Showing $f^{-1}$ exists where $f(x) = \frac{x+2}{x-3}$ Let $f(x) = \dfrac{x + 2 }{x - 3}$.
There's three parts to this question:


*

*Find the domain and range of the function $f$.

*Show $f^{-1}$ exists and find its domain and range.

*Find $f^{-1}(x)$.


I'm at a loss for #2, showing that the inverse function exists.  I can find the inverse by solving the equation for $x$, showing that it exists without just solving for the inverse. Can someone point me in the right direction? 
 A: Can you graph it?  Passing the horizontal line test would show it has an inverse.  That may be what your teacher is looking for.
A: It is a valid way to find the inverse by solving for x first and then verify that $f^{-1}(f(x))=x$ for all $x$ in your domain. It is quite preferable to do it here because you need it for 3. anyways.
A: A function will be invertible (on a proper domain) if and only if it is injective, that is, if it never takes the same value twice.  If your function is continuous and is defined on an interval, this forces it to be increasing or decreasing.  Even without being defined on an interval we can use calculus to determine if the function is increasing/decreasing, which lets us solve the problem fairly simply.
We have that $f'(x)=\frac{(x-3)-(x+2)}{(x-3)^2}=\frac{-5}{(x-3)^2}$.  This is negative except where it is undefined, and so the function is decreasing, at least locally.  Unfortunately, because we have the discontinuity at $x=3$, we can't say immediately that the function is injective.  However, we do know that it never takes the same value twice on either side of the discontinuity.  This combined with limits, is all we need:
Since $\displaystyle \lim_{x\to -\infty}f(x)=\lim_{x\to \infty}f(x)=1$, we must have that $f(x)<1$ if $x<3$ and $f(x)>1$ if $x>3$.  Since no number is both less than 1 and greater than 1, we can't have $f(x_1)=f(x_2)$ for $x_1<3$ and $x_2>3$.  Therefore $f(x)$ is injective, and hence invertible on a suitable domain (namely on $(-\infty,3)\cup (3,\infty)$.)
