1)设y=f(x)=x^2-mx+1=(x-m/2)^2+1-m^2/4m/2>=2,m>=4,Ymax=f(-1)=2+m,Ymin=f(2)=5-2m1/2<=m/2<2,1<=m<=4,Ymax=f(-1)=2+m,Ymin=f(m/2)=1-m^2/4-1<=m/2<1/2,-2<=m<1,Ymax=f(2)=5-2m,Ymin=f(m/2)=1-m^2/4m/2<-1,m<-2,Ymax=f(2)=5-2m,Ymin=f(-1)=2+m 2)解: 将bx代入f(x)得: f(bx)=(bx)^2+2(bx)+a整理得:f(bx)=b^2X^2+2bx+a 与f(bx)比较得到 a=2 b=-3 |