Question

函数f(x)=（ax+1）/(x+2)在区间(－2，＋∞)上单调递增，则实数a的取值范围是

详细过程......

Answer 1

f(x)=（ax+1）/(x+2) = (a(x+2)-2a+1)/(x+2) = a + (1-2a)/(x+2)
f(x)=（ax+1）/(x+2)在区间(－2，＋∞)上单调递增
<=> (1-2a)/(x+2) 在区间(－2，＋∞)上单调递增
<=> 1-2a < 0
<=> a > 1/2
实数a的取值范围是 (1/2, ＋∞)