Parallel Tangent Line to y=(x-1)/(x+1)

[tink]

Find the equations of the tangent line to the curve y= (x-1)/(x+1) that are parallel to the line x-2y=2

help!

 

[HallsofIvy]

What is the slope of the line x- 2y= 2?

For what values of x is the derivative of (x-1)/(x+1) equal to that slope?

 

[wais]

To find the equations of parallel tangent lines to the curve y= (x-1)/(x+1), we first need to find the derivative of the curve. Using the quotient rule, we get:

y' = [(x+1)(1) - (x-1)(1)] / (x+1)^2
y' = 2 / (x+1)^2

Now, we know that parallel lines have the same slope. So, the equation of the parallel line must also have a slope of 2. Using the point-slope form of a line, we get:

y - y1 = m(x - x1)
y - y1 = 2(x - x1)

Since the line is parallel to the curve, it must also pass through the point (x1, y1). Substituting the coordinates of this point into the equation, we get:

y - y1 = 2(x - x1)
y - (x1-1)/(x1+1) = 2(x - x1)

Simplifying the equation, we get:

y = 2x - (2x1 + 1)/(x1+1)

This is the equation of the parallel tangent line to the curve y= (x-1)/(x+1). However, we also know that this line must be parallel to the line x-2y=2. So, we can set these two equations equal to each other and solve for x1:

2x - (2x1 + 1)/(x1+1) = x - 2y + 2
2x - (2x1 + 1)/(x1+1) = x - 2((x-1)/(x+1)) + 2
2x - (2x1 + 1)/(x1+1) = (x+1) - 2x + 2
2x - (2x1 + 1)/(x1+1) = 3
2x - 2x1 - 1 = 3(x1+1)
2x - 2x1 - 1 = 3x1 + 3
2x - 3 = 5x1
x1 = (2x-3)/5

Therefore, the equation of the parallel tangent line to the curve y= (x-1)/(x+1) that