题目内容
(本小题满分12分)探究函数
的最小值,并确定取得最小值时x的值.列表如下:
请观察表中y值随x值变化的特点,完成以下的问题.
(1)函数
在区间(0,2)上递减;函数
在区间 上递增.当
时,
.
(2)证明:函数
在区间(0,2)递减.
(3)思考:函数
时,有最值吗?是最大值还是最小值?此时x为何值?(直接回答结果,不需证明)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002712912979.png)
x | … | 0.5 | 1 | 1.5 | 1.7 | 1.9 | 2 | 2.1 | 2.2 | 2.3 | 3 | 4 | 5 | 7 | … |
y | … | 16 | 10 | 8.34 | 8.1 | 8.01 | 8 | 8.01 | 8.04 | 8.08 | 8.6 | 10 | 11.6 | 15.14 | … |
(1)函数
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002712912943.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002712912943.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002712959283.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002712959522.png)
(2)证明:函数
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002712912943.png)
(3)思考:函数
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002712912943.png)
(1)
;当
(2)证明:设
是区间,(0,2)上的任意两个数,且![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713083436.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240027130992945.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713130685.png)
又![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240027131461821.png)
函数在(0,2)上为减函数.
(3)思考:
。
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713037548.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713052791.png)
(2)证明:设
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713068422.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713083436.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240027130992945.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713115966.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713130685.png)
又
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240027131461821.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713161195.png)
(3)思考:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240027131771542.png)
试题分析:(1)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713037548.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713052791.png)
(2)证明:设
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713068422.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713083436.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240027130992945.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713115966.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713130685.png)
又
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240027131461821.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824002713161195.png)
(3)思考:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240027131771542.png)
点评:典型题,“对号函数”是高考常常考查的一类函数,其单调性及取得最值的情况又具有一般性,因此,学习中应倍加关注。
![](http://thumb.zyjl.cn/images/loading.gif)
练习册系列答案
相关题目