Cauchy's Mean Value Theorem

Comparing the Changes of Two Functions

The Mean Value Theorem relates the change of one function across an interval to its derivative at some interior point. Sometimes the question involves two functions at once: how do their changes compare, and can that comparison be expressed using their derivatives at the same point? Cauchy’s Mean Value Theorem answers this by producing one common point for both functions.

The theorem is useful even when a ratio of endpoint changes cannot be formed. Its basic conclusion is an identity, not a quotient. When the denominator factors are nonzero, that identity can be rearranged as a comparison of ratios. Keeping these two forms distinct prevents division by zero and makes the full scope of the theorem clear.

Theorem (Cauchy’s Mean Value Theorem): Let $a<b$, and let $f,g:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists $c\in(a,b)$ such that $$ \bigl(f(b)-f(a)\bigr)g'(c)=\bigl(g(b)-g(a)\bigr)f'(c). $$

The point $c$ is shared: both derivatives in the identity are evaluated at that same point. The theorem does not say that the derivative of either function equals its own secant slope at $c$. Instead, it says the two increments and the two derivatives satisfy a cross-multiplied relation.

If $g(b)\ne g(a)$ and $g'(c)\ne0$, division gives

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

This ratio form is especially useful when $g$ is strictly increasing, since then $g(b)-g(a)\ne0$, and when $g'$ is positive, since then division by $g'(c)$ is valid. Neither nonvanishing condition is required for the theorem’s original identity.

Reading the Identity Correctly

The ordinary Mean Value Theorem is recovered by choosing $g(x)=x$. Then $g(b)-g(a)=b-a$ and $g'(c)=1$, so the identity becomes

$$ f(b)-f(a)=(b-a)f'(c). $$

The function $g$ in Cauchy’s theorem can instead serve as a scale for measuring change. If $g$ changes rapidly in some region and slowly in another, the ratio $f'/g'$ compares the local rates of change relative to that scale. The theorem says that the ratio of total changes agrees with this derivative ratio at at least one interior point, provided the relevant denominators are nonzero.

Here are two forms to keep in view:

Always available: the cross-multiplied identity in Cauchy’s Mean Value Theorem.
Available with nonzero denominators: the equality of ratios of increments and derivatives.

The first form is more general. For example, if $g$ has equal values at the endpoints, the increment ratio is undefined, but the cross-multiplied identity still carries information.

Worked Example: Finding the Common Point for Two Powers

Take $f(x)=x^2$ and $g(x)=x^3$ on $[1,2]$. Both functions are continuous on this closed interval and differentiable on its interior. Their endpoint increments are

f(2)-f(1)=4-1=3,\qquad g(2)-g(1)=8-1=7.

Their derivatives are $f'(x)=2x$ and $g'(x)=3x^2$. Cauchy’s identity requires a point $c\in(1,2)$ satisfying

$$ 3(3c^2)=7(2c). $$

Since $c\in(1,2)$, it is nonzero, so solving gives $9c^2=14c$, hence $c=14/9$, which is indeed in $(1,2)$. Substitution verifies the identity:

3\left(3\left(\frac{14}{9}\right)^2\right) =3\left(\frac{196}{27}\right) =\frac{196}{9}, \qquad 7\left(2\left(\frac{14}{9}\right)\right) =\frac{196}{9}.

In ratio form, the endpoint-change ratio is $3/7$, and at $c=14/9$ the derivative ratio is $\frac{f'(c)}{g'(c)}=\frac{2c}{3c^2}=\frac{2}{3c}=\frac{3}{7}$.

Worked Example: When the Denominator Increment Is Zero

Let $f(x)=x$ and $g(x)=(x-1)^2$ on $[0,2]$. The functions meet the continuity and differentiability hypotheses, and their endpoint increments are

f(2)-f(0)=2,\qquad g(2)-g(0)=1-1=0.

The derivatives are $f'(x)=1$ and $g'(x)=2(x-1)$. The theorem’s identity becomes

2\bigl(2(c-1)\bigr)=0\cdot1.

This equation holds at $c=1$, which lies in $(0,2)$. In fact, the equation forces $c=1$. The ratio $\frac{f(2)-f(0)}{g(2)-g(0)}$ is not defined, so a proof or application that starts by dividing by $g(2)-g(0)$ would fail. The cross-multiplied conclusion remains valid and identifies an interior point where $g'$ vanishes.

Bounding One Increment by Another

A useful consequence turns bounds on the ratio of derivatives into bounds on the ratio of endpoint changes. This is a two-function counterpart to the secant-slope bounds established earlier in the course. The positivity assumption on $g'$ ensures that both the derivative ratio and the increment ratio are well-defined with positive denominators.

