If `f(x)=(1)/(x),xne0 and f^(n+1)(x)=f^(n)(f(x))`, find the product of `f^(11)(1).f^(33)(1).f^(55)(1)f^(77)(1).f^(99)(1)`:

Options

A. 110
B. 88
C. 275
D. 1

Answer

To solve the problem, we need to find the product of \( f^{11}(1) \cdot f^{33}(1) \cdot f^{55}(1) \cdot f^{77}(1) \cdot f^{99}(1) \), where \( f(x) = \frac{1}{x} \) and the recursive function definition is given by \( f^{(n+1)}(x) = f^{(n)}(f(x)) \).

### Step-by-Step Solution:

1. **Understanding the Function**:
We start with the function \( f(x) = \frac{1}{x} \). We will compute the values of \( f^n(1) \) for different values of \( n \).

2. **Calculating \( f(1) \)**:
\[ f(1) = \frac{1}{1} = 1 \]

3. **Calculating \( f^2(1) \)**:
Using the recursive definition:
\[ f^2(1) = f(f(1)) = f(1) = 1 \]

4. **Calculating \( f^3(1) \)**:
\[ f^3(1) = f(f^2(1)) = f(1) = 1 \]

5. **Generalizing**:
From the calculations, we can see a pattern emerging:
\[ f^n(1) = 1 \quad \text{for all } n \geq 1 \]

6. **Calculating Higher Orders**:
Continuing this pattern, we find:
\[ f^{11}(1) = 1, \quad f^{33}(1) = 1, \quad f^{55}(1) = 1, \quad f^{77}(1) = 1, \quad f^{99}(1) = 1 \]

7. **Finding the Product**:
Now we can calculate the product:
\[ f^{11}(1) \cdot f^{33}(1) \cdot f^{55}(1) \cdot f^{77}(1) \cdot f^{99}(1) = 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 = 1 \]

### Final Answer:
The product is \( \boxed{1} \).