設(shè)函數(shù)f(x)=|x2-4x-5|.
(1)在區(qū)間[-2,6]上畫出函數(shù)f(x)的圖象;
(2)設(shè)集合A={x|f(x)≥5},?B=(-∞,-2]∪[0,4]∪[6,+∞).試判斷集合A和B之間的關(guān)系(要寫出判斷過程);
(3)當(dāng)k>2時,求證:在區(qū)間[-1,5]上,y=kx+3k的圖象位于函數(shù)f(x)圖象的上方.
【答案】
分析:(1)當(dāng)x
2-4x-5>0時,f(x)=x
2-4x-5;當(dāng)x
2-4x-5<0時,f(x)=x
2-4x-5,進而畫出圖象.
(2)先求出f(x)≥5的解集,再判斷集合A和B的關(guān)系.
(3)設(shè)函數(shù)g(x)=kx+3k-f(x),只要證明g(x)>0恒成立即可.
解答:![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302713962/SYS201310251245513027139020_DA/images0.png)
解:(1)設(shè)-2≤x≤6,當(dāng)x
2-4x-5≥0時,
即6≥x≥5或-1≥x≥-2時,f(x)=x
2-4x-5=(x-2)
2-9
當(dāng)x
2-4x-5<0時,即-1<x<5時,f(x)=-(x
2-4x-5)=-(x-2)
2+9
故作圖如下.
(2)方程f(x)=5的解分別是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302713962/SYS201310251245513027139020_DA/0.png)
和
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302713962/SYS201310251245513027139020_DA/1.png)
,由于f(x)在(-∞,-1]和[2,5]上單調(diào)遞減,
在[-1,2]和[5,+∞)上單調(diào)遞增,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302713962/SYS201310251245513027139020_DA/2.png)
.
由于2+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302713962/SYS201310251245513027139020_DA/3.png)
<6,2-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302713962/SYS201310251245513027139020_DA/4.png)
>-2
∴B?A.
(3)當(dāng)x∈[-1,5]時,f(x)=-x
2+4x+5.
g(x)=k(x+3)-(-x
2+4x+5)=x
2+(k-4)x+(3k-5)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302713962/SYS201310251245513027139020_DA/5.png)
,
∵k>2,∴?
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302713962/SYS201310251245513027139020_DA/6.png)
.又-1≤x≤5,
①當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302713962/SYS201310251245513027139020_DA/7.png)
,即2<k≤6時,
取
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302713962/SYS201310251245513027139020_DA/8.png)
,g(x)
min=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302713962/SYS201310251245513027139020_DA/9.png)
.
∵?16≤(k-10)
2<64,?
∴?(k-10)
2-64<0,則g(x)
min>0.
②當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302713962/SYS201310251245513027139020_DA/10.png)
,即k>6時,取x=-1,g(x)
min=2k>0.
由①、②可知,當(dāng)k>2時,g(x)>0,x∈[-1,5].
因此,在區(qū)間[-1,5]上,y=k(x+3)的圖象位于函數(shù)f(x)圖象的上方.
點評:本題主要考查了函數(shù)圖象的應(yīng)用.注意數(shù)形結(jié)合.