题目内容
9.函数f(x)=$\left\{\begin{array}{l}{(a-1)x+\frac{5}{2},x≤1}\\{\frac{2a+1}{x},x>1}\end{array}\right.$,在定义域R上满足对任意实数x1≠x2都有$\frac{f({x}_{1})-f({x}_{2})}{{x}_{1}-{x}_{2}}$<0,则a的取值范围是(-$\frac{1}{2}$,$\frac{1}{2}$].分析 由已知可得函数f(x)=$\left\{\begin{array}{l}{(a-1)x+\frac{5}{2},x≤1}\\{\frac{2a+1}{x},x>1}\end{array}\right.$,在定义域R上为减函数,则$\left\{\begin{array}{l}a-1<0\\ 2a+1>0\\ a-1+\frac{5}{2}≥2a+1\end{array}\right.$,解得a的取值范围.
解答 解:若在定义域R上满足对任意实数x1≠x2都有$\frac{f({x}_{1})-f({x}_{2})}{{x}_{1}-{x}_{2}}$<0,
则函数f(x)=$\left\{\begin{array}{l}{(a-1)x+\frac{5}{2},x≤1}\\{\frac{2a+1}{x},x>1}\end{array}\right.$,在定义域R上为减函数,
则$\left\{\begin{array}{l}a-1<0\\ 2a+1>0\\ a-1+\frac{5}{2}≥2a+1\end{array}\right.$,
解得:a∈(-$\frac{1}{2}$,$\frac{1}{2}$],
故答案为:(-$\frac{1}{2}$,$\frac{1}{2}$]
点评 本题考查的知识点是分段函数的应用,正确理解分段函数的单调性,是解答的关键.
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