= Solution
Since $x^2+1\geq2|x|$, the <image of a function> is contained in $[-1/2,1/2]$. Both endpoints occur, at $x=-1$ and $x=1$. By <continuity> and the <intermediate value theorem> applied on $[-1,1]$, every intermediate value occurs too. Hence
$$
\boxed{f(\mathbb R)=[-1/2,1/2],\qquad f\text{ is neither onto nor one-to-one}.}
$$
It is not <surjective> onto $\mathbb R$ because, for example, $1$ is not attained. It is not <injective> because $f(2)=f(1/2)=2/5$ although $2\ne1/2$.
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