已知函數(shù)f(x)是定義在R上的奇函數(shù),當(dāng)x>0時(shí),f(x)=2x3+mx2+(1-m)x.
(I)當(dāng)m=2時(shí),求f(x)的解析式;
(II)設(shè)曲線y=f(x)在x=x0處的切線斜率為k,且對(duì)于任意的x0∈[-1,1]-1≤k≤9,求實(shí)數(shù)m的取值范圍.
分析:(I)f(x)是定義在R上的奇函數(shù),設(shè)x<0,則-x>0應(yīng)用f(x)=2x
3+mx
2+(1-m)x求解,再由f(0)=0得解.
(Ⅱ)因?yàn)榍y=f(x)在x=x
0處的切線斜率為k所以由(I)求導(dǎo)
f′(x)= | 6x2+2mx+(1-m),(x≥0) | 6x2-2mx+(1-m),(x<0) |
| |
再由對(duì)任意的x
0∈[-1,1],總能-1≤k≤9,則在任意x
0∈[-1,1]時(shí),-1≤f'(x)≤9恒成立,又因?yàn)閒'(x)是偶函數(shù)∴對(duì)任意x
0∈(0,1]時(shí),-1≤f'(x
0)≤9恒成立即可.
解答:解:(I)∵f(x)是定義在R上的奇函數(shù),∴f(0)=0.
當(dāng)x>0時(shí),f(x)=2x
3+mx
2+(1-m)x.
當(dāng)x<0時(shí),∵f(x)=-f(-x)∴f(x)=-(-2x
3+mx
2-(1-m)x)=2x
3-mx
2+(1-m)x∴
f(x)= | 2x3+mx2+(1-m)x(x≥0) | 2x3-mx2+(1-m)x(x<0) |
| |
.
當(dāng)m=2時(shí),∴
f(x)= | 2x3+2x2-x,(x≥0) | 2x3-2x2-x(x<0) |
| |
(Ⅱ)由(I)得:∴
f′(x)= | 6x2+2mx+(1-m),(x≥0) | 6x2-2mx+(1-m),(x<0) |
| |
曲線y=f(x)在x=x
0處的切線斜率,對(duì)任意的x
0∈[-1,1],總能不小于-1且不大于9,
則在任意x
0∈[-1,1]時(shí),-1≤f'(x)≤9恒成立,
∵f'(x)是偶函數(shù)
∴對(duì)任意x
0∈(0,1]時(shí),-1≤f'(x
0)≤9恒成立
1
0當(dāng)
-≤0時(shí),由題意得
∴0≤m≤2
2
0當(dāng)
0<-≤1時(shí)
∴
∴-6≤m<0
3
0當(dāng)
->1時(shí)∴
∴-8≤m<-6
綜上:-8≤m≤2
∴實(shí)數(shù)m的取值范圍是{m|-8≤m≤2}.
點(diǎn)評(píng):本題主要考查利用奇偶性求對(duì)稱區(qū)間上的解析式和二次函數(shù)研究最值解決恒成立問(wèn)題.