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
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 02:31:25
![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](/uploads/image/z/1645084-28-4.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BAD%3D%7Bx%7Cx%E4%B8%8D%E7%AD%89%E4%BA%8E%E9%9B%B6%7D%2C%E4%B8%94%E6%BB%A1%E8%B6%B3%E5%AF%B9%E4%BA%8E%E4%BB%BB%E6%84%8Fx1%2Cx2%E2%88%88D%2C%E6%9C%89f%28x1x2%29%3Df%28x1%29%2Bf%28x2%29%3B%281%29%E6%B1%82f%281%29%E7%9A%84%E5%80%BC%3B%282%29%E5%88%A4%E6%96%ADf%28x%29%E7%9A%84%E5%A5%87%E5%81%B6%E6%80%A7%E5%B9%B6%E8%AF%81%E6%98%8E%3B%283%29%E5%A6%82%E6%9E%9Cf%284%29%3D1%2Cf%283x%2B1%29%2Bf%282x-6%29%E5%B0%8F%E4%BA%8E%E6%88%96%E7%AD%89%E4%BA%8E3%2C%E4%B8%94f%28x%29%E5%9C%A8%E9%9B%B6%E5%88%B0%E6%AD%A3%E6%97%A0%E7%A9%B7%E4%B8%8A%E6%98%AF%E5%A2%9E%E5%87%BD+%E6%95%B0%2C%E6%B1%82x%E7%9A%84)
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
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.
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
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))
1 f(1)=f(1*1)=2f(1) so f(1)=0
2 f(1)=f(-1*-1)=2f(1) so f(-1)=0
f(-x)=f(-1*x)=f(-1)+f(x)=f(x) 又f(x)的定义域为D={x|x不等于零} so偶函数
3