求过点(1,-1)与曲线f(x)=x3-2x相切的直线方程.

问题描述:

求过点(1,-1)与曲线f(x)=x3-2x相切的直线方程.

若直线与曲线切于点(x0,y0)(x0≠0),则k=

y0+1
x0-1
=
x 30
-2x0+1
x0-1
=
x 20
+x0-1

∵y′=3x2-2,∴y′|x=x0=3x02-2,
x 20
+x0-1
=3x02-2,
∴2x02-x0-1=0,∴x0=1,x0=-
1
2

∴过点A(1,-1)与曲线f(x)=x3-2x相切的直线方程为x-y-2=0或5x+4y-1=0.