设函数f(x)=(x+1)^2 x<1

x<1
f(x)=(x+1)^2>=1
x^2+2x+1>=1
x(x+2)>=0
x>=0,x<=-2
所以x<=-2,0<=x<1

x>=1
f(x)=4-√(x-1)>=1
√(x-1)<=3
0<=x-1<=9
1<=x<=10

综上
x<=-2,0<=x<=10

分x<1和x>=1讨论
当x<1
(x+1)^2>=1 x<-2或x>0
故x<=-2或0<=x<1

当x>=1
4-√x-1>=1 √x-1<=3 0<=x-1<=9
1<=x<=10

得f(x)>=1的自变量x的取值范围是_x<=-2或0<=x<10____