函数两题

1.求证y=x+1/2x在[√2/2,+∞)上是增函数
证明:
设x1>x2>=根号2/2
f(x1)-f(x2)
=x1+1/2x1-x2-1/2x2
=(x1-x2)+(x2-x1)/2x1x2
=[(x2-x1)(1-2x1x2)]/2x1x2
因为:
x1>x2>=根号2/2
x2-x1<0
2x1x2>1
x1x2>0
f(x1)>f(x2)
所以y=x+1/2x在[√2/2,+∞)上是增函数

2.y=-√5-4x-x^2单调区间和值域
依题意可知;
5-4x-x^2>=0
x^2+4x-5<=0
(x+5)(x-1)<=0
-5<=x<=1
定义域是[-5,1]

设g(x)=-x^2-4x+5
=-(x^2+4x+4)+9
=-(x+2)^2+9
x属于[-5,1]
所以当x=-2,g(x)有最大值是:9
当x=1,或者-5时,g(x)min=0
所以
-根号(9)<=y<=0
-3<=y<=0
值域是[-3,0]
g(x)在[-5,-2]递增
[-2,1]递减
所以
f(x)在[-5,-2]递减
[-2,1]递增

（1）
设√2/2≤x1<x2，则
f(x1)-f(x2)
=(x1+1/2x1)-(x2+1/2x2)
=(x1-x2)+(1/2x1-1/2x2)
=(x1-x2)+(x2-x1)/(2x1x2)
=(x1-x2)[2x1x2-1]/(2x1x2)
因为x1-x2<0，2x1x2-1>2×(√2/2)×(√2/2)-1=0，x1x2>0
所以(x1-x2)[2x1x2-1]/(