The identity on $[0,1]$ attains its maximum at $1$ and its minimum at $0$ with derivative $1$ at both, so Fermat's theorem genuinely needs the extremum to be at an interior point

Refuted claim: let $A \subseteq \mathbb{R}$, let $f : A \to \mathbb{R}$ and let $c \in A$ be a limit point of $A$ at which $f$ has a local extremum (Local (relative) maximum and minimum of $f : A \to \mathbb{R}$ at a point, the strict forms, and what it means for the point to be interior to $A$) and is differentiable (The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$, and differentiability on a set). Then $f'(c) = 0$.

That is Fermat's interior extremum theorem: if $f$ has a local extremum at a point $c$ interior to its domain and is differentiable at $c$, then $f'(c) = 0$ with the hypothesis "$c$ is interior to $A$" deleted and replaced by the weaker one needed for $f'(c)$ to be a defined symbol at all. It is false: the identity on $[0,1]$ attains a greatest and a least value, both at points of the domain that are not interior to it, and its derivative is $1$ everywhere.

• What the endpoint case would need instead. At $1$ the domain supplies points on the left only, and the difference quotient there is positive, so the most one can conclude is $f'(1) \ge 0$. That one-sided refinement is not stated at this point in the reading order, since nothing here uses it; the point of the witness is only that the two-sided conclusion $f'(c) = 0$ is unavailable.

• The witness is not delicate. Any function increasing on $[0,1]$ and differentiable there whose derivative vanishes at neither endpoint does the same job, and the identity is chosen for having a derivative that can be computed from The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$, and differentiability on a set in one line. The nonvanishing clause has to be said and is not automatic: $x \mapsto x^{3}$ is increasing on $[0,1]$ and differentiable there, and attains its least value at $0$, yet $f'(0) = 0$, so it refutes nothing at the left endpoint. What removing interiority destroys is the guarantee that the derivative vanishes, not the possibility. So the failure at an endpoint is the generic situation and not an artefact.

• Why this matters for Rolle's theorem. Rolle's theorem: if $a < b$, $f$ is continuous on $[a,b]$, differentiable at every point of $(a,b)$, and $f(a) = f(b)$, then $f'(c) = 0$ for some $c \in (a,b)$ produces an interior point precisely by ruling this case out: when both extrema sit at the endpoints, the hypothesis $f(a) = f(b)$ forces the function to be constant, and any interior point then serves. Without that hypothesis the endpoint case is exactly the one that survives, and the identity on $[0,1]$ is it.