令t=x^2+1>=1
則x^2=t-1
代入函數(shù)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
當且僅當t=4/t時,即t=2時取等號,此時x^2=1,x=±1
所以f的最小值為fmin=4+1=5,當x=±1時取得.