# Finding horizontal tangents on an interval

by noelwolfe
Tags: horizontal, interval, tangents
 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?
HW Helper
P: 928
 Quote by noelwolfe 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.)?
 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?
P: 42

## Finding horizontal tangents on an interval

Look up the arctan function (also called inverse tangent) and its uses.
 P: 3 Got it. Thanks!

 Related Discussions Calculus & Beyond Homework 3 Calculus & Beyond Homework 4 Calculus & Beyond Homework 4 Calculus 14 Calculus & Beyond Homework 30