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

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

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



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    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by lsnyder View Post
    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:

    (-\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:

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


    Can you figure out yours?
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    Quote Originally Posted by mollymcf2009 View Post
    But interval notation looks something like this:
    You can write the union symbol with \cup (and similarly, \cap gives intersection): (-\infty,-2)\cup(-2,\infty)\text. A minor thing, but I thought it might be helpful to point out.
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    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
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    Quote Originally Posted by lsnyder View Post
    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 \frac{x}{x^2 -9}, I would factor the denominator into \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 (-\infty, -3)\cup(-3, 3)\cup(3, \infty).
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