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

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