已知x满足不等式:2(log1/2 x)²+7log1/2 x +3≤0,则函数f(x)=(log2 x/4)
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已知x满足不等式:2(log1/2 x)²+7log1/2 x +3≤0,则函数f(x)=(log2 x/4)·(log2 x/2)的最大值和最小值分别为
令t=log1/2 x
所以
2(log1/2 x)²+7log1/2 x +3≤0
变为
2t^2+7t+3
再问: 最小不应该是3/4么。
再答: 为啥?代入x=2^(3/2)=(1/2)^(-3/2) (log2 x/4)·(log2 x/2) =(log2 x-log 2 4)(log 2 x- log 2 2) =(3/2-2)(3/2-1) =(-1/2)(1/2) =-1/4 2(log1/2 x)²+7log1/2 x +3 =2(-3/2)^2+7(-3/2)+3 =9/2-21/2+3 =-3
所以
2(log1/2 x)²+7log1/2 x +3≤0
变为
2t^2+7t+3
再问: 最小不应该是3/4么。
再答: 为啥?代入x=2^(3/2)=(1/2)^(-3/2) (log2 x/4)·(log2 x/2) =(log2 x-log 2 4)(log 2 x- log 2 2) =(3/2-2)(3/2-1) =(-1/2)(1/2) =-1/4 2(log1/2 x)²+7log1/2 x +3 =2(-3/2)^2+7(-3/2)+3 =9/2-21/2+3 =-3
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