Question

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

Hint:

Linear functions of the form \( f(x) = ax + b \) (where \( a \neq 0 \)) mapping from \( \mathbb{R} \) to \( \mathbb{R} \) are always bijective (both one-one and onto).

Answer 1

Step 1: Understanding the Concept:
A function is one-one (injective) if distinct elements in the domain have distinct images in the codomain.
A function is onto (surjective) if every element in the codomain is the image of at least one element in the domain.
Step 2: Detailed Explanation:
Checking for One-one:
Let \( x_1, x_2 \in \mathbb{R} \) such that \( f(x_1) = f(x_2) \).
\[ \frac{5x_1 + 3}{4} = \frac{5x_2 + 3}{4} \]
Multiply both sides by 4:
\[ 5x_1 + 3 = 5x_2 + 3 \]
Subtract 3 from both sides:
\[ 5x_1 = 5x_2 \]
Divide by 5:
\[ x_1 = x_2 \]
Since \( f(x_1) = f(x_2) \implies x_1 = x_2 \), the function is one-one.
Checking for Onto:
Let \( y \in \mathbb{R} \) (Codomain). We need to find \( x \in \mathbb{R} \) (Domain) such that \( f(x) = y \).
\[ y = \frac{5x + 3}{4} \]
\[ 4y = 5x + 3 \]
\[ 5x = 4y - 3 \]
\[ x = \frac{4y - 3}{5} \]
For every \( y \in \mathbb{R} \), the value \( x = \frac{4y - 3}{5} \) is a real number.
Thus, for every \( y \) in the codomain, there exists an \( x \) in the domain such that \( f(x) = y \).
Therefore, the function is onto.
Step 3: Final Answer:
The function \( f(x) \) is both one-one and onto.