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若二次函数满足f(x+1)-f(x)=2x,f(0)=1 求 f(x)=
人气:114 ℃ 时间:2019-10-19 16:15:39
解答
设f(x) = ax^2 + bx + c
f(0) = 1
则 c =1
f(x) = ax^2 + bx + 1
f(x+1) = a(x+1)^2 + b(x+1) + 1 = ax^2 + (b+2a)x + a + b + 1
f(x+1) - f(x)
= [ax^2 + (b+2a)x + a + b + 1] - [ ax^2 + bx + 1]
= 2ax + a + b
f(x+1) - f(x) = 2x 对任何x成立,则
2a = 2
a + b = 0
a = 1
b = -1
f(x) = x^2 - x + 1
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