Question

Prove that the function \( f : \mathbb{R} \to \mathbb{R} \), \( f(x) = \frac{5x + 3}{4} \), is one-one and onto.

Hint:

A function is one-one if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), and it is onto if every element in the target set has a corresponding element in the domain.

Answer 1

Step 1: Proving the function is one-one (Injective).
A function is one-one if, for all \( x_1, x_2 \in \mathbb{R} \), whenever \( f(x_1) = f(x_2) \), it follows that \( x_1 = x_2 \).
Let \( f(x_1) = f(x_2) \), so: \[ \frac{5x_1 + 3}{4} = \frac{5x_2 + 3}{4} \] Multiplying both sides by 4: \[ 5x_1 + 3 = 5x_2 + 3 \] Simplifying: \[ 5x_1 = 5x_2 \] \[ x_1 = x_2 \] Thus, the function is one-one. Step 2: Proving the function is onto (Surjective).
A function is onto if, for every element \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \).
Let \( y \) be an arbitrary element of \( \mathbb{R} \). We want to find \( x \) such that: \[ f(x) = y \] Substitute the expression for \( f(x) \): \[ \frac{5x + 3}{4} = y \] Multiplying both sides by 4: \[ 5x + 3 = 4y \] Solving for \( x \): \[ 5x = 4y - 3 \] \[ x = \frac{4y - 3}{5} \] Thus, for every \( y \in \mathbb{R} \), there exists an \( x = \frac{4y - 3}{5} \in \mathbb{R} \) such that \( f(x) = y \). Therefore, the function is onto. Step 3: Conclusion.
Since the function is both one-one and onto, it is bijective.