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

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

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

2. 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?

3. 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.

4. 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

5. 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 $\displaystyle (-\infty, -3)\cup(-3, 3)\cup(3, \infty)$.