Question

Check whether the function \[ f:\mathbb{R}\setminus\{3\}\rightarrow\mathbb{R} \] defined by \[ f(x)=\frac{x-2}{x-3} \] is onto (surjective) or not.

Hint:

For any rational function of the form \( f(x) = \frac{ax+b}{cx+d} \), the horizontal asymptote is always given by \( y = \frac{a}{c} \). The function will never be able to attain this specific value, meaning its range will automatically exclude \( \frac{a}{c} \). Here \( a=1, c=1 \), so \( y = 1 \) is excluded.

Answer 1

Concept: A function \( f : A \rightarrow B \) is defined as onto (surjective) if every element in the codomain set \( B \) has at least one corresponding pre-image element in the domain set \( A \).
• This means that the total Range of the function must be exactly identical to its given Codomain.
• To find the range, we set \( y = f(x) \), rearrange the algebraic terms to express \( x \) explicitly as a function of \( y \), and evaluate which real values of \( y \) produce valid real values of \( x \) lying inside the domain.

Step 1: Set up the equation \( y = f(x) \).
Let \( y \) be an arbitrary element belonging to the codomain \(\mathbb{R}\). We set: \[ y = \frac{x - 2}{x - 3} \]

Step 2: Rearrange to express \( x \) in terms of \( y \).
Cross-multiply to remove the fraction denominator: \[ y(x - 3) = x - 2 \] Expand the left side of the equation: \[ xy - 3y = x - 2 \] Bring all terms containing the variable \( x \) to the left side, and move all other terms to the right side: \[ xy - x = 3y - 2 \] Factor out \( x \) on the left-hand side: \[ x(y - 1) = 3y - 2 \] Dividing both sides by \((y - 1)\): \[ x = \frac{3y - 2}{y - 1} \quad \cdots (1) \]

Step 3: Analyze the range constraints and domain compatibility.
From equation (1), we can see that \( x \) is a well-defined real number for all values of \( y \) except when the denominator becomes zero: \[ y - 1 = 0 \quad \Rightarrow \quad y = 1 \] If we substitute \( y = 1 \) into our initial setup equation: \[ 1 = \frac{x - 2}{x - 3} \quad \Rightarrow \quad x - 3 = x - 2 \quad \Rightarrow \quad -3 = -2 \] This is a logical contradiction, which means there is absolutely no value of \( x \in \mathbb{R} - \{3\} \) that can map onto the value \( y = 1 \).

Step 4: Conclusion based on definitions.
The given codomain of the function is the entire set of real numbers \(\mathbb{R}\). However, the computed range of the function is: \[ \text{Range} = \mathbb{R} - \{1\} \] Since \(\text{Range} \neq \text{Codomain}\) (specifically, the element \(1 \in \text{Codomain}\) has no pre-image), the function \( f(x) \) is not onto.