摘要:即b2>2 (b+2c). --------------------------14分
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已知函数f(x)=
x3+
(b-1)x2+cx(b,c为常数).
(1)若f(x)在x=1和x=3处取得极值,试求b,c的值;
(2)若f(x)在x∈(-∞,x1)和x∈(x2,+∞)上单调递增,且在x∈(x1,x2)上单调递减,又满足x2-x1>1.求证:b2>2(b+2c).
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