题目
| 1+x |
| 1-x |
答案
∵f1(x)=f(x),fk+1(x)=f(fk(x)),k=1,2,…,
∴f1(x)=f(x),
f2(x)=f(f1(x))=-
| 1 |
| x |
f3(x)=f(f2(x))=
| x-1 |
| x+1 |
f4(x)=f(f3(x))=x;
f5(x)=f(f4(x))=
| 1+x |
| 1-x |
…
归纳出规律:fk(x)以周期T=4的周期数列,
∴f2012(x)=f4(x)=x,
则f2012(2012)=2012,
故答案为:2012.
| 1+x |
| 1-x |
| 1 |
| x |
| x-1 |
| x+1 |
| 1+x |
| 1-x |