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Math Help - Find the domain of this function?

  1. #1
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    Find the domain of this function?

    x/abs(x^2 - 4)

    Would there be a vertical asymptote at only x = -2?
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  2. #2
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    Re: Find the domain of this function?

    Quote Originally Posted by explodingtoenails View Post
    x/abs(x^2 - 4)

    Would there be a vertical asymptote at only x = -2?
    There are vertical asymptotes at x = 2 and x = -2, and the function is not defined at only these two values of x.
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  3. #3
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    Re: Find the domain of this function?

    Why x = 2? I thought the absolute value bars would negate the (x - 2)?
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    Re: Find the domain of this function?

    Quote Originally Posted by explodingtoenails View Post
    Why x = 2? I thought the absolute value bars would negate the (x - 2)?
    Why did you think that? Note that abs(2^2-4) = 0.
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  5. #5
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    Re: Find the domain of this function?

    I thought if you set abs(x - 2) to 0, it would be like this:

    abs(x - 2) = 0
    abs(x) = -abs(2)
    x = -2
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  6. #6
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    Re: Find the domain of this function?

    Quote Originally Posted by explodingtoenails View Post
    I thought if you set abs(x - 2) to 0, it would be like this:

    abs(x - 2) = 0
    abs(x) = -abs(2)
    x = -2
    abs(x^2-4)=0

    \implies x^2-4=0

    \implies (x+2)(x-2)=0

    \implies x=\pm 2
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  7. #7
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    Re: Find the domain of this function?

    Why do the absolute value bars not do anything?
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  8. #8
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    Re: Find the domain of this function?

    Quote Originally Posted by explodingtoenails View Post
    Why do the absolute value bars not do anything?
    The only number whose absolute value is 0, is 0.
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  9. #9
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    Re: Find the domain of this function?

    So when there are absolute value bars around a binomial, it does nothing?
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    Re: Find the domain of this function?

    Quote Originally Posted by explodingtoenails View Post
    So when there are absolute value bars around a binomial, it does nothing?
    Do you know what absolute value means?
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  11. #11
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    Re: Find the domain of this function?

    Yes. It's the value of the number, except it's positive no matter what. Correct?
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  12. #12
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    Re: Find the domain of this function?

    Quote Originally Posted by explodingtoenails View Post
    Yes. It's the value of the number, except it's positive no matter what. Correct?
    Correct.

    abs(x)=m\implies x=\pm m
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  13. #13
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    Re: Find the domain of this function?

    abs(x^2 - 4) = abs[(x - 2) (x + 2)]

    For (x - 2), I keep thinking the absolute value bars would make the - 2 into a + 2. So I keep doing this:

    abs(x - 2) = 5
    abs(x) + abs(-2) = 5
    abs(x) = 5 - 2

    Why would the abs bars not doing anything to (x - 2)?
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  14. #14
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    Re: Find the domain of this function?

    Quote Originally Posted by explodingtoenails View Post
    abs(x^2 - 4) = abs[(x - 2) (x + 2)]

    For (x - 2), I keep thinking the absolute value bars would make the - 2 into a + 2. So I keep doing this:

    abs(x - 2) = 5
    abs(x) + abs(-2) = 5
    abs(x) = 5 - 2

    Why would the abs bars not doing anything to (x - 2)?
    abs(a+b)\neq abs(a)+abs(b)

    For example, abs (2-1)=1, whereas abs(2)+abs(-1)=3.
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  15. #15
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    Re: Find the domain of this function?

    So when would abs bars actually affect the value of a binomial?
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