Question

求函数y=(1/4)^x-(1/2)^x+1,x属于[-3,2]的值域

Answer 1

设(1/2)^x=t

x∈[-3,2]

t∈[1/4,8]

y=t^2-t+1=[t-(1/2)]^2+(3/4)

ymin=3/4

ymax=[8-(1/2)]^2+(3/4)=225/4+(3/4)=57

y的值域：y∈[3/4,57]