Prove that $\frac{x^3 - x + 10}{x^2-9}$ is surjective

Question

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?

Answer 1

On the interval $(-3,3),$ the function $f(x)=\frac{x^3 - x + 10}{x^2-9}$ is continuous, and $x^2-9<0$.

As $x$ approaches $-3$ from above, the numerator approaches $-14$ and the denominator approaches $0$ from below, so the function approaches $\infty$.

As $x$ approaches $3$ from below, the numerator approaches $34$ and the denominator approaches $0$ from below, so the function approaches $-\infty$.

Now use the intermediate value theorem.

Answer 2

Every polynomial of degree $3$ has always a root in reals. Now take the polynomial $x^3-x+10-y(x^2-9)$ with $y$ as a parameter and deduce that there is at least one such $x$ .

Answer 3

To reduce from cubic polynomial solving to quadratic polynomial divide and simplify the function like below$$y=\frac{x^3 - x + 10}{x^2-9}=\frac{x^3 - 9x +8x+ 10}{x^2-9}=\\y=x+\frac{8x+ 10}{x^2-9}$$ now restrict $-3<x<3$ at this interval function is continuous, so you need to show $\frac{8x+ 10}{x^2-9}$ is surjective.in other hand you must show $-\infty<\frac{8x+ 10}{x^2-9}<\infty,-3<x<3$. hope it helps you.

Answer 4

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 $y$.