2 |
3 |
∴
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∴f(x)=ax3+2ax2-4ax,
由图象可知函数y=f(x)在(-∞,-2)上单调递减,在(−2,
2 |
3 |
2 |
3 |
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,
2 |
3 |
2 |
3 |
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}.