∵y=x^2-x+2,∴(y+3)/(x+2)=(x^2-x+5)/(x+2).
引入函数f(x)=(x^2-x+5)/(x+2),则:
f′(x)
=[(x^2-x+5)′(x+2)-(x^2-x+5)(x+2)′]/(x+2)^2
=[(2x-1)(x+2)-(x^2-x+5)]/(x+2)^2
=[(2x^2+3x-2)-(x^2-x+5)]/(x+2)^2
=(x^2+4x-7)/(x+2)^2.
令f′(x)<0,得:(x^2+4x-7)/(x+2)^2<0,∴x^2+4x-7<0,∴x^2+4x+4<11,
∴(x+2)^2<11,∴-√11<x+2<√11,∴-2-√11<x<-2+√11.
∵-1≦x≦1,∴f(x)在[-1,1]上单调递减,
∴f(x)在x=1时取得最小值.