【答案】
分析:(1)先求出函數(shù)的導(dǎo)數(shù),根據(jù)導(dǎo)數(shù)求函數(shù)的極值,再求出端點值,比較極值和端點值的大小,得出最值.
(2)由函數(shù)的定義域是正數(shù)知,s>0,故極值點(3,0)不在區(qū)間[s,t]上,討論st的取值范圍,最后兩式相減并整理得出結(jié)果
解答:解:(Ⅰ)f′(x)=3x
2+2ax+b.依題意則有:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/0.png)
所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/1.png)
解得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/2.png)
所以f(x)=x
3-6x
2+9x;
f′(x)=3x
2-12x+9=3(x-1)(x-3),由f′(x)=0可得x=1或x=3.
f′(x),f(x)在區(qū)間(0,4]上的變化情況為:
x | | (0,1) | 1 | (1,3) | 3 | (3,4) | 4 |
f′(x) | | + | | - | | + | |
f(x) | | 增函數(shù) | 4 | 減函數(shù) | | 增函數(shù) | 4 |
所以函數(shù)f(x)=x
3-6x
2+9x在區(qū)間[0,4]上的最大值是4,最小值是0.
(2)由函數(shù)的定義域是正數(shù)知,s>0,故極值點(3,0)不在區(qū)間[s,t]上;
①若極值點M(1,4)在區(qū)間[s,t]上,此時0<s≤1≤t<3,
故有(i)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/3.png)
或(ii)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/4.png)
(i)由k=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/5.png)
,1≤t<3知,k∈
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/6.png)
,當(dāng)且僅當(dāng)t=1時,k=4;
再由k=(s-3)
2,0<s≤1知,k∈[4,9),當(dāng)且僅當(dāng)s=1時,k=4.
由于s≠t,故不存在滿足要求的k值.
(ii)由s=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/7.png)
f(t)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/8.png)
f(t)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/9.png)
,及0<s≤1可解得2≤t<3,
所以k=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/10.png)
,2≤t<3知,k∈
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/11.png)
;
即當(dāng)k∈
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/12.png)
時,存在t=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/13.png)
∈[2,3),s=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/14.png)
f(t)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/15.png)
∈(0,1],
且f(s)≥4s=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/16.png)
f(t)>f(t),滿足要求.
②若函數(shù)f(x)在區(qū)間[s,t]上單調(diào)遞增,則0<s<t≤1或3<s<t,
且
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/17.png)
,故s,t是方程x
2-6x+9=k的兩根,
由于此方程兩根之和為3,故[s,t]不可能同在一個單調(diào)增區(qū)間內(nèi);
③若函數(shù)f(x)在區(qū)間[s,t]上單調(diào)遞減,則1<s<t<3,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/18.png)
,
兩式相減并整理得s
2(s-3)
2=t
2(t-3)
2,由1<s<t<3知s(s-3)=t(t-3),即s+t=3,
再將兩式相減并除以s-t得-k=(s
2+st+t
2)-6(s+t)+9=(s+t)
2-6(s+t)+9-st=-st,
即k=st,所以s,t是方程x
2-3x+k=0的兩根,
令g(x)=x
2-3x+k,
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/19.png)
解得2<k<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/20.png)
,即存在s=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/21.png)
,t=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/22.png)
滿足要求.
綜上可得,當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/23.png)
<k<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182422672705922/SYS201310241824226727059021_DA/24.png)
時,存在兩個不等正數(shù)s,t(s<t),使x∈[s,t]時,
函數(shù)f(x)=x
3-6x
2+9x的值域恰好是[ks,kt].
點評:該題考查函數(shù)的求導(dǎo)以及對st的討論,以及判別式的應(yīng)用,注意在討論函數(shù)單調(diào)性時要畫表格.