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Math Help - SAT HELP: finding range

  1. #1
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    SAT HELP: finding range

    Hello! I already mentioned this in a previous topic, but anyways, I bought an SAT practice book for getting ready for the big test. It was pretty well formated, except it only has the answers/explanations to random questions. I was just hoping that you guys could help me with a question here or there that I didn't know how to solve. Thanks!

    Find the range of the following function:

    y = (12x^2 + 48) / (27-3x^2)
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  2. #2
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    Quote Originally Posted by Mr_Green View Post

    Find the range of the following function:

    y = (12x^2 + 48) / (27-3x^2)
    The following method is my own, nobody seems to use it, thus if you do not understand I will try to find another one.

    We need to find all y such that,
    y=\frac{12x^2+48}{27-3x^2}
    Has a solution in the domain ( x\not = \pm 3).
    Multiply, by the denominator,
    y(27-3x^2)=12x^2+48
    27y-3x^2y=12x^2+48
    (12x^2+3x^2y)=27y-48
    3x^2(4+y)=27y-48
    x^2(4+y)=9y-16
    Note, y\not = -4 because otherwise,
    x^2(4-4)=9(-4)+16---> Contradiction.
    Thus, we can divide by it,
    x^2=\frac{9y-16}{y+4}
    In order to have a solution is x we require that,
    \frac{9y-16}{y+4}\geq 0
    Thus, that happens when,
    9y-16\geq 0 \mbox{ and }y+4>0
    Or,
    9y-16\leq 0 \mbox{ and }y+4<0
    Solving the pair of inequalities,
    y\geq 16/9 \mbox{ and } y>-4 (1)
    Or,
    y\leq 16/9 \mbox{ and } y<-4 (2)
    The first inequality is really saying that,
    y\geq 16/9
    The second inequality is really saying that,
    y<-4.
    Thus, the range is,
    (-\infty,-4)\cup [16/9,\infty)
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  3. #3
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    .......

    Thanks. I apreciate u going over it step by step.

    I like your method as well.
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  4. #4
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    Hello, Mr_Green!

    Another approach . . .


    Find the range of the following function: . y \:= \:\frac{12x^2 + 48}{27-3x^2}

    We see that the graph is symmetric to the y-axis.
    So we need consider only x \geq 0 and "reflect" the graph.

    There is a vertical asymptote at x = 3.

    We see that: . \lim_{x\to3^+}\frac{12(x^2 + 4)}{3(9 - x^2)} \:=\:-\infty

    . . .and that: . \lim_{x\to3^-}\frac{12(x^2+4)}{3(9-x^2)} \:=\:+\infty


    There is a y-intercept \left(0,\,\frac{16}{9}\right) and no x-intercepts.


    Since \lim_{x\to\infty}\frac{12(x^2 + 4)}{3(9 -x^2)} \;=\;\lim_{x\to\infty}\frac{4(1 + \frac{4}{x^2})}{\frac{9}{x^2} - 1} \:=\:\frac{4}{-1} \:=\:-4
    . . there is a horizontal asymptote: . y\,=\,-4


    The graph looks something like this:
    Code:
                      :       |       :
                      :*      |      *:
                      :       |       :
                      : *     |     * :
                      :  *    |    *  :
                      :    *  |  *    :
                      :       *       :
                      :       |16/9   :
                      :       |       :
    ------------------:-------+-------:--------------------
                    -3:       |       :3
                      :       |       :
    - - - - - - - - - : - - - + - - - : - - - - - - - - -
      *               :     -4|       :               *
             *        :       |       :        *
                 *    :       |       :    *
                   *  :       |       :  *
                    * :       |       : *
                      :       |       :
                     *:       |       :*

    The range appears to be below \text{-}4 and above \frac{16}{9}.

    As ThePerfectHacker said: . (\text{-}\infty,\,\text{-}4)\, \cup \left[\frac{16}{9},\,\infty\right)

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