Question

设x1、x2是f(x)=(a/3)x^3+(b-1)x^2/2+x(a>0)的两个极值点,f'(x)为f(x)导函数,求

Answer 1

(1) f(x)=ax^3/3+(b-1)x^2/2+x

f'(x)=ax^2+(b-1)x+1=a(x-x1)(x-x2)

由韦达定理x1*x2=1/a,x1+x2=(1-b)/a

a=1/(x1x2),b=1-(1/x1+1/x2)

f'(-2)=4a-2b+3=4/x1x2+2/x1+2/x2+1

=(2/x1+1)(2/x2+1)

如果0＜x1＜2＜x2＜4->2/x1＞1,2/x2＞2/4

则f'(-2)＞2×(3/2)=3

(2)f'(x)=ax^2+(b-1)x+1

如果0