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Calculate Sup (Supremum) and Inf (Infimum) of the following set

$A=\{x\in\mathbb{R}:x|x|>x+2\}$

My solution.

$A=\{x\in\mathbb{R}:x^2>x+2 \ \ (x>0)\ \cup\ -x^2>x+2\ \ (x<0) \}$

The inequality $-x^2>x+2$ is satisfied on the empty set, and the

$x^2>x+2 \ \ (x>0)$ is satisfied for $x>2$. Then $Inf(A)=2$ and $Sup(A)=+\infty$. Is my procedure right? I made ​​a mistake?

thank you very much

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Seems OK. Perhaps you could say why $-x^2>x+2$ has no solutions, maybe by showing the discriminant is negative. –  Alex R. Oct 2 '12 at 18:14
    
Looks fine to me. –  Rick Decker Oct 2 '12 at 23:36

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