## Precalculus (6th Edition)

See: Important Facts about Inverses, p.414, 1. If $f$ is one-to-one, then $f^{-1}$ exists. 2. The domain of $f$ is the range of $f^{-1}$, and the range of $f$ is the domain of $f^{-1}$. 3. If the point $(a, b)$ lies on the graph of $f$, then $(b, a)$ lies on the graph of $f^{-1}$. ------------------- $(-1,-1)$ is on the graph of f, BUT $(-1,1)$ is on the graph of g, not $(-1,-1).$ So, $g(f(-1))=1\neq-1$ meaning that f and are not inverses.