Question

设不等式2(log1/2^x)^2+9(log1/2^x)+9≤0的解集为M

求当x属于m时f(x)=（log2^x/2)(log2^x/8)的最大、最小值

Answer 1

(2(log1/2x) 3)((log1/2x) 3)<=0

-3<=log1/2x<=-3/2

换底公式 log1/2x=log(2)x/log2(1/2)=-log（2）x

∴3/2<=log(2)x<=3
令y=log（2）x 3/2<=y<=3
f(x)=(y-1)(y-3)=y²-4y 3=（y-2）²-1
y=2时，有最小值-1
y=3时，有最大值0