函數(shù)y=(3-x2)ex的單調(diào)遞增區(qū)是( )
A.(-∞,0)
B.(0,+∞)
C.(-∞,-3)和(1,+∞)
D.(-3,1)
【答案】分析:求導(dǎo)函數(shù),令其大于0,解不等式,即可得到函數(shù)的單調(diào)遞增區(qū)間.
解答:解:求導(dǎo)函數(shù)得:y′=(-x2-2x+3)ex
令y′=(-x2-2x+3)ex>0,可得x2+2x-3<0
∴-3<x<1
∴函數(shù)y=(3-x2)ex的單調(diào)遞增區(qū)間是(-3,1)
故選D.
點評:本題重點考查導(dǎo)數(shù)知識的運用,考查函數(shù)的單調(diào)性,解題的關(guān)鍵是求導(dǎo)函數(shù),令其大于0.