题目内容
9.已知函数f1(x)=$\frac{2x-1}{x+1}$,对于n∈N,定义fn+1(x)=f1(fn(x)),则f28(x)=$\frac{1}{1-x}$.分析 分别求出f1(x)到f6(x)的值,每6个一循环,得到函数具备周期性,利用周期性进行求解即可.
解答 解:∵函数对于n∈N*,定义fn+1(x)=f1[fn(x)],
∴f2(x)=f1[f1(x)]=f1($\frac{2x-1}{x+1}$)=$\frac{2•\frac{2x-1}{x+1}-1}{\frac{2x-1}{x+1}+1}$=$\frac{x-1}{x}$.
f3(x)=f1[f2(x)]=f1($\frac{x-1}{x}$)=$\frac{2•\frac{x-1}{x}-1}{\frac{x-1}{x}+1}$=$\frac{x-2}{2x-1}$,
f4(x)=f1[f3(x)]=f1($\frac{x-2}{2x-1}$)=$\frac{2•\frac{x-2}{2x-1}-1}{\frac{x-2}{2x-1}+1}$=$\frac{1}{1-x}$,
f5(x)=f1[f4(x)]=f1($\frac{1}{1-x}$)=$\frac{2•\frac{1}{1-x}-1}{\frac{1}{1-x}+1}$=$\frac{x+1}{2-x}$,
f6(x)=f1[f5(x)]=f1($\frac{x+1}{2-x}$)=$\frac{2•\frac{x+1}{2-x}-1}{\frac{x+1}{2-x}+1}$=x,
f7(x)=f1[f6(x)]=f1(x)=$\frac{2x-1}{x+1}$=f1(x).
∴从f1(x)到f6(x),每6个一循环.周期为6,
∵28=4×6+4,
∴f28(x)=f4(x)=$\frac{1}{1-x}$,
故答案为:$\frac{1}{1-x}$.
点评 本题考查函数的周期性,是基础题.解题时要认真审题,解题的关键是得到从f1(x)到f6(x),每6个一循环.考查学生的运算能力.
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