The inverse of the function `f:Rto` range of f, defined by `f(x)=(e^(x)-e^(-x))/(e^(x)+e^(-x))` is

Options

A. `(1)/(2)log""(1+x)/(1-x)`
B. `(1)/(2)log""(2+x)/(2-x)`
C. `(1)/(2)log""(1-x)/(1+x)`
D. none of these

Answer

To find the inverse of the function \( f: \mathbb{R} \to \text{range of } f \) defined by

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

we will follow these steps:

### Step 1: Set up the equation
Let \( y = f(x) \). Then we have:

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

### Step 2: Clear the fraction
Multiply both sides by \( e^x + e^{-x} \):

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

### Step 3: Rearrange the equation
Rearranging gives:

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

### Step 4: Collect like terms
Rearranging further, we can group the terms involving \( e^x \):

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

This simplifies to:

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

### Step 5: Eliminate \( e^{-x} \)
Multiply both sides by \( e^x \):

\[ e^{2x}(y - 1) = -(y + 1). \]

### Step 6: Solve for \( e^{2x} \)
Rearranging gives:

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

### Step 7: Take the natural logarithm
Taking the natural logarithm on both sides:

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

### Step 8: Solve for \( x \)
Dividing by 2 gives:

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

### Step 9: Express the inverse function
Since \( y = f(x) \), we can express the inverse function as:

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

### Step 10: Replace \( y \) with \( x \)
To write the inverse function in terms of \( x \):

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

### Final Result
Thus, the inverse of the function \( f \) is:

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

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