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Math problems

  1. Mar 22, 2004 #1
    Math problems!!

    I understand how to solve this inequality, but i do not know what sign to put (greater than or less than) in my answer. I would really appreciate it if someone could explain this concept to me. THanks

    Write a cubic inequality that has this solution:
    -2<X<0, x>3
    (X)(X+2)(X-3)
    (X^2 +2X) (X-3)
    X^3 - X^2 - 6X (< or >) 0
     
  2. jcsd
  3. Mar 22, 2004 #2
    Whoa, hold on. What's the inequality?

    x>3 into the expression x^3 - x^2 - 6x?

    What's the -2<x<0?

    cookiemonster
     
  4. Mar 22, 2004 #3
    teh cubic inequality is:
    -2<X<0 and x>3
     
  5. Mar 22, 2004 #4
    I've never heard of a cubic inequality before (other than a cubic polynomial used in an inequality), so I'll just butt out of this before I make a fool of myself again.

    cookiemonster
     
  6. Mar 23, 2004 #5
    you can always check if you have the inequality the right way round you do not need a method for checking if its "<" or ">" other than subbing in points

    for example
    when x=4, the statement x> 3 is true
    and the statement 4^3-4^2-6*4=24 > 0 is true so the answer better be
    x^3 - x^2 - 6x> 0
    not x^3 - x^2 - 6x <0.


    There is no mystery here just sketch the curve y=x^3 - x^2 - 6x
    and notice there are two regions above the x-axis, one region has the property that all x values are between -2 and 0 (-2< x< 0) and the other region is x>3
     
  7. Mar 23, 2004 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Apparently the problem is not to "solve an inequality" but to construct an inequality having all numbers between -2 and 0 and all numbers above 3 as solutions.

    Start by constructing an equation:

    (x-(-2))(x-0)(x-3)= (x+2)(x)(x-3)= x3+ 5x2+ 6x= 0 is 0 at exactly -2, 0, and 3. Since a polynomial is continuous, it can change from positive to negative and vice-versa only where it is 0.

    Now check what happens in between: at x= 1, for example,
    13+ 5(12)+ 6(1)= 1+ 5+ 6= 11> 0. Therefore, the inequality x3+ 5x2+ 6x> 0 has x<-2 and 0< x< 3 as solution set.

    IF the problem had asked for the complement set: -2< x< 0, x> 3, then we would have tried x= -1, say, and found that
    (-1)3+ 5(-1)2+ 6(-1)= -1+5-6= -2< 0.
    The inequality x3+ 5x2+ 6x< 0 has that solution set.
     
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