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求解函數題目

發問:

1. 已知 f(x) = x^2 - kx (a) 求 f(x+2) 和 f(x-2) 的值. (b) 若 f(x+2) - f(x-2) = kx -32 , 求 k 的值. (c) 由此, 若 f(x+2) = f(x-2) + 40, 求 x 的值.

最佳解答:

1. f(x) = x2 - kx (a) f(x + 2) = (x + 2)2 - k(x + 2) = x2 + 4x + 4 - kx - 2k = x2 + (4 - k)x + (4 - 2k) (b) f(x - 2) = (x - 2)2 - k(x - 2) = x2 - 4x + 4 - kx + 2k = x2 - (4 + k)x + (4 + 2k) ===== (b) f(x + 2) - f(x - 2) = kx - 32 [x2 + (4 - k)x + (4 - 2k)] - [x2 - (4 + k)x + (4 + 2k)] = kx - 32 x2 + 4x - kx + 4 - 2k - x2 + 4x + kx - 4 - 2k = kx - 32 8x - 4k = kx - 32 kx + 4k = 8x + 32 k(x + 4) = 8(x + 4) k = 8 ===== (c) f(x + 2) = f(x - 2) + 40 f(x + 2) - f(x - 2) = 40 ...... (1) 由(c)，k = 8 所以，f(x + 2) - f(x - 2) = 8x - 32 ...... (2) (2) = (1): 8x - 32 = 40 8x = 72 x = 9 =

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