求解函數題目 - Ans.fyi 參考答案

求解函數題目

2008-11-22 8:19 am

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.

回答 (1)

Uncle Michael

2008-11-22 9:29 am

✔ 最佳答案

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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