1 |
x |
x−1 |
x |
∴f(x)在区间[1,+∞)上是递增的
当0<x<1时f(x)=1-x-lnx f′(x)=-1-lnx<0
∴f(x)在区间(0,1)上是递减的
f(x)在(0,1)内单调递减,在【1,+∞)上单调递增,故当x=1时,f(x)有最小值f(1),且f(1)=0
(2)由(1)x>1时,有x-1-lnx>0即
lnx |
x |
1 |
x |
∴
ln22 |
22 |
ln32 |
32 |
lnn2 |
n2 |
1 |
22 |
1 |
32 |
1 |
n2 |
1 |
22 |
1 |
32 |
1 |
n2 |
1 |
2•3 |
1 |
3•4 |
1 |
n(n+1) |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
n |
1 |
n+1 |
1 |
2 |
1 |
n+1 |
(n−1)(2n+1) |
2(n+1) |