设三次多项式函数f(x)=ax3+bx2+cx+d满足
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[解析] 由题干f(t)dt=12x2+18x+1知(x+1)-f(x)=12x2+18x+1,
即:3ax2+(3a+2b)x+(a+b+c)=12x2+18x+1,
解方程组:,可得a=4,b=3,c=-6,即有:f(x)=4x3+3x2-6x+d,从而可知:f'(x)=12x2+6x-6,f"(x)=24x+6,所以有稳定点x1=,x2=-1,由f"(