Finding horizontal tangents on an interval

  1. 1. The problem statement, all variables and given/known data

    Determine all x in [-pi/2, pi/2] at which the graph has horizontal tangents.

    2. Relevant equations

    1.) f'(x)= 9cos(x)-2sin(x)
    2.) f'(x)= -5csc(x) (5cot(x)-csc(x))

    3. The attempt at a solution

    1.) 9cos(x)=2sin(x)
    9/2=sin(x)/cos(x)
    9/2=tan(x)
    and then I'm not sure what to do with 9/2?

    2.) 5cot(x)=csc(x)
    5= csc(x)/cot(x)
    5=1/cos(x)
    cos(x)=1/5
    same problem here... what do I do with 1/5?
     
  2. jcsd
  3. hotvette

    hotvette 931
    Homework Helper

    You are looking for the values of x such that tan(x) = 9/2. Remember inverse trigonometric functions (i.e. sine / arcsine, cos / arccos, tan / arctan, etc.)?
     
  4. I'm sorry, I still don't know--I'm not aware of a "simple" solution to tan(x)=9/2. How would I go about finding the solution?
     
  5. Look up the arctan function (also called inverse tangent) and its uses.
     
  6. Got it. Thanks!
     
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