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Math Help - help with multivariable domain

  1. #1
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    help with multivariable domain

    i want to find and sketch the domain of

    f(x,y) = 1/[(x+y)(2-x)]^

    so what I did

    2-x > 0, thus x <2

    x+y > 0, so x > -y

    that right?
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  2. #2
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    Re: help with multivariable domain

    Quote Originally Posted by Kuma View Post
    i want to find and sketch the domain of

    f(x,y) = 1/[(x+y)(2-x)]^

    so what I did

    2-x > 0, thus x <2

    x+y > 0, so x > -y

    that right?
    That is correct, x < 2, and x > -y .
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  3. #3
    A Plied Mathematician
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    Re: help with multivariable domain

    What about x > 2 and x < -y^{2}?
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    Re: help with multivariable domain

    Quote Originally Posted by Ackbeet View Post
    What about x > 2 and x < -y^{2}?
    How can you have x take on both those values at the same time?
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  5. #5
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    Re: help with multivariable domain

    Quote Originally Posted by Kuma View Post
    How can you have x take on both those values at the same time?
    It's impossible. y is ≥ 0 , so -y ≤ 0 . Clearly 2 > 0 .

    I don't know what Ackbeet was thinking.
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  6. #6
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    Re: help with multivariable domain

    Quote Originally Posted by Kuma View Post
    How can you have x take on both those values at the same time?
    It is necessary to express all the possible cases:

    \exists f(x,y)\Leftrightarrow [(x<2)\;\wedge \;(x>-y^2)]\;\vee\;[(x>2)\;\wedge\; (x<-y^2)]

    Another thing is that in this problem , the second case (x>2)\;\wedge\; (x<-y^2) represents the empty set. That is the reason of the Ackbeet's remark.
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  7. #7
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    Re: help with multivariable domain

    Quote Originally Posted by SammyS View Post
    It's impossible. y is ≥ 0 , so -y ≤ 0 . Clearly 2 > 0 .

    I don't know what Ackbeet was thinking.
    Yeah, all I wanted to do was point out that the product of two numbers can be positive if both of the multiplicands are negative. I hadn't gotten so far as to determine if there actually were any numbers in the set or not. So the answer is, I wasn't thinking as deeply as was necessary!
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