
定义在R上的函数f(x),满足f(x)=log2(1-x) x≤0; f(x)=f(x-1)-f(x-2) x>0,求f(2011)
2个回答
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f(x)=f(x-1)-f(x-2)
f(x-1)=f(x-2)-f(x-3)
f(x)=-f(x-3)
f(x-3)=f(x-4)-f(x-5)
f(x-4)=f(x-5)-f(x-6)
f(x)=f(x-6)
f(2011)=f(1+335*6)=f(1)=f(0)-f(-1)=log2(1-0)-log2(1-(-1))=0-1=-1
f(x-1)=f(x-2)-f(x-3)
f(x)=-f(x-3)
f(x-3)=f(x-4)-f(x-5)
f(x-4)=f(x-5)-f(x-6)
f(x)=f(x-6)
f(2011)=f(1+335*6)=f(1)=f(0)-f(-1)=log2(1-0)-log2(1-(-1))=0-1=-1
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