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

Options

A. Identity function
B. Even function
C. Odd function
D. None of these

Answer

To solve the problem, we need to find \( f(f^{-1}(x)) \) where \( f(x) = \frac{x-1}{x+1} \) and \( x \neq -1, 1 \).

### Step 1: Find the inverse function \( f^{-1}(x) \)

1. Start with the equation \( y = f(x) = \frac{x-1}{x+1} \).
2. Cross-multiply to eliminate the fraction:
\[ y(x + 1) = x - 1 \]
This simplifies to:
\[ xy + y = x - 1 \]
3. Rearrange the equation to isolate \( x \):
\[ xy - x = -1 - y \]
Factor out \( x \):
\[ x(y - 1) = -1 - y \]
4. Solve for \( x \):
\[ x = \frac{-1 - y}{y - 1} \]
5. Replace \( y \) with \( x \) to express the inverse:
\[ f^{-1}(x) = \frac{-1 - x}{x - 1} \]

### Step 2: Substitute \( f^{-1}(x) \) back into \( f(x) \)

Now we need to find \( f(f^{-1}(x)) \):
1. Substitute \( f^{-1}(x) \) into \( f(x) \):
\[ f\left(f^{-1}(x)\right) = f\left(\frac{-1 - x}{x - 1}\right) \]
2. Using the definition of \( f(x) \):
\[ f\left(\frac{-1 - x}{x - 1}\right) = \frac{\frac{-1 - x}{x - 1} - 1}{\frac{-1 - x}{x - 1} + 1} \]
3. Simplify the numerator:
\[ \frac{-1 - x - (x - 1)}{x - 1} = \frac{-1 - x - x + 1}{x - 1} = \frac{-2x}{x - 1} \]
4. Simplify the denominator:
\[ \frac{-1 - x + (x - 1)}{x - 1} = \frac{-1 - x + x - 1}{x - 1} = \frac{-2}{x - 1} \]
5. Now, combine the results:
\[ f(f^{-1}(x)) = \frac{\frac{-2x}{x - 1}}{\frac{-2}{x - 1}} = x \]

### Conclusion

Thus, we find that:
\[ f(f^{-1}(x)) = x \]

### Final Answer
The answer is \( x \).