Question

1. - / 1 points find equations of the tangent lines to the curve y = \frac{x - 1}{x + 1} that are parallel to the line x - 2y = 2. (enter your answers as a comma - separated list of equations.)

Explanation:

Step1: Rewrite the given line in slope - intercept form

Rewrite $x - 2y=2$ as $y=\frac{1}{2}x - 1$. The slope of this line is $m = \frac{1}{2}$.

Step2: Differentiate the curve function

Use the quotient rule. If $y=\frac{x - 1}{x + 1}$, and the quotient rule is $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u=x - 1$, $u^\prime = 1$, $v=x + 1$, $v^\prime=1$. Then $y^\prime=\frac{1\cdot(x + 1)-(x - 1)\cdot1}{(x + 1)^{2}}=\frac{x + 1-x + 1}{(x + 1)^{2}}=\frac{2}{(x + 1)^{2}}$.

Step3: Set the derivative equal to the slope of the given line

Set $\frac{2}{(x + 1)^{2}}=\frac{1}{2}$. Cross - multiply to get $(x + 1)^{2}=4$. Then $x+1=\pm2$.

• Case 1: If $x + 1 = 2$, then $x = 1$. When $x = 1$, $y=\frac{1-1}{1 + 1}=0$. Using the point - slope form $y - y_{1}=m(x - x_{1})$ with $(x_{1},y_{1})=(1,0)$ and $m=\frac{1}{2}$, we have $y-0=\frac{1}{2}(x - 1)$, or $y=\frac{1}{2}x-\frac{1}{2}$.
• Case 2: If $x + 1=-2$, then $x=-3$. When $x=-3$, $y=\frac{-3 - 1}{-3 + 1}=\frac{-4}{-2}=2$. Using the point - slope form $y - y_{1}=m(x - x_{1})$ with $(x_{1},y_{1})=(-3,2)$ and $m=\frac{1}{2}$, we have $y - 2=\frac{1}{2}(x + 3)$, or $y=\frac{1}{2}x+\frac{7}{2}$.

Answer:

$y=\frac{1}{2}x-\frac{1}{2},y=\frac{1}{2}x+\frac{7}{2}$