已知函數(shù)f(x)=x3-3x,x∈[-2,2]和函數(shù)g(x)=ax-1,x∈[-2,2],若對(duì)于?x1∈[-2,2],總?x∈[-2,2],使得g(x)=f(x1)成立,則實(shí)數(shù)a的取值范圍 .
【答案】
分析:根據(jù)對(duì)于?x
1∈[-2,2],總?x
∈[-2,2],使得g(x
)=f(x
1)成立,得到函數(shù)f(x)在[-2,2]上值域是g(x)在[-2,2]上值域的子集,下面利用導(dǎo)數(shù)求函數(shù)f(x)、g(x)在[-2,2]上值域,并列出不等式,解此不等式組即可求得實(shí)數(shù)a的取值范圍
解答:解:∵f(x)=x
3-3x,
∴f′(x)=3(x-1)(x+1),
當(dāng)x∈[-2,-1],f′(x)≥0,x∈(-1,1),f′(x)<0;x∈(1,2],f′(x)>0.
∴f(x)在[-2,-1]上是增函數(shù),(-1,1)上遞減,(1,2)遞增;
且f(-2)=-2,f(-1)=2,f(1)=-2,f(2)=2.
∴f(x)的值域A=[-2,2];
又∵g(x)=ax+1(a>0)在[-2,2]上是增函數(shù),
∴g(x)的值域B=[-2a-1,2a-1];
根據(jù)題意,有A⊆B
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124922416763552/SYS201310251249224167635015_DA/0.png)
⇒a≥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124922416763552/SYS201310251249224167635015_DA/1.png)
.
同理g(x)=ax+1(a<0)在[-2,2]上是減函數(shù),
可以求出a≤-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124922416763552/SYS201310251249224167635015_DA/2.png)
.
故實(shí)數(shù)a的取值范圍是:(-∞,-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124922416763552/SYS201310251249224167635015_DA/3.png)
]∪[-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124922416763552/SYS201310251249224167635015_DA/4.png)
,+∞).
故答案為:(-∞,-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124922416763552/SYS201310251249224167635015_DA/5.png)
]∪[-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124922416763552/SYS201310251249224167635015_DA/6.png)
,+∞).
點(diǎn)評(píng):此題是個(gè)中檔題.考查利用導(dǎo)數(shù)研究函數(shù)在閉區(qū)間上的最值問(wèn)題,難點(diǎn)是題意的理解與轉(zhuǎn)化,體現(xiàn)了轉(zhuǎn)化的思想.同時(shí)也考查了同學(xué)們觀(guān)察、推理以及創(chuàng)造性地分析問(wèn)題、解決問(wèn)題的能力,