Register to reply

Finding Critical Numbers

by cdhotfire
Tags: critical, numbers
Share this thread:
Nov8-05, 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.
Phys.Org News Partner Science news on
NASA team lays plans to observe new worlds
IHEP in China has ambitions for Higgs factory
Spinach could lead to alternative energy more powerful than Popeye
Nov8-05, 11:52 PM
P: 193
Nov9-05, 12:39 AM
Sci Advisor
HW Helper
P: 3,146
Haven't I seen this somewhere before? :)

Nov9-05, 12:42 AM
P: 193
Finding Critical Numbers

Quote Quote by Tide
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.

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