Finding Critical Numbers


by cdhotfire
Tags: critical, numbers
cdhotfire
cdhotfire is offline
#1
Nov8-05, 08:19 PM
P: 193
Well, I got this equation [latex]f(x)=\frac{2\sin{2x}}{x}[/latex] [latex] [-\pi,\pi][/latex]
So I took the 1st derivative, [latex]f'(x)=\frac{2(2x\cos{2x} - \sin{2x})}{x^2}[/latex]
Then I set that equal to 0, and got [latex]0=2x\cos{2x} - \sin{2x}[/latex]
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
cdhotfire is offline
#2
Nov8-05, 11:52 PM
P: 193
Please.
Tide
Tide is offline
#3
Nov9-05, 12:39 AM
Sci Advisor
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P: 3,149
Haven't I seen this somewhere before? :)

cdhotfire
cdhotfire is offline
#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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