If f (x) `=(x-1)/(x+1),` then prove that `f{f(x)}=-1/x.`

Explanation

To prove that \( f(f(x)) = -\frac{1}{x} \) for the function \( f(x) = \frac{x-1}{x+1} \), we will follow these steps:

### Step 1: Define the function
We start with the function:
\[ f(x) = \frac{x-1}{x+1} \]

### Step 2: Find \( f(f(x)) \)
To find \( f(f(x)) \), we need to substitute \( f(x) \) into itself:
\[ f(f(x)) = f\left(\frac{x-1}{x+1}\right) \]

### Step 3: Substitute \( f(x) \) into the function
Now we replace \( x \) in \( f(x) \) with \( \frac{x-1}{x+1} \):
\[ f\left(\frac{x-1}{x+1}\right) = \frac{\left(\frac{x-1}{x+1}\right) - 1}{\left(\frac{x-1}{x+1}\right) + 1} \]

### Step 4: Simplify the numerator
The numerator becomes:
\[ \frac{x-1}{x+1} - 1 = \frac{x-1 - (x+1)}{x+1} = \frac{x-1-x-1}{x+1} = \frac{-2}{x+1} \]

### Step 5: Simplify the denominator
The denominator becomes:
\[ \frac{x-1}{x+1} + 1 = \frac{x-1 + (x+1)}{x+1} = \frac{x-1+x+1}{x+1} = \frac{2x}{x+1} \]

### Step 6: Combine the results
Now we can write:
\[ f(f(x)) = \frac{\frac{-2}{x+1}}{\frac{2x}{x+1}} = \frac{-2}{2x} = -\frac{1}{x} \]

### Step 7: Conclusion
Thus, we have shown that:
\[ f(f(x)) = -\frac{1}{x} \]
This completes the proof.