函数的概念及其表示法设函数f(x)的定义域为R+,且满足条件f(4)=1,对于任意x1,x2∈R+,有f(x1*x2)=f(x1)+f(x2),当x1〉x2时,有f(x1)〉f(x2)1.求f(1)的值2.如果f(3x+1)+f(2x-6)《3,求x的取值范围
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![函数的概念及其表示法设函数f(x)的定义域为R+,且满足条件f(4)=1,对于任意x1,x2∈R+,有f(x1*x2)=f(x1)+f(x2),当x1〉x2时,有f(x1)〉f(x2)1.求f(1)的值2.如果f(3x+1)+f(2x-6)《3,求x的取值范围](/uploads/image/z/3511159-7-9.jpg?t=%E5%87%BD%E6%95%B0%E7%9A%84%E6%A6%82%E5%BF%B5%E5%8F%8A%E5%85%B6%E8%A1%A8%E7%A4%BA%E6%B3%95%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BAR%2B%2C%E4%B8%94%E6%BB%A1%E8%B6%B3%E6%9D%A1%E4%BB%B6f%284%29%3D1%2C%E5%AF%B9%E4%BA%8E%E4%BB%BB%E6%84%8Fx1%2Cx2%E2%88%88R%2B%2C%E6%9C%89f%28x1%2Ax2%29%3Df%28x1%29%2Bf%28x2%29%2C%E5%BD%93x1%E3%80%89x2%E6%97%B6%2C%E6%9C%89f%28x1%29%E3%80%89f%28x2%291.%E6%B1%82f%281%29%E7%9A%84%E5%80%BC2.%E5%A6%82%E6%9E%9Cf%283x%2B1%29%2Bf%282x-6%29%E3%80%8A3%2C%E6%B1%82x%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
函数的概念及其表示法设函数f(x)的定义域为R+,且满足条件f(4)=1,对于任意x1,x2∈R+,有f(x1*x2)=f(x1)+f(x2),当x1〉x2时,有f(x1)〉f(x2)1.求f(1)的值2.如果f(3x+1)+f(2x-6)《3,求x的取值范围
函数的概念及其表示法
设函数f(x)的定义域为R+,且满足条件f(4)=1,对于任意x1,x2∈R+,有f(x1*x2)=f(x1)+f(x2),当x1〉x2时,有f(x1)〉f(x2)
1.求f(1)的值
2.如果f(3x+1)+f(2x-6)《3,求x的取值范围
函数的概念及其表示法设函数f(x)的定义域为R+,且满足条件f(4)=1,对于任意x1,x2∈R+,有f(x1*x2)=f(x1)+f(x2),当x1〉x2时,有f(x1)〉f(x2)1.求f(1)的值2.如果f(3x+1)+f(2x-6)《3,求x的取值范围
1.
令x1=x2=1.
就有:f(1*1)=f(1)+f(1) =>f(1)=0.
2.
对于任意x1,x2∈R+,有f(x1*x2)=f(x1)+f(x2)
=>f(3x+1)+f(2x-6)=f[(3x+1)(2x-6)] (3x+1>0,2x-6>0 =>x>3).
因为3=3*f(4)=f(4)+f(4)+f(4)
又对于任意x1,x2∈R+,有f(x1*x2)=f(x1)+f(x2).
所以,就有:f(4)+f(4)+f(4)=f(16)+f(4)=f(64).
于是,就化为:
f[(3x+1)(2x-6)] 3)
而当x1〉x2时,有f(x1)〉f(x2).
于是就有:
(3x+1)(2x-6)3).
=>(3x+7)(x-5)3)
=>-7/3