已知函数f(x)的定义域为(0,+∞)且f(4)=1,对于任意x1,x2∈(0,+∞),

来源:百度知道 编辑:UC知道 时间:2024/05/20 07:34:08
有f(x1·x2)=f(x1)+f(x2),当x1>x2时有f(x1)>f(x2).
问:若f(3x+1)+f(2x-6)<=3,求x的取值范围

f(4)=1,f(4)+f(4)+f(4)=3=f(4*4*4)=f(64)
f(3x+1)+f(2x-6)=f[(3x+1)(2x-6)]=f(6x^2-16x-6)<=3
当x1>x2时有f(x1)>f(x2).
f(6x^2-16x-6)<=3=f(64)
6x^2-16x-6<=64
6x^2-16x-70<=0
3x^2-8x-35<=0
(3x+7)(x-5)<=0,f(x)的定义域为(0,+∞)
0<x<=5
x的取值范围(0,5]

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