已知√2≤x≤8,求函数f（x）=（log2 x/2).(log2 4/x)的最大值和最小值

因为√2≤x≤8
故：1/2≤log2^ x≤3,令t= log2^ x
故：1/2≤t≤3,
又：f（x）=（log2^ x/2).(log2^ 4/x)
=( log2^ x- log2^ 2)( log2^ 4- log2^ x)
=( log2^ x- 1)( 2- log2^ x)
=(t-1)(2-t)
=-t²+3t-2
=-(t-3/2)²+1/4
故：t=3/2时,取最大值1/4,此时log2^ x=3/2,x=2√2
当t=3时,取最小值-2,此时log2^ x=3,x=8