Finding Critical Numbers

  • Thread starter cdhotfire
  • Start date
  • #1
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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.:mad:
Any help, would be appreciated.:smile:
 

Answers and Replies

  • #2
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Please.:uhh:
 
  • #3
Tide
Science Advisor
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Haven't I seen this somewhere before? :)
 
  • #4
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Tide said:
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.:smile:

btw, i posted back on you reply.:tongue2:
 

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