Problem: Prove that the mapping $ f: \mathbb R \rightarrow \mathbb R , f(x) = x^3-x$ is surjective.
Let $y\in \mathbb R$ such that $f(x)=y$ for some $x\in \mathbb R$. Then $x^3-x=y$. If we can express $x$ in terms of $y$, then we can say something about surjectiveness. But I stuck at that point.
What is the way to solve this problem?
You have $\lim_{x\to\infty}f(x)=\infty$ and $\lim_{x\to-\infty} f(x)=-\infty$, thus $f$ is surjective by the intermediate value theorem.
Any odd-degree polynomial with real coefficients has at least one real root (in fact, it has an odd number of them, counted with multiplicity). Now, consider the polynomial $x^3-x-a$ for any $a \in \mathbb{R}$. A root of this polynomial then gives a real number $b$ for which $b^3-b=a$. Hence, $\forall a \in \mathbb{R} \ \exists b \in \mathbb{R} \ (b^3-b=a)$, so the function $f(x)=x^3-x$ is surjective.
More generally, any odd-degree polynomial with real coefficients is surjective (from the reals to the reals).
