第2个回答 2018-07-01
x^2-2x+2 >0
(x-1)^2+1 >0
对于所有实数x 都是解
y= 2 - 2/√(x^2-2x+2)
y' =-2(-1/2). (x^2-2x+2)^(-3/2) . ( 2x-2)
= (2x-2)/(x^2-2x+2)^(3/2)
y'=0
2x-2 =0
x=1
y'|x=1+ >0
y'|x=1- <0
x=1 (min)
y= 2 - 2/√(x^2-2x+2)
miny = y(1) = 2 - 2/√(1-2+2) = 2 - 2 =0
x->+∞ , y->2
x->-∞ , y->2
值域 =[0, 2)