# Inverse functions

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})##?

Dr. Courtney
Gold Member
2020 Award
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.

pasmith
Homework Helper
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})##?
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.

PFuser1232
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.
Perfect. Thanks!