定义法求单调性

1、
f(x)=x+2/x
令x1>x2>1
f(x1)-f(x2)=x1+2/x1-x2-2/x2
=(x1-x2)+2/x1-2/x2
=[x1x2(x1-x2)+2x2-2x1]/x1x2
=(x1x2-2)(x1-x2)/x1x2
显然x1x2>0,x1-x2>0
就看x1x2-2的符号
则x>√2,时，x1x2>2,1<x<√2,x1x2<2
所以
x1>x2>√2,(x1x2-2)(x1-x2)/x1x2>0,f(x1)>f(x2),增函数
1<x2<x1<√2,(x1x2-2)(x1-x2)/x1x2<0,f(x1)<f(x2),减函数

综上
1<x<√2,减函数
x>√2,增函数

2、
f(x)=x³-4x
令0<=x1<x2<=1
f(x1)-f(x2)=x1³-x2²-4x1+4x2
=(x1-x2)(x1²+x1x2+x2²)-4(x1-x2)
=(x1-x2)(x1²+x1x2+x2²-4)
x1<x2,x1-x2<0
0<x1<1,所以0<x1²<1
同理,0<x1x2<1
0<x2²<1
所以0<x1²+x1x2+x2²<3
所以-4<x1²+x1x2+x2²-4<-1
即x1²+x1x2+x2²-4<0
所以(x1-x2)(x1²+x1x2+x2²-4)>0
即0<=x1<x2<=1时，f(x1)>f(x2)
所以是减函数