LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-01

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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.

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Cauchy's mean-value theorem in quotient form when the denominator derivative is nonzero

Statement

Let $a < b$ . If $f, g$ are continuous on $[a, b]$ , differentiable on $(a, b)$ , and $g^{'} (x) \neq 0$ throughout $(a, b)$ , then $g (b) \neq g (a)$ and there is $c \in (a, b)$ such that $\frac{f (b) - f (a)}{g (b) - g (a)} = \frac{f^{'} (c)}{g^{'} (c)} .$

Facts & Assumptions

Given: The functions and hypotheses in the statement.

[L1]

Cauchy's mean value theorem gives $(f (b) - f (a)) g^{'} (c) = (g (b) - g (a)) f^{'} (c)$ for some $c \in (a, b)$ (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 $(f (b) - f (a)) g^{'} (c) = (g (b) - g (a)) f^{'} (c)$ ; no hypothesis on $g^{'}$ is needed in this product form).

[L2]

Rolle's theorem says equal endpoint values force an interior zero of the derivative (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 $c \in (a, b)$ ).

Proof

technique · direct

1.1L2given

If $g (a) = g (b)$ , Rolle gives $d \in (a, b)$ with $g^{'} (d) = 0$ , contrary to the hypothesis. Hence $g (b) - g (a) \neq 0$ .

1.2L1

Cauchy's theorem supplies $c \in (a, b)$ with the cross-product identity in [L1].

2.1step 1.1step 1.2algebra∎

Divide that identity by the two nonzero factors $g (b) - g (a)$ and $g^{'} (c)$ to obtain the quotient formula.

Cauchy's mean-value theorem in quotient form when the denominator derivative is nonzero

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources