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Math Help - Help with inequality involving cube

  1. #1
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    Help with inequality involving cube

    x^3 >= 6x - x^2

    Where >= : less than or equal to ..

    I am not sure at all about how to solve an inequality of this type but I can solve inequalities involving squares.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by struck View Post
    x^3 >= 6x - x^2

    Where >= : less than or equal to ..

    I am not sure at all about how to solve an inequality of this type but I can solve inequalities involving squares.
    x^3 \ge 6x - x^2

    x= 0 is an obvious answer
    ---------------------------------
    If x>0

    x^3 \ge 6x - x^2

    x^2 \ge 6 - x

    x^2 +x - 6 \ge 0

    (x-2)(x+3) \ge 0

    x \ge 2 , both terms positive hence x>3 is an answer

    2 > x > 0 One of the term is negative hence

    its not an answer

    -----------------------------------

    When x<0 , sign of inequality changes as , you can remember that we canceled an x initially

    0 \le x \le -3 , is an answer

    x< -3 is similarly not an answer
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  3. #3
    Senior Member pankaj's Avatar
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     <br />
x^3+x^2-6x\ge 0<br />

     <br />
x(x+3)(x-2)\ge 0<br />

    x\in[-3,0]\cup[2,\infty)
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