Question

consider the following curve.
$y = \frac{x - 1}{x + 1}$
find $y(x)$.
$y(x)=\frac{2}{(x + 1)^2}$
find equations of the tangent lines to the curve that are parallel to the line $x - 2y = 5$. (enter your answers as a comma - separated list of equations.)

Explanation:

Step1: Rewrite the given line in slope - intercept form

Rewrite $x - 2y=5$ as $y=\frac{1}{2}x-\frac{5}{2}$. The slope of this line is $m = \frac{1}{2}$.

Step2: Set the derivative equal to the slope

Since the tangent lines are parallel to the given line, we set $y'(x)=\frac{1}{2}$. So, $\frac{2}{(x + 1)^2}=\frac{1}{2}$.

Step3: Solve for x

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

• When $x + 1 = 2$, $x=1$.
• When $x + 1=-2$, $x=-3$.

Step4: Find the corresponding y - values

• When $x = 1$, $y=\frac{1 - 1}{1+1}=0$.
• When $x=-3$, $y=\frac{-3 - 1}{-3 + 1}=2$.

Step5: Use the point - slope form to find the tangent line equations

The point - slope form is $y - y_1=m(x - x_1)$.

• For the point $(1,0)$ with $m=\frac{1}{2}$, the equation is $y-0=\frac{1}{2}(x - 1)$, which simplifies to $y=\frac{1}{2}x-\frac{1}{2}$.
• For the point $(-3,2)$ with $m=\frac{1}{2}$, the equation is $y - 2=\frac{1}{2}(x + 3)$, which simplifies to $y=\frac{1}{2}x+\frac{7}{2}$.

Answer:

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