设f(x)=1+x1−x,又记f1(x)=f(x),fk+1(x)=f(fk(x)),k=1,2,…,则f2009(x)=( ) A.-1x B.x C.x−1x+1 D.1+x1−x
问题描述:
设f(x)=
,又记f1(x)=f(x),fk+1(x)=f(fk(x)),k=1,2,…,则f2009(x)=( )1+x 1−x
A. -
1 x
B. x
C.
x−1 x+1
D.
1+x 1−x
答
因为f(x)=
,且f1(x)=f(x),fk+1(x)=f(fk(x)),1+x 1−x
所以有:f2(x)=f(f1(x))=f(
)=1+x 1−x
=-1+
1+x 1−x 1−
1+x 1−x
;1 x
f3(x)=f(f2(x))=f(-
)=1 x
=1−
1 x 1+
1 x
;x−1 x+1
f4(x)=f(f3(x))=f(
)=x−1 x+1
=x.1+
x−1 x+1 1−
x−1 x+1
所以fk(x)的周期为4,又2009=4×1002+1
故f2009(x)=f1(x)=
1+x 1−x
故选D.