已知x满足不等式log1/2(x)≥log1/2(2-x),
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已知x满足不等式log1/2(x)≥log1/2(2-x),
求函数f(x)=[log2(x/4)]*[log2(x/2)]的最小值
求函数f(x)=[log2(x/4)]*[log2(x/2)]的最小值
![已知x满足不等式log1/2(x)≥log1/2(2-x),](/uploads/image/z/3126388-4-8.jpg?t=%E5%B7%B2%E7%9F%A5x%E6%BB%A1%E8%B6%B3%E4%B8%8D%E7%AD%89%E5%BC%8Flog1%2F2%EF%BC%88x%EF%BC%89%E2%89%A5log1%2F2%EF%BC%882-x%EF%BC%89%2C)
由log1/2(x)≥log1/2(2-x),得:x大于0小于等于1,(*)
f(x)=[log2(x/4)]*[log2(x/2)]=[log2(x)-2]*[log2(x)-1]=[log2(x)]^2-3log2(x)+2=[log2(x)-3/2]^2-3/4
由(*)得log2(x)小于等于0
所以f(x)的最小值=[0-3/2]^2-3/4=2
f(x)=[log2(x/4)]*[log2(x/2)]=[log2(x)-2]*[log2(x)-1]=[log2(x)]^2-3log2(x)+2=[log2(x)-3/2]^2-3/4
由(*)得log2(x)小于等于0
所以f(x)的最小值=[0-3/2]^2-3/4=2
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