用换元法求这个函数的值域. y=2x-根号下x-1

Question

如题所述

Answer 1

根号下x-1= t(t>=0) ,x=t^2+1
y=2x-根号下x-1
=2 t^2+2- t
=2 (t- 1/4)^2+2 - 1/8
=2 (t- 1/4)^2+15/8
>=15/8
t=1/4,y取最小值,＝15/8
值域{y| y>=15/8}