1.y=x²-x+1/x²-x-1,求值域2.f（x）是一次函数,且满足3f（x+1）-2f（x-1）=2x+17,求f（x）步骤祥细点,或大概思路

Question

1.y=x²-x+1/x²-x-1,求值域
2.f（x）是一次函数,且满足3f（x+1）-2f（x-1）=2x+17,求f（x）
步骤祥细点,或大概思路

Answer 1

(1)
y
= (x² - x + 1)/(x² - x - 1)
= (x² - x - 1 + 2)/(x² - x - 1)
= 1 + 2/(x² - x - 1)
x² - x - 1
= (x - 1/2)² - 5/4 ≥ -5/4
所以 -4/5 ≤ 1/(x² - x - 1) < 0 或 1/(x² - x - 1) > 0
所以 -8/5 ≤ 2/(x² - x - 1) < 0 或 2/(x² - x - 1) > 0
所以 -3/5 ≤ 1 + 2/(x² - x - 1) < 1 或 1 + 2/(x² - x - 1) > 1
所以值域是 [-3/5 ,1)∪(1 ,+∞)
(2)
设 f(x) = kx + b
则 3k(x + 1) + 3b - 2k(x - 1) - 2b = 2x + 17
kx + 5k + b = 2x + 17
所以
k = 2
5k + b = 17
解得:
k = 2
b = 7
所以f(x) = 2x + 7