发布时间:2021-02-18 11:30:48
(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).
(1)f′(x)=x2+(b-1)x+c,据题意知x=1和3是方程x2+(b-1)x+c=0的两根,
∴1-b=4,c=3,即b=-3,c=3.
(2)∵f(x)在x∈(-∞,x1)和x∈(x2,+∞)上单调递增,∴f′(x)>0,
又f(x)在(x1,x2)上单调递减,∴f′(x)<0.∴f′(x1)=f′(x2)=0,
∴x1,x2是方程x2+(b-1)x+c=0的两根,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).