题目内容
若函数y=
为奇函数,
(1)确定a的值;
(2)求函数的定义域;
(3)求函数的值域;
(4)讨论函数的单调性.
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(1)确定a的值;
(2)求函数的定义域;
(3)求函数的值域;
(4)讨论函数的单调性.
(1)由奇函数的定义,可得f(-x)+f(x)=0,即a-
=0,
∴2a+
=0.
∴a=-
.
(2)∵y=-
-
,
∴2x-1≠0.
∴函数y=-
-
的定义域为{x|x≠0}.
(3)方法一:(逐步求解法)
∵x≠0,
∴2x-1>-1.
∵2x-1≠0,
∴0>2x-1>-1或2x-1>0.
∴-
-
>
,-
-
<-
,
即函数的值域为{y|y>
或y<-
}.
方法二:(利用有界性)由y=-
-
≠-
,可得2x=
.
∵2x>0,∴
>0.可得y>
或y<-
,
即函数的值域为{y|y>
或y<-
}.
(4)当x>0时,设0<x1<x2,则y1-y2=
.
∵0<x1<x2,
∴1<
<
.
∴
-
<0,
-1>0,
-1>0.
∴y1-y2<0.
因此y=-
-
在(0,+∞)上递增.
同样可以得出y=-
-
在(-∞,0)上递增.
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∴2a+
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∴a=-
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(2)∵y=-
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∴2x-1≠0.
∴函数y=-
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(3)方法一:(逐步求解法)
∵x≠0,
∴2x-1>-1.
∵2x-1≠0,
∴0>2x-1>-1或2x-1>0.
∴-
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即函数的值域为{y|y>
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方法二:(利用有界性)由y=-
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∵2x>0,∴
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即函数的值域为{y|y>
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(4)当x>0时,设0<x1<x2,则y1-y2=
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∵0<x1<x2,
∴1<
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∴
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∴y1-y2<0.
因此y=-
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同样可以得出y=-
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先将函数
化简为y=a-
.
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