Mean Value Theorems (Lagrange and Cauchy)

Chapter 3 defined the derivative pointwise: $f'(a)$ is a number attached to a single point $a$. This chapter proves the theorems that let a single value of the derivative, at some unspecified point, control the global behaviour of $f$ across an entire interval — why $f'=0$ everywhere forces $f$ to be constant, why $f'>0$ everywhere forces $f$ to be increasing, and ultimately why L'Hopital's rule (next chapter) is valid at all. The Mean Value Theorem of Lagrange is the central result; Rolle's Theorem is its special case and also the engine of its proof; Cauchy's Mean Value Theorem is its two-function generalisation.

Have to hand: the Extreme Value Theorem (Chapter 2), the definition of the derivative (Chapter 3), and the fact that differentiability implies continuity.

The shape of every proof in this chapter. Each theorem below is proved by constructing an auxiliary function specifically engineered to have equal values at the two endpoints, then applying Rolle's Theorem to that auxiliary function. Learning to construct the right auxiliary function is the one transferable skill of this entire chapter.

1. Fermat's Interior Extremum Theorem

Theorem. Let $f$ be differentiable at $c$, and suppose $f$ has a local maximum or local minimum at $c$, where $c$ is an interior point of the domain being considered. Then $f'(c)=0$.

Proof (local maximum case). Since $f$ has a local maximum at $c$, there is $\delta>0$ such that $f(x)\le f(c)$ for all $x$ with $|x-c|<\delta$. For $0<h<\delta$, $f(c+h)-f(c)\le0$ and $h>0$, so $\frac{f(c+h)-f(c)}{h}\le0$; letting $h\to0^+$ gives $f'_+(c)\le0$. For $-\delta<h<0$, $f(c+h)-f(c)\le0$ and $h<0$, so $\frac{f(c+h)-f(c)}{h}\ge0$; letting $h\to0^-$ gives $f'_-(c)\ge0$. Since $f$ is differentiable at $c$, $f'_+(c)=f'_-(c)=f'(c)$, and a number that is simultaneously $\le0$ and $\ge0$ must be $0$. Hence $f'(c)=0$. $\blacksquare$

"Interior" is load-bearing. The theorem says nothing about an extremum at an endpoint of a closed interval: $f(x)=x$ on $[0,1]$ has a maximum at $x=1$, yet $f'(1)=1\ne0$. One-sided derivatives at an endpoint need not vanish, because only one of the two one-sided inequalities above is available there.

2. Rolle's Theorem

Theorem (Rolle). Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$, with $f(a)=f(b)$. Then there exists $c\in(a,b)$ with $f'(c)=0$.

Proof. By the Extreme Value Theorem (Chapter 2), $f$ attains a maximum $M$ and a minimum $m$ on $[a,b]$. Case 1: $M=m$. Then $f$ is constant on $[a,b]$, so $f'(c)=0$ for every $c\in(a,b)$. Case 2: $M>m$. Since $f(a)=f(b)$, at least one of $M,m$ is attained at an interior point $c\in(a,b)$ (otherwise both the max and min would be attained only at the two endpoints, forcing $M=f(a)=f(b)=m$, contradicting $M>m$). At that interior point $c$, $f$ has a local extremum, and $f$ is differentiable there by hypothesis, so Fermat's theorem gives $f'(c)=0$. $\blacksquare$

Every hypothesis is necessary. Continuity on $[a,b]$: $f(x)=x$ for $x\in[0,1)$, $f(1)=0$ is discontinuous at $1$, satisfies $f(0)=f(1)=0$, but $f'(x)=1\ne0$ everywhere it is defined. Differentiability on $(a,b)$: $f(x)=|x|$ on $[-1,1]$ has $f(-1)=f(1)=1$ but no point with $f'(c)=0$, because $f$ fails to be differentiable at the one candidate point $c=0$. $f(a)=f(b)$: $f(x)=x$ on $[0,1]$ has no $c$ with $f'(c)=0$ at all, consistent with $f(0)\ne f(1)$.

Worked example. Verify Rolle's Theorem for $f(x)=x^2-4x+3$ on $[1,3]$: $f(1)=1-4+3=0=f(3)=9-12+3$, and $f$ is a polynomial (continuous and differentiable everywhere). Rolle's Theorem guarantees some $c\in(1,3)$ with $f'(c)=0$. Indeed $f'(x)=2x-4=0$ gives $c=2\in(1,3)$, confirming the theorem.

