- #1

- 193

- 0

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.