Register to reply 
Finding Critical Numbers 
Share this thread: 
#1
Nov805, 08:19 PM

P: 193

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. 


#2
Nov805, 11:52 PM

P: 193

Please.



#3
Nov905, 12:39 AM

Sci Advisor
HW Helper
P: 3,144

Haven't I seen this somewhere before? :)



#4
Nov905, 12:42 AM

P: 193

Finding Critical Numbers
btw, i posted back on you reply. 


Register to reply 
Related Discussions  
Finding the critical numbers  Calculus & Beyond Homework  2  
Critical numbers  Calculus & Beyond Homework  2  
Finding Critical Numbers  Calculus & Beyond Homework  7  
Finding Critical numbers  Introductory Physics Homework  5 