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Is it possible to define a bijection $f:\mathbb{R}\setminus\mathbb{R}^{-}\rightarrow[0,1)$ such that $f$ is continuously differentiable on its entire domain?

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How could it be your function if it did not exist?! :) –  Mariano Suárez-Alvarez Mar 11 '11 at 15:24
    
And the ones that have been suggested have probably been patented by Amazon. –  Ross Millikan Mar 11 '11 at 15:32
    
@mar; Maybe if he finds a function that is "surprising", we will name it the mac function ?? –  The Chaz 2.0 Mar 11 '11 at 15:37
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@Mariano: "What about Hobbes?" -- Calvin. –  Aryabhata Mar 11 '11 at 15:46
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Actually, this notation is not strange. What I find strange is writing $\mathbb{R}$\$\mathbb{R}^{−}$ rather than $\mathbb{R}^{+}_{*}$, which made me want to clarify your question. –  Wok Mar 11 '11 at 18:10
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3 Answers

up vote 6 down vote accepted

Yes. For example, $f(x)=1-\frac{1}{1+x}$.

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If your domain is the reals greater than or equal to zero, $1-\exp(-x)$ seems to fill the bill.

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Let us consider $f\,:\,\left[0;+\infty\right[\rightarrow\left[0;1\right[$ defined as: $$f = \frac{2}{\pi} \arctan$$

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