# how do you write the domain of function in interval notation?

• Feb 15th 2009, 05:31 PM
lsnyder
how do you write the domain of function in interval notation?
how do you write the domain of function in interval notation?

• Feb 15th 2009, 05:39 PM
mollymcf2009
Quote:

Originally Posted by lsnyder
how do you write the domain of function in interval notation?

Find where the function exists first and find your asymptotes/holes. I didn't work this out, you need to do this. But interval notation looks something like this:

$\displaystyle (-\infty, -2) U (-2, \infty)$

Or, -infinity to -2, union, -2 to infinity. -2 would be a vertical asymptote or a hole. If you had more than one asymptote, let's say at x=6, you would just add it in as another interval like this:

$\displaystyle (-\infty, -2) U (-2, 6) U (6, \infty)$

Can you figure out yours?
• Feb 15th 2009, 05:45 PM
Reckoner
Quote:

Originally Posted by mollymcf2009
But interval notation looks something like this:

You can write the union symbol with \cup (and similarly, \cap gives intersection): $\displaystyle (-\infty,-2)\cup(-2,\infty)\text.$ A minor thing, but I thought it might be helpful to point out.
• Feb 15th 2009, 05:51 PM
lsnyder
I am not understanding. Mainly because my teacher is from India so its hard to understand him. So he might of taught this but he wasn't very clear. Mainly I don't understand what the hole is or acyetopes
• Feb 15th 2009, 07:18 PM
VENI
Quote:

Originally Posted by lsnyder
I am not understanding. Mainly because my teacher is from India so its hard to understand him. So he might of taught this but he wasn't very clear. Mainly I don't understand what the hole is or acyetopes

The expression in the denominator of your original problem may equal 0 for a given value of x. So you would be dividing by 0 which is illegal. Therefore you have to find out for what value(s) of x that expression equals 0 by factoring. For example, if the problem is $\displaystyle \frac{x}{x^2 -9}$, I would factor the denominator into $\displaystyle \frac{x}{(x-3)(x+3)}$. Clearly the denominator would be 0 when x = 3 and x = -3. Therefore, the domain for this example would be $\displaystyle (-\infty, -3)\cup(-3, 3)\cup(3, \infty)$.