If `f(x)=(x-2)/(x+2), x ne 2,` then what is `f^(-1)(x)` equal to?

Options

A. `(4(x+2))/(x-2)`
B. `(x+2)/(4(x-2))`
C. `(x+2)/(x-2)`
D. `(2(1+x))/(1-x)`

Answer

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

### Step 1: Set \( f(x) \) equal to \( y \)
We start by letting \( y = f(x) \):
\[ y = \frac{x-2}{x+2} \]

### Step 2: Solve for \( x \) in terms of \( y \)
Next, we will rearrange this equation to solve for \( x \). We can cross-multiply:
\[ y(x + 2) = x - 2 \]
Expanding both sides gives:
\[ yx + 2y = x - 2 \]

### Step 3: Collect all terms involving \( x \) on one side
Now, we will move all terms involving \( x \) to one side of the equation:
\[ yx - x = -2 - 2y \]
Factoring out \( x \) from the left side:
\[ x(y - 1) = -2 - 2y \]

### Step 4: Solve for \( x \)
Now, we can isolate \( x \):
\[ x = \frac{-2 - 2y}{y - 1} \]
We can simplify this expression:
\[ x = \frac{-2(1 + y)}{y - 1} \]

### Step 5: Write the inverse function
Now we have expressed \( x \) in terms of \( y \). The inverse function \( f^{-1}(x) \) is obtained by replacing \( y \) with \( x \):
\[ f^{-1}(x) = \frac{-2(1 + x)}{x - 1} \]
This can also be rewritten as:
\[ f^{-1}(x) = \frac{2(1 + x)}{1 - x} \]

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