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.
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.:tongue2: