Question

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

Hint:

For rational functions \( \frac{ax+b}{cx+d} \), the value \( y = a/c \) is usually excluded from the range.
If the codomain were changed to \( R - \{1\} \), then the function would be onto.

Answer 1

Concept:
• A function \( f: A \rightarrow B \) is said to be onto (surjective) if every element in the codomain \( B \) has at least one pre-image in the domain \( A \).
• To check for onto, we let \( y \in B \) and try to find \( x \in A \) such that \( f(x) = y \).
• If there is any value in the codomain for which no pre-image exists, the function is not onto.

Step 1: Set the function equal to \( y \) and solve for \( x \)
Let \( y = f(x) \), where \( y \in R \) (the codomain).
\[ y = \frac{x-2}{x-3} \]
Multiply both sides by \( (x-3) \):
\[ y(x-3) = x - 2 \]
\[ xy - 3y = x - 2 \]
Rearrange the terms to group \( x \) on one side:
\[ xy - x = 3y - 2 \]
\[ x(y-1) = 3y - 2 \]
Divide by \( (y-1) \):
\[ x = \frac{3y-2}{y-1} \]

Step 2: Check if every value of \( y \) in the codomain has a pre-image
From the expression \( x = \frac{3y-2}{y-1} \), we can see that \( x \) is defined for all real values of \( y \) except when the denominator is zero.
The denominator is zero when:
\[ y - 1 = 0 \implies y = 1 \]
This means that for the value \( y = 1 \) in the codomain \( R \), there is no corresponding real value of \( x \) in the domain.
Even if we check the domain restriction, \( x = 3 \) would require \( 3y-2 = 3y-3 \), which is impossible, so \( y=1 \) truly has no pre-image.

Step 3: Conclusion
Since there exists an element \( y = 1 \) in the codomain \( R \) which does not have a pre-image in the domain \( R - \{3\} \), the function \( f \) is not onto.