
   \[
   f'(x) = 3(x - 2)^2
   \]

2. **Set \( f'(x) = 0 \) to find critical points:**
   \[
   3(x - 2)^2 = 0
   \]
   This gives \( x = 2 \) as a critical point.

3. **Evaluate \( f(x) \) at critical points and endpoints:**
   - **At \( x = -7 \):**
     \[
     f(-7) = (-7 - 2)^3 + 4 = (-9)^3 + 4 = -729 + 4 = -725
     \]
   - **At \( x = 2 \):**
     \[
     f(2) = (2 - 2)^3 + 4 = 0 + 4 = 4
     \]
   - **At \( x = 7 \):**
     \[
     f(7) = (7 - 2)^3 + 4 = (5)^3 + 4 = 125 + 4 = 129
     \]

4. **Determine the absolute extrema:**
   - The absolute minimum value is \(-725\) at \( x = -7 \).
   - The absolute maximum value is \(129\) at \( x = 7 \).

Thus, the answer is:
- **Absolute maximum** at \( x = 7 \) with \( f(7) = 129 \).
- **Absolute minimum** at \( x = -7 \) with \( f(-7) = -725 \).

Do you want more details or have any questions?

### Related Questions
1. What are local maxima and minima, and how do they differ from absolute extrema?
2. How can we determine the critical points of more complex functions?
3. What is the role of derivatives in finding extrema?
4. Can we find absolute extrema on an unbounded interval?
5. How does concavity affect the shape of a function near its extrema?

**Tip:** Always check both the endpoints and the critical points within the interval to find absolute extrema.

Determine the location and value of the absolute extreme values of f on the interval [-7,7]. The function is f(x) = (x - 2)^3 + 4.

Find the absolute maximum and minimum values of the function f(x) = (x - 2)^3 + 4 on the interval [-7,7] using calculus. This includes finding the derivative, evaluating critical points, and comparing endpoint values. Ideal for college calculus students.

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