已知函数f(x)=x+
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102455613337449/SYS201311031024556133374019_ST/0.png)
有如下性质:如果常数a>0,那么该函数在(0,
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102455613337449/SYS201311031024556133374019_ST/1.png)
]上是减函数,在[
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102455613337449/SYS201311031024556133374019_ST/2.png)
,+∞)上是增函数.
(1)如果函数y=x+
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102455613337449/SYS201311031024556133374019_ST/3.png)
(x>0)在(0,4]上是减函数,在[4,+∞)是增函数,求b的值;
(2)证明:函数f(x)=x+
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102455613337449/SYS201311031024556133374019_ST/4.png)
(常数a>0)在(0,
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102455613337449/SYS201311031024556133374019_ST/5.png)
]上是减函数;
(3)设常数c∈(1,9),求函数f(x)=x+
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102455613337449/SYS201311031024556133374019_ST/6.png)
在x∈[1,3]上的最小值和最大值.
考点分析:
相关试题推荐
定义在[-1,1]上的偶函数f(x),已知当x∈[0,1]时的解析式为f(x)=-2
2x+a2
x (a∈R).
(1)求f(x)在[-1,0]上的解析式;
(2)求f(x)在[0,1]上的最大值h(a).
查看答案
已知函数y=
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102455613337449/SYS201311031024556133374017_ST/0.png)
的定义域为集合A,B={x|2<x<9}.
(1)分别求:∁
R(A∩B),(∁
RB)∪A;
(2)已知C={x|a<x<a+3},若C⊆B,求实数a的取值范围.
查看答案
设函数
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102455613337449/SYS201311031024556133374016_ST/0.png)
,若用m表示不超过实数m的最大整数,则函数[
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102455613337449/SYS201311031024556133374016_ST/1.png)
]+[
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102455613337449/SYS201311031024556133374016_ST/2.png)
]的值域为
.
查看答案
函数y=
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102455613337449/SYS201311031024556133374015_ST/0.png)
的定义域为
,值域为
.
查看答案
已知函数f(x)在R上为增函数,且满足f(4)<f(2
x),则x的取值范围是
.
查看答案