1、
f(2x+1)的定义域为[0,1]
所以0<=x<=1
1<=2x+1<=3
所以f(x)定义域是[1,3]
f(x)值域为[1,2]
所以f(x)的值域也是[1,2]
2、
f(x-1)中x-1满足1<=x-1<=3
2<=x<=4
所以定义域是[2,4]
显然1<=f(x-1)<=2
则3<=3f(x-1)<=6
所以值域是[3,6]