Darboux function

Remarks

Equivalently, $f$ maps every subinterval of $I$ onto an interval. Each continuous function on $I$ is a Darboux function, by the intermediate value theorem. Each derivative on an open interval is a Darboux function, by Darboux's theorem.

Example: let $f(x) = \sin(1/x)$ for $x \ne 0$ and $f(0) = 0$, on $\R$. On a subinterval not containing $0$, $f$ is continuous, so its image is an interval. A subinterval with more than one point that contains $0$ contains a full period of $\sin(1/x)$ on one side of $0$, so its image is $[-1, 1]$. So $f$ is a Darboux function, although it is not continuous at $0$.

Non-example: the floor function $\floor{x}$ on $\R$ has $\floor{0} = 0$ and $\floor{1} = 1$, but it never takes the value $1/2$.

The Conway base 13 function is a Darboux function that is continuous at no point.