Question

已知f(x)的定义域为(0,+∞)且满足f(2)=1,f(xy)=f(x)+f(y),又当x2>x1>0时，f(x2)>f(x1)

（1）求f(1),f(4),f(8)得值 （2）若有f(x)+f(x-2)小于等于3成立，求x的取值范围

Answer 1

f(2*1) = f(2)+f(1)
f(1) = 0;
f(4) = f(2*2) = f(2) + f(2) = 1+1 = 2
f(8) = f(4*2) = f(4) + f(2) = 2 + 1 = 3
f(x)+f(x-2) = f(x(x-2)) <= 3 = f(8)
因为当x2>x1>0时，f(x2)>f(x1)，
所以，f(x)在(0,+∞)单曾
则有x(x-2)<=8
x^2 - 2*x + 1 <= 9
(x-1)^2 <=3^2
-3 <= x-1 <= 3
-2 <= x <= 4
结合定义域，
0<x<=4