Given $f(x)$ its inverse function, domain and range $f(x) = \frac{{2x + 3}}{{x - 1}},\left[ {x \in {R},x > 1} \right]$
I've got the inverse function to be:
${f^{ - 1}}(x) = \frac{{x + 3}}{{x - 2}}$
How would I go about working out the range and domain of this function? The range is possibly $f^{-1}(x)>1$ if im not mistaken, but how do I figure out the domain?
 A: One quick note, if you write $f^{-1}$ as a function of $x$ then the variable on the right hand should be $x$. So either $f^{-1}(y)=\frac{y+3}{y-2}$, which is perfectly fine to write, or $f^{-1}(x)=\frac{x+3}{x-2}$, which is also perfectly fine to write.
To find the domain, you need to find the set of $x$ values such that there exists a value $y=f(x)$. Now, it seems that you're given a domain—you've written "$\left[x\in R,x>1\right]$". But if you weren't given that, you would want to find the largest possible such set of $x$ values. There's really only one real number $z$ such that we can't compute $f(z)$ for this $f$. Can you see what it is?
The range isn't always easy to find. The range is the set of $y$ values such that $y=f(x)$ for some $x$ in the range. Read that sentence over and over and over again until you understand it.
In this case the range is easy to find, since you've computed an inverse function. Then, the range of $f$ is just the domain of $f^{-1}$. Make sense?
A: Hint: Use that


*

*the domain of the inverse function is the range of the original function

*the range of the inverse function is the domain of the original function.

