令f′(x)=a(x+2)(x-1)=0得x=-2或x=1
x∈(-∞,-2)時f′(x)的符號與x∈(-2,1)時f′(x)的符號相反,x∈(-2,1)時f′(x)的符號與x∈(1,+∞)時f′(x)的符號相反
∴f(-2)=?
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∵圖象經(jīng)過四個象限
∴f(-2)?f(1)<0即(
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解得?
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16 |
故答案為B
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