The inverse of the function `f(x)=(e^(x)-e^(-x))/(e^(x)+e^(-x))`

Options

A. `1/2 ln ((1+x)/(1-x))`
B. `1/2 ln (2+x)/(2-x)`
C. `1/2 ln (1-x)/(1+x)`
D. `2 ln (1+x)`

Explanation

To find the inverse of the function \( f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \), we will follow these steps:

### Step 1: Set the function equal to \( y \)
Let \( y = f(x) \):
\[ y = \frac{e^x - e^{-x}}{e^x + e^{-x}} \]

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ y(e^x + e^{-x}) = e^x - e^{-x} \]
This simplifies to:
\[ ye^x + ye^{-x} = e^x - e^{-x} \]

### Step 3: Rearrange the equation
Rearranging the equation, we get:
\[ ye^x - e^x = -ye^{-x} - e^{-x} \]
Factoring out \( e^x \) and \( e^{-x} \):
\[ (1 - y)e^x = -(1 + y)e^{-x} \]

### Step 4: Multiply both sides by \( e^x \)
To eliminate \( e^{-x} \), we multiply both sides by \( e^x \):
\[ (1 - y)e^{2x} = -(1 + y) \]

### Step 5: Solve for \( e^{2x} \)
Rearranging gives:
\[ e^{2x} = -\frac{(1 + y)}{(1 - y)} \]

### Step 6: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ 2x = \ln\left(-\frac{(1 + y)}{(1 - y)}\right) \]

### Step 7: Solve for \( x \)
Dividing by 2 gives:
\[ x = \frac{1}{2} \ln\left(-\frac{(1 + y)}{(1 - y)}\right) \]

### Step 8: Replace \( y \) with \( x \) to find the inverse
Now we replace \( y \) with \( x \) to express the inverse function:
\[ f^{-1}(x) = \frac{1}{2} \ln\left(-\frac{(1 + x)}{(1 - x)}\right) \]

### Final Answer
Thus, the inverse of the function \( f(x) \) is:
\[ f^{-1}(x) = \frac{1}{2} \ln\left(-\frac{(1 + x)}{(1 - x)}\right) \]

---