问题
解答题
已知函数y=log2
(1)当x=4
(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
=22 3
时,log2x=4 3 4 3
∴y=log2
•log2x 4 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
•log2x 4 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=
为对称轴的抛物线3 2
又∵1≤t≤2
∴当t=
时,ymin=-3 2 1 4
当t=1或2时,ymax=0
故函数的值域是[-
,0]1 4