Final answer:
After applying each recursive rule to the base case f(0) = 1, the values for f(1), f(2), f(3), and f(4) were found for rules (a), (b), (c), and (d), demonstrating the process of recursive function evaluation.
Step-by-step explanation:
To find f(1), f(2), f(3), and f(4) for each of the recursive functions provided, we will use the initial condition f(0) = 1 and apply the recursive rule to find the subsequent values.
For rule (a): f(n+1) = f(n) + 2
- f(1) = f(0) + 2 = 1 + 2 = 3
- f(2) = f(1) + 2 = 3 + 2 = 5
- f(3) = f(2) + 2 = 5 + 2 = 7
- f(4) = f(3) + 2 = 7 + 2 = 9
For rule (b): f(n+1) = 3f(n)
- f(1) = 3f(0) = 3 × 1 = 3
- f(2) = 3f(1) = 3 × 3 = 9
- f(3) = 3f(2) = 3 × 9 = 27
- f(4) = 3f(3) = 3 × 27 = 81
For rule (c): f(n+1) = 2ˣⁿₙ⁴
- f(1) = 2¹ = 2
- f(2) = 2ⁿ₁ = 2² = 4
- f(3) = 2ⁿ₂ = 2⁴ = 16
- f(4) = 2ⁿ₃ = 2¹⁶ = 256
For rule (d): f(n+1) = f(n)² + f(n) + 1
- f(1) = f(0)² + f(0) + 1 = 1² + 1 + 1 = 3
- f(2) = f(1)² + f(1) + 1 = 3² + 3 + 1 = 13
- f(3) = f(2)² + f(2) + 1 = 13² + 13 + 1 = 183
- f(4) = f(3)² + f(3) + 1 = 183² + 183 + 1 = 33558