Question

已知函数f(x)对任意实数x,y∈R,总有f(x+y)=f(x)+f(y),且x＞0时,f(x)＜0,f(1)=-2.

Answer 1

取x＝y＝0,由“f(x)对任意实数x,y∈R,总有f(x+y)=f(x)+f(y)”得f(0)=f(0)+f(0),所以f(0)=0,

再取y＝－x,由“f(x)对任意实数x,y∈R,总有f(x+y)=f(x)+f(y)”得f(x)+f(－x) = f(0)=0,所以f(x)是奇函数,所以f(x2)+ f(-x1)= f〔x2+ (-x1)〕=f(x2-x1),因为“对任意实数x∈R ,x＞0时,f(x)＜0”,那么取x＝x2-x1＞0,则f(x)＝f(x2-x1) f(-2),f(11-5x)> f(-2),f（x)在R上为减函数,11-5x13/5