f'x(x,y)=e^x(x+2y+y^2+1)=0f'y(x,y)=2e^x(1+y)=0解得x=0 y=-1A=f''xx(x,y)=e^x(x+2y+y^2+2)=1B=f''xy(x,y)=2e^x(1+y)=0C=f''yy(x,y)=2e^x=2AC-B^2=2>0,A>0所以函数在(0,-1)处取得极小值=-1