The derivative of the function f is defined by *f*′(*x*)=(*x^*2+2)cos(*x*+5). What is the x-coordinate of the absolute maximum value of the function *f* on the closed interval [−4,2]? You may use a calculator and round your answer to the nearest thousandth.

To find the absolute maximum value of the function \( f(x) \) on the closed interval \([-4, 2]\), we will apply the **First Derivative Test** and evaluate the function at critical points and at the endpoints.

### Steps to solve:

1. **Find the critical points**:
   The critical points occur when \( f'(x) = 0 \). We're given \( f'(x) = (x^2 + 2)\cos(x + 5) \). To find critical points, set \( f'(x) = 0 \):
   \[
   (x^2 + 2)\cos(x + 5) = 0
   \]
   Since \( x^2 + 2 \neq 0 \) for any real \( x \) (because \( x^2 + 2 \geq 2 \)), the only way this product equals 0 is when:
   \[
   \cos(x + 5) = 0
   \]
   
2. **Solve \( \cos(x + 5) = 0 \)**:
   The cosine function is zero at \( \frac{\pi}{2} + n\pi \), for integer \( n \). Therefore:
   \[
   x + 5 = \frac{\pi}{2} + n\pi
   \]
   Solving for \( x \):
   \[
   x = \frac{\pi}{2} - 5 + n\pi
   \]
   This gives us a general solution for \( x \), but we are only interested in solutions within the interval \([-4, 2]\).

3. **Find specific solutions**:
   By substituting different integer values of \( n \) and solving for \( x \), we get the possible critical points within the interval \([-4, 2]\):
   - For \( n = 0 \), \( x = \frac{\pi}{2} - 5 \approx -3.429 \)
   - For \( n = 1 \), \( x = \frac{3\pi}{2} - 5 \approx -0.288 \)
   
   So the critical points are \( x \approx -3.429 \) and \( x \approx -0.288 \).

4. **Evaluate the function at the critical points and endpoints**:
   Since we do not have an explicit formula for \( f(x) \), we can only evaluate the function \( f(x) \) at the critical points and endpoints using the information from \( f'(x) \). 

   The endpoints are \( x = -4 \) and \( x = 2 \).

5. **Determine the absolute maximum**:
   Using the calculator, evaluate the function at the critical points and the endpoints (by analyzing the sign changes of \( f'(x) \) around those points) to find the \( x \)-coordinate where \( f(x) \) has an absolute maximum.

   After evaluating, the absolute maximum occurs at \( x \approx -3.429 \).

### Conclusion:
The \( x \)-coordinate of the absolute maximum value of the function \( f(x) \) on the closed interval \([-4, 2]\) is approximately \( \boxed{-3.429} \).

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Would you like further details or clarifications on any of these steps?

Here are some related questions:
1. How do we find critical points for a general function?
2. What is the First Derivative Test, and how does it work?
3. How can we evaluate trigonometric equations like \( \cos(x) = 0 \) in different intervals?
4. What is the significance of closed intervals in optimization problems?
5. How does the behavior of the derivative inform us about maximum and minimum points?

**Tip:** Always check both critical points and endpoints when looking for absolute extrema on a closed interval.

The derivative of the function f is defined by f′(x)=(x^2+2)cos(x+5). What is the x-coordinate of the absolute maximum value of the function f on the closed interval [−4,2]? You may use a calculator and round your answer to the nearest thousandth.

Explore how to find the absolute maximum of the function f(x) with f'(x) = (x^2 + 2)cos(x + 5) on the interval [-4, 2]. This detailed solution includes steps on using the First Derivative Test, solving trigonometric equations, and evaluating critical points. Ideal for college calculus students.

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