Well, I got this equation [itex]f(x)=\frac{2\sin{2x}}{x}[/itex] [itex] [-\pi,\pi][/itex](adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums - The Fusion of Science and Community**

# Finding Critical Numbers

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Finding Critical Numbers

Loading...

**Physics Forums - The Fusion of Science and Community**