题目内容

设n为正整数,规定:fn(x)=
f{f[…f(x)…]}
n个f
,已知f(x)=
2(1-x)(0≤x≤1)
x-1(1<x≤2)

(1)解不等式:f(x)≤x;
(2)设集合A={0,1,2},对任意x∈A,证明:f3(x)=x;
(3)求f2008(
8
9
)
的值.
(1)①当0≤x≤1时,由2(1-x)≤x得,x≥
2
3

2
3
≤x≤1.
②当1<x≤2时,因x-1≤x恒成立.
∴1<x≤2.
由①,②得,f(x)≤x的解集为{x|
2
3
≤x≤2}.

(2)∵f(0)=2,f(1)=0,f(2)=1,
∴当x=0时,f3(0)=f(f(f(0)))=f(-f(2))=f(1)=0;
当x=1时,f3(1)=f(f(f(1)))=f(f(0))=f(2)=1;
当x=2时,f3(2)=f(f(f(2)))=f(f(1))=f(0)=2.
即对任意x∈A,恒有f3(x)=x.

(3)f1(
8
9
)=2(1-
8
9
)=
2
9

f2(
8
9
)=f(f(
8
9
))=f(
2
9
)=
14
9

f3(
8
9
)=f(f2(
8
9
))=f(
14
9
)=
14
9
-1=
5
9

f4(
8
9
)=f(f3(
8
9
))=f(
5
9
)=2(1-
5
9
)=
8
9

一般地,f4k+r(
8
9
)=fr(
8
9
)
(k,r∈N).
f2008(
8
9
)=f0(
8
9
)=
8
9
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