Critical Numbers for f(x)=2sin(2x)/x on [-pi,pi]

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Homework Help Overview

The discussion revolves around finding critical numbers for the function f(x) = 2sin(2x)/x over the interval [-π, π]. Participants are exploring the implications of the first derivative and the conditions for critical points.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find critical numbers by setting the first derivative equal to zero, leading to the equation 0 = 2xcos(2x) - sin(2x). There is uncertainty about how to proceed from this point. Other participants express curiosity about the topic and the visibility of the mathematics section.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and expressing a need for assistance. There is no clear consensus or resolution yet, but engagement is present as participants interact with each other.

Contextual Notes

There is mention of difficulty in finding critical numbers and a reference to previous encounters with similar problems, indicating a potential familiarity with the topic among participants.

cdhotfire
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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.:mad:
Any help, would be appreciated.:smile:
 
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Please.:rolleyes:
 
Haven't I seen this somewhere before? :)
 
Tide said:
Haven't I seen this somewhere before? :)

ya, i know, i though no one visited the mathematics section, because i was only able to join it by using the search tool. Also, i put a note for someone to delete my post in the other section.:smile:

btw, i posted back on you reply.:-p
 

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