已知函數(shù)f(x)=x(x+2)(x-3).
(1)求導(dǎo)數(shù)f′(x);
(2)求f(x)的單調(diào)區(qū)間.
解:(1)由f(x)=x(x+2)(x-3),得:f(x)=x
3-x
2-6x,
∴f'(x)=3x
2-2x-6.
(2)令f'(x)<0,解得
,
令f'(x)>0,解得
或
,
所以f(x)的單調(diào)遞減區(qū)間為
,
單調(diào)遞增區(qū)間為
,
.
分析:(1)把給出的函數(shù)解析式利用多項(xiàng)式乘多項(xiàng)式展開(kāi),然后直接利用導(dǎo)數(shù)的加法法則運(yùn)算;
(2)求出導(dǎo)函數(shù)的零點(diǎn),由零點(diǎn)對(duì)定義域分段,由導(dǎo)函數(shù)在各段內(nèi)的符號(hào)判斷函數(shù)的單調(diào)區(qū)間.
點(diǎn)評(píng):本題主要考查導(dǎo)函數(shù)的正負(fù)與原函數(shù)的單調(diào)性之間的關(guān)系,即當(dāng)導(dǎo)函數(shù)大于0時(shí)原函數(shù)單調(diào)遞增,當(dāng)導(dǎo)函數(shù)小于0時(shí)原函數(shù)單調(diào)遞減,此題是中檔題..