令t=x^2+1>=1
则x^2=t-1
代入函数得：f=[(t-1)^2+3(t-1)+6]/t=[t^2-2t+1+3t-3+6]/t=[t^2+t+4]/t=t+4/t+1
t+4/t>=2√(t*4/t)=4
当且仅当t=4/t时,即t=2时取等号,此时x^2=1,x=±1
所以f的最小值为fmin=4+1=5,当x=±1时取得.