设x1＜x2＜0
f(x1)-f(x2)=(x1^2+1)-(x2^2+1)
=x1^2-x2^2
=(x1-x2)(x1+x2)
x1-x2＜0,x1+x2＜0
所以,f(x1)-f(x2)＞0
f(x1)＞f(x2)
所以,f(x)在（-无穷大,0）上是减函数