Positive derivative implies strictly increasing

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 strictly increasing on $[a,b]$.

Connection to parent theorem: Apply the mean value theorem to any $x<y$ in $[a,b]$ to get $f(y)-f(x)=f'(c)(y-x)$ for some $c\in(x,y)$. Since $f'(c)>0$ and $y-x>0$, one has $f(y)>f(x)$.