Trig graph problem

Your equation is showing up funny on my browser.

Is it supposed to be this:

[tex]
y=cos(3x-1)
[/tex]

or this:

[tex]
y=cos(\frac{1}{3x-1})
[/tex]

?

 

y=cos¶/3(x-1)???????????????????????????????????????????????????
Neither can I. Please type in the CORRECT formula.

 

I think it's something else!

 

Ok, so it is:
[tex]y=\cos(\frac{\pi}{3x-1})[/tex]

Hints:
1. Find the point of singularity, where y is not defined.
2. Determine the set of points were y is -1, 0, 1 respectively.
In particular, see what this implies for the function's behaviour very close to the singularity found in 1.

 

Still not OK, and I suspect that it's because you aren't placing your parentheses properly.

What you have written is this:

[tex]y=cos(\frac{\pi}{3x-1})[/tex].

Now that's a perfectly legitimate function, but I strongly suspect that it is not the one you are supposed to graph, simply because it's too difficult for a trigonometry course. I am inclined to think that you really meant this:

[tex]y=cos[\frac{\pi}{3}(x-1)][/tex].

Am I right?

 

I don't know what that means.

 

özge:
If Tom Mattson is right, here's how you could have written this properly:
y=cos(pi*(x-1)/3)

 

This can't be right. cos without an argument makes no sense.

 

Do you know what the cosine function is??
What is its argument?

 

Alrighty then!

Here are a couple of tips to get you started.

*Examine the graph of [itex]y=cos(x)[/itex].

*Note that if you have the graph of some function [itex]y=f(x)[/itex] then you can obtain the graph of [itex]y=f(x+a)[/itex] by shifting the graph of [itex]y=f(x)[/itex] horizontally by [itex]a[/itex] units. You move to the right if [itex]a<0[/itex] and to the left if [itex]a>0[/itex].

*The period [itex]T[/itex] of a cosine function [itex]y=A\cos[k(x+a)][/itex] is [itex]T=\frac{2\pi}{k}[/itex].

Give it a shot and see how you do, OK?