$\displaystyle x^3 >= 6x - x^2$

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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- Mar 24th 2009, 05:49 AMstruckHelp with inequality involving cube
$\displaystyle x^3 >= 6x - x^2$

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. - Mar 24th 2009, 06:07 AMADARSH
$\displaystyle x^3 \ge 6x - x^2$

x= 0 is an obvious answer

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If x>0

$\displaystyle x^3 \ge 6x - x^2$

$\displaystyle x^2 \ge 6 - x$

$\displaystyle x^2 +x - 6 \ge 0 $

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

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

$\displaystyle 2 > x > 0$ One of the term is negative hence

its not an answer

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

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

$\displaystyle x< -3$ is similarly not an answer - Mar 24th 2009, 09:29 AMpankaj
$\displaystyle

x^3+x^2-6x\ge 0

$

$\displaystyle

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

$

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