Question

设函数f(x)=大括号x+1(x<=0),-x+1(x>0)，则不等式f(x)>=2x^2的解集是

答案是，[-1/2，1/2]

Answer 1

分区间讨论：
x<=0
f(x)=x+1>=2x^2 即2x^2-x-1<=0 即 (2x+1)(x-1)<=0 -1/2<=x<=1
又x<=0 所以-1/2<=x<=0
x>0
f(x)=-x+1>=2x^2
2x^2+x-1<=0 即(2x-1)(x+1)<=0 -1<=x<=1/2
又x>0 所以0<x<=1/2
合并讨论的区间就是-1/2<=x<=1/2
解集[-1/2,1/2]

Answer 2

-1<=x<=1

追问

答案是，[-1/2，1/2]

追答

不好意思，看错了。。。

Answer 3

分区间讨论：
x<=0
f(x)=x+1>=2x^2
即2x^2-x-1<=0
即
(2x+1)(x-1)<=0
-1/2<=x<=1
又x<=0
所以-1/2<=x<=0
x>0
f(x)=-x+1>=2x^2
2x^2+x-1<=0
即(2x-1)(x+1)<=0
-1<=x<=1/2
又x>0
所以0<x<=1/2
合并讨论的区间就是-1/2<=x<=1/2
解集[-1/2,1/2]