Question

已知曲线C:y=x3 求过点(1,1)处的切线方程并求相应切线与曲线的交点坐标

Answer 1

y=x^3
y'=3x^2
y'(1)=3
切线为：y=3(x-1)+1=3x-2
代入曲线方程得：x^3=3x-2
x^3-3x+2=0
x^3-x-2x+2=0
x(x-1)(x+1)-2(x-1)=0
(x-1)(x^2+x-2)=0
x=1, 此交点为(1,1)
x^2+x-2=0, x1=(-1+√5)/2, x2=(-1-√5)/2，此两交点为（x1,3x1-2),(x2,3x2-2)

Answer 2

y'=3x^2
k=y'|(x=1)=3
切线方程 y=3(x-1)+1
y=3x-2
交点坐标
x^3-3x+2=0
x^3-x-2(x-1)=0
x(x+1)(x-1)-2(x-1)=0
x-1=0 x=1 y=1 (1,1)为切点
x^2+x-2=0
x1=1 y1=1
x2=-2 y2=-8
交点坐标(-2,-8)

Answer 3

y'=3x²
x=1 y'=3
切线方程：y=3(x-1)+1=3x-2
y=x^3
y=3x-2
联立得 x^3-3x+2=0
x^3-x-2x+2=0
x(x²-1)-2(x-1)=0
x(x-1)(x+1)-2(x-1)=0
(x-1)(x²+x-2)=0
(x-1)(x-1)(x+2)=0
x=1 or x=-2
交点(1,1) (-2,-8)