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Finding Critical Numbers

by cdhotfire
Tags: critical, numbers
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cdhotfire
#1
Nov8-05, 08:19 PM
P: 193
Well, I got this equation [itex]f(x)=\frac{2\sin{2x}}{x}[/itex] [itex] [-\pi,\pi][/itex]
So I took the 1st derivative, [itex]f'(x)=\frac{2(2x\cos{2x} - \sin{2x})}{x^2}[/itex]
Then I set that equal to 0, and got [itex]0=2x\cos{2x} - \sin{2x}[/itex]
But I do not see how to get the critical numbers, I also tried to do double angle, but that just resulted in more pain.
Any help, would be appreciated.
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cdhotfire
#2
Nov8-05, 11:52 PM
P: 193
Please.
Tide
#3
Nov9-05, 12:39 AM
Sci Advisor
HW Helper
P: 3,144
Haven't I seen this somewhere before? :)

cdhotfire
#4
Nov9-05, 12:42 AM
P: 193
Finding Critical Numbers

Quote Quote by Tide
Haven't I seen this somewhere before? :)
ya, i know, i though no one visited the mathematics section, because i was only able to join it by using the search tool. Also, i put a note for someone to delete my post in the other section.

btw, i posted back on you reply.


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