Question

16．若函數(shù)f（x）=$\left\{\begin{array}{l}{（3a+2）x-1，x≤1}\\{\frac{a}{x}，x＞1}\end{array}\right.$是R上的單調(diào)函數(shù)，則實(shí)數(shù)a的取值范圍為$（-\frac{2}{3}，-\frac{1}{2}]$．

Answer 1

分析 通過函數(shù)的單調(diào)性，列出不等式，化簡求解即可．

解答 解：當(dāng)函數(shù)f（x）=$\left\{\begin{array}{l}{（3a+2）x-1，x≤1}\\{\frac{a}{x}，x＞1}\end{array}\right.$是R上的單調(diào)增函數(shù)，
可得：$\left\{\begin{array}{l}{3a+2＞0}\\{a＜0}\\{a≥3a+1}\end{array}\right.$，解得a∈$（-\frac{2}{3}，-\frac{1}{2}]$．
當(dāng)函數(shù)f（x）=$\left\{\begin{array}{l}{（3a+2）x-1，x≤1}\\{\frac{a}{x}，x＞1}\end{array}\right.$是R上的單調(diào)減函數(shù)，
可得：$\left\{\begin{array}{l}{3a+2＜0}\\{a＞0}\\{3a+2-1＞a}\end{array}\right.$，解得a∈∅．

故答案為：$（-\frac{2}{3}，-\frac{1}{2}]$．

點(diǎn)評 本題考查分段函數(shù)的應(yīng)用，考查分類討論思想，轉(zhuǎn)化思想以及計算能力．