Question

设函数f(x)的定义域是(0,+∞),对任意正实数m,n恒有f(mn)=f(m)+f(n)，且当x＞1时，f(x)＜0，f(2)=-1，求证

f(x)在(0,+∞)上是减函数。（写过程）快！在线等！

Answer 1

设x1>x2>0,则有x1/x2>1,即f(x1/x2)<0
f(x1)=f(x1/x2*x2)=f(x1/x2)+f(x2)
故f(x1)-f(x2)=f(x1/x2)<0
即f(x1)<f(x2)
所以，函数在（0，＋无穷）上是减函数．

Answer 2

for x>y>0
x = ky ( where k >1)
f(x) = f(ky)
= f(k) + f(y)
< f(y) ( f(k) < 0)
=>f(x)在(0,+∞)上是减函数