Answer
\[
G^{-1}(y) \;=\; \frac{2 - y}{3}.
\]
Work Step by Step
Given the function
\[
G : \mathbb{R} \;\to\; \mathbb{R}
\quad\text{by}\quad
G(x) \;=\; 2 \;-\; 3x
\quad\text{for all real } x.
\]
We need to determine whether \(G\) is a one-to-one correspondence (i.e., bijection) on \(\mathbb{R}\). If it is, we also find its inverse.
---
## 1. Injectivity
Suppose \(G(x_1) = G(x_2)\). That is,
\[
2 - 3x_1 \;=\; 2 - 3x_2.
\]
Subtracting 2 from both sides gives
\[
-3x_1 \;=\; -3x_2
\;\;\Longrightarrow\;\;
x_1 \;=\; x_2.
\]
Hence, \(G\) is **injective**.
---
## 2. Surjectivity (Onto)
To check if \(G\) is onto \(\mathbb{R}\), let \(y\) be any real number. We want an \(x \in \mathbb{R}\) such that
\[
G(x) = y
\;\;\Longleftrightarrow\;\;
2 - 3x = y
\;\;\Longleftrightarrow\;\;
-3x = y - 2
\;\;\Longleftrightarrow\;\;
x = \frac{2 - y}{3}.
\]
Because \(\frac{2 - y}{3}\) is a real number for any real \(y\), **every** real \(y\) has a preimage under \(G\). Thus \(G\) is **surjective**.
---
## 3. Conclusion: One-to-One Correspondence
Since \(G\) is both injective and surjective on \(\mathbb{R}\), it is a **bijection** (one-to-one correspondence).
---
## 4. Finding the Inverse Function
To find \(G^{-1}\), solve for \(x\) in terms of \(y\):
\[
y = 2 - 3x
\quad\Longrightarrow\quad
2 - y = 3x
\quad\Longrightarrow\quad
x = \frac{2 - y}{3}.
\]
Hence, the inverse function \(G^{-1} : \mathbb{R} \to \mathbb{R}\) is
\[
G^{-1}(y) = \frac{2 - y}{3}.
\]
---
### Final Answer:
1. \(G(x) = 2 - 3x\) is a **one-to-one correspondence** \(\mathbb{R}\to \mathbb{R}\).
2. Its inverse is
\[
G^{-1}(y) \;=\; \frac{2 - y}{3}.
\]
