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Math Help - Confusing range of a Relation.

  1. #1
    Junior Member Kaloda's Avatar
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    Confusing range of a Relation.

    Okay. So our assignment is to get the domain, range, intercepts, etc. of the relation (also a function but not expressed as one)
    5y(x-3)(x-1)=2(5x+3). I didn't have a problem at all till I realize that I got (-\infty,-5]\cup[-1/5,0)\cup(0,+\infty)
    (I EXCLUDED ZERO because when I solved for x, y is in the DENOMINATOR) as the range and have an x-intercept (-3/5,0)
    which is a contradiction with the range.

    Now, it leaves me with a dilemma because I can't include zero in the range because x will be undefined (at least if you use quadratic formula by manipulating the given equation which I think alters its nature I presume), and I can't just simply remove the x-intercept because of the same reason.

    Here's my question, do I include zero in the range OR remove the x-intercept OR just leave my answer as it is.
    And please explain why.
    Last edited by Kaloda; June 28th 2012 at 04:34 PM.
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  2. #2
    Master Of Puppets
    pickslides's Avatar
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    Re: Confusing range of a Relation.

    It seems to me that y \in \mathbb{R} and x \in (-\infty, 1) \cup ( 1,3) \cup (3,\infty) , your x-intercept is correct. Include y=0 in the range as it gives a corresponding x-value.

    What about when  x=0 ?
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  3. #3
    Junior Member Kaloda's Avatar
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    Re: Confusing range of a Relation.

    Quote Originally Posted by pickslides View Post
    It seems to me that y \in \mathbb{R}
    Huh? Do you mean that the range is all real numbers? I think you must recheck it (or maybe I'm the one who is wrong).
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  4. #4
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    Re: Confusing range of a Relation.

    Hello, Kaloda!

    I think you're confusing yourself . . .


    Find the domain, range, intercepts, etc. of the relation: . 5y(x-3)(x-1)\:=\:2(5x+3)

    We have: . y \;=\;\frac{2}{5}\cdot\frac{5x+3}{(x-3)(x-1)}

    If y = 0\!:\;x = \text{-}\tfrac{3}{5}\;\hdots\;x\text{-intercept: }\left(\text{-}\tfrac{3}{5},\:0\right)

    If x = 0\!:\;y = \tfrac{2}{5}\;\hdots\;y\text{-intercept: }\left(0,\:\tfrac{2}{5}\right)


    Horizontal asymptote: . y = 0
    Vertical asymptotes: . x = 1,\;x = 3


    Domain: . x \ne 1,\:x \ne 3
    . . . . . . . (\text{-}\infty,\,1)\,\cup\,(1,\,3)\,\cup\,(3,\,\infty)

    Range: difficult to determine without using Calculus.


    The graph looks something like this:

    Code:
                            |
                            | :         :
                            | :         :*
                            |*:         :
                            | :         : *
                            | :         :  *
                            oB:         :     *
                          A | :         :          *
      - -------------------o+-:---------:--------------
            *             * | :1        :3
                  *     *   | :         :
                     o      | :         :
                     D      | :    C    :
                            | :    o    :
                            | :  *   *  :
                            | : *     * :
                            | :         :
                            | :*        *:
    A\text{ and }B\text{ are the }x\text{- and }y\text{-intercepts, respectively.}

    C\text{ has coordinates: }\,\left(\tfrac{9}{5},\text{-}5\right)
    D\text{ has coordinates: }\,\left(\text{-}3,\,\text{-}\tfrac{1}{5}\right)

    Therefore, the range is: . \left(\text{-}\infty,\,\text{-}5)\,\cup\,\leeft(\text{-}\tfrac{1}{5},\,\infty\right)
    Thanks from Kaloda
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