（1）设x10,因当x>0时f(x)<0，故f(x2-x1)<0,f(x2)=f[x1+(x2-x1)]=f(x1)+f(x2-x1)∴f(x)↓。（2）2f(0)=f(0),f(0)=0,f(3)=3f(1)=-2,f(-3)+f(3)=f(0)=0,∴f(-3)=-f(3)=2,∴f(x)|max=2，f(x)|min=-2.
a
f
