Precalculus (10th Edition)

$3$
Since f is one-to-one then for each $x$ there is a unique $f(x)$ and all the $f(x)$ values are distinct. Hence if $f(x)=y$, then $f^{-1}(y)=x$, hence with $f(3)=8$, then $f^{-1}(8)=3$