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📅 May 4, 2026 👤 scraface

May 04, 2026

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f(x+1) - f(x) = (ax^2 + 2ax + a + bx + b + c) - (ax^2 + bx + c) = 2ax...
f(x+1) - f(x) = 6x + 4
2a = 6 \Rightarrow a = 3, \quad a + b = 4 \Rightarrow 3 + b = 4 \Right...
Thus, the leading coefficient of $ f(x) $ is $ \boxed{3} $.
Question: A function $ f: \mathbb{R} \to \mathbb{R} $ satisfies $ f(a...
Solution: Let’s plug in $ a = b = 1 $ into the functional equation:
f(2) + f(0) = 2f(1) + 2f(1) = 4
We are given $ f(1) = 1 $, so:
f(2) + f(0) = 4 \quad \text{(1)}
Now plug in $ a = 0, b = 1 $:

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