Answer
\[
f^{-1}(x) \;=\; \frac{x+1}{x-1}, \quad x \neq 1.
\]
Work Step by Step
Given the function\[
f(x) \;=\; \frac{x + 1}{x - 1},
\quad\text{for real }x \neq 1.
\]
We will show that, with the domain \(\mathbb{R}\setminus\{1\}\) and the codomain \(\mathbb{R}\setminus\{1\}\), this \(f\) is a one-to-one correspondence (bijection), and we will find its inverse.
---
## 1. Injectivity
To check that \(f\) is injective, suppose
\[
f(x_1) \;=\; f(x_2).
\]
That is,
\[
\frac{x_1 + 1}{x_1 - 1} \;=\; \frac{x_2 + 1}{x_2 - 1},
\]
with \(x_1, x_2 \neq 1\). Cross-multiplying gives
\[
(x_1 + 1)(x_2 - 1) \;=\; (x_2 + 1)(x_1 - 1).
\]
Expand both sides:
- Left side: \((x_1 + 1)(x_2 - 1) = x_1x_2 - x_1 + x_2 - 1.\)
- Right side: \((x_2 + 1)(x_1 - 1) = x_1x_2 - x_2 + x_1 - 1.\)
Set them equal and cancel common terms \(x_1x_2\) and \(-1\):
\[
-x_1 + x_2 \;=\; -x_2 + x_1.
\]
Rearrange:
\[
-x_1 - x_1 \;=\; -x_2 - x_2
\;\;\Longrightarrow\;\;
-2x_1 \;=\; -2x_2
\;\;\Longrightarrow\;\;
x_1 \;=\; x_2.
\]
Hence \(f\) is **injective**.
---
## 2. Surjectivity onto \(\mathbb{R}\setminus\{1\}\)
We must see if every \(y \neq 1\) arises as \(f(x)\) for some \(x \neq 1\). So fix \(y \in \mathbb{R}\setminus\{1\}\) and solve
\[
\frac{x + 1}{x - 1} \;=\; y.
\]
Cross-multiply:
\[
x + 1 \;=\; y\,(x - 1).
\]
Distribute \(y\):
\[
x + 1 = yx - y.
\]
Rearrange to isolate \(x\):
\[
x - yx = -y - 1
\;\;\Longrightarrow\;\;
x(1 - y) = -\,(y + 1).
\]
Because \(y \neq 1\), \(1 - y \neq 0\). Thus,
\[
x = \frac{-(y + 1)}{1 - y}.
\]
Multiply numerator and denominator by \(-1\) for a cleaner look:
\[
x = \frac{y + 1}{y - 1}.
\]
We must check that this \(x\) is not 1. Indeed, if \(\frac{y+1}{y-1} = 1\), then \(y+1 = y-1\), forcing \(1=-1\), a contradiction. So \(x \neq 1\).
Hence, for **every** \(y \neq 1\), there is a valid \(x \neq 1\) with \(f(x) = y\). Therefore, \(f\) is **onto** \(\mathbb{R}\setminus\{1\}\).
---
## 3. Bijection and Inverse
Since \(f\) is both injective and surjective onto \(\mathbb{R}\setminus\{1\}\), it is a **one-to-one correspondence**. From the work above, the inverse is found by solving \(y = \tfrac{x+1}{x-1}\) for \(x\). We obtained
\[
x = \frac{y+1}{\,y - 1\,}.
\]
Hence, the inverse function \(f^{-1} : \mathbb{R}\setminus\{1\} \to \mathbb{R}\setminus\{1\}\) is
\[
f^{-1}(y) \;=\; \frac{y + 1}{y - 1}.
\]
Renaming the variable \(y\) back to \(x\) if desired,
\[
f^{-1}(x) \;=\; \frac{x + 1}{\,x - 1\,},
\quad x \neq 1.
\]
Notice this has exactly the same form as \(f\), meaning \(f\) is its own inverse.
---
### Final Answer
1. With domain \(\mathbb{R}\setminus\{1\}\) and codomain \(\mathbb{R}\setminus\{1\}\), the function
\[
f(x) \;=\; \frac{x+1}{x-1}, \quad x \neq 1,
\]
is a **one-to-one correspondence**.
2. Its inverse is
\[
f^{-1}(x) \;=\; \frac{x+1}{x-1}, \quad x \neq 1.
\]
(Indeed, \(f\) is an involution: \(f = f^{-1}\).)
