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
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![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](/uploads/image/z/12502521-9-1.jpg?t=%E5%B7%B2%E7%9F%A5M%3D%7B2log%26%23178%3B1%2F2%5Ex-11log2%5Ex%2B9%E2%89%A40%7D%2C%E6%B1%82x%E2%88%88M%E6%97%B6.f%28x%29%3Dlog2%5E%28x%2F2%29%C3%97log%281%2F2%29%5E%288%2Fx%EF%BC%89%E7%9A%84%E6%9C%80%E5%80%BC)
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
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
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
2log^2 (1/2)[x]—11log(2)[x]+9≤0
推出2log^2 (1/2)[x]+11log(1/2)[x]+9≤0
设log(1/2)[x]=t
则为2t^2+11t+9≤0
(2t+9)(t+1)≤0
解得-9/2≤ t ≤-1
f(x)=(log2[x/2])*(log(1/2) [8/x])
=(log(2)[x]-1)*{log(1/2)[8]+log(1/2)[1/x]}
=(-log(1/2)[x]-1)*(-3-log(1/2)[x])
因为log(1/2)[x]=t
所以
f(x)=(-t-1)(-3-t)
=t^2+4t+3=(t+2)^2-1
所以在t=-2时取得最小值f(x)min=-1