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.

所以f(-1)=2是極大值,f(1)=-2是極小值.

(2)曲線方程為y=x³-3x,點(diǎn)A(0,16)不在曲線上.

設(shè)切點(diǎn)為M(x₀,y₀),則點(diǎn)M的坐標(biāo)滿足y₀=x₀³-3x₀.

因f′(x₀)=3(x₀²-1),故切線的方程為y-y₀=3(x₀²-1)(x-x₀).

注意到點(diǎn)A(0,16)在切線上,有16-(x₀³-3x₀)=3(x₀²-1)(0-x₀),

化簡(jiǎn)得x₀³=-8,解得x₀=-2.

所以切點(diǎn)為M(-2,-2),

切線方程為9x-y+16=0.

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(shí),f′(x)=6x²,f(x)在(-∞,+∞)上遞增.

當(dāng)a＞1時(shí),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)	增	極大值	減	極小值	增

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(shí),函數(shù)f(x)沒(méi)有極值.

當(dāng)a＞1時(shí),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.