分析:先求函數(shù)導(dǎo)數(shù),令導(dǎo)數(shù)大于等于0,解得x的范圍就是函數(shù)的單調(diào)增區(qū)間.
解答:解:對函數(shù)y=3x-x3求導(dǎo),得,y′=3-3x2,
令y′≥0,即3-3x2≥0,解得,-1≤x≤1
∴函數(shù)y=3x-x3的遞增區(qū)間為[-1,1]
故答案為:[-1,1].
點評:本題主要考查了導(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時原函數(shù)單調(diào)遞增,當(dāng)導(dǎo)函數(shù)小于0時原函數(shù)單調(diào)遞減.