Find \(f^{-1}(x)\)

🎥 Video Explanation

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📝 Question

Let \( f:\mathbb{R} \to (-1,1) \) be defined by

\[ f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}} \]

Find \(f^{-1}(x)\).

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✅ Solution

🔹 Step 1: Recognize Form

\[ f(x)=\tanh x \]

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🔹 Step 2: Let \(y=f(x)\)

\[ y=\frac{e^x-e^{-x}}{e^x+e^{-x}} \]

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🔹 Step 3: Solve for \(x\)

Multiply:

\[ y(e^x+e^{-x})=e^x-e^{-x} \]

\[ ye^x+ye^{-x}=e^x-e^{-x} \]

\[ e^x(y-1)= -e^{-x}(y+1) \]

Divide:

\[ \frac{e^{2x}}{1}=\frac{1+y}{1-y} \]

\[ e^{2x}=\frac{1+y}{1-y} \]

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🔹 Step 4: Take log

\[ 2x=\ln\left(\frac{1+y}{1-y}\right) \]

\[ x=\frac{1}{2}\ln\left(\frac{1+y}{1-y}\right) \] —

🔹 Step 5: Replace \(y\) by \(x\)

\[ f^{-1}(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \] —

🔹 Final Answer

\[ \boxed{f^{-1}(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)} \]