(2)f(x)=x+
1 |
x |
1 |
x |
(3)函数f(x)=
1 |
x |
设x1、x2是(1,+∞)上的任意两个实数,且x1<x2,则
f(x1)-f(x2)=x1+
1 |
x1 |
1 |
x2 |
1 |
x1 |
1 |
x2 |
=x1-x2-
x1−x2 |
x1x2 |
x1x2−1 |
x1x2 |
当1<x1<x2时,x1x2>1,x1x2-1>0,从而f(x1)-f(x2)<0,
即f(x1)<f(x2).
∴函数f(x)=
1 |
x |
m |
x |
1 |
x |
1 |
x |
1 |
x |
1 |
x1 |
1 |
x2 |
1 |
x1 |
1 |
x2 |
x1−x2 |
x1x2 |
x1x2−1 |
x1x2 |
1 |
x |