`y'= 3x^2 -3`
`y'=0 ⇔ 3x^2 -3= 3(x^2 -1)=0 => x= 1 ; x=-1`
` A : x=1 => y(1)=1^3 -3.1+1=-1`
` => A ( 1;-1)`
` B : x=-1 => y(-1) =(-1)^3 -3.(-1)+1=3`
` => B ( -1;3)`
PT đường thẳng ` AB:`
$⇔\frac{x-x_A}{x_B-x_A}=$ $\frac{y-y_A}{y_B-y_A}$
$⇔\frac{x-1}{-1-1}=$ $\frac{y-(-1)}{3-(-1)}$
$⇔\frac{x-1}{-2}=$ $\frac{y+1}{4}$
`⇔ 4x-4=-2y-2 `
`⇒ 2y =-4x+2`
`⇒ y=-2x+1`
`⇒` Đáp án `: C`
