Question

find the maximum and minimum values of the function $g(\theta)=5\theta - 6\sin(\theta)$ on the interval $0,\pi$
minimum value $=-0.392$
maximum value $=15.71$

Explanation:

Step1: Find the derivative of the function

The derivative of \(g(\theta)=5\theta - 6\sin(\theta)\) is \(g'(\theta)=5 - 6\cos(\theta)\).

Step2: Find the critical points

Set \(g'(\theta) = 0\), so \(5-6\cos(\theta)=0\). Then \(\cos(\theta)=\frac{5}{6}\), and \(\theta=\cos^{-1}(\frac{5}{6})\approx 0.585\) (since \(\theta\in[0,\pi]\)).

Step3: Evaluate the function at critical and endpoints

• At \(\theta = 0\): \(g(0)=5\times0-6\sin(0)=0\).
• At \(\theta=\cos^{-1}(\frac{5}{6})\): \(g(\cos^{-1}(\frac{5}{6})) = 5\cos^{-1}(\frac{5}{6})-6\sin(\cos^{-1}(\frac{5}{6}))\). Using \(\sin(\cos^{-1}(x))=\sqrt{1 - x^{2}}\), we have \(g(\cos^{-1}(\frac{5}{6}))=5\cos^{-1}(\frac{5}{6})-6\sqrt{1 - (\frac{5}{6})^{2}}\approx5\times0.585-6\times\frac{\sqrt{11}}{6}\approx2.925 - 3.317=- 0.392\).
• At \(\theta=\pi\): \(g(\pi)=5\pi-6\sin(\pi)=5\pi\approx15.71\).

Answer:

Minimum value \(=-0.392\), Maximum value \(=15.71\)