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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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    $\begingroup$ Seems OK. Perhaps you could say why $-x^2>x+2$ has no solutions, maybe by showing the discriminant is negative. $\endgroup$
    – Alex R.
    Oct 2, 2012 at 18:14
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    $\begingroup$ Looks fine to me. $\endgroup$ Oct 2, 2012 at 23:36

1 Answer 1

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CW answer to push it from unanswered queue:

It is correct.

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