How do you solve $x^3>=9x^2$ ?

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Answer 1 · 2016-12-27T00:55:43

$x in [9, oo)uu{0}$

$x^3>=9x^2$

$=> x^3-9x^2>=0$

$=> x^2(x-9)>=0$

$x^2(x-9)$ has two roots: $x=0$ and $x=9$ . We will check what happens in the intervals on each side of them.

Case 1: $x in (-oo, 0)$

$=> x^2 > 0 and x-9 < 0$

$=> x^2(x-9) < 0$

Case 2: $x in (0, 9)$

$=> x^2>0 and x-9 < 0$

$=> x^2(x-9) < 0$

Case 3: $x in (9, oo)$

$=> x^2 > 0 and x-9 > 0$

$=> x^2(x-9) > 0$

Thus we have $x^2(x-9)>=0$ if $x=0$ or $x in [9, oo)$

How do you solve x^3>=9x^2x^3>=9x^2?