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Let us define a set $A = \{ x \in R : |x||x+1| < 2 \} $. Which is the $supA$?

My solution:

We know that: $|x||x+1| = |x(x+1)| = |x^2 +x|$

Then

$ -2 < x^2 + x < 2 $

Is $supA=2$ ?

$supA = 2$ if $\forall e > 0$ there is a $x \in A$ such that

$x^2 + x > 2 -e$

Let us define $e=|x|$ (First Question)

Then

$x^2 + x> 2 - |x| \Leftrightarrow ... \Leftrightarrow |x| > -x^2 - x + 2 \Leftrightarrow $

\begin{cases} x > - x^2 -x + 2 \Leftrightarrow x^2 + 2x > 2\\ x < x^2 + x - 2 \Leftrightarrow x^2 > 2 \end{cases}

So, we have two cases:

if $x^2>2$ then $ x > \sqrt 2$

We replace $x=\sqrt 2$ in the first inequality, so:

$|(\sqrt 2)^2 + \sqrt 2| = |2 + \sqrt 2| > 2$

On the other hand if $x^2 + 2x -2 >0$

We find the roots of $x^2 + 2x - 2 = 0$

and if we replace $x = -1 + \sqrt 3$ in the first inequality we have:

$|(\sqrt 3)^2 + \sqrt 3| > 2 $

So, there isn't any $x\in A$ when $e=|x|$. Consequently, there isn't $supA$

First Question: Is the selection of e correct ? Could we find anoother e ?

Second Question: Is my solution correct ? Is there any easier way to solve the exercise ?

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Hint: Use $$x^2+x<2 \iff (x+1/2)^2 < 9/4 \iff |x+1/2|<3/2$$ to prove that $\sup (A) = 1$.

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  • $\begingroup$ If A is bounded from top, ther must be a supA. It is bounded from top cause let say every X>10 is not in the group, so 10 is a good bound. So 10 doesnt have to the Sup, but there is one. $\endgroup$ – Shaq Jan 26 at 22:23
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Let's try and simplify the solution. You're correct that the condition translates to $$ -2<x^2+x<2 $$ The inequality $x^2+x-2<0$ is satisfied on the interval $(-2,1)$; the inequality $x^2+x+2>0$ is satisfied for every $x$. Thus your set $A$ is $A=(-2,1)$.

So no, your solution is incorrect, I'm afraid.

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  • $\begingroup$ Thank you ! it was very useful. So, I correctly find that $supA \neq 2 $, but I couldn't find the solution because I ignored that $A = (-2,1)$. $\endgroup$ – Dimitris Dimitriadis Jan 26 at 22:42

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