已知函数f(x)=x+2,x≤0和-x+2,x>0（大括号连接的）, 解不等式f（x）≥x^2

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

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

所以-1<=x<=1

解不等式f（x）≥x^2

(1)x<=0,则有x+2>=x^2
x^2-x-2<=0
(x-2)(x+1)<=0
-1<=x<=2
即有-1<=x<=0

(2)x>0,则有-x+2>=x^2
x^2+x-2<=0
(x+2)(x-1)<=0
-2<=x<=1
即有0<x=<1

综上所述,解是-1<=x=<1

x+2>=x^2
x^2-x-2<=0
-1<=x<=2
x<=0
-1<=x<=0

-x+2>=x^2
x^2+x-2<=0
-2<=x<=1
x>0
0<x<=1