问题 解答题
已知函数y=log2
x
4
log2
x
2
(2≤x≤4)
(1)当x=4
2
3
时,求y的值.
(2)令t=log2x,求y关于t的函数关系式.
(3)求该函数的值域.
答案

(1)x=4

2
3
=2
4
3
时,log2x=
4
3

∴y=log2

x
4
log2
x
2

=(log2x-log24)•(log2x-log22)

=(log2x-2)•(log2x-1)

=-

2
3
1
3
=-
2
9

(2)若t=log2x,(2≤x≤4)

则1≤t≤2,

则y=log2

x
4
log2
x
2

=(log2x-2)•(log2x-1)

=(t-2)•(t-1)

=t2-3t+2(1≤t≤2)

(3)∵y=t2-3t+2的图象是开口朝上,且以t=

3
2
为对称轴的抛物线

又∵1≤t≤2

∴当t=

3
2
时,ymin=-
1
4

当t=1或2时,ymax=0

故函数的值域是[-

1
4
,0]

判断题
判断题