If we know that f (x) is a function of degree and satisfies 3f (x + 1) = 2x + 17, then f (x) = () A. 23x+5B. 23x+1C. 2x-3D. 2x+5

Let f (x) = KX + B ∵ 3f (x + 1) = 2x + 17, ∵ 3 [K (x + 1) + b] = 2x + 17, i.e., 3kx + 3K + 3B = 2x + 17 ∵ 3K = 23K + 3B = 17, then we can solve the equation, k = 23, B = 5 ∵ f (x) = 23x + 5, so we choose a

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