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To find the maximum and minimum values of the function \( g(\theta) = 3\theta - 9 \sin(\theta) \) on the interval \( [0, \pi] \), let's go through the following steps:

### Step 1: Find the derivative of \( g(\theta) \)
We start by differentiating \( g(\theta) \) with respect to \( \theta \):
\[
g'(\theta) = \frac{d}{d\theta}(3\theta - 9\sin(\theta)) = 3 - 9\cos(\theta)
\]

### Step 2: Set the derivative equal to zero to find critical points
Set \( g'(\theta) = 0 \) and solve for \( \theta \):
\[
3 - 9\cos(\theta) = 0
\]
\[
9\cos(\theta) = 3
\]
\[
\cos(\theta) = \frac{1}{3}
\]
Now, solving \( \cos(\theta) = \frac{1}{3} \) within the interval \( [0, \pi] \) gives:
\[
\theta = \cos^{-1}\left(\frac{1}{3}\right)
\]

### Step 3: Evaluate \( g(\theta) \) at the endpoints and critical points
Now we need to evaluate \( g(\theta) \) at \( \theta = 0 \), \( \theta = \pi \), and \( \theta = \cos^{-1}\left(\frac{1}{3}\right) \).

1. **At \( \theta = 0 \):**
   \[
   g(0) = 3 \cdot 0 - 9 \sin(0) = 0
   \]

2. **At \( \theta = \pi \):**
   \[
   g(\pi) = 3\pi - 9\sin(\pi) = 3\pi
   \]

3. **At \( \theta = \cos^{-1}\left(\frac{1}{3}\right) \):**
   We substitute this value into \( g(\theta) \):
   \[
   g\left(\cos^{-1}\left(\frac{1}{3}\right)\right) = 3 \cos^{-1}\left(\frac{1}{3}\right) - 9 \sin\left(\cos^{-1}\left(\frac{1}{3}\right)\right)
   \]
   Since \( \sin(\cos^{-1}(1/3)) = \sqrt{1 - (1/3)^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \),
   \[
   g\left(\cos^{-1}\left(\frac{1}{3}\right)\right) = 3 \cos^{-1}\left(\frac{1}{3}\right) - 9 \cdot \frac{2\sqrt{2}}{3} = 3 \cos^{-1}\left(\frac{1}{3}\right) - 6\sqrt{2}
   \]

### Step 4: Determine the maximum and minimum values
Now, we compare the values:
- \( g(0) = 0 \)
- \( g(\pi) = 3\pi \)
- \( g\left(\cos^{-1}\left(\frac{1}{3}\right)\right) = 3 \cos^{-1}\left(\frac{1}{3}\right) - 6\sqrt{2} \)

Among these values, \( 3\pi \) is the maximum, and \( 3 \cos^{-1}\left(\frac{1}{3}\right) - 6\sqrt{2} \) is the minimum.

### Conclusion
- **Maximum value**: \( 3\pi \)
- **Minimum value**: \( 3 \cos^{-1}\left(\frac{1}{3}\right) - 6\sqrt{2} \)

Would you like a further breakdown of any part of this process?

### Related Questions
1. How do we find critical points for functions with trigonometric terms?
2. Why is it important to check endpoints when finding extrema on a closed interval?
3. How do we calculate the exact value of \( \cos^{-1}\left(\frac{1}{3}\right) \)?
4. How would the process change if the interval were \( (0, \pi) \) instead of \( [0, \pi] \)?
5. Can the minimum or maximum occur at more than one point in the interval?

### Tip
Always check both the critical points and endpoints when finding absolute extrema on a closed interval.

Find the maximum and minimum values of the function g(θ) = 3θ - 9 sin(θ) on the interval [0, π].

Learn how to find the maximum and minimum values of g(θ) = 3θ - 9 sin(θ) on the interval [0, π]. This solution includes differentiation, identifying critical points, and using the Extreme Value Theorem. Suitable for high school calculus students.

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