Finding horizontal tangents on an interval


by noelwolfe
Tags: horizontal, interval, tangents
noelwolfe
noelwolfe is offline
#1
Oct22-10, 01:54 PM
P: 3
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?
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hotvette
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#2
Oct22-10, 03:13 PM
HW Helper
P: 930
Quote Quote by noelwolfe View Post
9/2=tan(x)
and then I'm not sure what to do with 9/2?
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.)?
noelwolfe
noelwolfe is offline
#3
Oct22-10, 03:53 PM
P: 3
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?

einsteinoid
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#4
Oct22-10, 04:39 PM
P: 42

Finding horizontal tangents on an interval


Look up the arctan function (also called inverse tangent) and its uses.
noelwolfe
noelwolfe is offline
#5
Oct23-10, 04:52 PM
P: 3
Got it. Thanks!


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