# Find the supA of $A= \{ x \in R : |x||x+1| < 2 \}$

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 ?

Hint: Use $$x^2+x<2 \iff (x+1/2)^2 < 9/4 \iff |x+1/2|<3/2$$ to prove that $$\sup (A) = 1$$.
Let's try and simplify the solution. You're correct that the condition translates to $$-2 The inequality $$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)$$.
• 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 $A = (-2,1)$. – Dimitris Dimitriadis Jan 26 at 22:42