若-2<x<1,求函数y=(x2-2x+3)/(x-1)的最大值.
问题描述:
若-2<x<1,求函数y=(x2-2x+3)/(x-1)的最大值.
答
因x²-2x+3=(x-1)²+2.故y=(x²-2x+3)/(x-1)=(x-1)+[2/(x-1)].再令t=1-x,则由-2<x<1,===>0<t<3.且-y=t+(2/t).(0
若-2<x<1,求函数y=(x2-2x+3)/(x-1)的最大值.
因x²-2x+3=(x-1)²+2.故y=(x²-2x+3)/(x-1)=(x-1)+[2/(x-1)].再令t=1-x,则由-2<x<1,===>0<t<3.且-y=t+(2/t).(0