分类：教育/科学-理工学科-数学
问题：对数函数已知x满足不等式2(log1/2x)^2（以1/2为底x 
已知x满足不等式2(log1/2x)^2（以1/2为底x的的平方）+7log1/2x（以1/2为底x的对数）+3<=0求函数f(x)=（以2为底x/4的对数）(log2x/4).(log2x/2)（以2为底x/2的对数）的最大值和最小值
回答：2(log1/2x)^2+7log1/2x+3<=0
(2log1/2x+1)(log1/2x+3)<=0
2log1/2x+1<=0且log1/2x+3>=0或2log1/2x+1>=0且log1/2x+3<=0
-3<=log1/2x<=-1/2或者log1/2x>=-1/2且log1/2x<=-3(舍去)
√2<=x<=8
f(x)=(log2x/4)*(log2x/2)
=(log2x-log24)*(log2x-log22)
=(log2x-2)(log2x-1)
=(log2x-3/2)^2-9/4 +2
=(log2x-3/2)^2-1/4
因为√2<=x<=8，所以1/2<=log2x<=3
所以f(x)最小值为-1/4
f(x)最大值=(3-3/2)^2-1/4=2