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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
1  
Looks fine to me. – Rick Decker Oct 2 '12 at 23:36
up vote 1 down vote accepted

CW answer to push it from unanswered queue:

It is correct.

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