| 10x?10?x |
| 10x+10?x |
| 102x?1 |
| 102x+1 |
f（x）=
| 10?x?10x |
| 10?x+10x |
| 102x?1 |
| 102x+1 |
∴f（x）是奇函數(shù)
（2）f(x)＝
| 102x?1 |
| 102x+1 |
則f(x1) ?f(x2) ＝
| 102x1?1 |
| 102x1+1 |
| 102x2?1 |
| 102x2+1 |
| 2(102x1?102x2) |
| (102x1+1)(102x2+1) |
| 2(100x1?100x2) |
| (102x1+1)(102x2+1) |
因為x1＜x2，所以100x1＜100x2，所以f（x1）-f（x2）＜0，即f（x1）＜f（x2），
∴f（x）為增函數(shù)．