For `x in RR - {0, 1},` let `f_1(x) =1/x, f_2(x) = 1-x and f_3(x) = 1/(1-x)` be three given functions. If a function, `J(x)` satisfies `(f_2oJ_of_1)(x) = f_3(x)` then `J(x)` is equal to :

Options

A. `f_3(x)`
B. `f_1(x)`
C. `f_2(x)`
D. `1/x f_3(x)`

Answer

To solve the problem, we need to find the function \( J(x) \) such that:

\[ (f_2 \circ J \circ f_1)(x) = f_3(x) \]

where the functions are defined as follows:

- \( f_1(x) = \frac{1}{x} \)
- \( f_2(x) = 1 - x \)
- \( f_3(x) = \frac{1}{1 - x} \)

### Step-by-Step Solution:

1. **Substitute \( f_1(x) \) into the equation**:
We start with the left-hand side of the equation:

\[ (f_2 \circ J \circ f_1)(x) = f_2(J(f_1(x))) \]

Substituting \( f_1(x) \):

\[ f_1(x) = \frac{1}{x} \implies f_2(J(f_1(x))) = f_2(J(\frac{1}{x})) \]

2. **Express \( f_3(x) \)**:
The right-hand side of the equation is:

\[ f_3(x) = \frac{1}{1 - x} \]

So we need:

\[ f_2(J(\frac{1}{x})) = \frac{1}{1 - x} \]

3. **Find the inverse of \( f_2 \)**:
We need to find \( J(\frac{1}{x}) \). First, let's find the inverse of \( f_2 \):

\[ f_2(x) = 1 - x \implies y = 1 - x \implies x = 1 - y \implies f_2^{-1}(x) = 1 - x \]

4. **Apply the inverse function**:
Now we apply the inverse of \( f_2 \) to both sides:

\[ J\left(\frac{1}{x}\right) = f_2^{-1}\left(f_3(x)\right) \]

Substituting \( f_3(x) \):

\[ J\left(\frac{1}{x}\right) = f_2^{-1}\left(\frac{1}{1 - x}\right) = 1 - \frac{1}{1 - x} \]

5. **Simplify the right-hand side**:
Now simplify \( 1 - \frac{1}{1 - x} \):

\[ 1 - \frac{1}{1 - x} = \frac{(1 - x) - 1}{1 - x} = \frac{-x}{1 - x} \]

Thus, we have:

\[ J\left(\frac{1}{x}\right) = \frac{-x}{1 - x} \]

6. **Find \( J(x) \)**:
To find \( J(x) \), we replace \( \frac{1}{x} \) with \( x \):

Let \( y = \frac{1}{x} \) then \( x = \frac{1}{y} \):

\[ J(y) = \frac{-\frac{1}{y}}{1 - \frac{1}{y}} = \frac{-1}{y(1 - \frac{1}{y})} = \frac{-1}{\frac{y - 1}{y}} = \frac{-y}{y - 1} \]

Therefore, we can write:

\[ J(x) = \frac{-x}{x - 1} \]

### Final Answer:
Thus, the function \( J(x) \) is:

\[ J(x) = \frac{-x}{x - 1} \]