If `f(x)=(x+1)/(x-1)`, `x ne 1` , then `f^(-1)(x)` is

Options

A. `(x+1)/(x-1)`, `x ne 1`
B. `(x+1)/(1-x)`, `x ne 1`
C. `(1)/(1+x)`, `x ne -1`
D. `(1)/(x-1)`, `x ne 1`

Answer

To find the inverse of the function \( f(x) = \frac{x + 1}{x - 1} \) where \( x \neq 1 \), we can follow these steps:

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

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ y(x - 1) = x + 1 \]

### Step 3: Expand and rearrange the equation
Expanding the left side:
\[ yx - y = x + 1 \]
Now, rearranging to isolate terms involving \( x \):
\[ yx - x = y + 1 \]

### Step 4: Factor out \( x \)
Factoring \( x \) out from the left side:
\[ x(y - 1) = y + 1 \]

### Step 5: Solve for \( x \)
Now, solve for \( x \):
\[ x = \frac{y + 1}{y - 1} \]

### Step 6: Change the variable to find \( f^{-1}(x) \)
Since we initially set \( y = f(x) \), we can now replace \( y \) with \( x \) to express the inverse function:
\[ f^{-1}(x) = \frac{x + 1}{x - 1} \]

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

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