Question

已知函数f(x)的定义域为（0，+∞），且满足条件f(xy)=f(x)+f(y)；f(2)=1；当x>1时，f(x)>0.

1.讨论函数的单调性
2.求不等式f(x)+f(x-3)小于等于2的解集
“2.求不等式f(x)+f(x-3)小于等于2的解集”打错了应该是“2.求不等式f(x)+f(x-2）＜3的解集”

Answer 1

(2)
x>y>0
x = ky ( k >1)
f(x) = f(ky) = f(k)+f(y) > f(y)
函数:(0,+∞)增加
(2)
put x=y=1, => f(1)=0
f(x)+f(x-3) <=0
f(x) is defined for x>0
f(x-3) is defined for x>3
f(x)+f(x-3) ≤0
f(x(x-3))≤0 = f(1)
x(x-3)≤1
x^2-3x-1 ≤0
(3-√13)/2≤x≤(3+√13)/2
f(x)+f(x-3)小于等于2的解集
"x>0" and "x>3" and "(3-√13)/2≤x≤(3+√13)/2"
ie 3<x≤(3+√13)/2

Answer 2

1)函数在（0，+∞）上为单调增函数
2）解集为（2，4）