To find the intervals on which $f(x) = -2 \cos(x) - \sqrt{2} x$ is increasing or decreasing on the interval $[0, 2\pi]$, we need to follow these steps:

Step 1: Find the Derivative of $f(x)$

The derivative $f'(x)$ will help us determine where the function is increasing or decreasing.

$f(x) = -2 \cos(x) - \sqrt{2} x$

The derivative of $f(x)$ is: $f'(x) = 2 \sin(x) - \sqrt{2}$

Step 2: Set $f'(x) = 0$ to Find Critical Points

To find the points where $f(x)$ could change from increasing to decreasing or vice versa, set $f'(x) = 0$:

$2 \sin(x) - \sqrt{2} = 0$

Solving for $\sin(x)$: $\sin(x) = \frac{\sqrt{2}}{2}$

This equation has solutions within $[0, 2\pi]$ at: $x = \frac{\pi}{4} \quad \text{and} \quad x = \frac{3\pi}{4}$

Step 3: Determine the Sign of $f'(x)$ in Each Interval

We now analyze the sign of $f'(x)$ in the intervals determined by the critical points $\left(0, \frac{\pi}{4}\right)$, $\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)$, and $\left(\frac{3\pi}{4}, 2\pi\right)$.

1. Interval $\left(0, \frac{\pi}{4}\right)$: Choose a test point, say $x = \frac{\pi}{8}$.

   • $f'\left(\frac{\pi}{8}\right) = 2 \sin\left(\frac{\pi}{8}\right) - \sqrt{2}$
   • Since $\sin\left(\frac{\pi}{8}\right) < \frac{\sqrt{2}}{2}$, $f'(x) < 0$ in this interval.

2. Interval $\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)$: Choose a test point, say $x = \frac{\pi}{2}$.

   • $f'\left(\frac{\pi}{2}\right) = 2 \cdot 1 - \sqrt{2} > 0$, so $f'(x) > 0$ in this interval.

3. Interval $\left(\frac{3\pi}{4}, 2\pi\right)$: Choose a test point, say $x = \pi$.

   • $f'(\pi) = 2 \cdot 0 - \sqrt{2} = -\sqrt{2} < 0$, so $f'(x) < 0$ in this interval.

Step 4: Conclusion

• Increasing Interval: $\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)$
• Decreasing Intervals: $\left(0, \frac{\pi}{4}\right)$ and $\left(\frac{3\pi}{4}, 2\pi\right)$

Would you like further details on the steps or have any questions?

Related Questions

1. How do you find the derivative of trigonometric functions?
2. What are critical points, and how are they used to determine intervals of increase and decrease?
3. How can you determine the intervals where a function is concave up or concave down?
4. Why is it necessary to use test points in each interval?
5. What is the role of the derivative in identifying local maxima and minima?

Tip

When finding increasing or decreasing intervals, always check each interval created by the critical points by substituting test points back into the derivative.