2^x<=256且log2( x)>=1/2,求函数f(x)=log2 (x/2)*log√2(√x/2)的最大,最小值

2^x≤256得x≤8
log2( x)≥1/2得x≥√2
所以x∈[√2,8]
f(x)=log2 (x/2)*log√2(√x/2)
=log2 (x/2)*log2(x/4)
=log2 (x/2)*[log2(x/2)-1]
令log2 (x/2)=t,则t∈[-1/2,2]
t^2-t=(t-1/2)^2-1/4
当t=1/2,即x=2√2时,有最小值-1/4
当t=2,即x=8时,有最大值2

由2^x<=256得x<=8；由log2( x)>=1/2得x>=√2.
函数f(x)=log2 (x/2)*log√2(√x/2)
=(log2 x-1)(0.5*log2 x-1)/log2 √2
=(log2 x-1)(log2 x-2)
令log2 x=t，
则f(x)=(t-1)(t-2)=t^2-3t+2
=(t-3/2)^2-1/4=g(t)
另外由√2<=x<=8，知0.5<=t<=3
g(0.5)=0.75,g(3)=2,g(3/2)=-1/4
故f(x)的最大值为2,最小值为-1/4