m=n=1
f(1*1) = f(1)+f(1)
f(1) = 0
设x11,f(a) < 0
有f(x2) = f(a) + f(x1) < f(x1)
所以f(x)是减函数
f(2)=-1/2
f(4) = f(2)+f(2) = -1
f(x^2 -3x) > -1
x^2 - 3x < 4
(x+1)(x-4) < 0
-1 < x < 4