这道题求解答 第四题
1个回答
展开全部
4、从f(2+x)=f(2-x)可知,x=2是f(x)的对称轴。
0=<x<=2时,f(x)=2x-1
∵x是对称轴
∴2=<x<=4时,0=<x-2<=2
f(x)=f(x-2)
=2(x-2)-1
=2x-5
即:f(x)=2x-5 (2=<x<=4)
-2=<x<=0时,0=<-x<=2
f(-x)=2(-x)-1=-2x-1
∵f(x)是偶函数
∴f(x)=f(-x)=-2x-1
即:f(x)=-2x-1 (-2=<x<=0)
-4=<x<=-2时,
-2=<x+2<=0
f(x)=f(x+2)
=-2(x+2)-1
=-2x-5
即:f(x)=-2x-5 (-4=<x<=-2)
综上,f(x)在[-4,0]上的解析式如下:
f(x)=-2x-5 (-4=<x<=-2)
f(x)=-2x-1 (-2=<x<=0)
0=<x<=2时,f(x)=2x-1
∵x是对称轴
∴2=<x<=4时,0=<x-2<=2
f(x)=f(x-2)
=2(x-2)-1
=2x-5
即:f(x)=2x-5 (2=<x<=4)
-2=<x<=0时,0=<-x<=2
f(-x)=2(-x)-1=-2x-1
∵f(x)是偶函数
∴f(x)=f(-x)=-2x-1
即:f(x)=-2x-1 (-2=<x<=0)
-4=<x<=-2时,
-2=<x+2<=0
f(x)=f(x+2)
=-2(x+2)-1
=-2x-5
即:f(x)=-2x-5 (-4=<x<=-2)
综上,f(x)在[-4,0]上的解析式如下:
f(x)=-2x-5 (-4=<x<=-2)
f(x)=-2x-1 (-2=<x<=0)
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询