函数f(x)=（ax+1）/(x+2)在区间(-2,＋∞)上为增函数,则a的取值范围是?

方法一:f(x)=（ax+1）/(x+2)=[a(x+2)-2a+1]/(x+2)=a+(1-2a)/(x+2).令,Y=1/(x+2),而此函数,在x∈(-2,＋∞)上为减函数,现要使Y=(1-2a)/(x+2),在x∈(-2,＋∞)上为增函数,则须满足(1-2a)1/2.即,函数f(x)=（ax+1）/(x+2)...