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.