Show \(f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}\) is Invertible and Find \(f^{-1}\)
📝 Question
Let:
\[ f:\mathbb{R}\to(-1,1), \quad f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}} \]
Show that \(f\) is invertible and find \(f^{-1}\).
✅ Solution
🔹 Step 1: Prove that \(f\) is one-one
The function is strictly increasing since exponential functions are increasing.
Hence, \(f\) is one-one (injective).
🔹 Step 2: Range of the function
We know:
\[ -1 < \frac{e^x-e^{-x}}{e^x+e^{-x}} < 1 \]
Thus, range of \(f\) is \((-1,1)\).
So \(f:\mathbb{R}\to(-1,1)\) is onto.
Hence, \(f\) is bijective and invertible.
🔹 Step 3: Find the inverse
Let:
\[ y=\frac{e^x-e^{-x}}{e^x+e^{-x}} \]
Multiply numerator and denominator by \(e^x\):
\[ y=\frac{e^{2x}-1}{e^{2x}+1} \]
Solve for \(x\):
\[ y(e^{2x}+1)=e^{2x}-1 \]
\[ y e^{2x}+y=e^{2x}-1 \]
\[ e^{2x}(1-y)=1+y \]
\[ e^{2x}=\frac{1+y}{1-y} \]
Taking natural logarithm:
:contentReference[oaicite:0]{index=0}Interchanging \(x\) and \(y\):
\[ f^{-1}(x)=\frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right) \]
🎯 Final Answer
\[ \boxed{f^{-1}(x)=\frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right)} \]
🚀 Exam Shortcut
- Recognize this as \(\tanh x\) form
- Use identity: \( \tanh^{-1}x = \frac{1}{2}\ln\frac{1+x}{1-x} \) :contentReference[oaicite:1]{index=1}
- Convert into \(e^{2x}\) form for quick solving
- Apply logarithm and interchange variables