If `f(x)={(sqrt(2)x,xepsilonQ-{0}),(3x,xepsilonQ^(c)):}` then define `foff(x)` and hence define fofof………….`f(x)` where `f` is `n` times.

Explanation

To solve the problem, we need to define the function \( f(x) \) and then find \( f \circ f(x) \) and continue this process for \( n \) times.

### Step 1: Define the function \( f(x) \)

The function is defined as follows:
\[ f(x) = \begin{cases} \sqrt{2} x & \text{if } x \in \mathbb{Q} \setminus \{0\} \\ 3x & \text{if } x \in \mathbb{Q}^c \end{cases} \]

### Step 2: Find \( f(f(x)) \)

We need to evaluate \( f(f(x)) \) based on the input \( x \).

1. **Case 1**: If \( x \in \mathbb{Q} \setminus \{0\} \):
\[ f(x) = \sqrt{2} x \]
Now, we need to determine whether \( \sqrt{2} x \) is rational or irrational.
- Since \( \sqrt{2} \) is irrational, \( \sqrt{2} x \) will be irrational for any rational \( x \neq 0 \).
- Therefore, \( \sqrt{2} x \in \mathbb{Q}^c \).
\[ f(f(x)) = f(\sqrt{2} x) = 3(\sqrt{2} x) = 3\sqrt{2} x \]

2. **Case 2**: If \( x \in \mathbb{Q}^c \):
\[ f(x) = 3x \]
Here, \( 3x \) is also irrational (since multiplying an irrational number by a rational number results in an irrational number).
- Thus, \( 3x \in \mathbb{Q}^c \).
\[ f(f(x)) = f(3x) = 3(3x) = 9x \]

### Step 3: Generalize to \( f^n(x) \)

Now, we can generalize this process:

- If \( x \in \mathbb{Q} \setminus \{0\} \):
- \( f(x) = \sqrt{2} x \)
- \( f(f(x)) = 3\sqrt{2} x \)
- \( f(f(f(x))) = f(3\sqrt{2} x) = 9\sqrt{2} x \)
- Continuing this pattern, we see that:
\[ f^n(x) = 3^{n-1} \cdot \sqrt{2} x \quad \text{for } n \geq 1 \]

- If \( x \in \mathbb{Q}^c \):
- \( f(x) = 3x \)
- \( f(f(x)) = 9x \)
- \( f(f(f(x))) = 27x \)
- Continuing this pattern, we see that:
\[ f^n(x) = 3^n x \quad \text{for } n \geq 1 \]

### Final Result

Thus, we can summarize the results as follows:
\[ f^n(x) = \begin{cases} 3^{n-1} \cdot \sqrt{2} x & \text{if } x \in \mathbb{Q} \setminus \{0\} \\ 3^n x & \text{if } x \in \mathbb{Q}^c \end{cases} \]