3. The Mean Value Theorem (Lagrange)

Theorem (MVT). Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists $c\in(a,b)$ with

Proof. Define the auxiliary function

which subtracts from $f$ the secant line through $(a,f(a))$ and $(b,f(b))$. Then $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$ (a sum of $f$ and a linear function), and

So $g(a)=g(b)=0$, and Rolle's Theorem applies to $g$: there is $c\in(a,b)$ with $g'(c)=0$. Differentiating,

so $g'(c)=0$ gives exactly $f'(c) = \dfrac{f(b)-f(a)}{b-a}$. $\blacksquare$

Geometric meaning. $\dfrac{f(b)-f(a)}{b-a}$ is the slope of the secant line through the endpoints. MVT says some point strictly between $a$ and $b$ has a tangent line parallel to that secant. Rolle's Theorem is the special case of a horizontal secant ($f(a)=f(b)$), giving a horizontal tangent.

Worked example. For $f(x)=x^2$ on $[1,4]$, the secant slope is $\dfrac{f(4)-f(1)}{4-1} = \dfrac{16-1}{3} = 5$. MVT guarantees some $c\in(1,4)$ with $f'(c)=2c=5$, i.e. $c=2.5$, which indeed lies in $(1,4)$.

4. Consequences of the MVT

Corollary 1 (zero derivative implies constant). If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f'(x)=0$ for every $x\in(a,b)$, then $f$ is constant on $[a,b]$.

Proof. Let $x_1<x_2$ be any two points of $[a,b]$. Apply MVT to $f$ on $[x_1,x_2]$: there is $c\in(x_1,x_2)$ with $f(x_2)-f(x_1) = f'(c)(x_2-x_1) = 0$, since $f'(c)=0$. So $f(x_2)=f(x_1)$ for every such pair, hence $f$ is constant. $\blacksquare$

Corollary 2 (equal derivatives differ by a constant). If $f,g$ are both continuous on $[a,b]$, differentiable on $(a,b)$, and $f'(x)=g'(x)$ for every $x\in(a,b)$, then $f(x)-g(x)$ is constant on $[a,b]$.

Proof. Apply Corollary 1 to $h=f-g$: $h'(x)=f'(x)-g'(x)=0$ everywhere on $(a,b)$, so $h$ is constant. $\blacksquare$

Corollary 3 (sign of $f'$ controls monotonicity). If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f'(x)>0$ for every $x\in(a,b)$, then $f$ is strictly increasing on $[a,b]$. (Similarly $f'(x)<0$ gives strictly decreasing, $f'(x)\ge0$ gives non-decreasing.)

Proof. Let $x_1<x_2$ in $[a,b]$. MVT on $[x_1,x_2]$ gives $c\in(x_1,x_2)$ with $f(x_2)-f(x_1)=f'(c)(x_2-x_1)$. Since $f'(c)>0$ and $x_2-x_1>0$, the right side is positive, so $f(x_2)>f(x_1)$. $\blacksquare$

Corollary 4 (bounded derivative gives a Lipschitz-type bound). If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $|f'(x)|\le K$ for every $x\in(a,b)$, then $|f(x_2)-f(x_1)|\le K|x_2-x_1|$ for all $x_1,x_2\in[a,b]$.

Proof. For $x_1<x_2$ in $[a,b]$, MVT on $[x_1,x_2]$ gives $c\in(x_1,x_2)$ with $f(x_2)-f(x_1)=f'(c)(x_2-x_1)$. Taking absolute values, $|f(x_2)-f(x_1)| = |f'(c)|\,|x_2-x_1| \le K|x_2-x_1|$. $\blacksquare$

Worked example. Since $f(x)=\sin x$ has $f'(x)=\cos x$ with $|\cos x|\le1$ everywhere, Corollary 4 gives $|\sin b - \sin a| \le |b-a|$ for all real $a,b$ — a clean inequality obtained with no trigonometric identity at all, purely from a bound on the derivative.

This is the theorem that justifies every "increasing/decreasing" argument you have ever made with a derivative. Without MVT, "$f'>0$ on an interval" and "$f$ is increasing on that interval" are two separate, unconnected facts; MVT is precisely the bridge between them, and it is why Chapter 6's curve-sketching methods are valid at all.

5. Cauchy's Mean Value Theorem

Theorem (Cauchy MVT). Let $f,g$ be continuous on $[a,b]$ and differentiable on $(a,b)$, with $g'(x)\ne0$ for every $x\in(a,b)$. Then there exists $c\in(a,b)$ with

(Note $g(b)\ne g(a)$ is guaranteed: if they were equal, Rolle's Theorem applied to $g$ would give some point with $g'=0$, contradicting the hypothesis.)

