Find equations of the tangent lines to the curve \[ y=\frac{x-1}{x+1} \] that are parallel to the line \( x-2 y=4 \). (Enter your answers as a comma-separated list of equations.) F

We are given the curve \[ y=\frac{x-1}{x+1} \] and we need to find the equations of the tangent lines to this curve that are parallel to the line \[ x-2y=4. \] --- **Step 1. Find the slope of the given line.** Rewrite the line \( x-2y=4 \) in slope-intercept form: \[ x-2y=4 \implies -2y = 4-x \implies y=\frac{1}{2}x-2. \] Thus, the slope of the line is \[ m=\frac{1}{2}. \] --- **Step 2. Differentiate the curve.** The curve is given by \[ y=\frac{x-1}{x+1}. \] Differentiate using the quotient rule: \[ y' = \frac{(x+1)\cdot(1) - (x-1)\cdot(1)}{(x+1)^2} = \frac{x+1-x+1}{(x+1)^2} = \frac{2}{(x+1)^2}. \] --- **Step 3. Find the x-coordinates where the slope is \(\frac{1}{2}\).** Set the derivative equal to \(\frac{1}{2}\): \[ \frac{2}{(x+1)^2} = \frac{1}{2}. \] Cross-multiply: \[ 2\cdot2 = (x+1)^2 \implies (x+1)^2 = 4. \] Take the square root: \[ x+1 = \pm2. \] Solving for \(x\): - If \(x+1 = 2\), then \(x = 1\). - If \(x+1 = -2\), then \(x = -3\). --- **Step 4. Find the corresponding y-coordinates on the curve.** For \(x=1\): \[ y=\frac{1-1}{1+1}=\frac{0}{2}=0. \] For \(x=-3\): \[ y=\frac{-3-1}{-3+1}=\frac{-4}{-2}=2. \] Thus, the points of tangency are: \[ (1,0) \quad \text{and} \quad (-3,2). \] --- **Step 5. Find the equations of the tangent lines.** Using the point-slope form \( y-y_0=m(x-x_0) \) with \( m=\frac{1}{2} \): - For point \( (1,0) \): \[ y-0=\frac{1}{2}(x-1) \implies y=\frac{1}{2}x-\frac{1}{2}. \] - For point \( (-3,2) \): \[ y-2=\frac{1}{2}(x+3) \implies y=\frac{1}{2}x+\frac{3}{2}+2=\frac{1}{2}x+\frac{7}{2}. \] --- **Final Answer** \[ y=\frac{1}{2}x-\frac{1}{2}, \quad y=\frac{1}{2}x+\frac{7}{2}. \]