# Inverse functions

1. Aug 17, 2015

Is there a way to formally prove that if $f$ and $g$ are multiplicative inverses of each other, then $f^{-1} (x) = g^{-1} (\frac{1}{x})$?

2. Aug 17, 2015

### Dr. Courtney

I'd start by testing a number of special cases and see what approach the trickier special cases suggest.

Simple special cases almost never suggest approaches to proof, but the tricky special cases often do.

3. Aug 17, 2015

### pasmith

Let $h$ be the function which takes $x$ to $1/x$. Now if $f(x)g(x) = 1$ for all $x$ then $f = h \circ g$. Then $f^{-1} = g^{-1} \circ h^{-1}$. But $h = h^{-1}$ so $f^{-1} = g^{-1} \circ h$ as required.

4. Aug 17, 2015