已知函数f(X)=(ax+1)/(x+2)在(-2,+∞)上单调递增,则a的取值范围是f(X)=(ax+1)/(x+2)=a+[(1-2a)/(x+2)]这步怎么转换过来

Question

已知函数f(X)=(ax+1)/(x+2)在(-2,+∞)上单调递增,则a的取值范围是
f(X)=(ax+1)/(x+2)
=a+[(1-2a)/(x+2)]这步怎么转换过来

Answer 1

易知函数f(X)=(ax+1)/(x+2) =[(ax+2a)+1-2a]/(x+2)=a+[(1-2a)/(x+2)]
a+[(1-2a)/(x+2)] 其实就是配出来的.
因为函数在(-2,+∞)上单调递减,
而函数f(X)=(ax+1)/(x+2)=a+[(1-2a)/(x+2)]
在(-2,+∞)上单调递增
所以须(1-2a)1/2