If `f(x) = (x-1)/(x+1),` then `f(alphax)=`

Options

A. `(f(x)+alpha)/(1+alpha f(x))`
B. `((alpha-1)f(x) + alpha +1)/((alpha+1)f(x)+(alpha-1))`
C. `((alpha+1)f(x) +alpha-1)/((alpha-1) f(x) + (alpha+1))`
D. none of these

Explanation

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

### Step-by-Step Solution:

1. **Substituting \( \alpha x \) into the function:**
\[ f(\alpha x) = \frac{\alpha x - 1}{\alpha x + 1} \]

2. **Rearranging the expression:**
We can rewrite the expression for clarity:
\[ f(\alpha x) = \frac{\alpha x - 1}{\alpha x + 1} \]

3. **Finding a common form:**
To simplify, we can express \( f(\alpha x) \) in terms of \( f(x) \). We will use the relationship derived from the original function \( f(x) \):
\[ x = \frac{1 + f(x)}{1 - f(x)} \]
Thus, substituting \( \alpha x \) for \( x \):
\[ \alpha x = \frac{1 + f(\alpha x)}{1 - f(\alpha x)} \]

4. **Cross-multiplying to eliminate the fraction:**
We can cross-multiply:
\[ \alpha x (1 - f(\alpha x)) = 1 + f(\alpha x) \]
Expanding this gives:
\[ \alpha x - \alpha x f(\alpha x) = 1 + f(\alpha x) \]

5. **Rearranging the equation:**
Bringing all terms involving \( f(\alpha x) \) to one side:
\[ \alpha x - 1 = f(\alpha x)(\alpha x + 1) \]

6. **Solving for \( f(\alpha x) \):**
Now, we can isolate \( f(\alpha x) \):
\[ f(\alpha x) = \frac{\alpha x - 1}{\alpha x + 1} \]

7. **Final expression:**
Therefore, we conclude:
\[ f(\alpha x) = \frac{\alpha - 1}{\alpha + 1} \cdot f(x) + \frac{\alpha - 1}{\alpha + 1} \]

### Conclusion:
The expression for \( f(\alpha x) \) is:
\[ f(\alpha x) = \frac{\alpha - 1}{\alpha + 1} + \frac{2}{\alpha + 1} \]