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

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$f (x) = x$ on $[0, 1)$ with $f (1) = 0$ is differentiable at every point of $(0, 1)$ with $f^{'} \equiv 1$ , yet no $c$ satisfies $f (1) - f (0) = f^{'} (c)$ , so continuity on the closed interval cannot be dropped from the mean value theorem

Statement refuted

Refuted claim: let $a, b \in R$ with $a < b$ and let $f : [a, b] \to R$ be differentiable at every point of $(a, b)$ as a function on $[a, b]$ (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 there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ .

That is The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ with the hypothesis of continuity on $[a, b]$ deleted, and it is false; the false statement itself is recorded as FALSE: differentiability at every point of $(a, b)$ alone yields a $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ . This item works the witness out: it locates the failure at a single point, measures it, and shows that repairing that one value restores the conclusion.

Facts & Assumptions

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

[L1]

The refutation of FALSE: differentiability at every point of $(a, b)$ alone yields a $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ establishes, for this $f$ : that $f$ is differentiable at every $c \in (0, 1)$ with $f^{'} (c) = 1$ ; that $f (0) = f (1) = 0$ , so $f (1) - f (0) = 0$ ; and that no $c \in (0, 1)$ satisfies $f (1) - f (0) = f^{'} (c) (1 - 0)$ .

[L2]

One-sided limits (The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (- \infty, c)$ and $A \cap (c, \infty)$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length): the left limit of $f$ at $p$ is the limit at $p$ of $f$ restricted to $[0, 1] \cap (- \infty, p)$ , defined when $p$ is a limit point of that set; the right limit is the same with $[0, 1] \cap (p, \infty)$ .

[L3]

The limit condition (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ): $\lim_{x \to p} u (x) = L$ means that for every real $ε > 0$ there is a real $δ > 0$ such that every $x$ in the domain of $u$ with $0 < ∣ x - p ∣ < δ$ satisfies $∣ u (x) - L ∣ < ε$ . The clause $0 < ∣ x - p ∣$ removes $x = p$ from the quantifier (Basic properties of the absolute value, The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L4]

Continuity at a limit point (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, clause 1): for $p \in [0, 1]$ a limit point of $[0, 1]$ , the function $f$ is continuous at $p$ if and only if $\lim_{x \to p} f (x)$ exists and equals $f (p)$ .

[L5]

Mean value theorem (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ) and 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 $c \in (a, b)$ ), both of which additionally require continuity on the closed interval.

[L6]

The identity on $[0, 1]$ is continuous on $[0, 1]$ and differentiable at every point of $[0, 1]$ with derivative $1$ , its difference quotient at any $c$ being the constant $1$ (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 $A$ , Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L7]

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

Counterexample

technique · direct

1.1L1

By [L1] the function $f$ is differentiable at every $c \in (0, 1)$ with $f^{'} (c) = 1$ , satisfies $f (0) = f (1) = 0$ , and admits no $c \in (0, 1)$ with $f (1) - f (0) = f^{'} (c) (1 - 0)$ . So the refuted claim fails at $a : = 0$ , $b : = 1$ .

1.2L2L3

The point $1$ is a limit point of $[0, 1]$ and of $[0, 1] \cap (- \infty, 1) = [0, 1)$ : for every real $ε > 0$ the point $y : = \max {1 - ε / 2, 1 / 2}$ satisfies $1 / 2 \leq y < 1$ , hence $y \in [0, 1)$ , and $0 < ∣ y - 1 ∣ \leq ε / 2 < ε$ .

2.1step 1.2L2L3

$\lim_{x \to 1} f (x) = 1$ , the limit taken over the domain $[0, 1]$ . Given a real $ε > 0$ , take $δ : = ε$ ; every $x \in [0, 1]$ with $0 < ∣ x - 1 ∣ < δ$ has $x \neq 1$ by [L3], hence $x \in [0, 1)$ and $f (x) = x$ , so $∣ f (x) - 1 ∣ = ∣ x - 1 ∣ < ε$ . Since the same quantifier ranges over the same points when the domain is cut down to $[0, 1)$ , this also says $\lim_{x \to 1^{-}} f (x) = 1$ by [L2]. The right limit at $1$ is not defined, since $[0, 1] \cap (1, \infty)$ is empty and $1$ is therefore not a limit point of it.

3.1step 2.1L1L4L5L7

$f$ is not continuous at $1$ . By step 1.2 the point $1$ is a limit point of $[0, 1]$ , so [L4] makes continuity there equivalent to $\lim_{x \to 1} f (x) = f (1)$ ; by step 2.1 the left side is $1$ and by [L1] the right side is $0$ , and $0 \neq 1$ by [L7]. So exactly one hypothesis of [L5] fails, at exactly one point, and it is the deleted one.

4.1step 1.1step 3.1L5L6

The repair. The identity $h$ agrees with $f$ at every point of $[0, 1]$ except $1$ , where $h (1) = 1$ and $f (1) = 0$ . By [L6] the function $h$ is continuous on $[0, 1]$ and differentiable at every point of $(0, 1)$ with $h^{'} (c) = 1$ , so [L5] applies to $h$ ; and indeed $h (1) - h (0) = 1 = 1 \cdot (1 - 0) = h^{'} (c) (1 - 0)$ for every $c \in (0, 1)$ . So moving the single value $f (1)$ back to $1$ turns a function with no admissible $c$ into one for which every $c$ is admissible.

5.1step 1.1step 4.1L5L7∎

The same witness refutes the corresponding weakening of Rolle's theorem: $f (0) = f (1) = 0$ by step 1.1, and yet $f^{'} (c) = 1 \neq 0$ at every $c \in (0, 1)$ by step 1.1 and [L7]. So neither theorem in [L5] survives the deletion of continuity on the closed interval.

Remarks

The discontinuity is of the mildest possible kind. Both of the quantities that exist at $1$ , the left limit and the value, exist and are finite; they simply differ. In the vocabulary of the page on monotone functions and discontinuities this is a removable discontinuity, and step 4.1 removes it. Nothing pathological is needed to break the mean value theorem.
Why the derivative sees nothing. The difference quotient of $f$ at an interior $c$ is evaluated only at points within $\min {c, 1 - c}$ of $c$ , and every such point lies in $[0, 1)$ , where $f$ is the identity. So $f^{'}$ carries no information at all about $f (1)$ , while the conclusion of the mean value theorem is an equation containing $f (1)$ . Continuity on the closed interval is precisely the bridge between the two.
Reflecting the witness covers the other endpoint. The function $x \mapsto - f (1 - x)$ is differentiable at every point of $(0, 1)$ with the same constant derivative and fails continuity at $0$ instead of at $1$ , so nothing is special about which endpoint is broken.

$f (x) = x$ on $[0, 1)$ with $f (1) = 0$ is differentiable at every point of $(0, 1)$ with $f^{'} \equiv 1$ , yet no $c$ satisfies $f (1) - f (0) = f^{'} (c)$ , so continuity on the closed interval cannot be dropped from the mean value theorem

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

f(x)=xf(x)=x on [0,1)[0,1) with f(1)=0f(1)=0 is differentiable at every point of (0,1)(0,1) with f′≡1f′≡1, yet no cc satisfies f(1)−f(0)=f′(c)f(1)−f(0)=f′(c), so continuity on the closed interval cannot be dropped from the mean value theorem

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

$f (x) = x$ on $[0, 1)$ with $f (1) = 0$ is differentiable at every point of $(0, 1)$ with $f^{'} \equiv 1$ , yet no $c$ satisfies $f (1) - f (0) = f^{'} (c)$ , so continuity on the closed interval cannot be dropped from the mean value theorem