Answer
\[
G^{-1}(y) \;=\; \frac{y + 5}{4}.
\]
Work Step by Step
Given the function
\[
G : \mathbb{R} \;\to\; \mathbb{R}
\quad\text{by}\quad
G(x) = 4x \;-\; 5
\quad\text{for all real } x.
\]
We are to check whether \(G\) is onto (and, more broadly, whether it is a one-to-one correspondence).
---
## 1. Injectivity
Even though the exercise specifically asks about onto, let us first note injectivity. Suppose
\[
G(x_1) = G(x_2).
\]
That is,
\[
4x_1 - 5 = 4x_2 - 5
\;\;\Longrightarrow\;\;
4x_1 = 4x_2
\;\;\Longrightarrow\;\;
x_1 = x_2.
\]
Hence \(G\) is **injective**.
---
## 2. Surjectivity (Onto)
To check surjectivity, take an arbitrary real number \(y \in \mathbb{R}\). We want to find \(x \in \mathbb{R}\) such that
\[
G(x) = y \quad\Longleftrightarrow\quad 4x - 5 = y \quad\Longleftrightarrow\quad 4x = y + 5 \quad\Longleftrightarrow\quad x = \frac{y + 5}{4}.
\]
Since \(\frac{y+5}{4}\) is a real number for any real \(y\), we see that **every** \(y \in \mathbb{R}\) has a preimage \(x \in \mathbb{R}\). Hence \(G\) is **onto**.
---
## 3. Conclusion: One-to-One Correspondence
Because \(G\) is both injective and surjective on \(\mathbb{R}\), it is a **one-to-one correspondence** (bijection). Therefore, it has an inverse function.
---
## 4. Finding the Inverse
The inverse function \(G^{-1} : \mathbb{R} \to \mathbb{R}\) is given by solving \(y = 4x - 5\) for \(x\):
\[
x = \frac{y + 5}{4}.
\]
Thus,
\[
G^{-1}(y) = \frac{y + 5}{4}.
\]
---
### Final Answer for Exercise 47
1. \(G(x) = 4x - 5\) is a **one-to-one correspondence** \(\mathbb{R} \to \mathbb{R}\).
2. Its inverse is
\[
G^{-1}(y) \;=\; \frac{y + 5}{4}.
\]
