log1/2 x +3≤0有
log1/2 x ≤-3=log1/2 8
∵log1/2(x)是减函数,有
∴ X≥8.
令t=log(2) x （ t≥3）
此时函数f(x)等价于：
y=(log2 x/4)(log2 x/2)
=(log2 x-log2 4)(log2 x-log2 2)
=(log2 x-2)(log2 x-1)
=(t－2)(t－1)
=t²－3t＋2
=[t－(3/2)]²－(1/4)
其中,t≥3
则是当t=3,即log(2) x =3,x=8时y取得最小值2,
且f(x)无最大值