Show \(f(x)=\frac{4x}{3x+4}\) is Invertible and Find \(f^{-1}\)
📝 Question
Let:
\[ f:\mathbb{R}\setminus\left\{-\frac{4}{3}\right\}\to \mathbb{R}, \quad f(x)=\frac{4x}{3x+4} \]
Show that \(f:\mathbb{R}\setminus\{-\frac{4}{3}\}\to \text{Range}(f)\) is one-one and onto. Hence, find \(f^{-1}\).
✅ Solution
🔹 Step 1: Prove that \(f\) is one-one
Assume:
\[ f(x_1)=f(x_2) \]
\[ \frac{4x_1}{3x_1+4}=\frac{4x_2}{3x_2+4} \]
Cross-multiplying:
\[ 4x_1(3x_2+4)=4x_2(3x_1+4) \]
\[ 12x_1x_2+16x_1=12x_1x_2+16x_2 \]
\[ 16x_1=16x_2 \Rightarrow x_1=x_2 \]
Hence, \(f\) is one-one (injective).
🔹 Step 2: Prove that \(f\) is onto (Range as Codomain)
Let \(y \in \text{Range}(f)\). Then:
\[ y=\frac{4x}{3x+4} \]
Solve for \(x\):
\[ y(3x+4)=4x \]
\[ 3xy+4y=4x \]
\[ x(4-3y)=4y \]
\[ x=\frac{4y}{4-3y} \]
Thus, for every \(y\) in the range, there exists \(x\) in the domain.
Hence, \(f\) is onto.
Therefore, \(f\) is bijective and invertible.
🔹 Step 3: Find the inverse function
From above:
\[ x=\frac{4y}{4-3y} \]
Interchanging \(x\) and \(y\):
\[ f^{-1}(x)=\frac{4x}{4-3x} \]
🎯 Final Answer
\[ \boxed{f^{-1}(x)=\frac{4x}{4-3x}} \]
Hence, \(f:\mathbb{R}\setminus\{-4/3\}\to \text{Range}(f)\) is invertible.
🚀 Exam Shortcut
- Use cross-multiplication to prove injectivity
- Take range as codomain ⇒ onto automatically
- Solve \(y=f(x)\) for \(x\) to get inverse