Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 75

Determine the equations of both lines that are tangent to the curve $y = 1+x^3$ and are parallel to the line $12x -y = 1$


$\begin{equation} \begin{aligned} \text{Given:}&&& \text{Curve}\quad y = 1+x^3\\ \phantom{x}&&& \text{Line} \quad 12- y = 1 \end{aligned} \end{equation}$


The slope$(m)$ of the curve is equal to the slope$(m)$ of the line because there are parallel.
Using the formula $mx+b$, we take the equation of the line $12x-y=1$ or $y = 12x-1$ the slope is 12.


$\begin{equation} \begin{aligned} y &= 1 + x^3\\ \\ y'&= \frac{d}{dx} (1) + \frac{d}{dx}(x^3) && \text{Derive each terms}\\ \\ y'&= 0 + 3x^2 && \text{Simplify the equation}\\ \\ y'&= 3x^2\\ \end{aligned} \end{equation}$

Let $y' =$ slope$(m)$


$\begin{equation} \begin{aligned} m &= 12\\ \\ m &= 3x^2 && \text{Substitute the value of slope}(m)\\ \\ \frac{12}{3} &= \frac{3x^2}{3} && \text{Divide both sides by 3}\\ \\ x^2 &= 4 && \text{Take the square root of both sides}\\ \\ \sqrt{x^2} &= \pm \sqrt{4} && \text{Simplify the equation}\\ \\ x &= \pm 2 \end{aligned} \end{equation}$


Substitute the values of $x$ to the equation of the curve to solve for $y$


$\begin{equation} \begin{aligned} y &= 1 + x^3 &&& \phantom{x} && y &= 1 + x^3\\ \\ y &= 1 + (2)^3 &&& \Longleftarrow\text{(Simplify the equation)} \Longrightarrow && y &= 1 + (-2)^3\\ \\ y &= 9 &&& \phantom{x} && y &= -7\\ \\ \end{aligned} \end{equation}$


Using point slope form
@ $x = 2$ $y = 9$ $m = 12$


$\begin{equation} \begin{aligned} y - y_1 &= m(x-x_1) && \text{Substitute the value of } x, y \text{ and slope}(m)\\ \\ y - 9 &= 12(x-2) && \text{Distribute 12 in the equation}\\ \\ y - 9 &= 12x - 24 && \text{Add 9 to each sides}\\ \\ y &= 12x - 24 + 9 && \text{Combine like terms}\\ \end{aligned} \end{equation}$


The first equation of the tangent line is $y = 12x - 15$

@ $x = -2$ $y = -7$ $m = 12$

$\begin{equation} \begin{aligned} y - y_1 &= m(x-x_1) && \text{Substitute the value of }x,y\text{ and slope}(m)\\ \\ y+7 &= 12(x+2) && \text{Distribute 12 in the equation}\\ \\ y+7 &= 12x+24 && \text{Add -7 to each sides}\\ \\ y &= 12x+24-7 && \text{Combine like terms} \end{aligned} \end{equation}$


The second equation of the tangent line is $y = 12x + 17$