Question

已知函数f(x)=不等式组-x+1,x<0和x-1,x≥0，则不等式x+(x+1)f(x+1)≤1的解集是

求详细过程！谢谢~！急！

Answer 1

x+1<0
则f(x+1)=-(x+1)+1=-x
所以x-x(x+1)<=1
x²+1>=0
恒成立
则x<-1

x+1>=0
f(x)=(x+1)-1=x
所以x+x(x+1)<=1
x²+2x-1<=0
-2-√2<=x<=-2+√2
所以-1<=x<=-2+√2

综上
x≤-2+√2