已知函數(shù)f(x)=x3-ax2-3x.
(1)若f(x)在x∈[1,+∞)上是增函數(shù),求實(shí)數(shù)a的取值范圍;
(2)若x=3是f(x)的極值點(diǎn),求f(x)在x∈[1,a]上的最小值和最大值.
【答案】
分析:利用導(dǎo)數(shù)求閉區(qū)間上函數(shù)的極值、最值和對函數(shù)單調(diào)性的判定.
解答:解:(1)f′(x)=3x
2-2ax-3≥0在[1,+∞)恒成立.
∵x≥1.∴a≤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181503100767390/SYS201310241815031007673018_DA/0.png)
(x-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181503100767390/SYS201310241815031007673018_DA/1.png)
),
當(dāng)x≥1時,令g(x)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181503100767390/SYS201310241815031007673018_DA/2.png)
(x-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181503100767390/SYS201310241815031007673018_DA/3.png)
)是增函數(shù),g(x)
min=![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181503100767390/SYS201310241815031007673018_DA/4.png)
(1-1)=0.
∴a≤0.
(2)∵x=3是f(x)的極值點(diǎn)
∴f′(3)=0,即27-6a-3=0,∴a=4.
∴f(x)=x
3-4x
2-3x有極大值點(diǎn)x=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181503100767390/SYS201310241815031007673018_DA/5.png)
,極小值點(diǎn)x=3.
此時f(x)在x∈[-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181503100767390/SYS201310241815031007673018_DA/6.png)
,3]上時減函數(shù),在x∈[3,+∝)上是增函數(shù).
∴f(x)在x∈[1,a]上的最小值是:f(3)=-18,最大值是:f(1)=-6,(因f(a)=f(4)=-12).
點(diǎn)評:利用導(dǎo)數(shù)求函數(shù)的單調(diào)性和最值問題,先根據(jù)極值確定參數(shù)a的值,再求閉區(qū)間上的最值.