设三次函数方程为y=ax^3+bx^2+cx+d,把x=1,y=4,x=3,y=0,x=0,y=0代入方程,可得d=0,则a+b+c=4,27a+9b+3c=0.再算此函数的导数为y'=3ax^2+2bx+c,当x=1时有极大值,即y'=0,得方程3a+2b+c=0,解三次方程,得a=-1,b=6,c=-9 检验...