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 $f(b)-f(a)=f^{\mathrm{\prime }}(c)(b-a)$

Let $a,b\in \mathbb{R}$ with $a<b$ and let $f:[a,b]\to \mathbb{R}$ be continuous on $[a,b]$ (Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ 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 a function on $[a,b]$ (The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ 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

$f(b)-f(a)=f^{\mathrm{\prime }}(c)(b-a)\mathrm{.}$

Equivalently, since $b-a\mathrm{\ne }0$, there is $c\in (a,b)$ at which $f^{\mathrm{\prime }}(c)=(f(b)-f(a))\mathrm{/}(b-a)$: the derivative somewhere inside equals the average rate of change across the whole interval.

Continuity on the closed interval cannot be dropped. Differentiability at every point of $(a,b)$ alone does not suffice: a function on $[0,1]$, differentiable at every point of $(0,1)$ with derivative constantly $1$, for which no $c$ works, is exhibited later on this page as a false statement, and the companion page works the same witness out in full.

Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ 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

The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set

• The geometric reading, and what it is not. The conclusion says that some tangent line is parallel to the chord from $(a,f(a))$ to $(b,f(b))$. It does not say which one, it does not say there is only one, and it says nothing at all about $f$ between the endpoints beyond the hypotheses. Every use of the theorem on this page is a use of the equation, never of the picture.

• Why this is a corollary and not the primitive statement. 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 $(f(b)-f(a))g^{\mathrm{\prime }}(c)=(g(b)-g(a))f^{\mathrm{\prime }}(c)$; no hypothesis on $g^{\mathrm{\prime }}$ is needed in this product form is proved from 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 $f^{\mathrm{\prime }}(c)=0$ for some $c\in (a,b)$ with one auxiliary function, and this statement is that theorem at $g(x)=x$; deriving it the other way round, from this statement to Cauchy's, is also possible but needs an auxiliary function of its own, so nothing is saved. What matters is that both rest on Rolle, and Rolle on the extreme value theorem.

• The two hypotheses are on different sets on purpose. Continuity is asked for on the closed interval and differentiability only on the open one, so nothing at all is required of the difference quotients at $a$ and at $b$. Weakening the continuity in step with the differentiability, so that both are asked for on $(a,b)$ only, destroys the theorem, which is exactly what $f(x)=x$ on $[0,1)$ with $f(1)=0$ is differentiable at every point of $(0,1)$ with $f^{\mathrm{\prime }}\equiv 1$, yet no $c$ satisfies $f(1)-f(0)=f^{\mathrm{\prime }}(c)$, so continuity on the closed interval cannot be dropped from the mean value theorem ↗ shows on the companion page.