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Math Help - Finding the domain of a function

  1. #1
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    Finding the domain of a function

    f(x)=x/\sqrt{39x^2-7}

    For clarification purposes, in word form this would be: the fraction of x over the square root of 39 x squared minus seven.

    It would help alot if someone could help me find and write the answer Set Builder Notation (ex: [-5, infinity)). Thanks in advance for your assistance.
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  2. #2
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    Quote Originally Posted by currypuff View Post
    f(x)=x/\sqrt{39x^2-7}

    For clarification purposes, in word form this would be: the fraction of x over the square root of 39 x squared minus seven.

    It would help alot if someone could help me find and write the answer Set Builder Notation (ex: [-5, infinity)). Thanks in advance for your assistance.
    The domain is all values of x that the function takes on. In this scenario, not only must the square root not be negative, but it cannot be zero, so:

    39x^2-7>0

    39x^2>7

    x^2>\frac{7}{39}

    x>\sqrt{\frac{7}{39}}
    x<-\sqrt{\frac{7}{39}}

    x: \left(-\infty, -\sqrt{\frac{7}{39}}\right) \cup \left(\sqrt{\frac{7}{39}}, \infty \right)
    Last edited by colby2152; January 30th 2008 at 04:42 AM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by colby2152 View Post
    The domain is all values of x that the function takes on. In this scenario, not only must the square root not be negative, but it cannot be zero, so:

    39x^2-7>0

    39x^2>7

    x^2>\frac{7}{39}

    x>\sqrt{\frac{7}{39}}
    x<-\sqrt{\frac{7}{39}}

    x: (-\infty, -\sqrt{\frac{7}{39}}) \cap (\sqrt{\frac{7}{39}}, \infty)
    use \left( and \right) so that the brackets encompass everything. so you would get:

    x: \left(-\infty, -\sqrt{\frac{7}{39}} \right) \cap  \left(\sqrt{\frac{7}{39}}, \infty \right)

    which looks nicer

    ...well, sort of, that answer's ugly either way...
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    use \left( and \right) so that the brackets encompass everything. so you would get:

    x: \left(-\infty, -\sqrt{\frac{7}{39}} \right) \cap  \left(\sqrt{\frac{7}{39}}, \infty \right)

    which looks nicer

    ...well, sort of, that answer's ugly either way...
    it should be x: \left(-\infty, -\sqrt{\frac{7}{39}} \right) \cup  \left(\sqrt{\frac{7}{39}}, \infty \right) ?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by curvature View Post
    it should be x: \left(-\infty, -\sqrt{\frac{7}{39}} \right) \cup  \left(\sqrt{\frac{7}{39}}, \infty \right) ?
    yes, that's right. the union.

    thanks for spotting it, i didn't see that, i just copied the code Colby used
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    yes, that's right. the union.

    thanks for spotting it, i didn't see that, i just copied the code Colby used
    Cup and cap!

    How does it look now Jhevon?
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by colby2152 View Post
    Cup and cap!

    How does it look now Jhevon?
    looks good!

    (though i would use \in instead of : but the answer is understood, so no big deal)
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