If `f(x)=((x-1)/(x+1))`, then `f(f(ax))` in terms of `f(x)` is equal to:

Options

A. `(f(x)+1)/(a(f(x)+1))`
B. `(f(x)+1)/(a(f(x)-1))`
C. `(f(x)-1)/(a(f(x)+1))`
D. `(f(x)+1)/(a(f(x)-1))`

Answer

To solve the problem, we need to find \( f(f(ax)) \) in terms of \( f(x) \), where \( f(x) = \frac{x-1}{x+1} \).

### Step-by-Step Solution:

1. **Find \( f(ax) \)**:
\[ f(ax) = \frac{ax - 1}{ax + 1} \]

2. **Substitute \( f(ax) \) into \( f(x) \)**:
Now we need to find \( f(f(ax)) \). We will substitute \( f(ax) \) into the function \( f(x) \):
\[ f(f(ax)) = f\left(\frac{ax - 1}{ax + 1}\right) \]

3. **Use the definition of \( f(x) \)**:
We substitute \( \frac{ax - 1}{ax + 1} \) into \( f(x) \):
\[ f\left(\frac{ax - 1}{ax + 1}\right) = \frac{\left(\frac{ax - 1}{ax + 1}\right) - 1}{\left(\frac{ax - 1}{ax + 1}\right) + 1} \]

4. **Simplify the numerator**:
The numerator becomes:
\[ \frac{ax - 1}{ax + 1} - 1 = \frac{ax - 1 - (ax + 1)}{ax + 1} = \frac{ax - 1 - ax - 1}{ax + 1} = \frac{-2}{ax + 1} \]

5. **Simplify the denominator**:
The denominator becomes:
\[ \frac{ax - 1}{ax + 1} + 1 = \frac{ax - 1 + (ax + 1)}{ax + 1} = \frac{ax - 1 + ax + 1}{ax + 1} = \frac{2ax}{ax + 1} \]

6. **Combine the results**:
Now we can combine the simplified numerator and denominator:
\[ f(f(ax)) = \frac{\frac{-2}{ax + 1}}{\frac{2ax}{ax + 1}} = \frac{-2}{2ax} = \frac{-1}{ax} \]

7. **Express in terms of \( f(x) \)**:
We know that \( f(x) = \frac{x - 1}{x + 1} \). We can express \( x \) in terms of \( f(x) \):
\[ x = \frac{f(x) + 1}{f(x) - 1} \]
Therefore, substituting this into our expression:
\[ f(f(ax)) = \frac{-1}{a \cdot \frac{f(x) + 1}{f(x) - 1}} = \frac{-(f(x) - 1)}{a(f(x) + 1)} = \frac{1 - f(x)}{a(1 + f(x))} \]

### Final Answer:
Thus, \( f(f(ax)) \) in terms of \( f(x) \) is:
\[ f(f(ax)) = \frac{1 - f(x)}{a(1 + f(x))} \]