Derivative zero implies constant

Let $I\subseteq\mathbb{R}$ be an interval and let $f:I\to\mathbb{R}$ be differentiable on $I^\circ$.

Proposition: If $f'(x)=0$ for all $x\in I^\circ$, then $f$ is constant on $I$; i.e., for all $x,y\in I$ one has $f(x)=f(y)$.

This is a fundamental rigidity result: having zero instantaneous rate of change everywhere forces no global change.

Proof sketch: Take $x<y$ in $I$ with $[x,y]\subseteq I$. By the mean value theorem , there exists $c\in(x,y)$ such that $f(y)-f(x)=f'(c)(y-x)=0.$ Hence $f(y)=f(x)$. Since any two points in an interval can be connected by such a segment, $f$ is constant.