Question

6．已知函數(shù)f（x）=$\left\{\begin{array}{l}{{x}^{2}+2ax+1，x＜0}\\{（a-3）x+4a，x≥0}\end{array}\right.$滿足對(duì)任意x₁≠x₂，都有$\frac{f（{x}_{1}）-f（{x}_{2}）}{{x}_{1}-{x}_{2}}$＜0成立，則實(shí)數(shù)a的取值范圍是（-∞，0]．

Answer 1

分析 若對(duì)任意x₁≠x₂，都有$\frac{f（{x}_{1}）-f（{x}_{2}）}{{x}_{1}-{x}_{2}}$＜0成立，則函數(shù)f（x）=$\left\{\begin{array}{l}{{x}^{2}+2ax+1，x＜0}\\{（a-3）x+4a，x≥0}\end{array}\right.$為減函數(shù)，進(jìn)而根據(jù)分段函數(shù)單調(diào)性的定義，可得答案．

解答 解：若對(duì)任意x₁≠x₂，都有$\frac{f（{x}_{1}）-f（{x}_{2}）}{{x}_{1}-{x}_{2}}$＜0成立，
則函數(shù)f（x）=$\left\{\begin{array}{l}{{x}^{2}+2ax+1，x＜0}\\{（a-3）x+4a，x≥0}\end{array}\right.$為減函數(shù)，
則$\left\{\begin{array}{l}-\frac{2a}{2}≥0\\ a-3＜0\\ 1≥4a\end{array}\right.$，
解得：a∈（-∞，0]，
故答案為：（-∞，0]

點(diǎn)評(píng) 本題考查的知識(shí)點(diǎn)是分段函數(shù)的應(yīng)用，熟練掌握并正確理解分段函數(shù)單調(diào)性的定義，是解答的關(guān)鍵．