问题 解答题
已知函数f(x)=ln(x+1),g(x)=
x
x+1

(1)求h(x)=f(x)-g(x)的单调区间;
(2)求证:当-1<x1<0<x2时,f(x1)g(x2)-f(x2)g(x1)>0;
(3)求证:f2(x)-xg(x)≤0恒成立.
答案

(1)h(x)=f(x)-g(x)=ln(x+1)-

x
x+1
,x>-1,h/(x)=
1
x+1
-
1
(x+1)2
=
x
(x+1)2

令h/(x)<0,得:-1<x<0,则h(x)在(-1,0)上单调递减;

令h/(x)>0,得:x>0,则h(x)在(0,+∞)上单调递增.

故增区间为(0,+∞),减区间为(-1,0).

(2)由(1)知h(x)min=h(0)=0,

则当x>-1时f(x)≥g(x)恒成立.f/(x)=

1
x+1
>0,g/(x)=
1
(x+1)2
>0

则f(x)、g(x)在(-1,+∞)上均单调递增.

易知:0>f(x1)>g(x1),f(x2)>g(x2)>0,

则-f(x2)g(x1)>-f(x1)g(x2),

即:f(x1)g(x2)-f(x2)g(x1)>0.

(3)f2(x)-xg(x)=ln2(x+1)-

x2
x+1

F(x)=ln2(x+1)-

x2
x+1

F/(x)=

2ln(x+1)
x+1
-
x2+2x
(x+1)2
=
2(x+1)ln(x+1)-(x2+2x)
(x+1)2

令G(x)=2(x+1)ln(x+1)-(x2+2x),

则G/(x)=2ln(x+1)-2x,

令H(x)=2ln(x+1)-2x,

H/(x)=

2
x+1
-2=
-2x
x+1

当-1<x<0时,H/(x)>0,则H(x)在(-1,0)上单调递增;

当x>0时,H/(x)<0,则H(x)在(0,+∞)上单调递减,

故H(x)≤H(0)=0,即G/(x)≤0,

则G(x)在(-1,+∞)上单调递减.

当-1<x<0时,G(x)>G(0)=0,

即F/(x)>0,则F(x)在(-1,0)上单调递增;

当x>0时,G(x)<G(0)=0,

即F/(x)<0,则F(x)在(0,+∞)上单调递减,

故F(x)≤F(0)=0,

即f2(x)-xg(x)≤0.

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