函数f(x)=ax+1/x+2(a为常数) (1)若a=1,证明f(x)在(-2,+∞)上为单调递增函数

(1)当a=1时,f(x)=x+1/x+2
设-2<x1<x2,∴f(x1)-f(x2)=x1-x2/(x1+2)(x2+2)
∵-2<x1<x2,x1<x2∴(x1+2)(x2+2)>0,x1-x2<0
即f(x1)-f(x2)0即a