Question

已知函数y=log2x4•log2x2(2≤x≤4) (1)当x=423时,求y...

已知函数y=log2x4•log2x2(2≤x≤4) (1)当x=423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.

Answer 1

(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]