The inverse of the function `y=(e^(2x)-e^(-2x))/(e^(2x)+e^(-2x))` is/an

Options

A. `log_(e )((1+2x)/(1-2x))`
B. `1/4 log_(e )((1-x)/(1+x))`
C. `1/4 log_(e )((1+x)/(1-x))`
D. an odd function

Answer

To find the inverse of the function \( y = \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} \), we will follow these steps:

### Step 1: Rewrite the equation
Start by expressing the function in terms of \( y \):
\[ y = \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} \]

### Step 2: Cross-multiply
To eliminate the fraction, cross-multiply:
\[ y(e^{2x} + e^{-2x}) = e^{2x} - e^{-2x} \]
This leads to:
\[ ye^{2x} + ye^{-2x} = e^{2x} - e^{-2x} \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ ye^{2x} - e^{2x} = -ye^{-2x} - e^{-2x} \]
Factoring out \( e^{2x} \) and \( e^{-2x} \):
\[ (1 - y)e^{2x} = -(1 + y)e^{-2x} \]

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

### Step 5: Isolate \( e^{4x} \)
Rearranging gives:
\[ e^{4x} = -\frac{(1 + y)}{(1 - y)} \]

### Step 6: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ 4x = \ln\left(-\frac{(1 + y)}{(1 - y)}\right) \]

### Step 7: Solve for \( x \)
Now, solve for \( x \):
\[ x = \frac{1}{4} \ln\left(-\frac{(1 + y)}{(1 - y)}\right) \]

### Step 8: Express \( y \) in terms of \( x \)
To find the inverse function, we replace \( y \) with \( x \):
\[ f^{-1}(x) = \frac{1}{4} \ln\left(-\frac{(1 + x)}{(1 - x)}\right) \]

### Final Result
Thus, the inverse of the function is:
\[ f^{-1}(x) = \frac{1}{4} \ln\left(-\frac{(1 + x)}{(1 - x)}\right) \]