Question

已知函数f(x)=x-1(x大于0) -x+1（x小于0）,求不等式x+(x+1)f(x+1)小于等于1的解集.

Answer 1

这就讨论下x的取值,

1.

当x+1≥0,(x≥-1)

x+(x+1)f(x+1)≤1

x+(x+1)(x+1-1)≤1

x+x(x+1)≤1

x^2+2x-1≤0

-1≤x≤-1+√2

2.

x+1＜0,(x＜-1)

x+(x+1)f(x+1)≤1

x+(x+1)[-(x+1)+1]≤1

-x^2≤1

x+(x+1)f(x+1)