= Cauchy mean value theorem
{c}
= Cauchy's mean value theorem
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{synonym}
For $a<b$ and <real-valued functions> $f,g$ that are <continuous> on the <closed interval> $[a,b]$ and <differentiable> on the <open interval> $(a,b)$, there exists $\xi\in(a,b)$ with
$$
[f(b)-f(a)]g'(\xi)=[g(b)-g(a)]f'(\xi).
$$
Apply <Rolle's theorem> to the <linear combination> $h(x)=[f(b)-f(a)]g(x)-[g(b)-g(a)]f(x)$, which has equal endpoint values. The displayed product identity permits zero <derivatives>. For the ratio form, additionally assume $g'(x)\ne0$ for every $x\in(a,b)$. Then <Rolle's theorem> ensures $g(b)\ne g(a)$, so both denominators below are nonzero:
$$
\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(\xi)}{g'(\xi)}.
$$
Neither <derivative> needs to be <continuous>, and $f'$ may vanish. Choosing $g(x)=x$ recovers the <mean value theorem>; the ratio form also gives the zero-over-zero endpoint case of <L'Hôpital's rule>. https://www.jirka.org/ra/html/sec_mvt.html[Jiří Lebl, Basic Analysis I, Section 4.2, Theorem 4.2.5]{ref}
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