- #1
cdhotfire
- 193
- 0
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.
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.