CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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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

Statement refuted

Refuted claim: let $A \subseteq R$ , let $f : A \to R$ and let $c \in A$ be a limit point of $A$ at which $f$ has a local extremum (Local (relative) maximum and minimum of $f : A \to R$ 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 $f : A \to R$ 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.

Facts & Assumptions

Given: The set $A : = [0, 1]$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) and the function $f : A \to R$ , $f (x) : = x$ .

[L1]

Derivative of the identity (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ): every point of the order-convex set $[0, 1]$ , which has at least two elements, is a limit point of it (Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ); and the difference quotient of $f$ at any $c \in [0, 1]$ is $(x - c) / (x - c) = 1$ at every $x \in [0, 1]$ with $x \neq c$ , a constant function whose limit at $c$ is $1$ . So $f$ is differentiable at every $c \in [0, 1]$ with $f^{'} (c) = 1$ .

[L2]

Local extrema (Local (relative) maximum and minimum of $f : A \to R$ at a point, the strict forms, and what it means for the point to be interior to $A$ ): $f$ has a local maximum at $c \in A$ when $f (x) \leq f (c)$ for every $x \in A \cap N_{ε} (c)$ for some real $ε > 0$ , and a local minimum with the inequality reversed; a value that is a greatest value of $f$ over the whole of $A$ is a local maximum, and a least value is a local minimum (claim 4 of its body); and $c$ is interior to $A$ exactly when $N_{ε} (c) \subseteq A$ for some real $ε > 0$ (Interior, closure, boundary and exterior of a subset of $R$ , The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L3]

Maximum and minimum of a set (Maximum and minimum of a set): $m$ is a maximum of $S$ when $m \in S$ and $s \leq m$ for every $s \in S$ , and a minimum when $m \in S$ and $m \leq s$ for every $s \in S$ .

[L4]

Fermat's interior extremum theorem (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 $f^{'} (c) = 0$ ) additionally requires $c$ to be interior to $A$ .

[L5]

$0 \neq 1$ , since $0 < 1$ (The multiplicative identity is positive).

Counterexample

technique · direct

1.1

By [L1] the function $f$ is differentiable at every $c \in [0, 1]$ , and $f^{'} (c) = 1$ ; in particular $f^{'} (0) = f^{'} (1) = 1$ , and every point of $[0, 1]$ is a limit point of $[0, 1]$ .

1.2

Every $x \in A$ satisfies $0 \leq x \leq 1$ , so $f (x) = x \leq 1 = f (1)$ and $f (x) = x \geq 0 = f (0)$ ; and $0, 1 \in A$ . So $f (1)$ is a maximum of $f [A]$ and $f (0)$ is a minimum of $f [A]$ by [L3], and by [L2] the function $f$ has a local maximum at $1$ and a local minimum at $0$ , hence a local extremum at each.

L2L3

1.3

Neither $1$ nor $0$ is interior to $A$ : for every real $ε > 0$ the point $1 + ε / 2$ lies in $N_{ε} (1)$ and not in $[0, 1]$ , and the point $- ε / 2$ lies in $N_{ε} (0)$ and not in $[0, 1]$ . So no $N_{ε}$ around either point is contained in $A$ .

2.1

The refuted claim therefore fails at $c : = 1$ : the point $1$ lies in $A$ and is a limit point of $A$ by step 1.1, $f$ has a local extremum there by step 1.2 and is differentiable there by step 1.1, and yet $f^{'} (1) = 1 \neq 0$ by [L5]. The same holds at $c : = 0$ .

step 1.1step 1.2L5

3.1

Nothing in [L4] is contradicted. By step 1.3 neither $0$ nor $1$ is interior to $A$ , so the hypothesis of that theorem is not met at either point, and the deleted hypothesis is exactly the one that fails. Indeed no point of $A$ at all carries a vanishing derivative, and consistently with [L4] no interior point of $A$ carries a local extremum: by step 1.2 the only extrema of $f$ over $A$ sit at the two endpoints.

step 1.3step 2.1L4∎

Remarks

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 $f^{'} (1) \geq 0$ . 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 $f : A \to R$ 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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

27 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Fermat's theorem (stationary points) (Wikipedia) (standard reference, not scraped)
Maximum and minimum (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, Mean Value Theorem (standard reference, not scraped)

The identity on [0,1][0,1] attains its maximum at 11 and its minimum at 00 with derivative 11 at both, so Fermat's theorem genuinely needs the extremum to be at an interior point