
#1
Nov805, 08:19 PM

P: 193

Well, I got this equation [latex]f(x)=\frac{2\sin{2x}}{x}[/latex] [latex] [\pi,\pi][/latex]
So I took the 1st derivative, [latex]f'(x)=\frac{2(2x\cos{2x}  \sin{2x})}{x^2}[/latex] Then I set that equal to 0, and got [latex]0=2x\cos{2x}  \sin{2x}[/latex] 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. 



#2
Nov805, 11:52 PM

P: 193

Please.




#3
Nov905, 12:39 AM

Sci Advisor
HW Helper
P: 3,149

Haven't I seen this somewhere before? :)




#4
Nov905, 12:42 AM

P: 193

Finding Critical Numbersbtw, i posted back on you reply. 


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