I was looking for a rational function $f: \mathbb{R} \to \mathbb{R}$ that looks like $\arctan$, in that it is
- injective
- not surjective
- well-defined on all $x\in \mathbb{R}$ (no vertical asymptotes)
I believe I have proven that one can't exist.
My thinking is this: it can't be a polynomial, because an odd-degree polynomial is surjective and an even degree polynomial is not injective. Therefore it must have a nontrivial denominator. The denominator can't be an odd degree polynomial, or else it has a zero somewhere. If the numerator is a polynomial of degree greater than the denominator $N>D$, then we get surjectivity (if $N-D$ is odd), or we lose injectivity (if $N-D$ is even). If $N\leq D$, then we lose injectivity, since the function will approach the same limit as $x \to \pm \infty$ (I claim that together with continuity, this implies that we lose injectivity).
Do you agree? Is there a general result about such functions?
Best Answer
This looks good to me as well, though the various cases could be cleaned up a bit: the key point is that the range of an injective continuous function is the interval between its two limits, and that for a rational function $f(x)/g(x)$ the two limits can only differ when $\deg f > \deg g$ in which case the interval is $(- \infty, \infty)$.
It's hard to say how to generalize this result: in $\mathbb{C}$ for instance all continuous rational functions are polynomials, and all injective polynomials are surjective. Over any proper subinterval of $\mathbb{R}$, you can find an injective rational functions with whatever values you like on the two boundaries.