f(x)=(x+b+a-b)/(x+b)
=1+(a-b)/(x+b)
任意x1,x2∈(-∞,-b] ,x1>x2
f(x1)-f(x2)
=(a-b)/(x1+b)-(a-b)/(x2+b)
=(a-b)(x2-x1)(x1+b)(x2+b)
因为x1<-b x2<-b
所以x1+b<0 x2+b<0
所以f(x1)-f(x2)<0
任意x1,x2∈[-b,+∞) ,x1>x2
f(x1)-f(x2)
=(a-b)/(x1+b)-(a-b)/(x2+b)
=(a-b)(x2-x1)(x1+b)(x2+b)
因为x1>-b x2>-b
所以x1+b>0 x2+b>0
所以f(x1)-f(x2)<0
函数在(-∞,-b]和[-b,+∞)上单调递减
