I want to find an injective map $f\colon\mathbb{R}\to[0,1)$ that is not a transcendental function (I prefer a rational function). Is it possible to find such a function or do I need a transcendental function like $$f(x)=\frac1{1+e^x}$$
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It is not possible for rational functions. Here is why not:
Suppose such an $f$ exists. Write $f(x) = \frac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials with no common terms. First, $q(x)$ can not have any roots. Since $f$ is continuous on $\mathbb{R}$ and $f$ is injective, it must either be increasing or decreasing. To finish it off, for any rational function $\lim_{x \to \infty} f(x) = \lim_{x \to -\infty}f(x),$ if either limit is finite. This can't happen if $f$ is increasing or decreasing.
D Poole
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A recent question asked for a proof that $f(x)=\frac{1}{x}+\frac{1}{x-1}$ defines a bijection $(0,1) \to \mathbb{R}$.
The inverse of $f$ gives a bijection $ \mathbb{R} \to (0,1)$ and so an injection $ \mathbb{R} \to [0,1)$.
This inverse function has an expression using square roots.
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How about $$ f(x) = \frac 12\left(1 + \frac{x}{|x| + 1}\right) $$ not quite a rational function, but it is continuously differentiable.
Ben Grossmann
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Might be worth noting explicitly that $(-1)|x|^4+(x^2)|x|^2=0$, so if I'm understanding the definitions correctly, this is not transcendental over $\Bbb R$ . – Eric Stucky Nov 18 '14 at 00:14