题目
(1)函数y=
| ax-a-x |
| 2 |
(2)函数f(x)=
| (ax+1)x |
| ax-1 |
(3)若f(x)=3x,则f(x+y)=f(x)f(y);
(4)若f(x)=ax(a>0,a≠1),且x1≠x2,则
| 1 |
| 2 |
| x1+x2 |
| 2 |
答案
| ax-a-x |
| 2 |
| a-x-ax |
| 2 |
对于(2)f(-x)=
| (a-x+1)(-x) |
| a-x-1 |
| (ax+1)x |
| ax-1 |
对于(3)f(x+y)=3x+y,f(x)f(y)=3x•3y=3x+y,有f(x+y)=f(x)f(y);
对于(4)
| 1 |
| 2 |
| 1 |
| 2 |
| x1+x2 |
| 2 |
| x1+x2 |
| 2 |
故答案为(4)