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.

• Maybe a more analytic argument is easier. You can observe the limit $x\to\pm 1$, and then argue with the intermediate value theorem. Mar 23 at 19:57

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$$\frac{1+\sqrt{1+4b^2}}{2b}\times\frac{1-\sqrt{1+4b^2}}{2b}=-1.$$
• I've edited my answer. Concerning your question, I would say that you should take $a=0$ if $b=0$ and $a=\frac{1-\sqrt{1+4b^2}}{2b}$ otherwise. Mar 23 at 20:14
• @blacknapkins7 Suppose that $b>0$. Then $1+\sqrt{1+4b^2}>\sqrt{4b^2}=2b$, and therefore $\frac{1+\sqrt{1+4b^2}}{2b}>1$. Since the product of those two numbers is equal to $-1$, the other number must belong to $(-1,0)\subset(-1,1)$. A similar argument works if $b<0$. Mar 27 at 12:37
You can also just argue that $$f$$ is continuous on $$(-1,1)$$ and that $$\lim_{x \to -1^+} f(x)=+\infty, \quad \lim_{x\to 1^-} f(x) = -\infty.$$