# Help with inequality involving cube

• Mar 24th 2009, 05:49 AM
struck
Help 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 AM
Quote:

Originally Posted by struck
$\displaystyle x^3 >= 6x - x^2$

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.

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

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

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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 AM
pankaj
$\displaystyle x^3+x^2-6x\ge 0$

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

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