Question

Find any critical numbers of the function. $$ \begin{array}{l} f(\theta)=2 \sec \theta+\tan \theta \\ 0<\theta<2 \pi \end{array} $$

Short answer

The critical numbers of the function \(f(\theta)=2 \sec \theta+\tan \theta\) on the interval \(0<\theta<2 \pi\) are \(\theta=\pi/2\) and \(\theta=3\pi/2\).

Step-by-step solution

Compute the derivative of the function

The function \(f(\theta)=2 \sec \theta+\tan \theta\) involves the functions \(\sec \theta\) and \(\tan \theta\). The derivative of \(\sec x\) is \(\sec x \tan x\) and the derivative of \(\tan x\) is \(\sec^2 x\). Using those, the derivative of \(f(\theta)\) becomes \(f'(\theta)= 2 \sec \theta \tan \theta+\sec^2 \theta\).

Locate where the derivative is undefined

The function \(\sec \theta\) is undefined at \(\pi/2 + k\pi\), where \(k\) is an integer. Within the given interval \(0<\theta<2 \pi\), \(\sec \theta\) will be undefined at \(\theta=\pi/2\) and \(\theta=3\pi/2\). Hence these are critical numbers.

Find out when the derivative equals zero

Setting the derivative equal to zero, we get \(2 \sec \theta \tan \theta + \sec^2 \theta = 0\). From the equation, we can deduce that \(\sec \theta\) cannot equal \(0\) because sec is never equal to \(0\) by definition, which means the critical points occur when both \(2\tan \theta = -\sec \theta\), this leads to the equation \(\tan \theta = -1/2\). Solving for \(\theta\), we get \(\theta = \arctan(-1/2)\). However, this value is not within the given range. Hence, \(0<\theta<2 \pi\) has no solution.

Combine all critical numbers

The critical numbers of the function are the values where the derivative is either zero or undefined. From our calculations, we find that \(\theta=\pi/2\) and \(\theta=3\pi/2\) are critical numbers.