Question

已知函數(shù)f(x)=x³+(b-1)x²+cx(b,c為常數(shù)).

(1)若f(x)在x=1和x=3處取得極值，試求b,c的值;

(2)若f(x)在x∈(-∞,x₁)和x∈(x₂,+∞)上單調(diào)遞增，且在x∈(x₁,x₂)上單調(diào)遞減，又滿足x₂-x₁＞1.求證：b²＞2(b+2c).

Answer 1

(1)f′(x)=x²+(b-1)x+c，據(jù)題意知x=1和3是方程x²+(b-1)x+c=0的兩根，

∴1-b=4，c=3，即b=-3，c=3.                                              

(2)∵f(x)在x∈(-∞,x₁)和x∈(x₂,+∞)上單調(diào)遞增，∴f′(x)＞0，

又f(x)在(x₁,x₂)上單調(diào)遞減，∴f′(x)＜0.∴f′(x₁)=f′(x₂)=0,

∴x₁,x₂是方程x²+(b-1)x+c=0的兩根，故有                   

∵x₂-x₁＞1，∴(x₂-x₁)²＞1,

∴(x₂-x₁)²=(x₁+x₂)²-4x₁x₂=(b-1)²-4c=b²-2(b+2c)+1＞1,

∴b²＞2(b+2c).