Question

函数f（x）的定义域为D=﹛x／x≠0﹜且满足对于任意x1，x2∈0，有f（x1乘x2）=f（x1）+f(x2)

⑴。求f（1）值
⑵判断f（x）的奇偶性
⑶如果f（4）=1，f（3x+1）+f（2x-6）≤3且f（x）在（0，+∞）上是增函数，求x的取值范围

Answer 1

1) 令x1 = x2 = 1;带入f(x1x2)=f(x1)+f(x2);
f(1) = f(1) + f(1) ===> f(1) = 0;

2) 令x1 = x2 = -1;带入f(x1x2)=f(x1)+f(x2);
0 = f(1) = f(-1) + f(-1) ===> f(-1) = 0;
令x1 = -1;带入f(x1x2)=f(x1)+f(x2);
f(-x2) = f(-1) + f(x2) ===> f(-x2) = f(x2);===> f(x)是偶函数

3) f(4*4) = f(4) + f(4) = 2
f(4*4*4) = f(4*4)+f(4)=2+1=3
所以 f(64) = 3 , f(x)是偶函数，所以，f(-64) = 3
f(3x+1)+f(2x-6) = f( (3x + 1)(2x - 6)) <= 3 = f(64)
又f(x)在零到正无穷上是增函数；
f( (3x + 1)(2x - 6)) <= 3 = f(64)
x>0时；(3x + 1)(2x - 6) <= 64 ====> 0 < x <= 5

因为f（x）是偶函数，所以f(x)在负无穷到零上是减函数；
f( (3x + 1)(2x - 6)) <= 3 = f(-64)
x<0时；(3x + 1)(2x - 6) >= -64 ====> x < 0

综上， 0<x <=5 或 x<0。