Question

y=2x+根号下(3-x)的值域.

Answer 1

设 根号下3-x＝t t>=0
3-t^2=x
y=6-2t^2+t
y=-2(t^2-1/2t+1/16)+6+1/8=-2(t-1/4)^2+49/8
所以 1/4时区最大值 49/8
(-无穷,49/8)