Question

函数f(x)=x三次方+bx平方+cx+d在(-无穷,0)上是递增在[0,2]上是递减f(x)=0的三根是m,2,n,求

Answer 1

f'(x)=3x^2+2bx+c

由已知得

x=0,2是3x^2+2bx+c=0两根

-2b/3=2

c=0,b=-3

f(x)=x^3-3x^2+d

f(0)=d=2

f(x)=x^3-3x^2+2

所以

m^3-3m^2+2=0

(m^3-1)-3(m^2-1)=0

(m-1)(m^2+m+1)-3(m+1)(m-1)=0

m^2-2m=2

m^2-2m+1=3

(m-1)^2=3

1-√3≤m≤1+√3

同理1-√3≤n≤1+√3

所以|m-n|≤2√3