题目内容
解:通过观察图象可设三次函数f(x)的解析式为f(x)=ax(x-1)(x-2)=ax3-3ax2+2ax,即ax3+bx2+cx+d=ax3-3ax2+2ax,于是有:b=-3a,c=2a,d=0.
由图象知f(3)>0,而f(3)=6a>0,
∴a>0,
即有c=2a>a>0=d>-3a=b,
故c>a>d>b.