Question

已知函数f(x)={-x+1,x<0;x-1,x>=0 则不等式x+(x+1)f(x+1)<=1的解集是多少

需要祥细过程

Answer 1

当x<0时，x+(x+1)f(x+1)<=1可化为：x+(x+1)[-(x+1)+1]=-x�0�5≤1∵x�0�5≥0∴-x�0�5≤0<1∴x是任何实数当x≥0时，x+(x+1)[(x+1)-1]=x�0�5+2x≤1(x+1)�0�5≤2∴-√2≤x+1≤√2∴-√2-1≤x≤√2-1