求函数f(x)=-(log1/2 2x)²-log1/4 x+5在2≤x≤4时的值域

换元法即可
t=log2(x)
∵ 2≤x≤4
∴ 1≤t≤2
又 log(1/2) 2x=log(1/2)2+log(1/2) x=-1-t
log(1/4)x=-log4(x)=-(1/2)log2(x)=-(1/2)t
∴ f(x)=-(-1-t)²+(1/2)t+5
=-(t+1)²+(1/2)t+5
=-t²-(3/2)t+4
=-(t+3/4)²+73/16
∴ 当t=1时,有最大值3/2
t=2时,有最小值-3
∴ 函数f(x)=-(log1/2 2x)²-log1/4 x+5在2≤x≤4时的值域[-3,3/2]