已知y=f(x)的定义域是[-1,1],求下列函数的定义域: (1)y=f(x-3) (2)y=f(1/x) (3)y=f(IX+1I)+f(x^2-1).

(1)
-1<=x-3<=1
解得 2<=x<=4

y=f(x-3)的定义域是[2,4]
(2)
1/x<=1
(1-x)/x<=0
x<0 or x>=1

1/x>=-1
(1+x)/x>=0
x<=-1 or x>0
所以x<=-1 or x>=1

y=f(1/x)的定义域是(-∞,-1]∪[1,+∞]
(3)
-1<=|x+1|<=1
即|x+1|<=1
-1<=x+1<=1
-2<=x<=0

-1<=x^2-1<=1
0<=x^2<=2
-√2<=x<=√2

y=f(|x+1|)+f(x^2-1)的定义域是[-√2,0]

1. -1<=x-3<=1 ,2<=x<=4
2.-1<=1/x<=1,x>=1 或x<=-1
3.-1<=Ix+1I<=1
-1<=x^2-1<=1解得 -√2<=x<=0

(1)y=f(x-3) [2,4]
(2)y=f(1/x) (-∞,-1]或[1,+∞)
(3)y=f(IX+1I)+f(x^2-1) [-√2,0]