Question

A function \( f \) is defined from \( R \to R \) as \( f(x) = ax + b \), such that \( f(1) = 1 \) and \( f(2) = 3 \). Find the function \( f(x) \). Hence, check whether the function \( f(x) \) is one-one and onto.

Hint:

\textbf{Part (b):} For a function \(f(x) = ax + b\), use the given points to form and solve simultaneous equations for \(a\) and \(b\). Check one-to-one by verifying \(f'(x) \neq 0\) and onto by solving \(f(x) = y\) for all \(y \in \mathbb{R}\).

Answer 1

Given the conditions, we establish two equations: \( f(1) = a(1) + b = 1 \), which simplifies to \( a + b = 1 \) (Equation 1), and \( f(2) = a(2) + b = 3 \), which simplifies to \( 2a + b = 3 \) (Equation 2). To solve these equations simultaneously, we first express \( b \) from Equation 1 as \( b = 1 - a \). Substituting this expression for \( b \) into Equation 2 yields \( 2a + (1 - a) = 3 \). This simplifies to \( 2a + 1 - a = 3 \), which further reduces to \( a = 2 \). Substituting \( a = 2 \) back into Equation 1 gives \( 2 + b = 1 \), leading to \( b = -1 \). Therefore, the function is determined as \( f(x) = 2x - 1 \).

The function is one-one (injective) if distinct inputs produce distinct outputs. As \( f(x) = 2x - 1 \) is a linear function with a non-zero slope, it satisfies this condition.

The function is onto (surjective) if for every \( y \in R \), there exists an \( x \in R \) such that \( f(x) = y \). For \( f(x) = 2x - 1 \), we can solve \( y = 2x - 1 \) for \( x \), obtaining \( x = \frac{y + 1}{2} \). Since for any \( y \in R \), a corresponding \( x \in R \) can be found, the function is onto.

Answer: The function \( f(x) = 2x - 1 \) is both one-one and onto. \bigskip