已知函數(shù)f(x)=x2+bx+c(b,c∈R),對任意的x∈R,恒有f′(x)≤f(x).
(Ⅰ)證明:當(dāng)x≥0時(shí),f(x)≤(x+c)2;
(Ⅱ)若對滿足題設(shè)條件的任意b,c,不等式f(c)-f(b)≤M(c2-b2)恒成立,求M的最小值.
【答案】
分析:(Ⅰ)f′(x)≤f(x)轉(zhuǎn)化為x
2+(b-2)x+c-b≥0恒成立,找到b和c之間的關(guān)系,再對f(x)和(x+c)
2作差整理成關(guān)于b和c的表達(dá)式即可.
(Ⅱ)對c≥|b|分c>|b|和c=|b|兩種情況分別求出對應(yīng)的M的取值范圍,再綜合求M的最小值即可.
解答:解:(Ⅰ)易知f'(x)=2x+b.由題設(shè),對任意的x∈R,2x+b≤x
2+bx+c,
即x
2+(b-2)x+c-b≥0恒成立,所以(b-2)
2-4(c-b)≤0,從而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277961595/SYS201310241805192779615019_DA/0.png)
.
于是c≥1,且
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277961595/SYS201310241805192779615019_DA/1.png)
,因此2c-b=c+(c-b)>0.
故當(dāng)x≥0時(shí),有(x+c)
2-f(x)=(2c-b)x+c(c-1)≥0.
即當(dāng)x≥0時(shí),f(x)≤(x+c)
2.
(Ⅱ)由(Ⅰ)得,c≥|b|
當(dāng)c>|b|時(shí),有M≥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277961595/SYS201310241805192779615019_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277961595/SYS201310241805192779615019_DA/3.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277961595/SYS201310241805192779615019_DA/4.png)
,
令t=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277961595/SYS201310241805192779615019_DA/5.png)
則-1<t<1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277961595/SYS201310241805192779615019_DA/6.png)
=2-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277961595/SYS201310241805192779615019_DA/7.png)
,
而函數(shù)g(t)=2-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277961595/SYS201310241805192779615019_DA/8.png)
(-1<t<1)的值域(-∞,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277961595/SYS201310241805192779615019_DA/9.png)
)
因此,當(dāng)c≥|b|時(shí)M的取值集合為[
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277961595/SYS201310241805192779615019_DA/10.png)
,+∞).
當(dāng)c=|b|時(shí),由(Ⅰ)知,b=±2,c=2.
此時(shí)f(c)-f(b)=-8或0,c
2-b
2=0,
從而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277961595/SYS201310241805192779615019_DA/11.png)
恒成立.
綜上所述,M的最小值為
點(diǎn)評:本題是對二次函數(shù)的恒成立問題和導(dǎo)函數(shù)的求法的綜合考查.二次函數(shù)的恒成立問題一般分兩類,一是大于0恒成立須滿足開口向上,且判別式小于0,二是小于0恒成立須滿足開口向下,且判別式小于0.