Cauchy mean value theorem

Rolle's theorem says a smooth function that returns to its starting value must, somewhere in between, have zero slope. The Lagrange mean value theorem (LMVT) generalizes this to functions that do not return to the same value: some tangent slope equals the average slope over the interval. Cauchy's mean value theorem pushes one step further, comparing two functions $f$ and $g$ at once. It says that at some common point $c$ , the ratio of their instantaneous rates of change, $f'(c)/g'(c)$ , equals the ratio of their total changes over $[a,b]$ . This is not "LMVT applied twice" — it guarantees a single shared $c$ that works for both functions simultaneously. That extra strength is exactly what is needed to justify L'Hôpital's rule for indeterminate limits, which is why this theorem sits directly between LMVT and L'Hôpital's rule in the syllabus.

The core insight

For two functions $f,g$ continuous on $[a,b]$ , differentiable on $(a,b)$ , with $g'\neq0$ throughout, some $c\in(a,b)$ makes the ratio of derivatives equal the ratio of total changes, $f'(c)/g'(c) = \big(f(b)-f(a)\big)/\big(g(b)-g(a)\big)$ .

The core idea

Cauchy's mean value theorem needs the same continuity and differentiability hypotheses as LMVT, applied to both functions, plus one extra condition on $g$ .

Statement. Let $f,g:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$ , with $g'(x)\neq 0$ for every $x\in(a,b)$ . Then there exists $c\in(a,b)$ such that

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

where: $f,g$ are the two functions being compared, $a,b$ are the interval endpoints, $c$ is the guaranteed interior point, and $g'(x)\neq 0$ on $(a,b)$ is the extra hypothesis that keeps the denominator on the left from vanishing. Two structural points matter more than memorizing the formula.

First, the denominator on the right is automatically safe. If $g(a)$ happened to equal $g(b)$ , Rolle's theorem applied to $g$ alone would force some point where $g'$ vanishes — but $g'\neq 0$ throughout $(a,b)$ is exactly the hypothesis ruling that out. So $g(b)\neq g(a)$ is guaranteed, not an extra thing to check.

Second, LMVT is the special case $g(x)=x$ . Then $g'(x)=1\neq0$ everywhere and $g(b)-g(a)=b-a$ , and the theorem collapses to $f'(c) = \dfrac{f(b)-f(a)}{b-a}$ , which is exactly LMVT. Cauchy's theorem is therefore the genuinely more general fact; LMVT is a corollary of it, not a separate result to remember independently. A useful picture: trace the parametric curve $\big(g(x), f(x)\big)$ as $x$ runs from $a$ to $b$ . The chord joining its two endpoints has slope $\dfrac{f(b)-f(a)}{g(b)-g(a)}$ , and the theorem says the curve has a tangent parallel to that chord at some parameter value $c$ — the tangent's slope there is $f'(c)/g'(c)$ , since $dy/dx = (dy/dt)/(dx/dt)$ along the curve.

Figure

A parametric curve traced by g of x on the horizontal axis and f of x on the vertical axis, with a straight chord joining its two endpoints and a tangent line at an interior point drawn parallel to that chord

Verify Cauchy's MVT for two power functions

The theorem is an existence result, so the standard verification task is: check the three hypotheses, compute the right-hand ratio, and solve for the $c$ that the theorem promises exists.

Derivation AProving Cauchy's mean value theorem from Rolle's theorem

The goal is to derive Cauchy's mean value theorem for $f,g$ continuous on $[a,b]$ , differentiable on $(a,b)$ , with $g'\neq0$ on $(a,b)$ , by building a single auxiliary function whose Rolle's-theorem zero-derivative point turns out to be exactly the point $c$ the theorem claims exists, rather than treating the ratio formula as something pulled out of thin air.

Step 1. Build the auxiliary function that combines $f$ and $g$ with the target ratio baked in. Define

\phi(x) = f(x) - f(a) - \frac{f(b)-f(a)}{g(b)-g(a)}\big(g(x)-g(a)\big).

This is legal because $g(b)\neq g(a)$ , guaranteed by $g'\neq0$ on $(a,b)$ as argued above. Note $\phi$ is a combination of continuous, differentiable functions, so $\phi$ is itself continuous on $[a,b]$ and differentiable on $(a,b)$ — Rolle's theorem is available for $\phi$ once its endpoint values match.

Step 2. Check the endpoint values are equal. At $x=a$ : $\phi(a) = f(a)-f(a) - \frac{f(b)-f(a)}{g(b)-g(a)}(g(a)-g(a)) = 0$ . At $x=b$ : $\phi(b) = f(b)-f(a) - \frac{f(b)-f(a)}{g(b)-g(a)}(g(b)-g(a)) = f(b)-f(a) - \big(f(b)-f(a)\big) = 0$ . So $\phi(a)=\phi(b)=0$ , and Rolle's theorem applies directly to $\phi$ on $[a,b]$ .

Step 3. Apply Rolle's theorem and differentiate $\phi$ . Rolle's theorem gives some $c\in(a,b)$ with $\phi'(c)=0$ . Differentiating the definition of $\phi$ ,

\phi'(x) = f'(x) - \frac{f(b)-f(a)}{g(b)-g(a)}\,g'(x),

so $\phi'(c)=0$ means $f'(c) = \dfrac{f(b)-f(a)}{g(b)-g(a)}\,g'(c)$ .

