## Precalculus (10th Edition)

$f^{-1}$ is also increasing on $(0,5)$.
If for a function $f(x)=y$ then for the inverse function $f^{-1}(y)=x.$ Since $f$ is one-to-one the domain of $f$ is the range of $f^{-1}$ and the range of $f$ is the domain of $f^{-1}$. Hence if $f$ is increasing on $(0,5)$, it means that for $a\lt b$, $f(a)\lt f(b)$. Hence for $f^{-1}$ if $f(a)\lt f(b)$, then $f^{-1}f(a)=a\lt b=f^{-1}f(b)$, hence it is also increasing on $(0,5)$.