If `f(x)=(x)/(x-1)`,then`underset( 19 "times")( (fofofo…of))` (x) is equal to

Options

A. `(x)/(x-1)`
B. `((x)/(x-1))^(19)`
C. `(19x)/(x-1)`
D. `x`

Explanation

To solve the problem, we need to evaluate the function \( f(x) = \frac{x}{x-1} \) applied 19 times, denoted as \( f^{(19)}(x) \).

### Step-by-step Solution:

1. **Define the Function**:
We start with the function:
\[ f(x) = \frac{x}{x-1} \]

2. **Calculate \( f(f(x)) \)**:
We need to find \( f(f(x)) \):
\[ f(f(x)) = f\left(\frac{x}{x-1}\right) \]
Substitute \( \frac{x}{x-1} \) into the function:
\[ f\left(\frac{x}{x-1}\right) = \frac{\frac{x}{x-1}}{\frac{x}{x-1} - 1} \]
Simplifying the denominator:
\[ \frac{x}{x-1} - 1 = \frac{x - (x-1)}{x-1} = \frac{1}{x-1} \]
Now substituting back:
\[ f(f(x)) = \frac{\frac{x}{x-1}}{\frac{1}{x-1}} = x \]

3. **Calculate \( f(f(f(x))) \)**:
Now we find \( f(f(f(x))) \):
\[ f(f(f(x))) = f(x) \]
Since we already calculated \( f(f(x)) = x \), we have:
\[ f(f(f(x))) = f(x) = \frac{x}{x-1} \]

4. **Identify the Pattern**:
We observe the pattern:
- \( f(x) = \frac{x}{x-1} \)
- \( f(f(x)) = x \)
- \( f(f(f(x))) = f(x) = \frac{x}{x-1} \)
- \( f(f(f(f(x)))) = f(f(x)) = x \)

This pattern shows that:
- For odd \( n \), \( f^{(n)}(x) = f(x) \)
- For even \( n \), \( f^{(n)}(x) = x \)

5. **Evaluate \( f^{(19)}(x) \)**:
Since 19 is odd, we find:
\[ f^{(19)}(x) = f(x) = \frac{x}{x-1} \]

### Final Answer:
Thus, the value of \( f^{(19)}(x) \) is:
\[ \frac{x}{x-1} \]