Theorem (Bounds for a Ratio of Increments): Let $a<b$, and let $f,g:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose $g'(x)>0$ for every $x\in(a,b)$, and suppose $m\leq f'(x)/g'(x)\leq M$ for every $x\in(a,b)$, where $m\leq M$. Then $$ m\leq\frac{f(b)-f(a)}{g(b)-g(a)}\leq M. $$

Proof. By the Mean Value Theorem, applied to $g$, there is $d\in(a,b)$ such that

$$ g(b)-g(a)=g'(d)(b-a)>0. $$

Thus the increment ratio in the conclusion is defined. By Cauchy’s Mean Value Theorem, there is $c\in(a,b)$ such that

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

Since $g'(c)>0$, division by $g'(c)$ and by the positive quantity $g(b)-g(a)$ gives

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

The assumed derivative-ratio bounds hold at this $c$, so $m\leq f'(c)/g'(c)\leq M$. Substitution proves the desired inequalities. $\square$

Worked Example: Bounding a Secant Ratio

On $[1,3]$, let $f(x)=x^2$ and $g(x)=x$. On the interior, $g'(x)=1>0$, while

\frac{f'(x)}{g'(x)}=2x, \qquad 2\leq2x\leq6\quad (1<x<3).

The theorem therefore bounds the ratio of increments between $2$ and $6$. Direct calculation confirms the resulting estimate:

\frac{f(3)-f(1)}{g(3)-g(1)} =\frac{9-1}{3-1} =\frac{8}{2} =4, \qquad 2\leq4\leq6.

The conclusion does not claim that the increment ratio equals either endpoint value of the derivative ratio. It guarantees that it lies between the stated bounds; Cauchy’s theorem gives the stronger fact that it equals $2c$ for some $c\in(1,3)$.

Proportional Derivatives Give Proportional Increments

The ratio bounds have an equality case that is often useful. If the derivative of $f$ is a fixed multiple of the derivative of $g$, then their changes across the interval have that same relationship, provided $g'$ is positive. The result follows directly from the common-point identity.

Proposition (Proportional Derivatives Give Proportional Increments): Let $a<b$, and let $f,g:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose $g'(x)>0$ and $f'(x)=\lambda g'(x)$ for every $x\in(a,b)$, where $\lambda\in\mathbb{R}$. Then $$ f(b)-f(a)=\lambda\bigl(g(b)-g(a)\bigr). $$

Proof. Cauchy’s Mean Value Theorem supplies $c\in(a,b)$ such that

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

By the assumed derivative relation, $f'(c)=\lambda g'(c)$. Therefore

\bigl(f(b)-f(a)\bigr)g'(c) =\lambda\bigl(g(b)-g(a)\bigr)g'(c).

Since $g'(c)>0$, division by $g'(c)$ gives the claimed equality. $\square$

Worked Example: Checking Proportional Changes

Let $f(x)=3x^2+1$ and $g(x)=x^2$ on $[1,2]$. On $(1,2)$, $g'(x)=2x>0$, and $f'(x)=6x=3g'(x)$. The proposition predicts that the change in $f$ is three times the change in $g$. Indeed,

f(2)-f(1)=(13-4)=9, \qquad 3\bigl(g(2)-g(1)\bigr)=3(4-1)=9.

Both functions satisfy the required endpoint continuity and interior differentiability conditions. The equality of increments holds across the whole interval, not merely at a selected point.

Hypotheses and Common Pitfalls

Continuity on the closed interval and differentiability on its interior are essential parts of the theorem’s setup. Differentiability at the endpoints is not required. Conversely, having both derivatives at an interior point is not enough: the theorem needs the functions to satisfy its interval-wide hypotheses.

A second common pitfall is to assume that the theorem gives a quotient even when a denominator may vanish. If $g(b)=g(a)$, the increment quotient is undefined. If $g'(c)=0$, the derivative quotient is undefined. In either situation the cross-multiplied identity is still the correct conclusion. Check the relevant nonzero conditions before dividing.

Finally, do not choose separate points for the two functions. The strength of Cauchy’s theorem is that the two derivatives are evaluated at one common $c$. This makes it possible to turn a bound on $f'/g'$ into a bound on the ratio of increments, as above. The proof of Cauchy’s Mean Value Theorem is the next step in developing this tool.

Check Your Understanding

Use the theorem and its consequences to answer the following questions.

State the continuity and differentiability hypotheses in Cauchy’s Mean Value Theorem.
Why is the cross-multiplied identity more general than the ratio form?
When $g(x)=x$, what familiar theorem follows from Cauchy’s Mean Value Theorem?
Which assumptions ensure that the ratio-of-increments bound has positive denominators?
If $f'(x)=\lambda g'(x)$ throughout an interval and $g'(x)>0$, what relation holds between the endpoint increments?

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