Question

设三次函数f(x)的图像与x轴交于不同的三点A、B、C,且在点A、B处的切线互相平行.求证：A、B、C三点的横坐

Answer 1

设y=ax^3+bx^2+cx=0, a0,

α、Υ、β为方程的三个根:x1, x2, x3

x1+x2+x3=-b/a

y'=3ax^2+2bx+c

由题意有:y'(x1)=y'(x2)

3ax1^2+2bx1+c=3ax2^2+2bx2+c

3a(x1^2-x2^2)+2b(x1-x2)=0

3a(x1+x2)+2b=0

x1+x2=-2b/(3a)

x3=-b/a-(x1+x2)=-b/a+2b/(3a)=-b/(3a)=(x1+x2)/2

因此x1, x3, x2成等差数列.