Question

f(x)=(ax+1)/(x+2)在（-2,+∞）上是增函数,则a的取值范围是（ ）

Answer 1

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.