求函数fx=log以2为底数2x×log以2为底数4分之x,x∈[二分之一,4]的最大值和最小值

f(x) = log2 (2x) * log2 (x/4),x∈[1/2,4],整理可得f(x) = (log2 2 + log2 x)*(log2 x – log2 4) = (1 + log2 x)*(-2 + log2 x) = log2 2x – log2 x – 2 = (log2 x – 1/2)2 – 9/4,因为x∈[1/2,4] => log2 x∈[-1,2],联系二次函数性质可得当log2 x = 1/2,即x = √2时,f(x)最小值为-9/4 ；
当log2 x = -1或者2,即x = 1/2或者4时,f(x)最大值为0 .