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为简便计,将log已a为底b的对数记为log(a,b),则原题即:若[log (1/2,x)]^2-6log(1/2,x)+8≤0,求f(x)=9[log(4,x)]^2-[log(2,x)]^2的平方的最大值与最小值.
[log (1/2,x)]^2-6log(1/2,x)+8=[log (2,x)]^2+6log(2,x)+8=[log (2,x)+2][log (2,x)+4]≤0得
-4≤log (2,x)≤-2,得1/16≤x≤1/4
于是f(x)=9[log(4,x)]^2-[log(2,x)]^2=9[1/2*log(2,x)]^2-[log(2,x)]^2=5/4*[log(2,x)]^2
则5/4*(-2)^2≤f(x)≤5/4*(-4)^2
5≤f(x)≤20
也即f(x)最大值为20,最小值为5
1年前
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