已知a∈R,函數(shù)f(x)=x2(x-a).
(Ⅰ)當(dāng)a=3時,求f(x)的零點;
(Ⅱ)求函數(shù)y=f (x)在區(qū)間[1,2]上的最小值.
【答案】
分析:(1)將a=3代入求出函數(shù)f(x)的解析式,然后令f(x)=0求出x的值,即得到答案.
(2)對函數(shù)f(x)進(jìn)行求導(dǎo)然后對a的值進(jìn)行分析:當(dāng)a≤0時,f′(x)>0,f(x)是區(qū)間[1,2]上的增函數(shù)進(jìn)而可得到最小值;
當(dāng)a>0時,根據(jù)導(dǎo)函數(shù)的正負(fù)對函數(shù)區(qū)間[1,2]上的單調(diào)性進(jìn)行討論,從而確定最小值.
解答:解:(Ⅰ)由題意f(x)=x
2(x-3),
由f(x)=0,解得x=0,或x=3;
(Ⅱ)設(shè)此最小值為m.,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181001097779533/SYS201310241810010977795020_DA/0.png)
,
(1)當(dāng)a≤0時,f′(x)>0,x∈(1,2),
則f(x)是區(qū)間[1,2]上的增函數(shù),所以m=f(1)=1-a
(2)當(dāng)a>0時,
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181001097779533/SYS201310241810010977795020_DA/1.png)
時,f'(x)>0,從而f(x)在[
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181001097779533/SYS201310241810010977795020_DA/2.png)
,+∞)上是增函數(shù);
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181001097779533/SYS201310241810010977795020_DA/3.png)
時,f'(x)<0,從而f(x)在區(qū)間[0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181001097779533/SYS201310241810010977795020_DA/4.png)
]上是單調(diào)減函數(shù)
①當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181001097779533/SYS201310241810010977795020_DA/5.png)
,即a≥3時,m=f(2)=8-4a
②當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181001097779533/SYS201310241810010977795020_DA/6.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181001097779533/SYS201310241810010977795020_DA/7.png)
時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181001097779533/SYS201310241810010977795020_DA/8.png)
③當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181001097779533/SYS201310241810010977795020_DA/9.png)
時,m=f(1)=1-a
綜上所述,所求函數(shù)的最小值
點評:本題主要考查函數(shù)的零點的求法、函數(shù)的單調(diào)性與其導(dǎo)函數(shù)的正負(fù)之間的關(guān)系、函數(shù)在閉區(qū)間的最值的求法.導(dǎo)數(shù)時高等數(shù)學(xué)下放到高中的內(nèi)容,是高考的必考內(nèi)容,要給予重視.