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.