Let `W = N cup {0}`. Suppose `f : W to W` is a function such that `f(0)=0, f(1) = 1,f(2) =2` and `f(x) = f(x-2) + f(x-3)` for x=3,4,……….. Then `f(9)` =……………..

Answer

To solve for \( f(9) \) using the function defined in the problem, we will follow these steps:

### Step 1: Identify the base cases
We know from the problem statement:
- \( f(0) = 0 \)
- \( f(1) = 1 \)
- \( f(2) = 2 \)

### Step 2: Use the recursive formula for \( x = 3 \)
For \( x = 3 \):
\[ f(3) = f(3-2) + f(3-3) = f(1) + f(0) = 1 + 0 = 1 \]

### Step 3: Use the recursive formula for \( x = 4 \)
For \( x = 4 \):
\[ f(4) = f(4-2) + f(4-3) = f(2) + f(1) = 2 + 1 = 3 \]

### Step 4: Use the recursive formula for \( x = 5 \)
For \( x = 5 \):
\[ f(5) = f(5-2) + f(5-3) = f(3) + f(2) = 1 + 2 = 3 \]

### Step 5: Use the recursive formula for \( x = 6 \)
For \( x = 6 \):
\[ f(6) = f(6-2) + f(6-3) = f(4) + f(3) = 3 + 1 = 4 \]

### Step 6: Use the recursive formula for \( x = 7 \)
For \( x = 7 \):
\[ f(7) = f(7-2) + f(7-3) = f(5) + f(4) = 3 + 3 = 6 \]

### Step 7: Use the recursive formula for \( x = 8 \)
For \( x = 8 \):
\[ f(8) = f(8-2) + f(8-3) = f(6) + f(5) = 4 + 3 = 7 \]

### Step 8: Finally, calculate \( f(9) \)
For \( x = 9 \):
\[ f(9) = f(9-2) + f(9-3) = f(7) + f(6) = 6 + 4 = 10 \]

Thus, the value of \( f(9) \) is \( \boxed{10} \).

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