Prove that $\frac{x^3 - x + 10}{x^2-9}$ is surjective I'm looking for an elegant way to prove that this function is surjective. One way to solve this would be to set $\frac{x^3-x+10}{x^2-9} = y$ and solve for $x$ in terms of $y$ but this would require using the cubic formula to solve. Is there any other nice way?

One can also approach the problem by way of Descartes' Rule of Signs.
render $\frac{x^3-x+10}{x^2-9} = y$ as $x^3-x^2y-x+(10+9y)=0$ for $|x|\ne 3$ and let
\begin{eqnarray}
f(x)&=&x^3-x^2y-x+(10+9y)\\
f(-x)&=&-x^3-x^2y+x+(10+9y)
\end{eqnarray}
Then for $y$ in the interval $\left(\infty, -\frac{10}{9}\right)$ Descartes' signs for $f(x)$ are $+\,+\,-\,-$, so there is one positive $x$ which maps to $y$.
Clearly, $x=0$ maps to $y=-\frac{10}{9}$.
In the interval $\left(-\frac{10}{9},0\right)$ Descartes' signs for $f(-x)$ are $-\,+\,+\,+$ so there is one negative $x$ which maps to $y$.
For $y=0$ a root of $x^3-x+10$ maps to $y$.
For $y$ in the interval $(0,\infty)$ Descartes' signs for $f(-x)$ are $-\,-\,+\,+$ so one negative value of $x$ maps to $y$.
Thus for each value of $y$ in $(-\infty,\infty)$ there is an $x$ which maps to $y$.
