# Finding Critical Numbers

1. Nov 8, 2005

### cdhotfire

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

edit: Can an mod please remove this post, seems I posted in the wrong section. :(

Last edited: Nov 8, 2005
2. Nov 9, 2005

### Tide

Either x = 0 or $\tan 2x = 2x$. You won't solve the latter analytically.

3. Nov 9, 2005

### cdhotfire

Thats what I got, but I couldnt figure out how to solve the tan 2x = 2x. How would I be able to solve it?

I tried to graph it, but the points that came out, did not work in the equation.:yuck:

So, do I go with x=0 only?

edit: my teacher said, I was gonna use decimals on this problem.

edit: meh, the graph shows more crit points, it seems i need to solve the 2x=tan2x

Last edited: Nov 9, 2005
4. Nov 9, 2005

### TD

As Tide said, tan(2x) = 2x isn't solvable analytically which is probably why your teacher told you to use decimals. Of course, due to the periodicy of the tangent there will be an infinite number of solutions but you only have to consider the ones in the interval given.

5. Nov 9, 2005

6. Nov 9, 2005

### Tide

Graphing will work but you're limited severely in the accuracy unless you have a good graphing program that will zoom in as deeply as you need to pick off the numbers.

A good Newton-Raphson algorithm will give you the numbers you need.

Or, you could take a shortcut and get a copy of a freebie program called "deadline" which will graph your functions AND give the numerical solutions to a high degree of accuracy! :)

7. Nov 9, 2005

### cdhotfire

hmmm, ive heard of the newton-raphson algorithm, but we never learned it in class. ill check out that prog.

thank you.

8. Nov 9, 2005

### dextercioby

$$f(x)\rightarrow f(u)=4 \ \mbox{sinc} \ u$$
with $u\in [-2\pi,2\pi]$.