## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$f[g(x)]=g[f(x)]=x$. This means that $f(x)$ and $g(x)$ are inverses of each other. We know that $x=0$ has to be excluded from both domains because the denominator cannot be $0$ for the function to be undefined.
We wish to plug $f(x)$ into $g(x)$ to obtain: $$\displaystyle g[f(x)]=g (\dfrac{1}{x}) \\ =\dfrac{1}{(1/x)} \\=x$$ We wish to plug $g(x)$ into $f(x)$ to obtain: $$f[(g(x)]=f (\dfrac{1}{x}) \\ =\dfrac{1}{(1/x)}\\=x$$ We see that $f[g(x)]=g[f(x)]=x$. This means that $f(x)$ and $g(x)$ are inverses of each other. We know that $x=0$ has to be excluded from both domains because the denominator cannot be $0$ for the function to be undefined.