3x−2−x |
3x+2−x |
2x•3x−1 |
2x•3x+1 |
6x−1 |
6x+1 |
∴f(−x)=
6−x−1 |
6−x+1 |
1−6x |
1+6x |
(2)f(x)=
6x−1 |
6x+1 |
(6x+1)−2 |
6x+1 |
2 |
6x+1 |
证明如下:任意取x1,x2,
使得:x1>x2∴6x1>6x2>0
则f(x1)−f(x2)=
2 |
6x2+1 |
2 |
6x1+1 |
2(6x1−6x2) |
(6x1+1)(6x2+1) |
∴f(x1)>f(x2),
则f(x)在R上是增函数.
∵0<
2 |
6x+1 |
∴f(x)=1−
2 |
6x+1 |
则f(x)的值域为(-1,1).