Questions: Find the maximum and minimum values of the function g(θ)=4 θ-8 sin (θ) on the interval [0, π/2] Round to 4 decimal places if needed.

Solution Steps

To find the maximum and minimum values of the function \( g(\theta) = 4\theta - 8\sin(\theta) \) on the interval \([0, \frac{\pi}{2}]\), we need to:

1. Compute the derivative of the function \( g'(\theta) \).
2. Find the critical points by setting \( g'(\theta) = 0 \) and solving for \( \theta \).
3. Evaluate the function \( g(\theta) \) at the critical points and at the endpoints of the interval.
4. Compare these values to determine the maximum and minimum values.

To find the critical points of the function \( g(\theta) = 4\theta - 8\sin(\theta) \), we first compute its derivative: \[ g'(\theta) = 4 - 8\cos(\theta) \]

Next, we set the derivative equal to zero to find the critical points: \[ 4 - 8\cos(\theta) = 0 \implies \cos(\theta) = \frac{1}{2} \] This gives us the critical point: \[ \theta = \frac{\pi}{3} \approx 1.0472 \]

We evaluate \( g(\theta) \) at the critical point and the endpoints of the interval \([0, \frac{\pi}{2}]\):

• At \( \theta = 0 \): \[ g(0) = 4(0) - 8\sin(0) = 0 \]
• At \( \theta = \frac{\pi}{3} \): \[ g\left(\frac{\pi}{3}\right) = 4\left(\frac{\pi}{3}\right) - 8\sin\left(\frac{\pi}{3}\right) = \frac{4\pi}{3} - 8\left(\frac{\sqrt{3}}{2}\right) = \frac{4\pi}{3} - 4\sqrt{3} \approx -2.7394 \]
• At \( \theta = \frac{\pi}{2} \): \[ g\left(\frac{\pi}{2}\right) = 4\left(\frac{\pi}{2}\right) - 8\sin\left(\frac{\pi}{2}\right) = 2\pi - 8 \approx -1.7168 \]

Now we compare the values obtained:

• \( g(0) = 0 \)
• \( g\left(\frac{\pi}{3}\right) \approx -2.7394 \)
• \( g\left(\frac{\pi}{2}\right) \approx -1.7168 \)

The minimum value is: \[ \min(g) \approx -2.7394 \] The maximum value is: \[ \max(g) = 0 \]

Final Answer