F(x)=1/3x^3-1/2(m+3)x^2+(m+6)x
F'(x)=x^2-(m+3)x+(m+6)
F'(x)为开口向上的抛物线,顶点为[(m+3)/2,(m+6)-(m+3)^2)/4]
要使F(x)在[1,∞]内有2个极点,只要F'(x)在[1,∞]内有2个零点,
所以F'(x)在[-∞,(m+3)/2]内单调递减,(m+3)/2＞1且顶点[(m+3)/2,(m+6)-(m+3)^2)/4]在x轴下方,
即m＞-1
F'(1)=1-(m+3)+(m+6)=4＞0
又(m+6)-(m+3)^2)/4＜0
m^2+2m-15＞0
m＞3或m＜-5
所以-1＜m＜-5