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Finding Critical Numbers

  1. Nov 8, 2005 #1
    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:
  2. jcsd
  3. Nov 8, 2005 #2
  4. Nov 9, 2005 #3


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    Haven't I seen this somewhere before? :)
  5. Nov 9, 2005 #4
    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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