求函数y=(1⼀4)^x-(1⼀2)^x+1(x∈[-3,2])的值域

2024-11-27 21:13:47
推荐回答(1个)
回答1:

y=f(x)=(1/4)^x-(1/2)^x+1(x∈[-3,2])
t=(1/2)^x>0
y=(t-1/2)^2+3/4
t=1/2,x=1
f(-3)=64-8+1=57
f(1)=3/4
f(2)=1/16-1/4+1=13/16
3/4<=y<=57