(本小题满分16分,每小题8分) 求下列函数的值域:(1) ;(2) ,.

(本小题满分16分,每小题8分) 求下列函数的值域:(1) ;(2) ,.

题型:解答题难度:简单来源:不详
(本小题满分16分,每小题8分)
求下列函数的值域:(1) ;(2)
答案

(1)
(2)
解析


即值域为……………………………………………………………8分
(2) 解:
,则y=t2-3t+2,·························································· 2分
x∈[-2,2],∴,················································ 4分
,····································································· 5分
时,,当t=4时,ymax=6,
········································································· 8分
举一反三
函数的定义域为(    )
A.B.
C.D.

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函数的图像是两条直线的一部份,如下图所示,其定义域为,则不等式的解集为(    )
A.{x|-1≤x<或0<x≤1}
B.{x|-1≤x≤1,且x≠0}
C.{x|≤x<或0<x≤}
D {x|-1≤x<0或<x≤1}

题型:单选题难度:一般| 查看答案
已知函数,则函数的定义域为(    )
A:           B:  
C:     D:
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下列函数中,值域为的是(     )
A:   B:   C:   D:
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函数的定义域是,则函数的定义域是
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