Question

設(shè)函數(shù)f(x)=2x³-3(a-1)x²+1,其中a≥1.

(1)求f(x)的單調(diào)區(qū)間;

(2)討論f(x)的極值.

Answer 1

解：由已知,得f′(x)=6x［x-(a-1)］.令f′(x)=0,解得x₁=0,x₂=a-1.

（1）當(dāng)a=1時,f′(x)=6x²,f(x)在(-∞,+∞)上單調(diào)遞增;

當(dāng)a＞1時,f′(x)=6x［x-(a-1)］,f′(x),f(x)隨x的變化情況如下表:

x	(-∞,0)	0	(0,a-1)	a-1	(a-1,+∞)
f′(x)	+	0	-	0	+
f(x)		極大值		極小值

 

從上表,可知函數(shù)f(x)在(-∞,0)上單調(diào)遞增;在(0,a-1)上單調(diào)遞減;在(a-1,+∞)上單調(diào)遞增.

(2)由(1),知當(dāng)a=1時,函數(shù)f(x)沒有極值;當(dāng)a＞1時,函數(shù)f(x)在x=0處取得極大值1,在x=a-1處取得極小值1-(a-1)³.