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Anyone can explain me how I find x-es that coresponds to this inequality $x^3-4x^2-4x\leq 0$ ?

I know that I must transform into $x(x^2-4x-4)\leq 0$ . But after this I'm lost.

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  • $\begingroup$ No clue how to formaly find this x-es basically $\endgroup$ – Matjaž Jerman Nov 22 '15 at 19:57
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$x^3-4x^2-4x=x\left(x^2-4x-4\right)=x\left((x-2)^2-8\right)\le 0$.

This is true if and only if either $x=0$ or $(x-2)^2=8$ or $\begin{cases}x>0\\ (x-2)^2<8\end{cases}$ or $\begin{cases}x<0\\ (x-2)^2>8\end{cases}$.

I.e. if and only if either $x\in\{0,2\pm 2\sqrt{2}\}$ or $\begin{cases}x>0\\|x-2|<2\sqrt{2}\end{cases}$ or $\begin{cases}x<0\\|x-2|>2\sqrt{2}\end{cases}$

I.e. if and only if either $x\in\{0,2\pm 2\sqrt{2}\}$ or $\begin{cases}x>0\\x\in\left(2-2\sqrt{2},2+2\sqrt{2}\right)\end{cases}$ or

$\begin{cases}x<0\\x\in\left(-\infty,2-2\sqrt{2}\right)\cup\left(2+2\sqrt{2},+\infty\right)\end{cases}$

I.e. if and only if either $x\in\{0,2\pm 2\sqrt{2}\}$ or $x\in\left(0,2+2\sqrt{2}\right)$ or $x\in\left(-\infty,2-2\sqrt{2}\right)$.

I.e. if and only if $x\in\left(-\infty,2-2\sqrt{2}\right]\cup\left[0,2+2\sqrt{2}\right]$.

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  • $\begingroup$ Thanks, I did it with your help! $\endgroup$ – Matjaž Jerman Nov 22 '15 at 20:23

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