Prove that $f : \mathbb R \smallsetminus \{−1\} \to\mathbb R \smallsetminus \{1\}$ given by $f(x) = \frac{x − 3}{x + 1}$ is bijective

Question

I know for a function to be bijective it must be one to one and onto. Here's what I have

Take by cases
Case 1 (one to one)

$$ \begin{align*} \frac{x-3}{x+1} &= \frac{y-3}{y+1} \\[1ex] (x-3)(y+1)&=(x+1)(y-3) \\[1ex] xy+x-3y-3&=xy-3x+y-3 \end{align*} $$

cleaning up we get

$$4x=4y$$

so

$$x=y$$

The function is one to one.

Case two (onto) $$ \begin{align*} y&=f(x)\\[1ex] y&= \frac{x-3}{x+1}\\[1ex] y(x+1)&=x-3 \\[1ex] yx+y&=x-3 \\[1ex] yx-x&=-3-y \\[1ex] x(y-1)&=-3-y \\[1ex] x&= \frac{-3-y}{y-1} \end{align*} $$

The function is onto.

Are we done here? Anything I should add?

Answer 1

Alternatively, simply note that $$f(x) = 1 - \frac{4}{x+1}$$

The fraction takes on all values except $0$ hence $f$ takes on all values except $1$. It is injective clearly since $\frac{4}{a} = \frac{4}{b} \Longleftrightarrow a=b$ where neither $a$ nor $b$ are $0$.