令2<=t<=4,且t=x+2,则0<=x<=2,
f(t)=f(2+x)=f(2-x),0<=(2-x)<=2
所以f(t)=f(2-x)=2(2-x)-1,
=3-2x(又x=t-2)
=-2t+7.（2<=t<=4）
因为f(x)是偶函数.可令-4<=a<-2,则
2<=-a<=4,f(a)=f(-a)=2a+7,(-4<=a<-2)
令-2<=k<=0,0<=-k<2,
f(k)=f(-k)=-2k-1,(-2<=k<=0)
综上,f(x)=2x+7,(-4<=x<-2)
f(x)=-2x-1,(-2<=x<=0)