y=1/3 x^3+x∧2+ax-5,求导得:y’= x^2+2x+a,若在(-∞,+∞)中是单调函数,则需导数恒大于等于0或恒小于等于0,y’= x^2+2x+a只能恒大于等于0,即x^2+2x+a>=0, a>=-x^2-2x,-x^2-2x=-(x+1)^2+1有最大值1,所以a>=1.若在[1,+∞)中是单调函数,y’= x^2+2x+a只能恒大于等于0,即x^2+2x+a>=0, a>=-x^2-2x,当x>=1时,-x^2-2x=-(x+1)^2+1有最大值-3,所以a>=-3.若函数在区间(-3,1)上单调递减,f’(x) = x^2+2x+a则f’(-3)<=0,且...
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