考點(diǎn):復(fù)合函數(shù)的單調(diào)性
專題:函數(shù)的性質(zhì)及應(yīng)用
分析:先求出函數(shù)的定義域,利用復(fù)合函數(shù)的單調(diào)性之間的關(guān)系進(jìn)行求解即可.
解答:
解:要使函數(shù)有意義,則2x-x2>0,即0<x<2.
設(shè)t=2x-x2=-(x-1)2+1,則當(dāng)0<x≤1時(shí),函數(shù)t=2x-x2單調(diào)遞增,
當(dāng)1≤x<2時(shí),函數(shù)t=2x-x2單調(diào)遞減.
∵函數(shù)y=log0.5t在定義域上為單調(diào)遞減函數(shù),
∴根據(jù)復(fù)合函數(shù)的單調(diào)性之間的關(guān)系可知,
當(dāng)0<x≤1時(shí),函數(shù)f(x)單調(diào)遞減,
即函數(shù)f(x)的遞減區(qū)間為(0,1],
故答案為:(0,1].
點(diǎn)評(píng):本題主要考查復(fù)合函數(shù)單調(diào)性的判斷,利用復(fù)合函數(shù)同增異減的原則進(jìn)行判斷即可,注意要先求出函數(shù)的定義域.