We are tasked with finding the equation of the tangent to the curve $y = \frac{x-1}{x+1}$ that is parallel to the line $x - 2y = 2$.

Step 1: Determine the slope of the given line.

The line $x - 2y = 2$ can be rewritten in slope-intercept form (i.e., $y = mx + b$):

$x - 2y = 2 \quad \Rightarrow \quad -2y = -x + 2 \quad \Rightarrow \quad y = \frac{1}{2}x - 1$

Thus, the slope of the line is $m = \frac{1}{2}$.

Step 2: Find the slope of the tangent to the curve.

To find the slope of the tangent line to the curve $y = \frac{x-1}{x+1}$ at any point, we need to take the derivative of the curve with respect to $x$.

We use the quotient rule for differentiation:

$\frac{dy}{dx} = \frac{(x+1)\cdot 1 - (x-1)\cdot 1}{(x+1)^2}$

Simplifying the numerator:

$\frac{dy}{dx} = \frac{x + 1 - x + 1}{(x+1)^2} = \frac{2}{(x+1)^2}$

Thus, the slope of the tangent at any point is $\frac{2}{(x+1)^2}$.

Step 3: Set the derivative equal to the slope of the given line.

Since the tangent line is parallel to the given line, their slopes must be equal. Therefore, we set the slope of the tangent to $\frac{1}{2}$ (the slope of the given line):

$\frac{2}{(x+1)^2} = \frac{1}{2}$

Step 4: Solve for $x$.

Multiply both sides of the equation by $2(x+1)^2$:

$4 = (x+1)^2$

Now, take the square root of both sides:

$x+1 = \pm 2$

Thus, the two possible solutions for $x$ are:

$x = 1 \quad \text{or} \quad x = -3$

Step 5: Find the corresponding $y$-coordinates.

We now find the corresponding $y$-coordinates for these values of $x$ by substituting them into the original equation $y = \frac{x-1}{x+1}$.

• For $x = 1$:

$y = \frac{1-1}{1+1} = 0$

• For $x = -3$:

$y = \frac{-3-1}{-3+1} = \frac{-4}{-2} = 2$

Step 6: Write the equations of the tangent lines.

Now we have two points on the curve where the tangents have a slope of $\frac{1}{2}$:

1. For $x = 1$ and $y = 0$, the tangent line has the equation $y - 0 = \frac{1}{2}(x - 1)$, or

   $y = \frac{1}{2}(x - 1)$

2. For $x = -3$ and $y = 2$, the tangent line has the equation $y - 2 = \frac{1}{2}(x + 3)$, or

   $y = \frac{1}{2}(x + 3) + 2 = \frac{1}{2}x + \frac{7}{2}$

Final Answer:

The two possible tangent lines are:

1. $y = \frac{1}{2}(x - 1)$
2. $y = \frac{1}{2}x + \frac{7}{2}$

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Would you like more details on any part of the solution? Or do you have any follow-up questions?

Related Questions:

1. How do we use the quotient rule to differentiate rational functions?
2. What does it mean for two lines to be parallel in terms of their slopes?
3. How can we verify if a given point is on a curve?
4. What are other methods for finding tangents to curves?
5. Can we solve for the tangent to a curve at a specific point without using derivatives?

Tip:

To find tangents to curves, always start by differentiating the curve equation to find the general slope formula. Then, you can set the derivative equal to the slope of the desired tangent.