Finding horizontal tangents on an interval

Click For Summary
SUMMARY

The discussion focuses on finding the values of x in the interval [-π/2, π/2] where the graph has horizontal tangents, specifically using the derivatives f'(x) = 9cos(x) - 2sin(x) and f'(x) = -5csc(x)(5cot(x) - csc(x)). The critical points are determined by solving the equations 9cos(x) = 2sin(x) and 5cot(x) = csc(x). The solution involves using the arctan function to find x such that tan(x) = 9/2, which is not a simple solution but can be approached using inverse trigonometric functions.

PREREQUISITES
  • Understanding of trigonometric functions and their derivatives
  • Knowledge of inverse trigonometric functions, specifically arctan
  • Familiarity with solving equations involving trigonometric identities
  • Basic calculus concepts, particularly differentiation
NEXT STEPS
  • Learn how to solve trigonometric equations using inverse functions
  • Study the properties and applications of the arctan function
  • Explore the concept of horizontal tangents in calculus
  • Review techniques for finding critical points in functions
USEFUL FOR

Students studying calculus, particularly those focusing on trigonometric functions and their applications in finding horizontal tangents, as well as educators looking for examples to illustrate these concepts.

noelwolfe
Messages
3
Reaction score
0

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?
 
Physics news on Phys.org
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!
 

Similar threads

Replies
3
Views
2K
  • · Replies 27 ·
Replies
27
Views
4K
Replies
3
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K