Question

已知函數(shù)f(x)=x3-x2+ax-a(a∈R).

(1)當(dāng)a=-3時(shí),求函數(shù)f(x)的極值.

(2)若函數(shù)f(x)的圖象與x軸有且只有一個(gè)交點(diǎn),求a的取值范圍.

 

Answer 1

(1) 當(dāng)x=-1時(shí),函數(shù)f(x)取得極大值為f(-1)=--1+3+3=,

當(dāng)x=3時(shí),函數(shù)f(x)取得極小值為f(3)=×27-9-9+3=-6.

(2) (0,+∞)

【解析】(1)當(dāng)a=-3時(shí),f(x)=x3-x2-3x+3.

f'(x)=x2-2x-3=(x-3)(x+1).

令f'(x)=0,得x1=-1,x2=3.

當(dāng)x<-1時(shí),f'(x)>0,

則函數(shù)在(-∞,-1)上是增函數(shù),

當(dāng)-1<x<3時(shí),f'(x)<0,

則函數(shù)在(-1,3)上是減函數(shù),

當(dāng)x>3時(shí),f'(x)>0,

則函數(shù)在(3,+∞)上是增函數(shù).

所以當(dāng)x=-1時(shí),函數(shù)f(x)取得極大值為f(-1)=--1+3+3=,

當(dāng)x=3時(shí),函數(shù)f(x)取得極小值為f(3)=×27-9-9+3=-6.

(2)因?yàn)?/span>f'(x)=x2-2x+a,

所以Δ=4-4a=4(1-a).

①當(dāng)a≥1時(shí),則Δ≤0,∴f'(x)≥0在R上恒成立,所以f(x)在R上單調(diào)遞增.

f(0)=-a<0,f(3)=2a>0,所以,當(dāng)a≥1時(shí)函數(shù)的圖象與x軸有且只有一個(gè)交點(diǎn).

②a<1時(shí),則Δ>0,∴f'(x)=0有兩個(gè)不等實(shí)數(shù)根,不妨設(shè)為x1,x2(x1<x2),∴x1+x2=2,x1·x2=a,

則

x	(-∞,x1)	x1	(x1,x2)	x2	(x2,+∞)
f'(x)	+	0	-	0	+
f(x)	↗	極大值	↘	極小值	↗

∵-2x1+a=0,∴a=-+2x1,

∴f(x1)=-+ax1-a

=-+ax1+-2x1

=+(a-2)x1

=x1[+3(a-2)],

同理f(x2)=x2[+3(a-2)].

∴f(x1)·f(x2)=x1x2[+3(a-2)][+3(a-2)]=a(a2-3a+3).

令f(x1)·f(x2)>0,解得a>0.

而當(dāng)0<a<1時(shí),f(0)=-a<0,f(3)=2a>0.

故0<a<1時(shí),函數(shù)f(x)的圖象與x軸有且只有一個(gè)交點(diǎn).

綜上所述,a的取值范圍是(0,+∞).