Question

函数的奇偶性和单调性求证：f(1)=0 f(xy)=f(x)+f(y)

Answer 1

令x=y=0,有

f(0)=f(0)+f(0) =>f(0)=0

令x=-1,y=0,有

f(0)=f(-1)+f(0) =>f(-1)=0

令x=-2,y=-1有

f(2)=f(-2)+f(-1)

令x=-2,y=0,有

f(0)=f(-2)+f(0)=>f(-2)=0

既是 f(x)=0

偶函数,