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Critical Numbers

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Pupil
Karma Points: 95
(Thomas A. Edison State College)
Respect (100%):
Date Posted: 7/24/2008 9:56:00 AM  Status: Closed
Critical Numbers
Course Textbook Chapter Problem
Calculus Calculus: Early Transcendental Functions (4th) by Edwards, Hostetler, Larson N/A N/A
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Mentor
Karma Points: 600
Date Posted: 7/24/2008 10:20:44 AM  Status: Live
Asker's Rating: Not Helpful   
Response:
OK, let's start by differentiating...
f'(x) = 2*sin(3x)*cos(3x)*3 - 3*sin(3x)
=6sin(3x)cos(3x)-3sin(3x)
Equate the derivative to zero to find critical numbers.
6sin(3x)cos(3x)-3sin(3x)=0 ==> 6sin(3x)cos(3x)=3sin(3x). Cancel 3sin(3x) under the condition that sin(3x) is not equal to zero, i.e. x is not 0 or nπ/3. Now, 2cos(3x) = 1, cos(3x) = 1/2, 3x = π/3 + 2nπ/3 = (2n+1)π/3 (Note the period division by 3 from trig.) Since  nπ/3 is not a solution, (2n+1)π/3 can't be a solution either. So, there are no critical numbers except where the derivative does not exist, which is at no given point in the domain.
Enjoy...

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Pupil
Karma Points: 95
(Thomas A. Edison State College)
Date Posted: 7/24/2008 10:25:17 AM  Status: Live
Asker's Rating: N/A-Posted by Person Asking Question   
Response:
well i am told there are at  3 critical numbers, and I know is one of them

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Oracle
Karma Points: 55,117
Date Posted: 7/24/2008 10:52:39 AM  Status: Live
Asker's Rating: Lifesaver   
Response:
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Testing Critical Numbers at , , , in the domain from
Intervals 0<x<pi/9 pi/9<x<5pi/9 5pi/9<x<2pi
Test Values x=0.2 x=1 x=3
Sign of F'(x) F'(x)=1.10>0 F'(x)=-1.126<0 F'(x)= -3.489<0
Conclusion Increasing Decreasing Decreasing
Decreasing Intervals: (pi/9<x<5pi/9), (5pi/9<x<2pi)
Increasing Interval: (0<x<pi/9)
Relative Maximum at x= pi/9
Relative Minimum: None
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
KorTig's Comment:
Thanks - Just what I was looking for

 



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