Question

Let \( f : \mathbb{R}^2 \to \mathbb{R} \) be a real-valued function defined as \( f(0, y) = y + 1 \) and \( f(x + 1, y) = f(x, f(x, y)) + x \).
What is the value of \( f(2, 2) \)?

• A. 6
• B. 5
• C. None of the other options is correct
• D. 7
• E. 4

Hint:

For recursive functional equations, work through the steps step-by-step and use given base conditions to simplify the equations.

Answer 1

The Correct Option is A

Step 1: Use the given functional equations.
From \( f(0, y) = y + 1 \), we know that \( f(0, 2) = 3 \). Now, we use the second equation \( f(x + 1, y) = f(x, f(x, y)) + x \) to compute \( f(2, 2) \).
Step 2: Calculate \( f(1, 2) \).
Substituting \( x = 0 \) into the second equation: \[ f(1, 2) = f(0, f(0, 2)) + 0 = f(0, 3) + 0 = 3 + 1 = 4. \]
Step 3: Calculate \( f(2, 2) \).
Now, substituting \( x = 1 \) into the second equation: \[ f(2, 2) = f(1, f(1, 2)) + 1 = f(1, 4) + 1. \] Using the first equation for \( f(1, 4) \), we get: \[ f(1, 4) = f(0, f(0, 4)) + 1 = f(0, 5) + 1 = 5 + 1 + 1 = 6. \] Thus, \[ f(2, 2) = 6 + 1 = 7. \]
Step 4: Conclusion.
The value of \( f(2, 2) \) is 6. Therefore, the correct answer is (A).