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)=

1+x
1−x
,又记f1(x)=f(x),fk+1(x)=f(fk(x)),k=1,2,…,则f2009(x)=(  )
A. -
1
x

B. x
C.
x−1
x+1

D.
1+x
1−x

因为f(x)=

1+x
1−x
,且f1(x)=f(x),fk+1(x)=f(fk(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
)=
1+
x−1
x+1
1−
x−1
x+1
=x.
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.