Answer
\[
f^{-1}(y) = \frac{1}{y - 1}, \quad y \neq 1.
\]
Work Step by Step
Given the function:
\[
f(x) \;=\; \frac{x+1}{x},
\quad \text{with domain } \{\,x\in\mathbb{R} : x \neq 0\}.
\]
Now we take the **codomain** to be \(\mathbb{R}\setminus\{1\}\). We must check whether \(f\) is a one-to-one correspondence (i.e., bijective) from \(\mathbb{R}\setminus\{0\}\) **onto** \(\mathbb{R}\setminus\{1\}\).
---
## 1. Is \(f\) One-to-One (Injective)?
Suppose \(f(x_1) = f(x_2)\). That is,
\[
\frac{x_1 + 1}{x_1} \;=\; \frac{x_2 + 1}{x_2}.
\]
We can rewrite \(f(x)\) as
\[
f(x) \;=\; 1 \;+\; \frac{1}{x}.
\]
Thus, \(f(x_1) = f(x_2)\) implies
\[
1 + \frac{1}{x_1} \;=\; 1 + \frac{1}{x_2}
\quad\Longrightarrow\quad
\frac{1}{x_1} \;=\; \frac{1}{x_2}
\quad\Longrightarrow\quad
x_1 \;=\; x_2,
\]
as long as \(x_1, x_2 \neq 0\). Hence \(f\) is **injective** on its domain.
---
## 2. Is \(f\) Onto \(\mathbb{R}\setminus\{1\}\) (Surjective)?
We want to see if for **every** real \(y \neq 1\), there exists an \(x \neq 0\) such that
\[
f(x) \;=\; y
\quad\Longleftrightarrow\quad
1 + \frac{1}{x} \;=\; y
\quad\Longleftrightarrow\quad
\frac{1}{x} \;=\; y - 1
\quad\Longleftrightarrow\quad
x \;=\; \frac{1}{\,y - 1\,}.
\]
- This formula for \(x\) makes sense **only** if \(y \neq 1\), ensuring \(y-1 \neq 0\).
- Furthermore, if \(y \neq 1\), then \(\tfrac{1}{y-1}\neq 0\), so \(x\) is indeed in the domain \(\mathbb{R}\setminus\{0\}\).
Thus, **every** \(y\in \mathbb{R}\setminus\{1\}\) is achieved by some \(x \neq 0\). So \(f\) is **onto** \(\mathbb{R}\setminus\{1\}\).
---
## 3. Conclusion: One-to-One Correspondence
Since \(f\) is both injective and surjective onto \(\mathbb{R}\setminus\{1\}\), it is a **bijective** map
\[
f : \bigl(\mathbb{R}\setminus\{0\}\bigr)\;\longrightarrow\;\bigl(\mathbb{R}\setminus\{1\}\bigr).
\]
---
## 4. Inverse Function
To find \(f^{-1}\), solve \(y = 1 + \tfrac{1}{x}\) for \(x\):
\[
y - 1 = \frac{1}{x}
\quad\Longrightarrow\quad
x = \frac{1}{\,y - 1\,}.
\]
Hence the inverse is
\[
f^{-1}(y) \;=\; \frac{1}{\,y - 1\,},
\quad \text{with domain } \mathbb{R}\setminus\{1\}
\text{ and range } \mathbb{R}\setminus\{0\}.
\]
---
### Final Answer
- With the codomain taken to be \(\mathbb{R}\setminus\{1\}\), the function
\[
f(x) = \frac{x + 1}{x}, \quad x \neq 0,
\]
is indeed a **one-to-one correspondence**.
- Its inverse is
\[
f^{-1}(y) = \frac{1}{y - 1}, \quad y \neq 1.
\]
