Find the intervals on which f is increasing and the intervals on which it is decreasing.
​f(x)equals=negative 3 cosine squared x−3cos2x on ​[minus−piπ​,piπ​

We are tasked with determining the intervals on which the function \( f(x) = -3\cos^2(x) - 3\cos(2x) \) is increasing and decreasing on the interval \([- \pi, \pi]\).

### Step 1: Differentiate the function
To determine where the function is increasing or decreasing, we need to first find its derivative \( f'(x) \). 

Using the chain rule and the derivative of trigonometric functions:

- Derivative of \( -3\cos^2(x) \):
  \[
  \frac{d}{dx}\left(-3\cos^2(x)\right) = -3 \cdot 2\cos(x) \cdot (-\sin(x)) = 6\cos(x)\sin(x) = 3\sin(2x)
  \]
  (using the double-angle identity \( 2\sin(x)\cos(x) = \sin(2x) \)).

- Derivative of \( -3\cos(2x) \):
  \[
  \frac{d}{dx} \left(-3\cos(2x)\right) = -3 \cdot (-2\sin(2x)) = 6\sin(2x)
  \]

Now, summing up both derivatives:
\[
f'(x) = 3\sin(2x) + 6\sin(2x) = 9\sin(2x)
\]

### Step 2: Set the derivative equal to zero
To find critical points, we solve:
\[
9\sin(2x) = 0
\]
This simplifies to:
\[
\sin(2x) = 0
\]

The general solution for \( \sin(2x) = 0 \) is \( 2x = n\pi \), where \( n \) is an integer. Dividing by 2, we get:
\[
x = \frac{n\pi}{2}
\]

Since we are working on the interval \( [-\pi, \pi] \), the solutions for \( x \) are:
\[
x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi
\]

### Step 3: Analyze the sign of \( f'(x) \)
Now, we analyze the sign of \( f'(x) = 9\sin(2x) \) in the intervals determined by the critical points \( x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi \).

- On \( (-\pi, -\frac{\pi}{2}) \), \( 2x \in (-2\pi, -\pi) \), and \( \sin(2x) < 0 \), so \( f'(x) < 0 \) (decreasing).
- On \( (-\frac{\pi}{2}, 0) \), \( 2x \in (-\pi, 0) \), and \( \sin(2x) > 0 \), so \( f'(x) > 0 \) (increasing).
- On \( (0, \frac{\pi}{2}) \), \( 2x \in (0, \pi) \), and \( \sin(2x) > 0 \), so \( f'(x) > 0 \) (increasing).
- On \( (\frac{\pi}{2}, \pi) \), \( 2x \in (\pi, 2\pi) \), and \( \sin(2x) < 0 \), so \( f'(x) < 0 \) (decreasing).

### Step 4: Conclusion
- The function is **increasing** on \( (-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}) \).
- The function is **decreasing** on \( (-\pi, -\frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi) \).

Would you like further details or explanations on any part of this solution?

Here are some related questions you might consider:
1. How do you determine the concavity of the function \( f(x) \)?
2. What happens at the critical points in terms of local maxima or minima?
3. How do you find inflection points for this function?
4. Can you apply this process to other trigonometric functions?
5. How does the double-angle identity help simplify the derivative?

**Tip**: Always check the endpoints of the interval when analyzing increasing/decreasing behavior over a restricted domain.

Find the intervals on which f is increasing and the intervals on which it is decreasing. f(x) = -3cos^2(x) - 3cos(2x) on [-π, π].

Find the intervals where f(x) = -3cos²(x) - 3cos(2x) is increasing and decreasing on [-π, π]. Using derivatives and the double-angle identity, this problem provides a step-by-step guide to understanding the behavior of the trigonometric function.

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