题目

函数f(x)的定义域为D={x|x不等于零},且满足对于任意x1,x2∈D,有f(x1x2)=f(x1)+f(x2);
(1)求f(1)的值;
(2)判断f(x)的奇偶性并证明;
(3)如果f(4)=1,f(3x+1)+f(2x-6)小于或等于3,且f(x)在零到正无穷上是增函 数,求x的取值范围.

提问时间：2020-08-26

答案

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))