(1)∵f"(x)=3ax2+2bx+c,且y=f"(x)的图象经过点(-2,0),(,0), ∴⇒ ∴f(x)=ax3+2ax2-4ax, 由图象可知函数y=f(x)在(-∞,-2)上单调递减,在(-2,)上单调递增,在(,+∞)上单调递减, 由f(x)极小值=f(-2)=a(-2)3+2a(-2)2-4a(-2)=-8,解得a=-1 ∴f(x)=-x3-2x2+4x (2)要使对x∈[-3,3]都有f(x)≥m2-14m恒成立, 只需f(x)min≥m2-14m即可. 由(1)可知函数y=f(x)在[-3,2)上单调递减,在(-2,)上单调递增,在(,3]上单调递减 且f(-2)=-8,f(3)=-33-2×32+4×3=-33<-8 ∴f(x)min=f(3)=-33(11分)-33≥m2-14m⇒3≤m≤11 故所求的实数m的取值范围为{m|3≤m≤11}. |