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 $y = 12x + 17$