Show \(f(x)=3x+5\) is Invertible on \(\mathbb{Q}\) and Find \(f^{-1}\)
📺 Video Explanation
📝 Question
Show that:
\[ f:\mathbb{Q}\to\mathbb{Q},\qquad f(x)=3x+5 \]
is invertible. Also find:
\[ f^{-1}(x) \]
✅ Solution
🔹 Step 1: Show that \(f\) is one-one
Assume:
\[ f(x_1)=f(x_2) \]
Then:
\[ 3x_1+5=3x_2+5 \]
Subtract 5:
\[ 3x_1=3x_2 \]
Divide by 3:
\[ x_1=x_2 \]
Therefore:
\[ f \text{ is one-one} \]
🔹 Step 2: Show that \(f\) is onto
Let:
\[ y\in\mathbb{Q} \]
We must show there exists:
\[ x\in\mathbb{Q} \]
such that:
\[ f(x)=y \]
So:
\[ 3x+5=y \]
Solve for \(x\):
\[ 3x=y-5 \]
\[ x=\frac{y-5}{3} \]
Since \(y\in\mathbb{Q}\), and rationals are closed under subtraction and division by non-zero numbers:
\[ x\in\mathbb{Q} \]
Hence:
\[ f \text{ is onto} \]
🔹 Step 3: Find inverse function
Let:
\[ y=3x+5 \]
Solve for \(x\):
\[ x=\frac{y-5}{3} \]
Replace \(y\) by \(x\):
\[ \boxed{f^{-1}(x)=\frac{x-5}{3}} \]
🎯 Final Answer
Since \(f\) is both one-one and onto, it is invertible.
Therefore:
\[ \boxed{f^{-1}(x)=\frac{x-5}{3}} \]
🚀 Exam Shortcut
- Linear function with non-zero coefficient is one-one
- Solve \(y=f(x)\) for \(x\)
- Replace \(y\) by \(x\) to get inverse