Inequalities Interval Notation

Solve: (2 + x)(5 + x)(4 - x) < 0 (write answer in interval notation)

Which I think is xe (-infinity, -5) or (-2, 4)

and

Solve: (4x + 1)(3 - x) > or equal to 0 (write answer in interval notation)

Which im unsure

asked by Sasha
  1. #1.
    (2 + x)(5 + x)(4 - x) < 0
    Equate each term inside the parenthesis to zero:
    2 + x < 0
    x < -2

    5 + x < 0
    x < -5

    4 - x < 0
    4 < x
    x > 4

    When x > 4, only the term (4 - x) is negative while the others are positive, so it satisfies the inequality.

    When x < -5, say x = -6,
    (2 - 6)(5 - 6)(4 + 6) < 0
    -4 * -1 * 10 < 0
    40 < 0
    which is false. So x cannot be less than or equal to -5.

    When x < -2 but greater than -5, say x = -3
    (2 - 3)(5 - 3)(4 + 3) < 0
    -1 * 2 * 7 < 0
    -14 < 0
    which is true. So -5 < x < -2. Combine it with x > 4, we'll have
    (-5, -2) U (4, +infinity)


    #2.
    (4x + 1)(3 - x) ≥ 0
    Same procedure we'll do. Equate each term to zero:
    4x + 1 ≥ 0
    4x ≥ -1
    x ≥ -1/4

    3 - x ≥ 0
    3 ≥ x

    We get the intersection. So x must be greater than or equal to -1/4 but less than or equal to 3:
    [-1/4 , 3]

    hope this helps~ `u`

    posted by Jai
  2. After you get a bit of practice with this kind of stuff, you can feel confident using what you know about the graphs of polynomials.

    (2 + x)(5 + x)(4 - x) < 0

    You know this is a cubic, with three roots. The graph crosses the x-axis at each root. So, pick a very large negative value for x.

    The product (-)(-)(+) is positive
    So, you know that the graph crosses below the x-axis at -5 back up at -2 and back down at 4. So, f(x) < 0 on

    (-5,-2)U(4,∞)

    posted by Steve

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