题目
| x |
| 1+x |
答案
| -1 |
| x+1 |
理由如下:
任取x1,x2∈[0,3],且x1<x2,
则f(x2)-f(x1)=1+
| -1 |
| x2+1 |
| -1 |
| x1+1 |
| x2-x1 |
| (x2+1)(x1+1) |
∵x2+1>0,x1+1>0,(8分)
又∵x2-x1>0,
∴f (x2)-f (x1)=
| x2-x1 |
| (x2+1)(x1+1) |
∴f (x) 在[0,3]上递增.(10分)
| x |
| 1+x |
| -1 |
| x+1 |
| -1 |
| x2+1 |
| -1 |
| x1+1 |
| x2-x1 |
| (x2+1)(x1+1) |
| x2-x1 |
| (x2+1)(x1+1) |