设函数f(x)=(ax+1)/(x+2a)在区间(-2,+∞)上是增函数,那么a的取值范围是多少
人气：251 ℃ 时间：2019-10-19 23:48:43
解答
f'(x)=[a(x+2a)-(ax+1)]/[(x+2a)^2]=(2a^2-1)/[(x+2a)^2]>0
所以2a^2-1>0且-2a=1
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