Question

已知x满足不等式-3≤log1/2x≤-1/2,求函数f(x)=log2x/4·log2x/2的值域

Answer 1

解由-3≤log1/2x≤-1/2,
得-3≤-log2x≤-1/2,
即1/2≤log2x≤3
令t=log2（x），则t属于[1/2,3]
故f(x)=log2x/4·log2x/2=(log2(x)+log2（1/4))(log2(x)+log2（1/2))
=(log2(x)+（-2）)(log2(x)+（-1))
故原函数变为y=（t-2)(t-1）=t^2-3t+2 (t属于[1/2,3])
故y=t^2+3t+3
=(t-3/2)^2-1/4
故当t=3/2时，y有最小值-1/4
当t=3时，y有最大值21
故函数的值域为[-1/4,21]。

Answer 2

x满足不等式-3≤log1/2x≤-1/2
x满足不等式1/2≤log(2)x≤3
log2x/4·log2x/2=[log(2)x-2]*[log(2)x-1]=[log(2)x-1.5]²-0.25

函数f(x)=log2x/4·log2x/2的值域[-0.25，2】

Answer 3

-3≤log1/2x≤-1/2,
1/2≤log2(x)≤3,
令t=log2(x)
1/2≤
t
≤3,
所求函数=[log2(x)-2]×[log2(x)-1]
=(t-2)(t-1)
=t²-3t+2
=(t-3/2)²-1/4
当
t=3/2时，函数有最小值
-1/4