False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

How statement and proof provenance work

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.

FALSE: differentiability at every point of $(a,b)$ alone yields a $c \in (a,b)$ with $f(b) - f(a) = f'(c)(b-a)$

Statement

False claim: let $a, b \in \mathbb{R}$ with $a < b$ and let $f : [a,b] \to \mathbb{R}$ 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 $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length). Then there is $c \in (a,b)$ with

$f(b) - f(a) \;=\; f'(c)\,(b - a) .$

This 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 " $f$ is continuous on $[a,b]$ " deleted, everything else left as it stands. It is false.

Why it is tempting. The conclusion mentions $f'$ only at interior points, and the hypothesis of continuity on the closed interval looks like a technical condition guaranteeing nothing the differentiability does not already give. It is not: the values $f(a)$ and $f(b)$ appear on the left-hand side of the conclusion, and nothing in a hypothesis about $(a,b)$ alone connects them to the behaviour of $f$ inside. A single unrelated value at one endpoint breaks the identity outright.

Facts & Assumptions

Given: The interval $[0,1]$ and the function $f : [0,1] \to \mathbb{R}$ defined by $f(x) := x$ for $x \in [0,1)$ and $f(1) := 0$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L1]

Derivative (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): for $c \in [0,1]$ a limit point of $[0,1]$ , the difference quotient $q(x) := (f(x)-f(c))/(x-c)$ is a function on $[0,1] \setminus \{c\}$ , and $f$ is differentiable at $c$ with $f'(c) = L$ exactly when for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that every $x \in [0,1]$ with $0 < |x-c| < \delta$ satisfies $|q(x) - L| < \varepsilon$ (The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ ).

[L2]

Every point of $[0,1]$ is a limit point of $[0,1]$ , that set being order-convex with at least two elements (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, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ ).

[L3]

Absolute value and order (Basic properties of the absolute value): $|u| \ge 0$ ; $|u| = 0$ exactly when $u = 0$ ; and for $\delta > 0$ the condition $|y - c| < \delta$ is $c - \delta < y < c + \delta$ .

[L4]

$0 \ne 1$ in $\mathbb{R}$ , since $0 < 1$ (The multiplicative identity is positive).

[L5]

Continuity at a point (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point): $f$ is continuous at $p \in [0,1]$ when for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that every $x \in [0,1]$ with $|x - p| < \delta$ satisfies $|f(x) - f(p)| < \varepsilon$ .

[L6]

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 $c \in (a,b)$ ) carries the same continuity hypothesis on the closed interval as 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 $f(b) - f(a) = f'(c)(b-a)$ does.

Refutation

technique · direct

1.1

$f$ is a well-defined function on $[0,1]$ : every $x \in [0,1]$ either equals $1$ or does not, exclusively, so exactly one of the two clauses applies to it.

given

1.2

$f(0) = 0$ , since $0 \in [0,1)$ , and $f(1) = 0$ by the second clause. Hence $f(1) - f(0) = 0$ , and also $f(0) = f(1)$ .

given

2.1

The derivative inside. Let $c \in (0,1)$ and put $\delta := \min\{c,\ 1-c\}$ , a positive real. Every $x \in [0,1]$ with $0 < |x - c| < \delta$ satisfies $x < c + \delta \le c + (1-c) = 1$ by [L3], so $x \in [0,1)$ and $f(x) = x$ ; and $c \in [0,1)$ , so $f(c) = c$ . Therefore $q(x) = (x - c)/(x - c) = 1$ for every such $x$ .

step 1.1L3

2.2

$f$ is not continuous at $1$ . Take $\varepsilon := 1/2$ and let a real $\delta > 0$ be given. Put $x := \max\{1 - \delta/2,\ 1/2\}$ ; then $1/2 \le x < 1$ , so $x \in [0,1)$ and $f(x) = x$ , while $|x - 1| \le \delta/2 < \delta$ . Yet $|f(x) - f(1)| = |x - 0| = x \ge 1/2 = \varepsilon$ . So no $\delta$ witnesses the condition of [L5] at $p = 1$ for this $\varepsilon$ .

step 1.1L3L5

3.1

Let $c \in (0,1)$ and let a real $\varepsilon > 0$ be given. The $\delta$ of step 2.1 satisfies: every $x \in [0,1]$ with $0 < |x-c| < \delta$ has $|q(x) - 1| = |1 - 1| = 0 < \varepsilon$ by [L3]. Since $c$ is a limit point of $[0,1]$ by [L2], this is exactly the condition of [L1] with $L := 1$ . So $f$ is differentiable at $c$ with $f'(c) = 1$ .

step 2.1L1L2L3

4.1

The claim fails on this witness. By step 3.1 the function $f$ is differentiable at every point of $(0,1)$ , so it satisfies the hypothesis of the false claim with $a := 0$ and $b := 1$ . For every $c \in (0,1)$ one has $f'(c)(b-a) = 1 \cdot (1 - 0) = 1$ , while $f(b) - f(a) = f(1) - f(0) = 0$ by step 1.2. By [L4] these are different, so no $c \in (0,1)$ satisfies the asserted identity, and the claim is false.

step 1.2step 3.1L4

5.1

The same witness refutes the corresponding weakening of Rolle's theorem: by step 1.2 one has $f(0) = f(1)$ , and by step 3.1 one has $f'(c) = 1 \ne 0$ at every $c \in (0,1)$ , so no interior point carries a vanishing derivative. What is missing in both cases is exactly the hypothesis deleted, continuity on the closed interval, and step 2.2 shows it fails at the single point $1$ .

step 1.2step 2.2step 3.1step 4.1L6∎

Remarks

The witness is as small as it can be. It agrees with the identity except at one endpoint, and it is differentiable at every interior point with the constant derivative $1$ . Nothing about the interior is disturbed; only the value $f(1)$ is moved, and the conclusion of the mean value theorem is a statement about that value.
Where the true proof would break. 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 $f(b) - f(a) = f'(c)(b-a)$ runs through 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 $c \in (a,b)$ , and Rolle runs through Extreme value theorem: a continuous real function on a nonempty compact subset of $\mathbb{R}$ attains a greatest and a least value, which needs $f$ continuous on the compact set $[a,b]$ . With continuity failing at $1$ the supremum of $f$ over $[0,1]$ is $1$ and is not attained, so the extreme value theorem has nothing to hand back and every later step is unavailable.
Continuity at the endpoints is the only thing deleted. In particular the witness is continuous at every point of $(0,1)$ , being differentiable there; it is even continuous at $0$ . So the false claim cannot be repaired by asking for continuity on $[a,b)$ , or on $(a,b]$ after reflecting the witness, and it is the closed interval that is needed.

Depends on

Used by

f(x) = x on [0,1) with f(1) = 0 is differentiable at every point of (0,1) with f' ≡ 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 Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 62 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Mean value theorem (Wikipedia) (standard reference, not scraped)
Rolle's theorem (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, Mean Value Theorem (standard reference, not scraped)

FALSE: differentiability at every point of (a,b)(a,b)(a,b) alone yields a c∈(a,b)c \in (a,b)c∈(a,b) with f(b)−f(a)=f′(c)(b−a)f(b) - f(a) = f'(c)(b-a)f(b)−f(a)=f′(c)(b−a)