已知f1(x)=(2x-1)／(x+1),对于n=1,2,…,定义fn+1(x)=f1(fn(x)),若f35(x)=f5(x),则f28(x)=?

题目

已知f1(x)=(2x-1)／(x+1),对于n=1,2,…,定义fn+1(x)=f1(fn(x)),若f35(x)=f5(x),则f28(x)=?

答案

f2(x)={2[(2x-1)/(x+1)]-1}/{[(2x-1)/(x+1)]+1}
=(x-1)/x
f3(x)={2[(x-1)/x]-1}/{[(x-1)/x]+1}
=(x-2)/(2x-1)
f4(x)={2[(x-2)/(2x-1)]-1}/{[(x-2)/(2x-1)]+1}
=-1/(x-1)
f5(x)={2[-1/(x-1)]-1}/{[-1/(x-1)]+1}
=(-x-1)/(x-2)
f6(x)={2[(-x-1)/(x-2)]-1}/{[(-x-1)/(x-2)]+1}
=x
f7(x)=(2x-1)/(x+1)=f1(x)
所以从f1(x)到f6(x)每6个一循环
28=4*6+4
所以f28(x)=-1/(x-1)