Let $S=$ {$x^3 +x< 1 \:|\:x\in \mathbb R$}
The goal is to find the supremum, i.e. the least upper bound, of S, in $\mathbb R$
I have been attempting to solve it for awhile now and I honestly have yielded little. So far I have tried understanding it geometrically in terms of cubes and lines, have tried thinking on the derivatives of the elements. Use Cardano's formula on the general element in S, $x^3 +x - 1< 0$ but it didn't speak to me.
I don't think the supremum is 1. I'm not really sure whether this would be expressable concretely, like a number, or more abstractly, like some kind of general cubic equation (which would be more specific than the equations that define S.)
Could direction or hints be provided? I would prefer some hints but full answers are fine too, I just want to understand this, so if explanation could accompanied it would be appreciated.
Thank you for reading!
If you really mean $S=\{x^3+x<1:x\in\Bbb R\}$, the set of numbers in the form $x^3+x$ with $x\in\Bbb R$ such that they are strictly less that $1$, then the supremum is indeed $1$.
To avoid any confusion, we can express the set $S$ as $S=\{y\in\Bbb R:y=x^3+x\text{ for some $x\in\Bbb R$}, y<1\}$. Now, $f(x)=x^3+x$ takes the value $0$ for $x=0$ and takes the value $2$ for $x=1$, so, since $f$ is continuous, by Rolle's theorem this means that $f$ has to take every possible value in between while $x$ is in $(0,1)$. Then $f$ takes every value greater than $0$ and less than $2$, so we can take $y=f(x)$ for those values such that $y=f(x)<1$. Therefore every value greater than $0$ and less than $2$ such that it is also less than $1$ is in $S$, so $[0,1)\subseteq S$. Consequently $\sup\big([0,1)\big)\le\sup(S)$, but $\sup\big([0,1)\big)=1$ and $S$ is bounded by $1$, so we conclude $\sup(S)=1$.