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Math Help - Finding the range of a relation

  1. #1
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    Finding the range of a relation

    I have troubles in working out the range of the following relations:

    y=x-a, x<0, a>0

    y=(a/x)+a, a>0

    May you please list out the steps because I really want to understand the way~

    Also:

    y=square root x, x>_0 ---> Shouldn't the range be "all real numbers"? I saw the answer is [0,infinity[

    (Because for example, assuming that x=81, so y=square root 81, which equals either (-9) or (9), meaning that negative numbers are also accepted, so I think it is "all real numbers") Am I right?
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  2. #2
    Super Member TheChaz's Avatar
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    Re: Finding the range of a relation

    As for the square root one, there is a major misunderstanding.

    If you had y^2 = x, then yes, your range of y values would be all reals.

    But you have the y = \sqrt x, which is understood to be THE positive square root (singular) of x.

    So you will never get a negative output.

    The following is also true, and might help:

    If a^2 = 4 then a = +/- 2

    If b = \sqrt 9, then b = 3.
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  3. #3
    Member anonimnystefy's Avatar
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    Re: Finding the range of a relation

    hi kevinlam2490

    you have thewrong idea.square root of 81 is just 9,because its value is always positive.

    it is much different than solving the equation x^2=81.
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