Question

2．函數(shù)y=2log${\;}_{\frac{1}{2}}$²x-2log${\;}_{\frac{1}{2}}$x+3的單調(diào)遞增區(qū)間為[$\frac{\sqrt{2}}{2}$，+∞）．

Answer 1

分析 根據(jù)復(fù)合函數(shù)的單調(diào)性規(guī)律求出．

解答 解：令f（x）=log${\;}_{\frac{1}{2}}$x，則f（x）在（0，+∞）上單調(diào)遞減．
令g（x）=2x²-2x+3，則g（x）在（-∞，$\frac{1}{2}$]上單調(diào)遞減，在（$\frac{1}{2}$，+∞）上單調(diào)遞增，
令log${\;}_{\frac{1}{2}}$x$≤\frac{1}{2}$，得x≥$\frac{\sqrt{2}}{2}$，
∴函數(shù)g（f（x））的單調(diào)遞增區(qū)間為[$\frac{\sqrt{2}}{2}$，+∞）．
故答案為[$\frac{\sqrt{2}}{2}$，+∞）．

點評 本題考查了復(fù)合函數(shù)的單調(diào)性，將函數(shù)分成兩個基本初等函數(shù)是關(guān)鍵．