Let $f = \frac{x}{x-1}$. What is ${f}^{-1}$? Show that f and ${f}^{-1}$ are symmetric about $y = x$.

Question

Let $f = \frac{x}{x-1}$. What is ${f}^{-1}$? Show that f and ${f}^{-1}$ are symmetric about $y = x$.

Finding the inverse of $f$ is easy enough. It actually turns out to be the same as $f$ itself.

But how do I show that they are symmetric about $y = x$?

Thanks!

Answer 1

We have to show that if $f$ passes through the point $(x,y)$, then $f^{-1}$ passes through the point $(y,x)$. This is pretty easy to see (for all $f$, $f^{-1}$, not just yours). If $f(x)=y$, then $f^{-1}(y)=f^{-1}(f(x))=x$.

Answer 2

Given some function $f(x)$, to show that another function $g(x) = f^{-1}(x),$ one must show that $(f \circ g) (x) = x$ and $(g \circ f)(x) = x$. Then one can conclude $g(x) = f^{-1}(x)$.

Also, your domain and codomain here are both $\mathbb{R}$ \ $\{1\}$.