 Solution

Since $x^2+1\ge 2|x|$, 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. HenceIt 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 $2\ne 1/2$.