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三次函数及其相关的问题,近年来在各级各类考查试卷中经常出现,其中大部分题型都可利用导数法来求解.本文介绍几种常见类型的求解方法,供参考.一、三次函数的切线例1已知函数f(x)=x3-x+2,试求过点P(1,2)的曲线y=f(x)的切线方程.解析设切点P0(x0,y0),由f’(x)=3x2-1,则f’(x0)=3x20-1,过点P0的方程为y-y0=f’(x0)(x-x0),即y-(x30-x0+2)=(3x20-1)(x-x0).又切线过点P(1,2),则2-(x30-x0+2)=(3x20-1)(1-x0),分解因式得(x0-1)2(2x0+1)=0,解之得x0=1或x0=-12.则f’(-12)=-14,f’(1)=2.故所求的切线方程为y-2=-14(x-1)和y-2=2(x-1).
Third-order functions and their related problems often appear in exam papers of various kinds at various levels in recent years, most of them can be solved by the derivative method.This paper introduces several common types of solving methods for reference.One, three times Tangential Function Example 1 The tangent equation of the curve y = f (x) of the point P (1,2) is obtained by using the function f (x) = x3-x + ), F ’(x0) = 3x20-1 from f’ (x) = 3x2-1, the equation of the over point P0 is y-y0 = f ’(x0) (x-x0) 2- (x30-x0 + 2) = (3x20-1) (1-x0), then tangentially points P (1, 2) Factorization (x0-1) 2 (2x0 + 1) = 0, the solution of x0 = 1 or x0 = -12. Then f ’(- 12) = - 14, f’ (1) = 2. Therefore The tangent equations are y-2 = -14 (x-1) and y-2 = 2 (x-1).