Question

Find any critical numbers of the function. 
f(θ) = 6sec θ + 3tan θ, 0 < θ < 2π 
f(θ) = 6sec θ + 3tan θ, 0 < θ < 2πθ =?
Enter your answers as a comma-separated list.

Accepted Answer

You need to remember that the function f(theta)  has critical values if the equation f'(theta)=0  has solutions, hence, you neded to evaluate f'(theta)  such that:
f'(theta) = (6sec theta + 3tan theta)'
You should remember that sec theta = 1/cos theta , hence, you need to use quotient rule to evaluate (sec theta)' such that:
(sec theta)' = (1'*cos theta - 1*(cos theta)')/(cos^2 theta)
(sec theta)' = -(-sin theta)/(cos^2 theta)=> (sec theta)' = (sin theta)/(cos^2 theta)
f'(theta) = 6sin theta/(cos^2 theta) + 3/(cos^2 theta)
Factoring out 3/(cos^2 theta)  yields:
f'(theta) = 3/(cos^2 theta)(2sin theta + 1)
You need to solve for theta the equation f'(theta) = 0  such that:
3/(cos^2 theta)(2sin theta + 1) = 0
Since 3/(cos^2 theta)!=0 => 2sin theta + 1 = 0  such that:
2sin theta + 1 = 0 => 2sin theta=- 1 => sin theta = -1/2
You need to remember that the sine function is negative in quadrants 3 and 4, hence, you will have the following solutions to the equation sin theta = -1/2  such that:
sin theta = -1/2 => theta = pi+pi/6 => theta = 7pi/6 (quadrant 3)
theta = 2pi-pi/6 => theta = 11pi/6  (quadrant 4)
Hence, evaluating the critical values of the function yields theta = 7pi/6  and theta = 11pi/6 .

Math Questions and Answers

Find any critical numbers of the function. f(θ) = 6sec θ + 3tan θ, 0 < θ < 2π f(θ) = 6sec θ + 3tan θ, 0 < θ < 2πθ =? Enter your answers as a comma-separated list.

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