Question

已知三次函数y=f （x）有三个零点x1,x2,x3,且在点（xi,f（xi））处的切线斜率为ki（i=1,2,3）,则

Answer 1

记f(x)=a(x-x1)(x-x2)(x-x3)

f'(x)=a(x-x1)(x-x2)+a(x-x2)(x-x3)+a(x-x1)(x-x3),记p=a(x1-x2)(x2-x3)(x1-x3)

k1=f'(x1)=a(x1-x2)(x1-x3)=p/(x2-x3)

k2=f'(x2)=a(x2-x1)(x2-x3)=p/(x3-x1)

k3=f'(x3)=a(x3-x1)(x3-x2)=p/(x1-x2)

1/k1+1/k2+1/k3=(x2-x3+x3-x1+x1-x2)/p=0