If `f(x)=(3x-1)/(x+1),xne`____________, then fof (x) is____________

Answer

To solve the problem, we need to find \( f(f(x)) \) given the function \( f(x) = \frac{3x - 1}{x + 1} \).

### Step-by-Step Solution:

1. **Identify the function**:
We have \( f(x) = \frac{3x - 1}{x + 1} \).

2. **Determine the domain**:
The function is defined for all \( x \) except where the denominator is zero.
\[ x + 1 \neq 0 \implies x \neq -1 \]

3. **Find \( f(f(x)) \)**:
We need to substitute \( f(x) \) back into itself. First, we compute \( f(x) \):
\[ f(x) = \frac{3x - 1}{x + 1} \]
Now, we will substitute this expression into \( f \):
\[ f(f(x)) = f\left(\frac{3x - 1}{x + 1}\right) \]

4. **Substituting into the function**:
Replace \( x \) in \( f(x) \) with \( \frac{3x - 1}{x + 1} \):
\[ f\left(\frac{3x - 1}{x + 1}\right) = \frac{3\left(\frac{3x - 1}{x + 1}\right) - 1}{\left(\frac{3x - 1}{x + 1}\right) + 1} \]

5. **Simplifying the numerator**:
\[ \text{Numerator: } 3\left(\frac{3x - 1}{x + 1}\right) - 1 = \frac{9x - 3}{x + 1} - 1 = \frac{9x - 3 - (x + 1)}{x + 1} = \frac{9x - 3 - x - 1}{x + 1} = \frac{8x - 4}{x + 1} \]

6. **Simplifying the denominator**:
\[ \text{Denominator: } \left(\frac{3x - 1}{x + 1}\right) + 1 = \frac{3x - 1 + (x + 1)}{x + 1} = \frac{3x - 1 + x + 1}{x + 1} = \frac{4x}{x + 1} \]

7. **Combining the results**:
Now we can write \( f(f(x)) \):
\[ f(f(x)) = \frac{\frac{8x - 4}{x + 1}}{\frac{4x}{x + 1}} = \frac{8x - 4}{4x} \]

8. **Simplifying the expression**:
\[ f(f(x)) = \frac{8x - 4}{4x} = \frac{4(2x - 1)}{4x} = \frac{2x - 1}{x} \]

9. **Final result**:
Thus, we have:
\[ f(f(x)) = 2 - \frac{1}{x} \]

### Conclusion:
The final answer is:
\[ f(f(x)) = 2 - \frac{1}{x} \]