The inverse of the function `f(x)=(e^(x)-2e^(-x))/(e^(x)+2e^(-x))+1` is

Options

A. `log_(10)((2x)/(2-x))`
B. `log_(10)((x)/(2-x))`
C. `log_(e )((2x)/(2-x))^(1//2)`
D. `In((2x)/(2-x))^(1//2)`

Answer

To find the inverse of the function \( f(x) = \frac{e^x - 2e^{-x}}{e^x + 2e^{-x}} + 1 \), we will follow these steps:

### Step 1: Set the function equal to \( y \)
Let \( y = f(x) \). This gives us the equation:
\[ y = \frac{e^x - 2e^{-x}}{e^x + 2e^{-x}} + 1 \]

### Step 2: Rearrange the equation
To eliminate the fraction, we can rearrange the equation:
\[ y - 1 = \frac{e^x - 2e^{-x}}{e^x + 2e^{-x}} \]
Multiplying both sides by \( e^x + 2e^{-x} \) gives:
\[ (y - 1)(e^x + 2e^{-x}) = e^x - 2e^{-x} \]

### Step 3: Expand and collect terms
Expanding the left side:
\[ (y - 1)e^x + 2(y - 1)e^{-x} = e^x - 2e^{-x} \]
Rearranging gives:
\[ (y - 1)e^x - e^x = -2(y - 1)e^{-x} - 2e^{-x} \]
This simplifies to:
\[ (y - 2)e^x = -2(y + 1)e^{-x} \]

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

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

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

### Step 7: Express \( y \) in terms of \( x \)
Now, we can express \( y \) in terms of \( x \) to find the inverse function:
\[ f^{-1}(x) = \frac{1}{2} \ln\left(\frac{-2(x + 1)}{x - 2}\right) \]

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

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