I am trying to show that the function $f:(-1,1) \rightarrow \mathbb{R}$, $f(x) = \frac{x}{x^2-1}$ is bijective. I have the following two questions:
1) Proving it using calculus: First I show that $f$ is injective. Since $f'(x) = -\frac{x^2+1}{(x^2-1)^2} < 0$ for all $x \in (-1, 1)$, then $f$ is monotonic and hence injective.
I am having trouble showing that it is surjective using the Intermediate Value Theorem (IVT). The IVT states that if $f: [a,b] \rightarrow \mathbb{R}$ is continuous and $L$ is a real number satisfying $f(a) < L < f(b)$ or $f(b) < L < f(a)$, then there exists a $c \in (a,b)$ where $f(c) = L$. Clearly the function at hand is continuous on $(-1, 1)$, but to use the aforementioned theorem, we need to have a closed interval $[a,b]$ (whereas $(-1, 1)$ is not a closed interval…). I "sort of" have an idea: Since $f$ is continuous on $(-1, 1)$, $f(-1)$ isn't defined but it is basically $+\infty$ and $f(1)$ is "like" $-\infty$, so if $L$ is a real number $-\infty< L < +\infty$, i.e., $L \in \mathbb{R}$, then there exists a $c \in (-1, 1)$ such that $f(c) = L$, hence $f$ is surjective. But I feel this is wrong because $f(-1)$ and $f(1)$ aren't defined and the IVT needs a closed interval.
2) How can I prove bijection without using calculus? E.g., to show injection, let $\frac{x_1}{x_1^2-1} = \frac{x_2}{x_2^2-1}$ and I want to reach the conclusion that $x_1 = x_2$, but I am stuck. Also, how can I prove surjection?
Best Answer
You can show it is surjective by solving the quadratic equation. Given $y \in \Bbb R$ we want to find $x$ such that $\frac x{x^2-1}=y$, or $yx^2-x-y=0, x=\frac 1{2y}(1\pm \sqrt{1+4y^2})$ and we want the signs of $x$ and $y$ to be opposite, so we take the minus sign. The fact that you get a unique answer shows it is injective, so answers your second part as well.