已知x满足不等式-3≤log1/2x≤-1/2,求函数f(x)=log2x/4·log2x/2的值域

解由-3≤log1/2x≤-1/2,
得-3≤-log2x≤-1/2,
即1/2≤log2x≤3
令t=log2（x）,则t属于[1/2,3]
故f(x)=log2x/4·log2x/2=(log2(x)+log2（1/4))(log2(x)+log2（1/2))
=(log2(x)+（-2）)(log2(x)+（-1))
故原函数变为y=（t-2)(t-1）=t^2-3t+2 (t属于[1/2,3])
故y=t^2+3t+3
=(t-3/2)^2-1/4
故当t=3/2时,y有最小值-1/4
当t=3时,y有最大值21
故函数的值域为[-1/4,21].

x满足不等式-3≤log1/2x≤-1/2
x满足不等式1/2≤log(2)x≤3
log2x/4·log2x/2=[log(2)x-2]*[log(2)x-1]=[log(2)x-1.5]²-0.25

函数f(x)=log2x/4·log2x/2的值域[-0.25，2】