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Find the equations of the tangent line(s) to $\large{\frac{x-1}{x+1}}$ that are parallel to the line $y=2(x-5)$.

I've found the slope at $2$, and set the derivative equal to it, but I cant think how to solve for $x$. It seams simple but i'm drawing a blank.

So far I'm at $\large{\frac{2}{(x+1)^2}} = 2$.

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  • $\begingroup$ Since $ \ x \ $ cannot equal -1 , you can multiply both sides of the equation by $ \ ( x + 1)^2 \ $ (so that you are not multiplying through by zero). $\endgroup$ Commented Jun 19, 2015 at 4:27
  • $\begingroup$ so then x would = -2 because after multiplying it out to get 2 = 2x squared + 4x +2, subtracting the two makes 0 = 2x squared +4x, subtract 4x and divide by 2x which makes x= -2 $\endgroup$
    – Alec
    Commented Jun 19, 2015 at 4:43
  • $\begingroup$ The equation becomes $$ \ ( x + 1 )^2 \ = \ 1 \ \ \Rightarrow \ \ x + 1 \ = \ \pm 1 \ \ , $$ which has two solutions. as also discussed below. So there are two possible tangent lines. $\endgroup$ Commented Jun 19, 2015 at 4:55

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Following what you have(in fact, you already did the important part), $x=-2, 0$.

For $x=-2$, $\frac{x-1}{x+1}=3$, so the line is $y-3=2(x+2)$, so $y=2x+7$;

For $x=0$, $\frac{x-1}{x+1}=-1$, so the line is $y+1=2(x-0)$, so $y=2x-1$.

It would be helpful to draw the plot on paper.

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