已知√2≤x≤8 求函数y=(log₂x/4)*(log₂x/2)的最大值和最小值
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2011-02-09 · 知道合伙人教育行家
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y=(log₂x/4)*(log₂x/2)
= (log 2 x - log 2 4) * (log 2 x - log 2 2)
= (log 2 x - 2) * (log 2 x - 1)
=(log 2 x)^2 - 3log 2 x +2
=(log 2 x - 3/2)^2 -9/4+1
=(log 2 x - 3/2)^2 -1/4
当x=2^(3/2)=2根号2时,有最小值ymin=0-1/4=-1/4
当根号2≤x<2根号2时单调减
当2根号2≤x≤8时,单调增
f(根号2)=(log2 根号2 - 3/2)^2-1/4 = (1/2-3/2)^2-1/4=1-1/4=3/4
f(8)=(log2 8 -3/2)-1/4=(3-3/2)^2-1/4=2
∴x=8时,最大值ymax=2
= (log 2 x - log 2 4) * (log 2 x - log 2 2)
= (log 2 x - 2) * (log 2 x - 1)
=(log 2 x)^2 - 3log 2 x +2
=(log 2 x - 3/2)^2 -9/4+1
=(log 2 x - 3/2)^2 -1/4
当x=2^(3/2)=2根号2时,有最小值ymin=0-1/4=-1/4
当根号2≤x<2根号2时单调减
当2根号2≤x≤8时,单调增
f(根号2)=(log2 根号2 - 3/2)^2-1/4 = (1/2-3/2)^2-1/4=1-1/4=3/4
f(8)=(log2 8 -3/2)-1/4=(3-3/2)^2-1/4=2
∴x=8时,最大值ymax=2
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