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

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With $f(x) = x^{3}$ and $g(x) = x^{2}$ on $[-1,1]$ the quotient form $\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)}$ is meaningless because $g(b) = g(a)$ , while the product form of Cauchy's theorem still holds

Statement refuted

Refuted claim: let $a, b \in \mathbb{R}$ with $a < b$ and let $f, g : [a,b] \to \mathbb{R}$ be continuous on $[a,b]$ (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) and differentiable at every point of $(a,b)$ (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). Then there is $c \in (a,b)$ with

$\frac{f(b)-f(a)}{g(b)-g(a)} \;=\; \frac{f'(c)}{g'(c)} .$

This is the shape in which Cauchy's mean value theorem is usually remembered, and it is not what Cauchy's mean value theorem: for $f, g$ continuous on $[a,b]$ with $a<b$ and differentiable on $(a,b)$ there is $c \in (a,b)$ with $\bigl(f(b)-f(a)\bigr)g'(c) = \bigl(g(b)-g(a)\bigr)f'(c)$ ; no hypothesis on $g'$ is needed in this product form says. It is false as stated, because under the hypotheses given neither quotient need be a real number at all. The witness below makes both denominators vanish.

Facts & Assumptions

Given: The reals $a := -1$ and $b := 1$ and the functions $f, g : [-1,1] \to \mathbb{R}$ with $f(x) := x^{3}$ and $g(x) := x^{2}$ (Integer powers $a^m$ , Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length); numerals denote canonical naturals (The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

[L1]

Power rule and restriction (For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, claim 2, and 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): $x \mapsto x^{n}$ on $\mathbb{R}$ is differentiable at every real $c$ with derivative $\iota(n)c^{\,n-1}$ ; every point of the order-convex set $[-1,1]$ , which has at least two elements, is a limit point of it (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ , Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length); and a function differentiable at such a point stays differentiable there after restriction, with the same derivative.

[L2]

Continuity (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 5, 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): $x \mapsto x^{n}$ is continuous at every point of its domain.

[L3]

Cauchy's mean value theorem (Cauchy's mean value theorem: for $f, g$ continuous on $[a,b]$ with $a<b$ and differentiable on $(a,b)$ there is $c \in (a,b)$ with $\bigl(f(b)-f(a)\bigr)g'(c) = \bigl(g(b)-g(a)\bigr)f'(c)$ ; no hypothesis on $g'$ is needed in this product form), in its product form: under the hypotheses above there is $c \in (a,b)$ with $\bigl(f(b)-f(a)\bigr)g'(c) = \bigl(g(b)-g(a)\bigr)f'(c)$ .

[L4]

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 $c \in (a,b)$ ): a function continuous on $[a,b]$ , differentiable at every point of $(a,b)$ and taking equal values at the endpoints has a vanishing derivative somewhere in $(a,b)$ .

[L5]

Signs and powers (Integer powers $a^m$ , Sign rules for products and monotonicity of multiplication, Monotonicity of $x \mapsto x^n$ and of $n \mapsto a^n$ ): the recursion $a^{n+1} = a^{n}a$ with $a^{0} = 1$ gives $(-1)^{2} = (-1)(-1) = 1$ , the product of two negatives being positive (Sign rules for products and monotonicity of multiplication), and $(-1)^{3} = (-1)^{2}(-1) = -1$ ; that $1^{n} = 1$ for every natural $n$ is claim 4 of Monotonicity of $x \mapsto x^n$ and of $n \mapsto a^n$ and is not read off Integer powers $a^m$ .

[L6]

Canonical naturals (The canonical natural $\iota(n) = n \cdot 1_F$ of a field, Canonical naturals are positive and strictly increasing): $\iota(1) = 1$ , $\iota(m+n) = \iota(m)+\iota(n)$ and $\iota(mn) = \iota(m)\iota(n)$ for $m, n \ge 1$ ; in particular $\iota(2)\iota(2) = \iota(4) > 0$ , so $\iota(4) \ne 0$ .

[L7]

Division by $0$ is not defined: $0$ has no multiplicative inverse in a field (Field).

