已知fx是二次函数,且f(0)=0,f(x+1)=f(x)+x+1,求f(x)
人气：148 ℃ 时间：2019-09-26 01:24:40
解答
由f（0）=0
设f(x)=ax²+bx
f(x+1)
=a(x+1)²+b(x+1)
=ax²+(2a+b)x+(a+b)
=ax²+(b+1)x+1
对应项系数相等
2a+b=b+1
a+b=1
解得a=b=0.5
f(x)=0.5x²+0.5x
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