Answer
\[
f^{-1}(x) = \frac{2(x+1)}{1-x}, \quad \text{for } x \in \mathbb{R}\setminus\{1\}.
\]
Work Step by Step
> Define
\[
f : \mathbb{R}\setminus\{-2\} \to \mathbb{R}\setminus\{1\} \quad\text{by}\quad f(x) = \frac{x-2}{x+2}.
\]
Show that \(f\) is a one‐to‐one correspondence (bijective) and find its inverse.
We proceed as follows.
---
### 1. Prove \(f\) Is Injective
Suppose that for some \(x_1,x_2\in \mathbb{R}\setminus\{-2\}\) we have
\[
f(x_1) = f(x_2).
\]
That is,
\[
\frac{x_1-2}{x_1+2} = \frac{x_2-2}{x_2+2}.
\]
**Cross–multiply:** (since \(x_1+2\) and \(x_2+2\) are nonzero)
\[
(x_1-2)(x_2+2) = (x_2-2)(x_1+2).
\]
**Expand both sides:**
\[
x_1 x_2 + 2x_1 - 2x_2 - 4 = x_1 x_2 + 2x_2 - 2x_1 - 4.
\]
Subtract \(x_1x_2\) and \(-4\) from both sides:
\[
2x_1 - 2x_2 = 2x_2 - 2x_1.
\]
Bring like terms together:
\[
2x_1 - 2x_2 - 2x_2 + 2x_1 = 0 \quad\Longrightarrow\quad 4x_1 - 4x_2 = 0.
\]
Dividing by 4 gives
\[
x_1 = x_2.
\]
Thus, \(f\) is **injective**.
---
### 2. Prove \(f\) Is Surjective
Let \(y\) be an arbitrary element of the codomain, i.e. \(y \in \mathbb{R}\setminus\{1\}\). We must show there exists an \(x \in \mathbb{R}\setminus\{-2\}\) such that
\[
f(x) = y.
\]
That is, we need to solve
\[
\frac{x-2}{x+2} = y \quad\Longrightarrow\quad x-2 = y(x+2).
\]
**Solve for \(x\):**
1. Distribute on the right:
\[
x-2 = yx + 2y.
\]
2. Collect \(x\)-terms on one side:
\[
x - yx = 2y + 2.
\]
3. Factor \(x\) on the left:
\[
x(1-y) = 2(y+1).
\]
4. Provided that \(y\neq 1\) (which is ensured since \(y\in\mathbb{R}\setminus\{1\}\)), we have
\[
x = \frac{2(y+1)}{1-y}.
\]
We must check that the computed \(x\) is not \(-2\). Assume for contradiction that
\[
\frac{2(y+1)}{1-y} = -2.
\]
Multiply both sides by \(1-y\) (which is nonzero):
\[
2(y+1) = -2(1-y).
\]
Expanding both sides:
\[
2y+2 = -2 + 2y.
\]
Subtract \(2y\) from both sides:
\[
2 = -2,
\]
a contradiction. Hence, \(x \neq -2\).
Thus, for every \(y\in\mathbb{R}\setminus\{1\}\) there exists an \(x\in\mathbb{R}\setminus\{-2\}\) with \(f(x)=y\); i.e., \(f\) is **surjective**.
---
### 3. Conclusion: \(f\) Is Bijective
Since \(f\) is both injective and surjective, it is a **one‐to‐one correspondence** between \(\mathbb{R}\setminus\{-2\}\) and \(\mathbb{R}\setminus\{1\}\).
---
### 4. Find the Inverse Function \(f^{-1}\)
We found in the surjectivity step that if \(y = f(x)\) then
\[
x = \frac{2(y+1)}{1-y}.
\]
Thus, the inverse function \(f^{-1} : \mathbb{R}\setminus\{1\} \to \mathbb{R}\setminus\{-2\}\) is given by
\[
f^{-1}(y) = \frac{2(y+1)}{1-y}.
\]
Replacing the dummy variable \(y\) with \(x\) for conventional notation,
\[
f^{-1}(x) = \frac{2(x+1)}{1-x}.
\]
---
### Final Answer
For
\[
f : \mathbb{R}\setminus\{-2\} \to \mathbb{R}\setminus\{1\},\quad f(x) = \frac{x-2}{x+2},
\]
we have shown that:
- \(f\) is **bijective**.
- Its inverse is
\[
f^{-1}(x) = \frac{2(x+1)}{1-x}, \quad \text{for } x \in \mathbb{R}\setminus\{1\}.
\]
