Finding horizontal tangents on an interval

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Homework Help Overview

The problem involves finding all values of x in the interval [-π/2, π/2] where the graph has horizontal tangents, specifically through the analysis of the derivative functions provided.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss setting the derivatives equal to zero to find horizontal tangents, with attempts to solve equations involving trigonometric functions like tan(x) and cot(x). Questions arise about how to proceed after obtaining values like 9/2 and 1/5.

Discussion Status

Some participants have suggested using inverse trigonometric functions to find solutions, while others express uncertainty about the simplicity of solving tan(x) = 9/2. Guidance has been offered regarding the arctan function, indicating a potential direction for further exploration.

Contextual Notes

Participants note the challenge of finding explicit solutions within the given interval and the limitations of their current understanding of inverse trigonometric functions.

noelwolfe
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Homework Statement



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

Homework Equations



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

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

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