已知函数y=(2≤x≤4)
(1)当x=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.
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解:(1)x==时,log2x=
∴y=
=(log2x-log24)?(log2x-log22)
=(log2x-2)?(log2x-1)
=-?=-
(2)若t=log2x,(2≤x≤4)
则1≤t≤2,
则y=
=(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
又∵1≤t≤2
∴当时,
当t=1或2时,ymax=0
故函数的值域是
解析分析:(1)当)x=时,log2x=,代入y==(log2x-2)?(log2x-1)可得