x<0,-x>0
f(-x)=(-x-1)^2=(x+1)^2=-f(x)
f(x)=-(x+1)^2
x=0
f(0)=f(-0)=-f(0)
f(0)=0
(x-1)^2, x>0;
f(x)={ 0, x=0;
-(x+1)^2, x<0.
(-1,0)和（0,1）单减
f>0,x>0
f>4,(x-1)^2>4,
x>3
f<0,x<0
-f>4,f<-4,-(x+1)^2<-4,(x+1)^2>4,
x<-3
不等式|f(x)|>4的解是x<-3 or x>3