Step 4. Divide by $g'(c)$ , which is nonzero by hypothesis, to isolate the ratio: $\dfrac{f'(c)}{g'(c)} = \dfrac{f(b)-f(a)}{g(b)-g(a)}$ , exactly the statement of Cauchy's mean value theorem. Note the proof never needed $f,g$ individually to satisfy Rolle's theorem — only the specific combination $\phi$ did, which is why the theorem is genuinely about the pair $f,g$ together, not two separate applications of Rolle's or Lagrange's theorem.

Worked example

Verify Cauchy's MVT and find $c$ for $f(x)=x^2$ , $g(x)=x^3$ on $[1,2]$ .

Step 1. Check the hypotheses. Both $f$ and $g$ are polynomials, so both are continuous on $[1,2]$ and differentiable on $(1,2)$ . Also $g'(x)=3x^2$ , which vanishes only at $x=0$ , so $g'(x)\neq0$ for every $x\in(1,2)$ . All three hypotheses hold.

Step 2. Compute the target ratio, the right-hand side of the theorem. $f(2)-f(1) = 4-1 = 3$ and $g(2)-g(1) = 8-1 = 7$ , so

\frac{f(2)-f(1)}{g(2)-g(1)} = \frac{3}{7}.

Step 3. Compute the left-hand ratio symbolically. $f'(x)=2x$ and $g'(x)=3x^2$ , so

\frac{f'(c)}{g'(c)} = \frac{2c}{3c^2} = \frac{2}{3c}.

Step 4. Solve $\dfrac{2}{3c}=\dfrac{3}{7}$ for $c$ : cross-multiplying gives $2\cdot7 = 3\cdot3c$ , so $14 = 9c$ , giving $c = \dfrac{14}{9}\approx1.556$ . Since $1 < 14/9 < 2$ , this $c$ lies in the open interval $(1,2)$ as promised. Answer: $c = 14/9 \approx 1.556$ .

Competitive stretch

A second classic pair sharpens the picture: $f(x)=\sin x$ , $g(x)=\cos x$ on $[0,\pi/2]$ . Here $g'(x)=-\sin x\neq0$ on $(0,\pi/2)$ , the right-hand ratio is $\dfrac{1-0}{0-1}=-1$ , and the left-hand ratio simplifies to $\dfrac{\cos c}{-\sin c}=-\cot c$ . Setting $-\cot c = -1$ gives $\cot c = 1$ , so $c=\pi/4$ — an exact standard-angle answer, unlike the irrational $c=14/9$ above. Both cases confirm the same theorem; the ease of pinning down $c$ exactly depends on the specific $f,g$ , not on the theorem itself, which only ever promises existence.

Common traps

Forgetting the extra hypothesis $g'(x)\neq0$ on $(a,b)$ — without it, $g'(c)$ in the denominator could vanish and the ratio $f'(c)/g'(c)$ would be undefined.
Treating Cauchy's MVT as "apply LMVT to $f$ , apply LMVT to $g$ , then divide the two results" — that produces two generally different points $c_1,c_2$ , not the single shared $c$ that Cauchy's theorem actually guarantees.
Skipping the check that $g(a)\neq g(b)$ before writing the ratio $\frac{f(b)-f(a)}{g(b)-g(a)}$ — it follows automatically from $g'\neq0$ , but state it rather than assume it silently.
Mixing this up with an "average value" computation — Cauchy's MVT is an existence statement about matching a derivative ratio to a secant ratio, not a formula for computing an average of a function.

Ask the tutor: if the auxiliary function $\phi(x)$ in the proof feels arbitrary, ask the AI tutor to show step by step why exactly that combination of $f$ , $g$ , and their endpoint values forces $\phi(a)=\phi(b)=0$ — seeing the motivation makes the whole proof easy to reconstruct from scratch.

Why do we need a whole new theorem instead of just using LMVT on f and on g separately

Using LMVT twice gives $f'(c_1) = \frac{f(b)-f(a)}{b-a}$ for some $c_1$ and $g'(c_2) = \frac{g(b)-g(a)}{b-a}$ for some, generally different, $c_2$ . Dividing those two equations does give a true statement, but it mixes $c_1$ and $c_2$ , not a single common point. Cauchy's theorem is a genuinely stronger claim: one point $c$ makes both derivatives match the ratio simultaneously. That single-point guarantee is exactly the ingredient L'Hôpital's rule proof needs, so it cannot be replaced by two separate LMVT applications.

Does Cauchy's mean value theorem tell us how to find c, or just that it exists

Just that it exists — like Rolle's theorem and LMVT before it, this is a pure existence statement, not a construction. In the worked example, $c=14/9$ could be solved for exactly only because the algebra after substituting $f'$ and $g'$ happened to be simple. For more complicated $f,g$ pairs, $c$ may only be describable as a root of some equation, possibly with no closed form at all. The theorem's value is guaranteeing that such a $c$ exists in $(a,b)$ , which is precisely the logical step the proof of L'Hôpital's rule needs, independent of ever computing $c$ explicitly.

How exactly does this theorem connect to L'Hopital's rule

L'Hôpital's rule evaluates limits of the form $0/0$ or $\infty/\infty$ by replacing $f(x)/g(x)$ with $f'(x)/g'(x)$ near the limit point. The rigorous justification applies Cauchy's mean value theorem on a shrinking interval between the limit point and a nearby $x$ : it produces a point $c$ between them with $\frac{f(x)-f(a)}{g(x)-g(a)} = \frac{f'(c)}{g'(c)}$ , and as $x$ approaches the limit point, $c$ is squeezed along with it, so the ratio of derivatives at $c$ tracks the ratio of function values. Without the single shared point Cauchy's theorem guarantees, this squeeze argument would not go through.