已知函數(shù)f(x)=x3-3ax(a∈R)
(1)當(dāng)a=1時(shí),求f(x)的極小值;
(2)若直線x+y+m=0對(duì)任意的m∈R都不是曲線y=f(x)的切線,求a的取值范圍;
(3)設(shè)g(x)=|f(x)|,x∈[-1,1],求g(x)的最大值F(a)的解析式.
分析:(1)由f(x)=x
3-3ax,得f′(x)=3x
2-3a,當(dāng)f′(x)>0,f′(x)<0時(shí),分別得到f(x)的單調(diào)遞增區(qū)間、單調(diào)遞減區(qū)間,由此可以得到極小值為f(1)=-2.
(2)要使直線x+y+m=0對(duì)任意的m∈R都不是曲線y=f(x)的切線,只需令直線的斜率-1小于f(x)的切線的最小值即可,也就是-1<-3a.
(3)由已知易得g(x)為[-1,1]上的偶函數(shù),只需求在[0,1]上的最大值F(a).有必要對(duì)a進(jìn)行討論:①當(dāng)a≤0時(shí),f′(x)≥0,得F(a)=f(1)=1-3a;②當(dāng)a≥1時(shí),f(x)≤0,且f(x)在[0,1]上單調(diào)遞減,得g(x)=-f(x),則F(a)=-f(1)=3a-1;當(dāng)0<a<1時(shí),得f(x)在[0,
]上單調(diào)遞減,在[
,1]上單調(diào)遞增.當(dāng)f(1)≤0時(shí),f(x)≤0,所以得g(x)=-f(x),F(xiàn)(a)=-f(
)=2a
,當(dāng)f(1)>0,需要g(x)在x=
處的極值與f(1)進(jìn)行比較大小,分別求出a的取值范圍,即綜上所述求出F(a)的解析式.
解答:解:(1)∵當(dāng)a=1時(shí),f′(x)=3x
2-3,令f′(x)=0,得x=-1或x=1,當(dāng)f′(x)<0,即x∈(-1,1)時(shí),f(x)為減函數(shù);當(dāng)f′(x)>0,即x∈(-∞,-1],或x∈[1,+∞)時(shí),f(x)為增函數(shù).∴f(x)在(-1,1)上單調(diào)遞減,在(-∞,-1],[1,+∞)上單調(diào)遞增∴f(x)的極小值是f(1)=-2
(2)∵f′(x)=3x
2-3a≥-3a,∴要使直線x+y+m=0對(duì)任意的m∈R都不是曲線y=f(x)的切線,當(dāng)且僅當(dāng)-1<-3a時(shí)成立,∴
a<(3)因g(x)=|f(x)|=|x
3-3ax|在[-1,1]上是偶函數(shù),故只要求在[0,1]上的最大值
①當(dāng)a≤0時(shí),f′(x)≥0,f(x)在[0,1]上單調(diào)遞增且f(0)=0,∴g(x)=f(x),F(xiàn)(a)=f(1)=1-3a.
②當(dāng)a>0時(shí),
f′(x)=3x2-3a=3(x+)(x-),
(ⅰ)當(dāng)
≥1,即a≥1時(shí),g(x)=|f(x)|=-f(x),-f(x)在[0,1]上單調(diào)遞增,此時(shí)F(a)=-f(1)=3a-1
(ⅱ)當(dāng)
0<<1,即0<a<1時(shí),當(dāng)f′(x)>0,即x>
或x<-
時(shí),f(x)單調(diào)遞增;當(dāng)f′(x)<0,即-
<x<
時(shí),f(x)單調(diào)遞減.所以
f(x)在[0,]上單調(diào)遞減,在
[,1]單調(diào)遞增.
1°當(dāng)
f(1)=1-3a≤0,即≤a<1時(shí),
g(x)=|f(x)|=-f(x),-f(x)在[0,]上單調(diào)遞增,在[,1]上單調(diào)遞減,
F(a)=-f()=2a;
2°當(dāng)
f(1)=1-3a>0,即0<a<(。┊(dāng)
-f()≤f(1)=1-3a,即0<a≤時(shí),F(xiàn)(a)=f(1)=1-3a(ⅱ)當(dāng)
-f()>f(1)=1-3a,即<a<時(shí),F(xiàn)(a)=-f()=2a綜上所述
F(x)= | 1-3a,(a≤) | 2a,(<a<1) | 3a-1,(a≥1) |
| |
點(diǎn)評(píng):本題綜合性較強(qiáng),主要考查導(dǎo)數(shù)的單調(diào)性、極值、最值等函數(shù)基礎(chǔ)知識(shí),尤其第三小題,考查帶有參數(shù)的函數(shù)題型,更是值得推敲,希望在平時(shí),多加練習(xí),掌握其要領(lǐng).