Equation of perpendicular line to tangent lines

Question

I'm completely stuck on a question about equation of perpendicular lines to tangent lines. I figured out my tangent lines equations, I know graphically what I should get for my perpendicular lines (x=-1 and x=1) but how can I prove that mathematically?

Thanks a lot for your help

Answer 1

A horizontal tangent line must have a slope of $0$. So that will be a line in the form of $y = 0*x + c = c$.

So at which points on the curve $y=x^3 -3x + 1$ is the slope $0$?

Well the derivative is the formula for the slope. (Assuming the function is differentiable).

So we must solve for $y' =3x^2 - 3 = 0$. Let $x_0, x_1$ be the two solutions to that.

So the equations of your tangent lines will be: $y= (x_0^3 -3x + 1)$ and $y = (x_1^3 -3x_1 + 1)$

2) A line perpendicular would be completely vertical. So they will have the form $x = c$ for some constant. In this case those lines are at $x=x_0$ and $x=x_1$. So those are you equations.

Answer 2

The equation of the curve is given by $y=x^3-3x+1$ Therefore the slope of the tangent is given by: $$\dot{y}=3x^2-3$$ Where the tangent is horizontal is when the slope of the tangent is zero $\implies \dot{y}=0$ or, $x=\pm 1$. The points where this happens are given by $(1,-1)$ and $(-1,3)$.

Now the when the tangent is horizontal the normal will be vertical $\implies$ it will have a form $x=c$. At $(1,-1)$ it is $x=1$ and at $(-1,3)$ it is $x=-1$.
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