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queennae
Oct 14, 2009, 11:16 AM
how to use interval notation to find the domain?
1.f(x) =
1
___________
(x+2) (x+1)

Unknown008
Oct 15, 2009, 12:01 AM
Well, if you draw the graph, you'll find that there are two asymptotes, one at the line x = -2 and the other one at x = -1.

Domain is therefore:
f(x) < -2
-2 < f(x) < -1
f(x) > -1

Clough
Oct 15, 2009, 12:54 AM
Hi, Unknown008!

Was the answer that you gave a direct answer to the question or a way to help queennae find the answer on their own, please?

I'm just curious and really have no idea about these sort of things.

Thanks!

Unknown008
Oct 15, 2009, 01:01 AM
I would have posted the shape of the graph... but that's a little difficult for me just now. My graphing program is on my 'Windows XP' section of my computer, and not on my Linux section, with which I connect to the internet. Perhaps galactus or ebaines will be able to post the graph before I do... :o

teferi
Oct 15, 2009, 01:03 AM
The domain of this question is R/(-2,-1)

Clough
Oct 15, 2009, 01:08 AM
the domain of this question is R/(-2,-1)

Hi, teferi!

Is that also a direct answer to the question, please?

Thanks!

Unknown008
Oct 15, 2009, 01:41 AM
Ok, let me try this:

25583

Do you see that the graph exists for all values of x, except for x = -2 and x = -1?

This is because when you put x = -2 in the equation, your denominator becomes zero, and 1/0 gives infinity. The same happens if you put x = -1.