1 |
x |
證明如下:設(shè)0<x1<x2<+∞,
則有f(x2)?f(x1)=x2?
1 |
x2 |
1 |
x1 |
1 |
x1 |
1 |
x2 |
1 |
x2 |
1 |
x1 |
=(x2?x1)+(
x2?x1 |
x1?x2 |
1 |
x1?x2 |
x1x2+1 |
x1?x2 |
1+x1x2 |
x1x2 |
∵0<x1<x2<+∞,x2-x1>0且x1x2+1>0,x1x2>0,
所以f(x2)-f(x1)>0,即f(x1)<f(x2).
所以函數(shù)y=f(x)在區(qū)間(0,+∞)上單調(diào)遞增.
1 |
x |
1 |
x |
1 |
x2 |
1 |
x1 |
1 |
x1 |
1 |
x2 |
1 |
x2 |
1 |
x1 |
x2?x1 |
x1?x2 |
1 |
x1?x2 |
x1x2+1 |
x1?x2 |
1+x1x2 |
x1x2 |