Question

若函数f(x)=(ax+1)/(x+2)在区间（－2，+∞,)上是增函数，求实数a的取值范围

Answer 1

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)<0,
a>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.