已知M={2log²1/2^x-11log2^x+9≤0},求x∈M时。f(x)=log2^(x/2)×log(1/2)^(8/x)的最值
1个回答
2012-10-28
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M={2log²1/2^x-11log2^x+9≤0},
2log²1/2^x-11log2^x+9≤0
2 -9
X
1 -1
(2lg²1/2^x-9)(log²1/2^x-1)<=0
1<=log1/2^x<=9/2
log2^(x/2)×log(1/2)^(8/x)
f(x)=1/2*log2^x*1/3*log2^x
=1/6*lg²2^x
1/6<=1/6*log²1/2^x<=27/8
即
1/6<=f(x)<=27/8
2log²1/2^x-11log2^x+9≤0
2 -9
X
1 -1
(2lg²1/2^x-9)(log²1/2^x-1)<=0
1<=log1/2^x<=9/2
log2^(x/2)×log(1/2)^(8/x)
f(x)=1/2*log2^x*1/3*log2^x
=1/6*lg²2^x
1/6<=1/6*log²1/2^x<=27/8
即
1/6<=f(x)<=27/8
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