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

Options

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

Answer

To find \( f(ax) \) in terms of \( f(x) \) given that \( f(x) = \frac{x-1}{x+1} \), we can follow these steps:

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

### Step 2: Substitute \( ax \) into the function
Now, we need to find \( f(ax) \):
\[ f(ax) = \frac{ax - 1}{ax + 1} \]

### Step 3: Express \( x \) in terms of \( f(x) \)
From the original function \( f(x) = \frac{x-1}{x+1} \), we can rearrange it to express \( x \) in terms of \( f(x) \):
\[ f(x)(x + 1) = x - 1 \]
This simplifies to:
\[ xf(x) + f(x) = x - 1 \]
Rearranging gives:
\[ x - xf(x) = f(x) + 1 \]
Factoring out \( x \):
\[ x(1 - f(x)) = f(x) + 1 \]
Thus, we can express \( x \) as:
\[ x = \frac{f(x) + 1}{1 - f(x)} \]

### Step 4: Substitute \( x \) into \( f(ax) \)
Now, we substitute \( ax \) into the expression we found:
\[ f(ax) = \frac{a \cdot \frac{f(x) + 1}{1 - f(x)} - 1}{a \cdot \frac{f(x) + 1}{1 - f(x)} + 1} \]

### Step 5: Simplify the numerator and denominator
The numerator becomes:
\[ a \cdot \frac{f(x) + 1}{1 - f(x)} - 1 = \frac{a(f(x) + 1) - (1 - f(x))}{1 - f(x)} = \frac{af(x) + a - 1 + f(x)}{1 - f(x)} = \frac{(a + 1)f(x) + (a - 1)}{1 - f(x)} \]

The denominator becomes:
\[ a \cdot \frac{f(x) + 1}{1 - f(x)} + 1 = \frac{a(f(x) + 1) + (1 - f(x))}{1 - f(x)} = \frac{af(x) + a + 1 - f(x)}{1 - f(x)} = \frac{(a - 1)f(x) + (a + 1)}{1 - f(x)} \]

### Step 6: Combine the results
Thus, we have:
\[ f(ax) = \frac{(a + 1)f(x) + (a - 1)}{(a - 1)f(x) + (a + 1)} \]

### Final Result
So, we can express \( f(ax) \) in terms of \( f(x) \) as:
\[ f(ax) = \frac{(a + 1)f(x) + (a - 1)}{(a - 1)f(x) + (a + 1)} \]