Cauchy's Mean Value Theorem — Project Hematite

Suppose you record both position and fuel consumption as you drive from city $A$ to city $B$ . At some moment during the trip, the ratio of your instantaneous speed to your instantaneous fuel-burn rate must equal the ratio of total distance to total fuel used. Cauchy’s theorem is the precise form of this observation: it handles two functions varying together and compares their rates of change in tandem.

Statement

Theorem (Cauchy’s Mean Value Theorem). Let $f, g : [a, b] \to \mathbb{R}$ . If

$f$ and $g$ are continuous on $[a, b]$ ,
$f$ and $g$ are differentiable on $(a, b)$ , and
$g'(x) \neq 0$ for all $x \in (a, b)$ ,

then there exists $c \in (a, b)$ such that

\frac{f(b) - f(a)}{g(b) - g(a)} \;=\; \frac{f'(c)}{g'(c)}. \tag{1}

Note. Condition 3 implies $g(b) \neq g(a)$ : if $g(b) = g(a)$ , then Lagrange’s MVT applied to $g$ would give some interior point where $g' = 0$ , contradicting condition 3. So the left side of $(1)$ is well-defined.

Proof

The trick is to build a single auxiliary function whose boundary values coincide, then apply Lagrange’s MVT.

Define

h(x) \;\coloneqq\; [f(b) - f(a)]\,g(x) \;-\; [g(b) - g(a)]\,f(x).

Since $f$ and $g$ are continuous on $[a, b]$ and differentiable on $(a, b)$ , so is $h$ . Compute:

h(a) = [f(b)-f(a)]\,g(a) - [g(b)-g(a)]\,f(a),

h(b) = [f(b)-f(a)]\,g(b) - [g(b)-g(a)]\,f(b).

Their difference is

h(b) - h(a) = [f(b)-f(a)][g(b)-g(a)] - [g(b)-g(a)][f(b)-f(a)] = 0,

so $h(a) = h(b)$ . By Lagrange’s MVT applied to $h$ , there exists $c \in (a, b)$ with $h'(c)(b-a) = h(b) - h(a) = 0$ , hence $h'(c) = 0$ . Differentiating $h$ :

h'(x) = [f(b) - f(a)]\,g'(x) - [g(b) - g(a)]\,f'(x).

Setting $h'(c) = 0$ gives $[f(b)-f(a)]\,g'(c) = [g(b)-g(a)]\,f'(c)$ . Dividing by $g'(c) \neq 0$ and $g(b)-g(a) \neq 0$ yields $(1)$ . $\square$

Relation to Lagrange’s theorem

Set $g(x) = x$ . Then $g'(x) = 1$ and $g(b) - g(a) = b - a$ , so $(1)$ becomes

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

which is exactly Lagrange’s MVT. Cauchy’s theorem is the strict generalisation: every instance of Lagrange’s theorem is a special case, obtained by taking the second function to be the identity.

Parametric interpretation

When $x$ is a parameter tracing a curve $(g(x),\, f(x))$ for $x \in [a, b]$ , the slope of the chord from $(g(a), f(a))$ to $(g(b), f(b))$ is $\frac{f(b)-f(a)}{g(b)-g(a)}$ . Theorem $(1)$ says the tangent to the curve at the parameter value $c$ has the same slope $f'(c)/g'(c)$ . It is the Mean Value Theorem for parametric curves.

Summary

Cauchy’s theorem: if $f, g$ are continuous on $[a,b]$ , differentiable on $(a,b)$ , and $g' \neq 0$ on $(a,b)$ , then there exists $c \in (a,b)$ with $\dfrac{f(b)-f(a)}{g(b)-g(a)} = \dfrac{f'(c)}{g'(c)}$ .
Proof: define $h(x) = [f(b)-f(a)]\,g(x) - [g(b)-g(a)]\,f(x)$ ; it satisfies $h(a) = h(b)$ , so Lagrange’s MVT gives a zero of $h'$ .
Lagrange’s MVT is the special case $g(x) = x$ .
For parametric curves, the theorem says the tangent slope equals the chord slope at some interior parameter value.