Find the absolute maximum and minimum values of the following function on the given interval. If there are multiple points in a single category list the points in increasing order in x value and enter N in any blank that you don't need to use. f(x) = 2x -4 cos(x), [-π,π[

To find the absolute maximum and minimum values of the function \( f(x) = 2x - 4\cos(x) \) on the interval \( [-\pi, \pi[ \), we need to follow these steps:

### 1. Find the critical points:
The critical points occur where the derivative \( f'(x) \) is zero or undefined.

First, let's compute the derivative of \( f(x) \):

\[
f'(x) = \frac{d}{dx}(2x - 4\cos(x)) = 2 + 4\sin(x)
\]

Now, set \( f'(x) = 0 \) to find the critical points:

\[
2 + 4\sin(x) = 0
\]

\[
\sin(x) = -\frac{1}{2}
\]

The solutions to \( \sin(x) = -\frac{1}{2} \) on the interval \( [-\pi, \pi[ \) are:

\[
x = -\frac{\pi}{6}, \frac{7\pi}{6}
\]

However, \( \frac{7\pi}{6} \) is not in the interval \( [-\pi, \pi[ \), so the only relevant critical point is \( x = -\frac{\pi}{6} \).

### 2. Evaluate \( f(x) \) at the critical point and the endpoints of the interval:
We now evaluate the function \( f(x) = 2x - 4\cos(x) \) at the critical point \( x = -\frac{\pi}{6} \) and at the endpoints \( x = -\pi \) and the right endpoint just before \( \pi \).

#### At \( x = -\frac{\pi}{6} \):

\[
f\left(-\frac{\pi}{6}\right) = 2\left(-\frac{\pi}{6}\right) - 4\cos\left(-\frac{\pi}{6}\right) = -\frac{\pi}{3} - 4\left(\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} - 2\sqrt{3}
\]

#### At \( x = -\pi \):

\[
f(-\pi) = 2(-\pi) - 4\cos(-\pi) = -2\pi - 4(-1) = -2\pi + 4
\]

#### At \( x = \pi^{-} \) (just before \( \pi \), but still within the interval):

The function behaves similarly to \( f(\pi) \), so:

\[
f(\pi^{-}) = f(\pi) = 2\pi - 4\cos(\pi) = 2\pi - 4(-1) = 2\pi + 4
\]

### 3. Compare the values:

- \( f\left(-\frac{\pi}{6}\right) = -\frac{\pi}{3} - 2\sqrt{3} \)
- \( f(-\pi) = -2\pi + 4 \)
- \( f(\pi^{-}) = 2\pi + 4 \)

Now, comparing these values, we find:

- \( f(\pi^{-}) = 2\pi + 4 \) is the **absolute maximum**.
- \( f(-\pi) = -2\pi + 4 \) is **larger** than \( f\left(-\frac{\pi}{6}\right) \), so \( f\left(-\frac{\pi}{6}\right) \) is the **absolute minimum**.

### 4. Conclusion:
- Absolute maximum occurs at \( x = \pi^{-} \).
- Absolute minimum occurs at \( x = -\frac{\pi}{6} \).

### Final Answer:
- Absolute Maximum Value: \( f(\pi^{-}) = 2\pi + 4 \)
- Absolute Minimum Value: \( f\left(-\frac{\pi}{6}\right) = -\frac{\pi}{3} - 2\sqrt{3} \)

Do you need further clarifications or details on any part of the process?

### Relative Questions:
1. How do you find critical points for a general function?
2. What is the significance of endpoints in absolute maximum/minimum problems?
3. How does the derivative help in determining critical points?
4. Can a function have multiple critical points within a given interval?
5. What is the role of symmetry in finding extrema for trigonometric functions?

### Tip:
Always remember to check the endpoints when finding absolute extrema on a closed or semi-open interval, as they could yield the maximum or minimum value.

Find the absolute maximum and minimum values of the following function on the given interval. If there are multiple points in a single category list the points in increasing order in x value and enter N in any blank that you don't need to use. f(x) = 2x - 4 cos(x), [-π,π[

Find the absolute maximum and minimum values of the function f(x) = 2x - 4cos(x) over the interval [-π, π). This solution includes step-by-step derivative calculation, critical points identification, and endpoint evaluation. Suitable for university-level calculus students.

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