<s>[INST] 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?
 [/INST] 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 $x \not = 1$.
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