已知函數(shù)f(x)=ax3+bx2+c(a,b,c∈R,a≠0).
(1)若函數(shù)y=f(x)的圖象經(jīng)過點(diǎn)(0,0),(-1,0),求函數(shù)y=f(x)的單調(diào)區(qū)間;
(2)若a=b=1,函數(shù)y=f(x)與直線y=2的圖象有兩個(gè)不同的交點(diǎn),求c的值.
解:(1)把點(diǎn)P(-1,0)代入y=f(x)得-a+b+c=0,又c=0,故a=b
由f’(x)=3ax
2+2ax=ax(3x+2)=0得,x
1=0,x
2=-
,
故當(dāng)a>0時(shí),f(x)的單調(diào)遞增區(qū)間是(-∞,-
),(0,+∞)
單調(diào)遞減區(qū)間是(-
,0)
當(dāng)a<0時(shí),f(x)的單調(diào)遞減區(qū)間是(-∞,-
),(0,+∞)
單調(diào)遞增區(qū)間是(-
,0)(6分)
(2)當(dāng)a=b=1時(shí),f(x)的單調(diào)遞增區(qū)間是(-∞,-
),(0,+∞),
單調(diào)遞減區(qū)間是(-
,0)
故當(dāng)x=-
時(shí),f(x)取極大值為f(-
)=-
+
+c,
當(dāng)x=0時(shí),f(x)的極小值為f(0)=c
要使函數(shù)y=f(x)與直線y=2的圖象有兩個(gè)不同的交點(diǎn),則必須滿足-
+
+c=2或c=2
故c=
或2.(6分)
分析:(1)先由“函數(shù)y=f(x)的圖象經(jīng)過點(diǎn)(0,0),(-1,0)”,求得函數(shù)f(x),再求導(dǎo),由f′(x)≥0求得單調(diào)增區(qū)間,由f′(x)≤0求得單調(diào)減區(qū)間,要注意討論.
(2)當(dāng)a=b=1時(shí),分別求得函數(shù)的極大值和極小值,再由“函數(shù)y=f(x)與直線y=2的圖象有兩個(gè)不同的交點(diǎn)”求解.
點(diǎn)評(píng):本題主要考查用導(dǎo)數(shù)法研究函數(shù)的單調(diào)性,基本思路是:當(dāng)函數(shù)為增函數(shù)時(shí),導(dǎo)數(shù)大于等于零;當(dāng)函數(shù)為減函數(shù)時(shí),導(dǎo)數(shù)小于等于零,還考查了導(dǎo)數(shù)法研究曲線的相對(duì)位置,基本思路是:求導(dǎo),明確極值點(diǎn),再動(dòng)靜結(jié)合求解參數(shù)的范圍.