Question

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Answer 1

2(log1/2x)^2+7log1/2x+3≤0,
令,t=log1/2(x),有
2t^2+7t+3≤0,
(2t+1)(t+3)≤0.
-3≤t≤-1/2.
当t=-3时,log1/2(x)=-3,x=(1/2)^-3=8.
当t=-1/2时,log1/2(x)=-1/2,x=(1/2)^(-1/2)=√2.
∵log1/2(x),是减函数,有
√2≤X≤8.
又∵log2(x)为增函数,
∴当X=√2时,f(x)有最小值,
f(x)最小值=(log2x/4)(log2x/2)=log2(√2/4)*log2(√2/2)=[log2(√2)-log2(4)]*[log2(√2)-log2(2)]=(1/2-2)(1/2-1)=3/4.
当X=8时,f(x)有最大值,f(x)最大值=[log2(8/4)][log2(8/2)]=log2(2)*log2(4)=1*2=2.
函数f(x)=(log2x/4)(log2x/2)的最大值和最小值分别是:2和3/4.

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Answer 2