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Math Help - range of x

  1. #1
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    range of x

    \frac{(x+2)(x+1/2)}{(x-1)(x-3)}<0.

    Find the range of values of x that satisfy the inequality .
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  2. #2
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    Quote Originally Posted by mathaddict View Post
    \frac{(x+2)(x+1/2)}{(x-1)(x-3)}<0.

    Find the range of values of x that satisfy the inequality .
    Case 1: Solve (x + 2)(x + 1/2) < 0 and (x - 1)(x - 3) > 0 simultaneously.

    Case 2: Solve (x + 2)(x + 1/2) > 0 and (x - 1)(x - 3) < 0 simultaneously.
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  3. #3
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    Hello, mathaddict!

    Here is another (very primitive) approach . . .


    \frac{(x+2)(x+\frac{1}{2})}{(x-1)(x-3)} \:<\:0.

    Find the range of values of x that satisfy the inequality .

    We see that the "critical values" are: . \text{-}2,\:\text{-}\tfrac{1}{2},\:1,\:3


    The four values divide the number line into five intervals:

    . . \begin{array}{ccccccccc}<br />
--- &*& --- &*& --- &*& --- &*& --- \\<br />
& \text{-}2 & & \text{-}\tfrac{1}{2} && 1 && 3 \end{array}


    Test a value in each of the intervals.

    On (\text{-}\infty,\text{-}2)\!:\;\;f(\text{-}3) \:=\:\frac{(\text{-}1)(\text{-}\frac{5}{2})}{(\text{-}4)(\text{-}6)}  \:=\:+\quad\hdots \text{  no}

    On (\text{-}2,\text{-}\tfrac{1}{2})\!:\;\;f(\text{-}1) \:=\:\frac{(1)(\text{-}\frac{1}{2})}{(\text{-}2)(\text{-}4)} \:=\:{\color{red}-}\quad\hdots \text{ yes!}

    On (\text{-}\tfrac{1}{2}, 3)\!:\;\;f(0) \:=\:\frac{(2)(\frac{1}{2})}{(\text{-}1)(\text{-}3)} \:=\:+\quad\hdots \text{ no}

    On (1,3)\!:\;\;f(2) \:=\:\frac{(4)(\frac{5}{2})}{(1)(\text{-}1)} \:=\:{\color{red}-} \quad\hdots \text{ yes!}

    On (3,\infty)\!:\;\;f(4) \:=\: \frac{(6)(\frac{9}{2})}{(3)(1)}   \:=\: + \quad\hdots \text{ no}


    Therefore: . (\text{-}2, \text{-}\tfrac{1}{2}) \:\cup\: (1,3)

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