Examining the injectivity and surjectivity of $f(x) = x/(x^2+1)$ I'm beginning with my self study in math from the $11$th grade, so bear with the simplicity.
I've got to test the objectivity of $x/(x^2+1)$, function is from $\Bbb R$ to $\Bbb R$.
First issue is, I think it is injective, because my algebra after equating the output of two inputs leads to both inputs coming equal to each other. The book answer says I'm wrong (without explanation).
With surjectivity, I can't find a simple expression for the inverse of the function, so I thought of this:
Let $g(x) = x, t(x) = x^2+1$
Then $f(x) = [g(x)/t(x)]$. Now $g(x)$ is surjective and $t(x)$ isn't. Can I argue that the ratio however, will be surjective because the numerator is surjective, and thus even if it is being divided only by numbers greater than or equal to $1$, the numerator's surjectivity will make the whole function surjective?
If I can, is there a rigorous way of arguing this? If I can't, a counterexample or counter-proof would help.

In this case, I think there is a simpler way to check injectivity/surjectivity. Let $f(x)=c$. Surjective means that for any $c$ there is a solution. Injective means that for any solution, the solution is unique. $$\frac x{x^2+1}=c\\cx^2-x+c=0$$
This is a simple quadratic equation if $c\ne 0$. Then you have a real solution if the discriminant is positive. $$\Delta=1-4c^2$$
If the discriminant is negative you have no solution. In this case it means $|c|>\frac12$. So the function is not surjective. Also if $|c|<\frac12$ and $c\ne 0$ the quadratic has two distinct solution, so the function is not injective either.
