If `f(x)=(x-1)/(x+1),xne1`, then `f(f(x))`=__________.

Answer

To solve the problem of finding \( f(f(x)) \) for the function \( f(x) = \frac{x-1}{x+1} \) where \( x \neq 1 \), we will follow these steps:

### Step 1: Substitute \( f(x) \) into itself
We start by substituting \( f(x) \) into itself:
\[ f(f(x)) = f\left(\frac{x-1}{x+1}\right) \]

### Step 2: Apply the function definition
Now we need to apply the function \( f \) to \( \frac{x-1}{x+1} \):
\[ f\left(\frac{x-1}{x+1}\right) = \frac{\left(\frac{x-1}{x+1}\right) - 1}{\left(\frac{x-1}{x+1}\right) + 1} \]

### Step 3: Simplify the numerator
To simplify the numerator:
\[ \frac{x-1}{x+1} - 1 = \frac{x-1}{x+1} - \frac{x+1}{x+1} = \frac{x-1 - (x+1)}{x+1} = \frac{x-1-x-1}{x+1} = \frac{-2}{x+1} \]

### Step 4: Simplify the denominator
Now simplify the denominator:
\[ \frac{x-1}{x+1} + 1 = \frac{x-1}{x+1} + \frac{x+1}{x+1} = \frac{x-1 + (x+1)}{x+1} = \frac{x-1+x+1}{x+1} = \frac{2x}{x+1} \]

### Step 5: Combine the results
Now we can combine the results from the numerator and denominator:
\[ f(f(x)) = \frac{\frac{-2}{x+1}}{\frac{2x}{x+1}} = \frac{-2}{2x} = \frac{-1}{x} \]

### Final Result
Thus, we have:
\[ f(f(x)) = \frac{-1}{x}, \quad \text{where } x \neq 0 \]

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