Show \(f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}}\) is Invertible and Find \(f^{-1}\)
📝 Question
Let:
\[ f:\mathbb{R}\to(-1,1), \quad f(x)=\frac{10^x-10^{-x}}{10^x+10^{-x}} \]
Show that \(f\) is invertible and find \(f^{-1}\).
✅ Solution
🔹 Step 1: Prove that \(f\) is one-one
The function is strictly increasing because exponential functions are increasing.
Hence, \(f\) is one-one (injective).
🔹 Step 2: Range of the function
Since:
\[ -1 < \frac{10^x-10^{-x}}{10^x+10^{-x}} < 1 \]
So, range of \(f\) is \((-1,1)\).
Thus, \(f:\mathbb{R}\to(-1,1)\) is onto.
Hence, \(f\) is bijective and invertible.
🔹 Step 3: Find the inverse
Let:
\[ y=\frac{10^x-10^{-x}}{10^x+10^{-x}} \]
Multiply numerator and denominator by \(10^x\):
\[ y=\frac{10^{2x}-1}{10^{2x}+1} \]
Solve for \(x\):
\[ y(10^{2x}+1)=10^{2x}-1 \]
\[ y\cdot 10^{2x}+y=10^{2x}-1 \]
\[ 10^{2x}(1-y)=1+y \]
\[ 10^{2x}=\frac{1+y}{1-y} \]
Taking logarithm:
:contentReference[oaicite:0]{index=0}Interchanging \(x\) and \(y\):
\[ f^{-1}(x)=\frac{1}{2}\log_{10}\!\left(\frac{1+x}{1-x}\right) \]
🎯 Final Answer
\[ \boxed{f^{-1}(x)=\frac{1}{2}\log_{10}\!\left(\frac{1+x}{1-x}\right)} \]
🚀 Exam Shortcut
- Multiply by \(10^x\) to simplify expression
- Convert into \(10^{2x}\) form
- Solve algebraically and apply logarithm
- Always interchange variables at the end