已知函數(shù)f(x)=x5+ax3+bx+1在x=1和x=2處取得極值.
(1)求a和b的值;
(2)求f(x)的單調(diào)區(qū)間.
分析:(I)利用函數(shù)的導(dǎo)數(shù)在極值點(diǎn)處的值為0,列出方程組,求出a,b的值.
(Ⅱ)將a,b的值代入導(dǎo)函數(shù),令導(dǎo)函數(shù)大于0求出解集為遞增區(qū)間;令導(dǎo)函數(shù)小于0,求出解集為遞減區(qū)間.
解答:解:(1)因?yàn)閒'(x)=5x
4+3ax
2+b,…(2分)
由已知得:f'(1)=5+3a+b=0.f'(2)=2
4×5+2
2×3a+b=0,解得
a=-,b=20. …(5分)
(2)由(1)知f'(x)=5x
4+3ax
2+b=5(x
2-1)(x
2-4)
=5(x+1)(x+2)(x-1)(x-2).…(7分)
當(dāng)x∈(-∞,-2)∪(-1,1)∪(2,+∞)時(shí),f'(x)>0;
當(dāng)x∈(-2,-1)∪(1,2)時(shí),f'(x)<0.…(9分)
因此f(x)的單調(diào)增區(qū)間是(-∞,-2),(-1,1),(2,+∞),f(x)的單調(diào)減區(qū)間是(-2,-1),(1,2). …(10分)
點(diǎn)評(píng):本題考查函數(shù)的極值點(diǎn)處的導(dǎo)數(shù)值為0、考查函數(shù)的單調(diào)性與導(dǎo)函數(shù)的符號(hào)有關(guān):導(dǎo)函數(shù)大于0時(shí),函數(shù)遞增;導(dǎo)函數(shù)小于0時(shí),函數(shù)遞減.