text: # 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.
  
  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...
  
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  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` .
  
  Approved by eNotes Editorial Team
explanation: The document contains a math question about finding the critical numbers of a given function. The text also provides a detailed solution to the problem, including the steps to find the derivative of the function and solve for the critical points.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 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.
  
  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...
  
  Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
  
  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` .
  
  Approved by eNotes Editorial Team
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question about finding the critical numbers of a given function. The text also provides a detailed solution to the problem, including the steps to find the derivative of the function and solve for the critical points.
  
  Label: question-and-answer