Proof. Define

$h$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Direct computation gives

and these are equal: $h(a)=h(b)$. Rolle's Theorem gives $c\in(a,b)$ with $h'(c)=0$, i.e.

Since $g'(c)\ne0$ and $g(b)\ne g(a)$, dividing through gives $\dfrac{f'(c)}{g'(c)} = \dfrac{f(b)-f(a)}{g(b)-g(a)}$. $\blacksquare$

Why this generalises Lagrange's MVT. Taking $g(x)=x$ (so $g'(x)=1\ne0$ always, $g(b)-g(a)=b-a$) recovers exactly the ordinary MVT. Cauchy's version is what makes L'Hopital's Rule provable in the next chapter: it relates the ratio of two functions' increments to the ratio of their derivatives at a single shared point $c$, which an ordinary (single-function) MVT applied separately to $f$ and to $g$ cannot do, since the two separate applications would generally produce two different points $c_1\ne c_2$.

Common Exam Pitfalls

• Forgetting the two different interval conditions. Continuity is required on the closed interval $[a,b]$; differentiability only on the open interval $(a,b)$ — endpoints need not be differentiable points of a larger domain.
• Applying Rolle's Theorem without checking $f(a)=f(b)$. This is not automatic; verify it explicitly before invoking the theorem.
• Misusing Fermat's theorem at an endpoint. $f'(c)=0$ is only guaranteed at an interior local extremum; an extremum at $a$ or $b$ gives no such guarantee, as $f(x)=x$ on $[0,1]$ shows.
• Treating MVT as giving a specific, computable $c$ in general. Like IVT and EVT, MVT is an existence theorem; the worked example with $f(x)=x^2$ happened to allow solving for $c$ explicitly, but this is not guaranteed for every function.
• Applying Corollary 1 (constant function) on a set that is not an interval. The proof genuinely needs $[x_1,x_2]\subseteq[a,b]$ for every pair of points; on a domain with a gap (two disjoint intervals), $f'=0$ on each piece only forces $f$ to be constant on each piece separately, possibly with different constants.
• Forgetting Corollary 4 requires a uniform bound $K$ on $|f'|$ across the whole interval, not just at one point. A derivative bounded at a single point gives no Lipschitz-type bound on the function's increments over an interval.
• Confusing Cauchy's MVT with simply applying the ordinary MVT twice. Applying MVT to $f$ and separately to $g$ produces two generally different points $c_1,c_2$; Cauchy's theorem is the stronger statement that a single $c$ works for both simultaneously, which is exactly the extra content needed for L'Hopital's Rule.

Summary

• Fermat's theorem: at an interior local extremum where $f$ is differentiable, $f'(c)=0$ — proved by comparing one-sided derivatives, which must be $\le0$ and $\ge0$ respectively.
• Rolle's Theorem: $f$ continuous on $[a,b]$, differentiable on $(a,b)$, $f(a)=f(b)$ $\implies$ some $c\in(a,b)$ has $f'(c)=0$ — proved via EVT plus Fermat's theorem; every hypothesis is independently necessary.
• MVT (Lagrange): under the same continuity/differentiability hypotheses, some $c\in(a,b)$ has $f'(c)=\dfrac{f(b)-f(a)}{b-a}$ — proved by applying Rolle's Theorem to $f$ minus its secant line; geometrically, a tangent parallel to the secant.
• Consequences: $f'\equiv0$ on an interval $\implies$ $f$ constant; $f'\equiv g'$ $\implies$ $f-g$ constant; $f'>0$ (resp. $<0$) on an interval $\implies$ $f$ strictly increasing (resp. decreasing) — the rigorous justification for every monotonicity argument from a derivative's sign; $|f'|\le K$ on an interval $\implies$ $|f(x_2)-f(x_1)|\le K|x_2-x_1|$, a Lipschitz-type bound.
• Cauchy's MVT: for $f,g$ continuous on $[a,b]$, differentiable on $(a,b)$, $g'\ne0$ on $(a,b)$, some single $c\in(a,b)$ has $\dfrac{f'(c)}{g'(c)}=\dfrac{f(b)-f(a)}{g(b)-g(a)}$ — proved via Rolle's Theorem applied to a two-function auxiliary combination; reduces to Lagrange's MVT when $g(x)=x$, and is the theorem underlying L'Hopital's Rule.