Question

find the critical numbers of the function. (enter your answers as a comma - separated list. use n to denote any arbitrary integer values. if an answer does not exist, enter dne.)
( g(\theta)=16 \theta-4 \tan (\theta) )
( \theta= )

Explanation:

Step1: Differentiate the function

The derivative of \(g(\theta)=16\theta - 4\tan(\theta)\) is \(g'(\theta)=16-4\sec^{2}(\theta)\) (using the rules \(\frac{d}{d\theta}(a\theta)=a\) and \(\frac{d}{d\theta}(\tan\theta)=\sec^{2}\theta\)).

Step2: Set the derivative equal to zero

Set \(g'(\theta) = 0\), so \(16-4\sec^{2}(\theta)=0\).
First, divide both sides by \(4\): \(4-\sec^{2}(\theta)=0\).
Since \(\sec^{2}\theta=\frac{1}{\cos^{2}\theta}\), we have \(4-\frac{1}{\cos^{2}\theta}=0\).
Then, \(\frac{1}{\cos^{2}\theta}=4\), so \(\cos^{2}\theta=\frac{1}{4}\), and \(\cos\theta=\pm\frac{1}{2}\).

Step3: Solve for \(\theta\)

When \(\cos\theta=\frac{1}{2}\), \(\theta = 2n\pi\pm\frac{\pi}{3}\) (where \(n\in\mathbb{Z}\)).
When \(\cos\theta=-\frac{1}{2}\), \(\theta = 2n\pi\pm\frac{2\pi}{3}\) (where \(n\in\mathbb{Z}\)).

Answer:

\(\theta = 2n\pi\pm\frac{\pi}{3},2n\pi\pm\frac{2\pi}{3}\)