To determine where the function $f(x) = -2 \cos(x) - \sqrt{3}x$ is increasing or decreasing on the interval $[0, \pi]$, we need to find the critical points by calculating the derivative $f'(x)$ and analyzing its sign.

1. Find $f'(x)$: $f(x) = -2 \cos(x) - \sqrt{3}x$ Differentiating $f(x)$ with respect to $x$: $f'(x) = 2 \sin(x) - \sqrt{3}$

2. Set $f'(x) = 0$ to find critical points: $2 \sin(x) - \sqrt{3} = 0$ Solving for $\sin(x)$: $2 \sin(x) = \sqrt{3} \Rightarrow \sin(x) = \frac{\sqrt{3}}{2}$ This value of $\sin(x) = \frac{\sqrt{3}}{2}$ occurs when: $x = \frac{\pi}{3}$

3. Analyze the sign of $f'(x)$ around $x = \frac{\pi}{3}$:

   • For $x$ in $\left[ 0, \frac{\pi}{3} \right)$, choose a point, say $x = \frac{\pi}{6}$: $f'\left( \frac{\pi}{6} \right) = 2 \sin\left( \frac{\pi}{6} \right) - \sqrt{3} = 2 \cdot \frac{1}{2} - \sqrt{3} = 1 - \sqrt{3} < 0$ Therefore, $f'(x) < 0$ on $\left[ 0, \frac{\pi}{3} \right)$, indicating that $f(x)$ is decreasing on this interval.

   • For $x$ in $\left( \frac{\pi}{3}, \pi \right]$, choose a point, say $x = \frac{2\pi}{3}$: $f'\left( \frac{2\pi}{3} \right) = 2 \sin\left( \frac{2\pi}{3} \right) - \sqrt{3} = 2 \cdot \frac{\sqrt{3}}{2} - \sqrt{3} = \sqrt{3} - \sqrt{3} = 0$ Therefore, $f'(x) > 0$ on $\left( \frac{\pi}{3}, \pi \right]$, indicating that $f(x)$ is increasing on this interval.

Conclusion:

• $f(x)$ is decreasing on $\left[0, \frac{\pi}{3}\right]$.
• $f(x)$ is increasing on $\left(\frac{\pi}{3}, \pi\right]$.

Would you like further details on any step? Here are related questions that might help deepen your understanding:

1. How do we determine the intervals of increase and decrease for more complex trigonometric functions?
2. What happens if the derivative of $f(x)$ has more than one critical point?
3. How do critical points relate to local maxima and minima?
4. How would the answer change if we considered the entire real line instead of $[0, \pi]$?
5. What if $f(x)$ had a coefficient other than $-2$ in front of $\cos(x)$?

Tip: When analyzing a function's behavior, checking the sign of its derivative around critical points is crucial to identifying intervals of increase or decrease.