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 $\displaystyle y = \frac{x+7}{2}$