Show \(f(x)=\frac{x-2}{x-3}\) is Bijective and Find \(f^{-1}\)
📺 Video Explanation
📝 Question
Let:
\[ A=\mathbb{R}\setminus\{3\},\qquad B=\mathbb{R}\setminus\{1\} \]
and:
\[ f:A\to B,\qquad f(x)=\frac{x-2}{x-3} \]
Show that \(f\) is one-one and onto, and hence find:
\[ f^{-1}(x) \]
✅ Solution
🔹 Step 1: Show that \(f\) is one-one
Assume:
\[ f(x_1)=f(x_2) \]
Then:
\[ \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3} \]
Cross multiply:
\[ (x_1-2)(x_2-3)=(x_2-2)(x_1-3) \]
Expand:
\[ x_1x_2-3x_1-2x_2+6=x_1x_2-3x_2-2x_1+6 \]
Simplify:
\[ -x_1-x_2=-x_2-x_1 \]
This reduces to:
\[ x_1=x_2 \]
Hence:
\[ f \text{ is one-one} \]
🔹 Step 2: Show that \(f\) is onto
Let:
\[ y\in B=\mathbb{R}\setminus\{1\} \]
Need:
\[ f(x)=y \]
So:
\[ \frac{x-2}{x-3}=y \]
Cross multiply:
\[ x-2=y(x-3) \]
Expand:
\[ x-2=xy-3y \]
Bring terms together:
\[ x-xy=2-3y \]
Factor:
\[ x(1-y)=2-3y \]
Since:
\[ y\ne1 \]
we can divide:
\[ x=\frac{2-3y}{1-y} \]
Thus for every \(y\in B\), there exists:
\[ x=\frac{2-3y}{1-y}\in A \]
So \(f\) is onto.
🔹 Step 3: Find inverse
Let:
\[ y=\frac{x-2}{x-3} \]
Then:
\[ x=\frac{2-3y}{1-y} \]
Replace \(y\) by \(x\):
\[ \boxed{f^{-1}(x)=\frac{2-3x}{1-x}} \]
🎯 Final Answer
Since \(f\) is one-one and onto, it is bijective.
Therefore:
\[ \boxed{f^{-1}(x)=\frac{2-3x}{1-x}} \]
🚀 Exam Shortcut
- Use cross multiplication for injective proof
- Solve for \(x\) in terms of \(y\)
- Exclude restricted values carefully