Counterexample

technique · direct

1.1

By [L2] both $f$ and $g$ are continuous on $[-1,1]$ , and by [L1] both are differentiable at every $c \in [-1,1]$ , with $f'(c) = \iota(3)c^{2}$ and $g'(c) = \iota(2)c^{1} = \iota(2)c$ , using [L5] for $c^{1} = c$ . So the pair $(f,g)$ satisfies every hypothesis of the refuted claim, and of [L3], with $a = -1$ and $b = 1$ .

L1L2L5

1.2

By [L5], $f(1) = 1$ , $f(-1) = -1$ , $g(1) = 1$ and $g(-1) = 1$ . Hence $f(b)-f(a) = 1 - (-1) = \iota(2)$ and $g(b)-g(a) = 1 - 1 = 0$ .

L5L6

2.1

The left-hand side of the refuted claim names no real number: its denominator $g(b)-g(a)$ is $0$ by step 1.2, and $0$ has no inverse by [L7]. So there is no $c$ for which the asserted equation holds, since the equation cannot even be formed; the claim fails on this pair.

step 1.2L7

2.2

The right-hand side fails as well at one point of the interval: $g'(0) = \iota(2) \cdot 0 = 0$ by step 1.1, so the quotient $f'(c)/g'(c)$ is undefined at $c = 0$ , again by [L7].

step 1.1L7

2.3

The product form is untouched. By [L3] there is $c \in (-1,1)$ with $\bigl(f(b)-f(a)\bigr)g'(c) = \bigl(g(b)-g(a)\bigr)f'(c)$ , which by steps 1.1 and 1.2 reads $\iota(2)\,\iota(2)c = 0 \cdot \iota(3)c^{2}$ , that is $\iota(4)c = 0$ ; since $\iota(4) \ne 0$ by [L6], this forces $c = 0$ . And $c = 0$ does lie in $(-1,1)$ and does satisfy the identity, both sides being $0$ . So [L3] holds on this pair, with $c = 0$ its only admissible point.

step 1.1step 1.2L3L6

3.1

The vanishing of $g(b)-g(a)$ is not an accident of the choice. By step 1.2 one has $g(-1) = g(1)$ , so [L4] already forces $g'$ to vanish at some point of $(-1,1)$ , and by step 2.3 that point is $c = 0$ , the same point the product form produces. So on this pair every quotient the refuted claim writes down is undefined, while [L3] is satisfied; the quotient form needs hypotheses the product form does not, and as stated it is false.

step 2.1step 2.2step 2.3L4∎

Remarks

What the quotient form would need. Two extra hypotheses, and they are of different kinds: $g(b) \ne g(a)$ , a condition on the endpoints, and $g'(c) \ne 0$ at the point produced, a condition on a point one does not choose. The second is the awkward one, since the theorem hands back a $c$ and says nothing about it. This is why Cauchy's mean value theorem: for $f, g$ continuous on $[a,b]$ with $a<b$ and differentiable on $(a,b)$ there is $c \in (a,b)$ with $\bigl(f(b)-f(a)\bigr)g'(c) = \bigl(g(b)-g(a)\bigr)f'(c)$ ; no hypothesis on $g'$ is needed in this product form is stated as a product identity in this library, with no hypothesis on $g'$ at all.
A cheaper repair than a hypothesis on $g'$ . If $g(b) \ne g(a)$ then, by 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 $c \in (a,b)$ read contrapositively, nothing forces $g'$ to vanish; and the product identity may then be divided by $g(b)-g(a)$ to give $\bigl(f(b)-f(a)\bigr)/\bigl(g(b)-g(a)\bigr) \cdot g'(c) = f'(c)$ , which is a true statement with no division by $g'(c)$ anywhere. That is the form worth remembering.
The witness is the smallest natural one. $g(x) = x^{2}$ is even and the interval is symmetric about $0$ , which is the whole of the mechanism; any even $g$ on a symmetric interval does the same. The choice $f(x) = x^{3}$ only makes $f(b) - f(a)$ nonzero, so that the failure is not hidden by both sides vanishing for a trivial reason.

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Sources

Mean value theorem (Wikipedia) (standard reference, not scraped)
Rolle's theorem (Wikipedia) (standard reference, not scraped)
J. Hunter, An Introduction to Real Analysis (standard reference, not scraped)