Proving a Rational Expression is Surjective I need to prove that the function $f:(-1,1)\rightarrow\mathbb{R}$ defined by $f(x) = \frac{x}{x^2 - 1}$ is surjective.
My work.
$b = \frac{a}{a^2 - 1} \iff b(a^2 -1)=a \iff ba^2 - b - a=0$
From here I did a few cases:
Case 1) $b=0$. Then $a=0$.
Using the quadratic formula: $a = \frac{1\pm \sqrt{1+4b^2}}{2b}$.
Case 2) $b \gt0$. Then $a = \frac{1+ \sqrt{1+4b^2}}{2b}$ $\notin (-1,1)$ $\forall$ b $\in R$.
Case 3) $b \gt0$. Then $a = \frac{1- \sqrt{1+4b^2}}{2b}$ $\in (-1,1)$ $\forall$ b $\in R$.
Using Case 1 and Case 3, f is subjective. Is this correct? I cannot use Calculus.

It is correct. But you should justify the assertions that $\frac{1+\sqrt{1+4b^2}}{2b}\notin(-1,1)$ and that $\frac{1-\sqrt{1+4b^2}}{2b}\in(-1,1)$. This follows easily from the fact that$$\frac{1+\sqrt{1+4b^2}}{2b}\times\frac{1-\sqrt{1+4b^2}}{2b}=-1.$$
