If f(x) = `(x)/(x-1)` then what is `f ( a)/(f(a+1))` equal to:

Options

A. `f(-a)/(a+1)`
B. `f(a^(2))`
C. `f(1)/(a)`
D. `f(-a)`

Answer

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

### Step-by-Step Solution:

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

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

3. **Set up the expression \( \frac{f(a)}{f(a+1)} \)**:
\[ \frac{f(a)}{f(a+1)} = \frac{\frac{a}{a-1}}{\frac{a+1}{a}} \]

4. **Simplify the expression**:
\[ \frac{f(a)}{f(a+1)} = \frac{a}{a-1} \times \frac{a}{a+1} = \frac{a^2}{(a-1)(a+1)} \]

5. **Recognize the denominator**:
\[ (a-1)(a+1) = a^2 - 1 \]

6. **Final expression**:
\[ \frac{f(a)}{f(a+1)} = \frac{a^2}{a^2 - 1} \]

7. **Relate to \( f(a^2) \)**:
Since \( f(x) = \frac{x}{x-1} \), we can see that:
\[ f(a^2) = \frac{a^2}{a^2 - 1} \]
Thus, we conclude:
\[ \frac{f(a)}{f(a+1)} = f(a^2) \]

### Final Answer:
\[ \frac{f(a)}{f(a+1)} = f(a^2) \]