Question

用定义证明f(x)=（ax+1）/(x+2),(a不等于0.5) 在（负无穷，-2）上的单调性

Answer 1

f(x)=（ax+1）/(x+2)
=(ax+2a-2a+1)/(x+2)
=(a(x+2)-2a+1)/(x+2)
=a-(2a-1)/(x+2)

在（-∞，-2）上任取x1<x2
f(x1)-f(x2)
=a-(2a-1)/(x1+2)-(a-(2a-1)/(x2+2))
=-(2a-1)/(x1+2)+(2a-1)/(x2+2)
=(2a-1)(1/(x2+2)-1/(x1+2))
=(2a-1)(x1+2-x2-2)/((x1+2)(x2+2))
=(2a-1)(x1-x2)/((x1+2)(x2+2))

因为x1,x2∈（-∞，-2）,x1<x2
所以(x1-x2)/((x1+2)(x2+2))<0

因此当(2a-1)>0,即a>1/2时
(2a-1)(x1-x2)/((x1+2)(x2+2))<0
f(x1)<f(x2),f(x)是增函数

因此当(2a-1)<0,即a<1/2时
(2a-1)(x1-x2)/((x1+2)(x2+2))>0
f(x1)>f(x2),f(x)是减函数