Real Analysis. Bounds, Infimum and Supremum Given that $S=\left\{x \mid 4x^2 > x^3 + x\right\}$.
(1) Determine whether $S$ is bounded.
(2) Determine their supremum and infimum.
I divided the equation by $x$ to have a quadratic. 
Then my roots are in decimals doesn't look correct.
Thank you!
 A: $$
4x^2 > x^3 + x
$$
Dividing both sides by $x$ and getting $4x > x^2 + 1$ is wrong.  You can divide both sides by a positive number and do that, or by a negative number and get "$<$" instead of "$>$", but $x$ is sometimes positive and sometimes negative.
You have
\begin{align}
x^3 - 4x^2 + x < 0 \\[10pt]
x(x^2 - 4x+1) < 0 \\[10pt]
(x-0)\Big(x-(2-\sqrt 3)\Big)\Big(x-(2+\sqrt 3)\Big) < 0
\end{align}
Thus there are three places where the sign changes: $0$, $2-\sqrt 3$, and $2+\sqrt 3$.  They divide the line into four intervals:


*

*numbers less than $0$,

*numbers between $0$ and $2-\sqrt3$,

*numbers between $2-\sqrt3$ and $2+\sqrt 3$

*numbers bigger than $2+\sqrt3$.

*On the first interval, all three of the factors are negative so the product is negative.

*On the second interval, the first factor is positive and the other two are negative, so the product is positive.

*On the third interval, the first two factors are positive and the third is negative, so the product is negative.

*On the fourth interval, all three factors are positive, so the product is positive.


So $S= (-\infty,0)\cup(2-\sqrt 3,\  2+\sqrt 3)$, and from that you get the infimum and supremum.
A: 1) $\forall x \in \mathbb{R} : x < 0 \Rightarrow x \in S$ (Why?).
So $S$ can't be bounded.
2) Can you figure it out yourself now?
