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)