若f(x)在(0,+∞)上的单调减函数,求a的取值范围
f(x) = (ax-1)/(x+1)
=(ax + a -a-1)/(x+1)
=[a(x+1) - (a+1)]/(x+1)
= a - (a+1)/(x+1)
为保证 f(x)在(0,+∞)上的单调递减,则要求 -(a+1)/(x+1) 递减
要求 (a+1)/(x+1) 递增
因此 a+1 < 0
a