已知函数f(x)=loga(ax^2-x+1/2)
问题描述：
已知函数f(x)=loga(ax^2-x+1/2)
(1)当a=3/8时,求函数f(x)的单调递减区间 （2）当0小于a小于1时,f(x)在1,2的闭区间上恒大于0,求实数a的取值范围
求详解 急
答
1)f(x) = log(a)[ax^2-x + 1/2 ]当a = 3/8f(x) = log(3/8)[3/8)x^2-x +1/2]令g(x) =(3/8)x^2-x +1/2因为0《3/8《1所以log(3/8)^x是单调减函数所以g'(x) = (3/4)x -1 0=> (3/8)x^2-x +1/2 > 03x^2-8x +4 > 0(3x-2)(x-...