Zero derivative implies constant

Let $f:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$.

Corollary: If $f'(x)=0$ for all $x\in(a,b)$, then $f$ is constant on $[a,b]$.

Connection to parent theorem: Apply the mean value theorem: for any $x<y$, there exists $c\in(x,y)$ with $f(y)-f(x)=f'(c)(y-x)=0$.