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embassy suites by hilton st louis st charles - Moon Smoking

📅 May 16, 2026 👤 scraface

May 16, 2026

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📌 f(x+1) = a(x+1)^2 + b(x+1) + c = a(x^2 + 2x + 1) + b(x + 1) + c = ax^2 + 2ax + a + bx + b + c

📌 f(x+1) - f(x) = (ax^2 + 2ax + a + bx + b + c) - (ax^2 + bx + c) = 2ax + a + b

📌 f(x+1) - f(x) = 6x + 4

📌 2a = 6 \Rightarrow a = 3, \quad a + b = 4 \Rightarrow 3 + b = 4 \Rightarrow b = 1

📌 Thus, the leading coefficient of $ f(x) $ is $ \boxed{3} $.

📌 Question: A function $ f: \mathbb{R} \to \mathbb{R} $ satisfies $ f(a + b) + f(a - b) = 2f(a) + 2f(b) $ for all real $ a, b $. If $ f(1) = 1 $, find $ f(2) $.

📌 Solution: Let’s plug in $ a = b = 1 $ into the functional equation:

📌 f(2) + f(0) = 2f(1) + 2f(1) = 4

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