函数f(x)=(ax+1)/(x+2)在区间(-2,+∞)上是递增的,求实数a的取值范围谢谢了,

题目

函数f(x)=(ax+1)/(x+2)在区间(-2,+∞)上是递增的,求实数a的取值范围谢谢了,

答案

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)在区间(-2,＋∞)上为增函数,则a的取值范围是:a>1/2.方法二:对f(x)求导,f(x)=（ax+1）/(x+2),f'(x)=[(ax+1)'(x+2)-(x+2)'(ax+1)]/(x+2)^2 =(2a-1)/(x+2)^2.要使f(x)在区间X∈(-2,＋∞)上为增函数,则f'(x)>0,即,(2a-1)/(x+2)^2>0,(2a-1)>0,a>1/2.则a的取值范围是:a>1/2.