已知定义在R上的函数y=f(x)满足f(1+x)=f(1-x),且f(x)是奇函数,当x∈[0,1]时,f(x)=2x,

-1<=x<0
0<-x<=1
f(-x)==-2x=-f(x)
f(x)=2x(-1<=x<0)
-2<=x<-1
f(1+x)=f(1-x)
x=x+1
f(x+2)=f(-x)=-f(x)
0<=x+2<1
f(x+2)=2(x+2)=-f(x)
f(x)=-2x-4(-2<=x<-1)
f(x)=2x(-1<=x<0)
f(x)=-2x-4(-2<=x<-1)
分两段