A function `f(x)` is such that `f(x+1)=sqrt(f(x)+2)`, where `x in I^+ uu {0}` and `f(0)=0` then
`f(x)` is a/an

Options

A. increasing function
B. decreasing function
C. constant function
D. none of these

Answer

To solve the problem, we will analyze the function defined by the recurrence relation \( f(x+1) = \sqrt{f(x) + 2} \) with the initial condition \( f(0) = 0 \).

### Step-by-Step Solution:

1. **Initial Condition**:
We start with the initial condition given:
\[ f(0) = 0 \]

2. **Calculate \( f(1) \)**:
Using the recurrence relation for \( x = 0 \):
\[ f(1) = \sqrt{f(0) + 2} = \sqrt{0 + 2} = \sqrt{2} \]

3. **Calculate \( f(2) \)**:
Now, using the recurrence relation for \( x = 1 \):
\[ f(2) = \sqrt{f(1) + 2} = \sqrt{\sqrt{2} + 2} \]

4. **Calculate \( f(3) \)**:
Next, for \( x = 2 \):
\[ f(3) = \sqrt{f(2) + 2} = \sqrt{\sqrt{\sqrt{2} + 2} + 2} \]

5. **General Pattern**:
We observe that each subsequent value of \( f(x) \) is derived from the previous value plus 2, followed by taking the square root. This indicates that \( f(x) \) is increasing.

6. **Show that \( f(x) \) is increasing**:
To confirm that \( f(x) \) is increasing, we can analyze the difference:
\[ f(x+1) - f(x) = \sqrt{f(x) + 2} - f(x) \]
We want to show that this difference is positive.

- Since \( f(x) \) is non-negative (as we start from \( f(0) = 0 \)), we can see that:
\[ \sqrt{f(x) + 2} > f(x) \quad \text{if} \quad f(x) < 2 \]
- As \( f(x) \) increases, it will approach a limit, but will always be less than \( 2 \) for small values of \( x \).

7. **Conclusion**:
Since \( f(x) \) is increasing for all \( x \) in the domain of positive integers including \( 0 \), we conclude that:
\[ f(x) \text{ is an increasing function.} \]

### Final Answer:
The function \( f(x) \) is increasing.