Question

由原點O向三次曲線y=x³-3x²引切線,切于異于原點的點P₁(x₁,y₁),再由P₁引此曲線的切線,切于異于點P₁的點P₂(x₂,y₂),如此繼續(xù)下去,得到點列{P_n(x_n,y_n)}.

(1)求x₁；

(2)求x_n與x_n+1滿足的關(guān)系式；

(3)求數(shù)列{x_n}的通項公式.

Answer 1

解:(1)x₁=.                                                                                                     ?

(2)過曲線上的點P_n+1(x_n+1,y_n+1)的切線方程為?

y-(x_n+1³-3x_n+1²)=(3x_n+1²-6x_n+1)(x-x_n+1),?

 

而此切線過點P_n(x_n,y_n)，則有?

x_n³-3x_n²-(x_n+1³-3x_n+1²)=(3x_n+1²-6x_n+1)(x_n-x_n+1)x_n+2x_n+1=3.                                              ?

(3)由x_n+2x_n+1=3x_n+1-1=-(x_n-1)x_n=1-(-)ⁿ.