Find the inverse of $f(x) = (x+1)/(x-8)$ Find the inverse of this function: 
I have gotten this far:
$x = y+1/y-8$
$x(y-8) = y+1$
$x(y-8)-1=y$
$xy-8x - 1 = y$
I think I went backwards?
 A: Since the question did not ask to find the formula for $f^{-1}$, but only $f^{-1}(6)$, we simply let $f(x) = 6$:
$$6 = \frac{x+1}{x-8}$$
$$6x - 48 = x + 1$$
$$5x = 49$$
$$x = \frac{49}{5}$$
Therefore $f^{-1}(6) = \frac{49}{5}$.

Of course, for educational purposes, it is often useful to find the formula describing $f^{-1}$. Here is one way to do it:
$$f(x) = \frac{x + 1}{x - 8} = 1 + \frac{9}{x - 8}$$
$$f(x) - 1 = \frac{9}{x-8}$$
$$\frac{1}{f(x)-1} = \frac{x-8}{9}$$
$$\frac{9}{f(x)-1} = x-8$$
$$x = \frac{9}{f(x)-1} + 8$$
So
$$f^{-1}(x) = \frac{9}{x  - 1} + 8$$
with $x \not = 1$.
A: If $f(x)= (x+1)/(x-8)$ and we let $y=f(x)$, we can use your method of switching the variables $x$ and $y$ and solving for $y$.
You were correct in reaching $xy-8x-1 = y$. Though we have "solved" for $y$, this solution is not useful for finding the inverse of $f(x)$, because it is a function of two variables.
So you want $y$ to only be in terms of $x$.
$-8x-1 = y-xy$
$-8x-1 = y(1-x)$
$(-8x-1)/(1-x) = y$. 
Remember this $y$ is not the $y=f(x)$ from the beginning, but rather we only named it $y$ so that we could solve for it. Once solved, you can give its proper name, $f^{-1}(y)$. This function takes a $y$, and spits back the $x$ used to form it.
I'll leave it to you to find $f^{-1}(6)$.
Also be careful with your manipulation.
$y+1/y-8 \neq (y+1)/(y-8)$.
