Question

A function \[ f:\mathbb{R}-\left\{\frac{3}{5}\right\} \to \mathbb{R}-\left\{\frac{3}{5}\right\} \] is defined as \[ f(x)=\frac{3x+2}{5x-3}. \] Show that \(f\) is one-one and onto.

Hint:

For rational functions \( \frac{ax+b}{cx+d} \), if \( ad - bc \neq 0 \), the function is always one-one.
The horizontal asymptote \( y = a/c \) is always excluded from the range of such functions.

Answer 1

Concept:
• One-one (Injective): \( f(x_1) = f(x_2) \implies x_1 = x_2 \).
• Onto (Surjective): For every \( y \) in the codomain, there exists an \( x \) in the domain such that \( f(x) = y \).

Step 1: Verify if the function is one-one
Let \( f(x_1) = f(x_2) \): \[ \frac{3x_1 + 2}{5x_1 - 3} = \frac{3x_2 + 2}{5x_2 - 3} \] Cross-multiplying: \[ (3x_1 + 2)(5x_2 - 3) = (3x_2 + 2)(5x_1 - 3) \] \[ 15x_1x_2 - 9x_1 + 10x_2 - 6 = 15x_1x_2 - 9x_2 + 10x_1 - 6 \] \[ -9x_1 + 10x_2 = -9x_2 + 10x_1 \] \[ 19x_2 = 19x_1 \implies x_1 = x_2 \] Thus, \( f \) is one-one.

Step 2: Verify if the function is onto
Let \( y = f(x) = \frac{3x + 2}{5x - 3} \). We need to express \( x \) in terms of \( y \): \[ y(5x - 3) = 3x + 2 \] \[ 5xy - 3y = 3x + 2 \] \[ 5xy - 3x = 3y + 2 \implies x(5y - 3) = 3y + 2 \] \[ x = \frac{3y + 2}{5y - 3} \] Since the codomain is \( R - \{3/5\} \), \( y \neq 3/5 \), so \( 5y - 3 \neq 0 \).
For any \( y \in R - \{3/5\} \), \( x \) is a real number.
Also, if \( x = 3/5 \), then \( 3(3/5) + 2 = y(5(3/5) - 3) \implies 1.8 + 2 = y(0) \), which is impossible. So \( x \in R - \{3/5\} \).
Thus, every \( y \) has a pre-image, and the function is onto.