Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 76

Determine the equations of the tangent lines to the curve $\displaystyle y = \frac{x-1}{x+1}$ that
are parallel to the line $x - 2y = 2$



$
\begin{equation}
\begin{aligned}
\text{Given:}&&& \text{Curve}\quad y = \frac{x-1}{x+1}\\
\phantom{x}&&& \text{Line} \quad x - 2y = 2
\end{aligned}
\end{equation}
$


The slope$(m)$ of the curve and the line are equal because they are parallel.

$
\begin{equation}
\begin{aligned}
x - 2y &= 2
&& \text{Solving for slope}(m) \text{ using the equation of the line}\\
\\
-2y & = 2-x
&& \text{Transpose } x \text{ to the other side}\\
\\
\frac{-2y}{-2} &= \frac{2-x}{-2}
&& \text{Divide both sides by -2}\\
\\
y &= \frac{x-2}{2}
&& \text{Simplify the equation}\\
\\
y &= \frac{1}{2}x - 1
&& \text{Use the formula of general equation of the line to get the slope}(m)\\
\\
y &= mx + b
&& \text{Slope}(m) \text{ is the numerical coefficient of $x$}\\
\\
m &= \frac{1}{2}\\
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
y &= \frac{x-1}{x+1}\\
\\
y' &= \frac{(x+1) \frac{d}{dx} (x-1) - \left[ (x-1) \frac{d}{dx} (x+1) \right]}{(x+1)^2}
&& \text{Using Quotient Rule}\\
\\
y' &= \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2}
&& \text{Simplify the equation}\\
\\
y' &= \frac{x+1-x+1}{(x+1)^2}
&& \text{Combine like terms}\\
\\
y' &= \frac{2}{(x+1)^2}
\end{aligned}
\end{equation}
$


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

$
\begin{equation}
\begin{aligned}
y' &= m = \frac{2}{(x+1)^2}
&& \text{Substitute value of slope}(m)\\
\\
\frac{1}{2} &= \frac{2}{(x+1)^2}
&& \text{Using cross multiplication}\\
\\
(x+1)^2 &= 4
&& \text{Take the square root of both sides}\\
\\
\sqrt{(x+1)^2} &= \pm \sqrt{4}
&& \text{Simplify the equation}\\
\\
x+1 &= \pm 2
&& \text{Solve for }x \\
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
x &= 2 -1 &&& x &= -2-1\\
x &= 1 &&& x &= -3
\end{aligned}
\end{equation}
$


Using the equation of the curve given,
@ $x= 1$

$
\begin{equation}
\begin{aligned}
y &= \frac{x-1}{x+1}
&& \text{Substitute the value of }x = 1\\
\\
y &= \frac{1-1}{1+1}
&& \text{Combine like terms}\\
\\
y &= \frac{0}{2}
&& \text{Simplify the equation}\\
\\
y &= 0
\end{aligned}
\end{equation}
$


@ $x=-3$

$
\begin{equation}
\begin{aligned}
y &= \frac{x-1}{x+1}
&& \text{Substitute the value of } x = -3\\
\\
y &= \frac{-3-1}{-3+1}
&& \text{Combine like terms}\\
\\
y &= \frac{-4}{-2}
&& \text{Simplify the equation}\\
\\
y &= 2
\end{aligned}
\end{equation}
$


Using point slope form to get the equations of the tangent line
@ $x=1$, $y=0$, $\displaystyle m = \frac{1}{2}$


$
\begin{equation}
\begin{aligned}
y - y_1 &= m(x-x_1)
&& \text{Substitute the value of } x,y \text{ and slope}(m)\\
\\
y - 0 &= \frac{1}{2} (x-1)
&& \text{Simplify the equation}
\end{aligned}
\end{equation}
$


The equation of the tangent line @ $x =1$, $y =0$ is $\displaystyle y = \frac{x-1}{2}$

@ $x = -3$, $y = 2$, $\displaystyle m = \frac{1}{2} $


$
\begin{equation}
\begin{aligned}
y -y_1 &= m(x-x_1)
&& \text{Substitute the value of } x,y \text{ and slope}(m)\\
\\
y - 2 & = \frac{1}{2} (x+3)
&& \text{Distribute } \frac{1}{2} \text{ in the equation}\\
\\
y - 2 & = \frac{x+3}{2}
&& \text{Transpose -2 to the right side}\\
\\
y & = \frac{x+3}{2} +2
&& \text{Get the LCD}\\
\\
y & = \frac{x+3+4}{2}
&& \text{Combine like terms}
\end{aligned}
\end{equation}
$


The equation of the tangent line @ $x=-3$, $y = 2$ is $\displaystyle y = \frac{x+7}{2}$