I have no idea how to do this problem at all.
Find equation of the tangent lines to the curve
$$y = \frac {x-1}{x+1}$$
that are parallel to the line $x-2y = 2$
I found the derivative of $y = \frac {x-1}{x+1}$ to be $\frac 2{x^2 + 2x +1}$ and the other line I rewrote as $\frac{-2+x}{2}$
From here I attempted to set them equal to each other and I got
$6 = x^3 – 3x$ which is incredibly difficult to solve and is probably not the answer. Where did I go wrong?
Best Answer
The line $x-2y=2$ has slope $\frac12$, so you want to set the derivative of $\frac{x-1}{x+1}$ equal to $\frac12$ and solve for $x$:
$$\frac2{x^2+2x+1}=\frac12\;.$$
This would actually be just a little easier if you hadn't multiplied out the denominator; then you'd have $$\frac2{(x+1)^2}=\frac12\;,\tag{1}$$ so $(x+1)^2=4$, $x+1=\pm 2$, ...
Added: Let's recap what's going on here. You want some lines that are to be parallel to the line $x-2y=2$. That line has slope $1/2$, so your lines have to have slope $1/2$: that's how you know that they're parallel to the line $x-2y=2$. (If they had a different slope, they'd eventually cross it somewhere.) The derivative $\frac2{(x+1)^2}$ gives the slope of the tangent at any point on the curve $y=\frac{x-1}{x+1}$; we want to know where that slope is $1/2$, so we set the derivative equal to $1/2$ in $(1)$ and solve for $x$. You found that $x=-3$ or $x=1$, so there are two points on the curve where the tangent is parallel to the line $x-2y=2$; one is $(-3,2)$, and the other is $(1,0)$. Now you just find the tangent lines to $y=\frac{x-1}{x+1}$ at those two points.
For the first, for instance, you have $y-2=\frac12\big(x-(-3)\big)=\frac12(x+3)=\frac12x+\frac32$, so $y=\frac12x+\frac72$; I'll leave the other to you.