1. f(1)=f(0)+f(1), f(0)=02. f(0)=f(x)+f(-x)=03. 对于任意x1,x2, x10, 设x2-x1=y, y>0, f(x2) = f(x1) + f(y), f(y)<0, 所以得证4. 既然是减函数，那么-3就是最大值了，3应该是最小值但是因为取不到所以不考虑，f(-3) = f(-1) + f(-1) + f(-1) = 3f(-1) = 6
