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Math Help - Inequalities

  1. #1
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    Inequalities

    how do i solve:
    1)
    4 - x
    ---------- < 0
    (x-3)(x+2)
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bigstarz
    how do i solve:
    1)
    4 - x
    ---------- < 0
    (x-3)(x+2)
    There are three critical points here, x = 4, 3, -2. (Critical points are found where the numerator or denominator are zero.) So the equation is possibly true for the following intervals: (-inf, -2), (-2, 3), (3, 4), and (4, inf). Check each of these possibilities.

    -Dan
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  3. #3
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    Quote Originally Posted by bigstarz
    how do i solve:
    1)
    4 - x
    ---------- < 0
    (x-3)(x+2)
    The method topsuark used is the preferred method. Here is another one.
    Whenever you have a faction it is negative when the signs or numerator and denominator do not match (one positive and one negative).

    Thus, there are two possibilities
    4-x<0 \mbox{ and } (x-3)(x+2)>0 (1)
    OR
    4-x>0 \mbox{ and } (x-3)(x+2)<0 (2)

    Solving (1),
    4-x<0 \rightarrow x>4
    The second inequality,
    (x-3)(x+2)>0 has solutions,
    x>3 \mbox{ or }x<-2
    Combining these solutions together (find their intersection in strict terms) and you have,
    x>4

    Solving (2),
    4-x>0 \rightarrow x<4
    The second inequality,
    (x-3)(x+2)<0 has solutions,
    -2<x<3
    Combining these solutions together (find their intersection in strict terms) and you have,
    -2<x<3 (cuz first inequlaity is contained in this one).

    Thus,
    \left\{ \begin{array}{c} x>4 \\ -2<x<3
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