已知函数f(x)=ln(x+1),g(x)=
(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. |
(1)h(x)=f(x)-g(x)=ln(x+1)-
,x>-1,h/(x)=x x+1
-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)=
>0,g/(x)=1 x+1
>0,1 (x+1)2
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
令F(x)=ln2(x+1)-
,x2 x+1
则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,
则H/(x)=
-2=2 x+1
.-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.