The problem you provided reads as follows (translated into English for clarity):

Exercise 1.8. Let $f: \mathbb{R} \to \mathbb{R}$ be a one-to-one and onto (bijective) function. Define $g: \mathbb{R} \to \mathbb{R}$ by $g(x) := a f(x) + b$, with $a \neq 0$ and $b$ a constant. Prove that $g$ is also a one-to-one and onto function. Then, find the formula for $g^{-1}(x)$ (the inverse of $g$). Apply the given formula when $f(x) = \frac{4x}{x-2}$.

Step-by-step solution:

1. Proving that $g$ is one-to-one (injective):

   We need to show that for $g(x_1) = g(x_2)$, it must follow that $x_1 = x_2$.

   $g(x_1) = a f(x_1) + b, \quad g(x_2) = a f(x_2) + b$

   If $g(x_1) = g(x_2)$, we get: $a f(x_1) + b = a f(x_2) + b$ Subtract $b$ from both sides: $a f(x_1) = a f(x_2)$ Since $a \neq 0$, divide by $a$: $f(x_1) = f(x_2)$ Because $f$ is injective (one-to-one), this implies $x_1 = x_2$. Thus, $g$ is injective.

2. Proving that $g$ is onto (surjective):

   We need to show that for every $y \in \mathbb{R}$, there exists an $x \in \mathbb{R}$ such that $g(x) = y$.

   Let $y = g(x) = a f(x) + b$. Solving for $f(x)$: $y - b = a f(x)$ Divide by $a$: $f(x) = \frac{y - b}{a}$ Since $f$ is surjective (onto), there exists an $x \in \mathbb{R}$ such that $f(x) = \frac{y - b}{a}$. Therefore, $g(x) = y$, meaning $g$ is surjective.

   Combining injectivity and surjectivity, we conclude that $g$ is bijective.

3. Finding the inverse $g^{-1}(x)$:

   To find $g^{-1}(x)$, we start with the equation: $y = g(x) = a f(x) + b$ Sol