How do you solve $x(x-3)^2>=0$ ?

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Answer 1 · 2016-12-05T14:47:43

The answer is $x in [0, +oo[$

Let $f(x)=x(x-3)^2$

The domain of $f(x)$ is $D_f(x)=RR$

and $(x-3)^2>=0$ , $AA x in RR$

So, the sign of $f(x)$ depends on the sign of $x$

Therefore, $f(x)>=0$ , when $x in [0, +oo[$

How do you solve x(x-3)^2>=0x(x-3)^2>=0?