己知二次函數(shù)y=f(x) 的圖象過點(diǎn)(1,-4),且不等式f(x)<0的解集是(O,5).
(I )求函數(shù)f(x)的解析式;
(II)設(shè)g(x)=x3-(4k-10)x+5,若函數(shù)h(x)=2f(x)+g(x)在[-4,-2]上單調(diào)遞增,在[-2,0]上單調(diào)遞減,求y=h(x)在[-3,1]上的最大值和最小值..
【答案】
分析:(1)根據(jù)函數(shù)零點(diǎn),方程根與不等式解集端點(diǎn)之間的關(guān)系,結(jié)合二次函數(shù)y=f(x) 的圖象過點(diǎn)(1,-4),可求出函數(shù)f(x)的解析式;
(II)由(I)可求出函數(shù)h(x)的解析式(含參數(shù)k),進(jìn)而由函數(shù)極大值點(diǎn)為-2,求出k值,結(jié)合導(dǎo)數(shù)法求最值的步驟,可得答案.
解答:解:(Ⅰ)由已知y=f (x)是二次函數(shù),且f (x)<0的解集是(0,5),
可得f (x)=0的兩根為0,5,
于是設(shè)二次函數(shù)f (x)=ax(x-5),
代入點(diǎn)(1,-4),得-4=a×1×(1-5),解得a=1,
∴f (x)=x(x-5). …(4分)
(Ⅱ)h(x)=2f (x)+g(x)=2x(x-5)+x
3-(4k-10)x+5=x
3+2x
2-4kx+5,
于是h′(x)=3x
2+4x-4k,
∵h(yuǎn)(x)在[-4,-2]上單調(diào)遞增,在[-2,0]上單調(diào)遞減,
∴x=-2是h(x)的極大值點(diǎn),
∴h′(2)=3×(-2)
2+4×(-2)-4k=0,解得k=1. …(6分)
∴h(x)=x
3+2x
2-4x+5,進(jìn)而得h′(x)=3x
2+4x-4.
令h′(x)=3x
2+4x-4=0,得x=-2,或x=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025123239402498432/SYS201310251232394024984019_DA/0.png)
.
由下表:
x | (-3,-2) | -2 | (-2, ) | ![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025123239402498432/SYS201310251232394024984019_DA/2.png) | ( ,1) |
h′(x) | + | | - | | + |
h(x) | ↗ | 極大 | ↘ | 極小 | ↗ |
可知:h(-2)=(-2)
3+2×(-2)
2-4×(-2)+5=13,h(1)=1
3+2×1
2-4×1+5=4,
h(-3)=(-3)
3+2×(-3)
2-4×(-3)+5=8,h(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025123239402498432/SYS201310251232394024984019_DA/4.png)
)=(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025123239402498432/SYS201310251232394024984019_DA/5.png)
)
3+2×(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025123239402498432/SYS201310251232394024984019_DA/6.png)
)
2-4×
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025123239402498432/SYS201310251232394024984019_DA/7.png)
+5=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025123239402498432/SYS201310251232394024984019_DA/8.png)
,
∴h(x)的最大值為13,最小值為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025123239402498432/SYS201310251232394024984019_DA/9.png)
.…(12分)
點(diǎn)評:本題考查的知識點(diǎn)是二次函數(shù)的性質(zhì),函數(shù)零點(diǎn),方程根與不等式解集端點(diǎn)的關(guān)系,導(dǎo)數(shù)法求函數(shù)的極值與最值,其中求出函數(shù)h(x)的解析式是解答的關(guān)鍵.