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I realize that this function has a horizontal asymptote $y=0$. And that the range of this function is $(0, 1]$

Is the function $f: \mathbb{R} \rightarrow \mathbb{R}$ since for every $x \in \mathbb{R}$ $\exists$ a $f(x) \in \mathbb{R}$.

i.e. can I say $f: \mathbb{R} \rightarrow \mathbb{R}$?

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Yes. The function maps real numbers to real numbers.

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Yes. All of the following statements are correct for $\,f(x) = 1 / (x^2+1)\,$:

  • $\;f : \mathbb{R} \to \mathbb{R}$ is a function

  • $\;f : \mathbb{R} \to (0,1]$ is a surjective function

  • $\;f : [0,\infty) \to \mathbb{R}$ is an injective function (and so is $\;f : (-\infty,0] \to \mathbb{R}$)

  • $\;f : [0,\infty) \to (0,1]$ is a bijection (and so is $\;f : (-\infty,0] \to (0,1]\,$)

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Yes, it's ok to say that. Technically, the definition of a function includes a description of its domain, so you are right to wonder. The second $\mathbb{R}$ is fine in any case, since it represents the codomain and not the range.

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OK. Yea, $f: \mathbb{R} \rightarrow \mathbb{R}$ where $D_f=\mathbb{R}\subseteq\mathbb{R}$ and $R_f=(0,1]\subseteq\mathbb{R}$.

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