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

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, Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) and differentiable at every point of $(a,b)$ as functions 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). Then there is $c \in (a,b)$ with

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

The statement is a product identity, and that is deliberate. The familiar quotient form

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

is not asserted here, and it is not equivalent: its left side needs $g(b) \ne g(a)$ and its right side needs $g'(c) \ne 0$, and neither follows from the hypotheses. The product form above needs neither, holds under exactly the hypotheses stated, and specialises to the quotient form whenever both denominators happen to be nonzero. The companion page exhibits an $f$ and a $g$ for which the quotient form is meaningless while the product form holds.

• Where the auxiliary function comes from. $h$ is built so that the two cross terms $f(a)g(a)$ and $f(b)g(b)$ cancel against themselves at the two endpoints, leaving the same antisymmetric expression $f(b)g(a) - f(a)g(b)$ at each. Nothing is optimised and nothing is guessed: the two coefficients are forced, up to a common scalar, by the requirement $h(a) = h(b)$.

• The ordinary mean value theorem is the case $g(x) = x$, and it is recorded as the next item rather than reproved. Cauchy's theorem is the more general statement and is proved first for that reason, not because it is harder: it costs one application of Rolle either way.

• What fails in the quotient form. If $g(a) = g(b)$ the left side is not a real number at all, and the theorem still says something: it says $\bigl(f(b)-f(a)\bigr)g'(c) = 0$ for some $c \in (a,b)$. That is the case worked out in 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 ↗.