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,+∞)

(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)閒'(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,+∞).