Show that exists $x \in \mathbb{R} $ such that $x^3+x=3$ Show that exists $x \in \mathbb{R} $ such that $x^3+x=3$
I did:
I take $S=\{x\in \mathbb{R} \mid x^3+x<3\}$
So, this set is bounded and by completeness of the real numbers exists $a=\sup(S)$
Then I want to prove that $a^3+a=3$
And then in order to get a contradiction I suppose that $a^3+a>3$
and how $a $ is the supremum then $\forall\epsilon\in\mathbb{R} $ exists $b \in S$ such that
$a-\epsilon<b$ then $(a-\epsilon)^3+b<b^3+b<3$
But I'm stuck from here. I'm not able to use derivatives, or continuity so I can't use IVT
 A: Hint 1:
Apply the intermediate value theorem (which does not require taking derivatives) to the continuous function $f(x)=x^3+x-3$ on some closed interval.
Hint 2:
For the function defined above, $f(-1)<0$ and $f(2)>0$.

I don't understand what you mean by "I'm not able to use derivates or continuity". If you wish, you can unwrap the proof of the intermediate value theorem and apply the steps to the function $f$ above.
Alternatively, you may use the fundamental theorem of algebra and the fact that all complex roots come up as conjugate pairs. (Well, you only have the real-analysis tag...)
A: Hint
If
$$a^3+a >3$$
and
$$0< \epsilon < \min \{ 1, \frac{a^3+a -3}{3a^2+3a+2} \}$$
[ I figured out the value by finding first the upperbound $3a^2\epsilon-3a\epsilon^2+\epsilon^3+\epsilon <3a^2\epsilon+3a\epsilon+2\epsilon$ below and then solving $3a^2\epsilon+3a\epsilon+2\epsilon< a^3+a-3$. I like to point out to my students that in many proofs in real Analysis you do the computations first to figure out the inequalities you need, but write the proof in reverse order]
Then
$$
3a^2\epsilon-3a\epsilon^2+\epsilon^3+\epsilon <3a^2\epsilon+3a\epsilon^2+\epsilon^3+\epsilon <3a^2\epsilon+3a\epsilon+2\epsilon< a^3+a -3
$$
This gives
$$(a-\epsilon)^3+a-\epsilon=a^3+a-3a^2\epsilon+3a\epsilon^2-\epsilon^3-\epsilon>3$$
Now, use the fact that there exists some $b \in S$ such that $b >a-\epsilon$. But this leads to the contradiction
$$
3< (a-\epsilon)^3+a-\epsilon \leq b^3+3 \leq 3
$$
A: If you cannot use the IVT or continuity, then the simplest method in my mind is to explicitly solve the polynomial equation. It is difficult to show that the following real number
$$x= \sqrt[3]{\frac{1}{18}(27+\sqrt{741})}-\sqrt[3]{\frac{2}{3(27+\sqrt{741})}}$$
is a solution to the equation $x^3+x-3=0$ (at least by hand). However, it is possible (again, not as easy as using the IVT) and suffices as a direct proof of the original claim.
A: $S = \{x| x^3 + x < 3\}$ is one set but $T= \{x|x^3 + x < 3\}$ is another.
If $s = \sup S$ and $t = \inf T$ it's easy to prove $\sup S \le \sup T$.
Now if you can prove $s^3 + s \ge 3$ then you can probably use a similar argument to prove $t^3 + t \le 3$.
That means $t^3 + t \le s^3 + s$.  It's easy to prove $x^3 + x$ is increasing so $t^3 + t \le s^3 + s\iff t\le s$.
so you have $t \le s$ and $s \le t$ and $t^3 +t \le 3 \le s^3 +s$ which altogether shows $s=t=3$.
so what argument can we use to prove that $s^3 + s \ge 3$?
Consider that if $u > s$ then $u \not \in S$ so $u^3 + u \ge 3$. If we assume $s^3 +s < 3$, the would mean for all $\epsilon > 0$ then $(s+\epsilon)^3 + (s+\epsilon) \ge 3$
If we can show there is an $\epsilon > 0$ so that $(s+\epsilon)^3 + (s + \epsilon) < 3$ that would be a contradiction and that would prove $s =\sup S \ge 3$.
Let's find such an epsilon we need
$(s+\epsilon)^3 + (s + \epsilon) < 3$
$s^3 + 3s^2\epsilon + 3s\epsilon^2 + \epsilon^3+s + \epsilon <3$.
$(3s^2+1)\epsilon + 3s\epsilon^2 + \epsilon^3 < 3 - (s^3 + s)$
Now we know if $s \ge 2$ then $s^3 + s \ge 10$ so $s< 2$ and if $\epsilon < 1$ we have:
$(3s^2+1)\epsilon + 3s\epsilon^2 + \epsilon^3 <13\epsilon + 6\epsilon + \epsilon =20 \epsilon$
So if we can find and $\epsilon < 1$ so that $20\epsilon < 3-(s^3 + s)$ we'd be done.
So as we are assuming $3-(x^3 + s) > 0$ we can just let $\epsilon < \min(1, \frac {3-(x^3 + s)}{20}$.
And we are done.  That's a a contradiction so $s^3 + s \ge 3$.
