【答案】
分析:①利用一次函數(shù)的單調(diào)性即可求得其值域;②先求函數(shù)的定義域,再利用函數(shù)的單調(diào)性求函數(shù)的值域;③利用分離常數(shù)法將函數(shù)變形,再利用反比例函數(shù)的圖象和性質(zhì)求函數(shù)的值域;④利用均值定理求函數(shù)的值域,注意均值定理成立的條件
解答:解:(1)∵一次函數(shù)f(x)=3x+2在[-1,1]上為增函數(shù),
∴f(-1)≤y≤f(1),即-1≤y≤5
∴函數(shù)y=3x+2(-1≤x≤1)的值域為[-1,5]
(2)∵函數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185224431674819/SYS201310241852244316748041_DA/0.png)
的定義域為(-∞,4]
且此函數(shù)在定義域上為單調(diào)減函數(shù),
∴f(x)≥f(4)=2
∴函數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185224431674819/SYS201310241852244316748041_DA/1.png)
的值域為[2,+∞)
(3)函數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185224431674819/SYS201310241852244316748041_DA/2.png)
=1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185224431674819/SYS201310241852244316748041_DA/3.png)
由反比例函數(shù)的圖象知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185224431674819/SYS201310241852244316748041_DA/4.png)
≠0
∴y≠1
∴函數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185224431674819/SYS201310241852244316748041_DA/5.png)
的值域為(-∞,1)∪(1,+∞)
(4)x>0時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185224431674819/SYS201310241852244316748041_DA/6.png)
≥2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185224431674819/SYS201310241852244316748041_DA/7.png)
=2
x<0時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185224431674819/SYS201310241852244316748041_DA/8.png)
=-(-x-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185224431674819/SYS201310241852244316748041_DA/9.png)
)≤-2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185224431674819/SYS201310241852244316748041_DA/10.png)
=-2
∴函數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024185224431674819/SYS201310241852244316748041_DA/11.png)
的值域為(-∞,-2]∪[2,+∞)
點評:本題考察了求函數(shù)的值域的方法:單調(diào)性法、函數(shù)圖象法、分離常數(shù)法、均值定理法