Mean value theorem

The overall average rate of change is attained at some instant; Rolle → Lagrange, and why no hypothesis can be dropped.

about 5 min

Start from a problem

Problem

A car covers 200 km in two hours: average speed 100 km/h. Can you assert that at some instant the speedometer read exactly 100?

Intuition says yes: slower than 100 the whole way and you never reach 200 km; faster the whole way and you overshoot. Speed varies continuously, so it must pass through 100. But "intuition says yes" is not a proof. We need a theorem that connects "overall average" to "some single instant" precisely.

Observe

Fix the point $x_0$ below and make $h$ large: the secant joins two endpoints. The mean value theorem says the tangent somewhere in between is parallel to that secant. Try a few functions and hunt for that place.

Interactive experimentSecant becomes tangent

Function

point x₀1.00

increment h1.200

\frac{f(x_0+h)-f(x_0)}{h}

secant slope -0.0275tangent slope f′(x₀) 0.5403gap 0.5678

Conjecture

If $f$ is "nice enough" on $[a,b]$ (continuous, differentiable), there is $\xi\in(a,b)$ with $f'(\xi)=\frac{f(b)-f(a)}{b-a}.$ Strategy for the proof: reduce to the special case of equal endpoint values, where all we need is an extremum.

Theorems and proofs

Theorem 3.1Rolle's theorem

Let $f$ be continuous on $[a,b]$ , differentiable on $(a,b)$ , with $f(a)=f(b)$ . Then there is $\xi\in(a,b)$ with $f'(\xi)=0$ .

Proof

A continuous function on a closed interval attains a maximum $M$ and a minimum $m$ (extreme value theorem). If $M=m$ , $f$ is constant and any $\xi$ works. Otherwise at least one of $M,m$ differs from $f(a)=f(b)$ and is attained at an interior point $\xi\in(a,b)$ ; say $f(\xi)=M$ . For small $h$ , $f(\xi+h)\le f(\xi)$ , so $\frac{f(\xi+h)-f(\xi)}{h}\le0\ (h>0),\qquad\frac{f(\xi+h)-f(\xi)}{h}\ge0\ (h<0).$ Taking limits (the one-sided limits agree because $f$ is differentiable at $\xi$ ) gives $f'(\xi)\le0$ and $f'(\xi)\ge0$ , hence $f'(\xi)=0$ .

The heart of this step: the derivative vanishes at an interior extremum. That is Fermat's lemma.

Theorem 3.2Lagrange's mean value theorem

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$ . Then there is $\xi\in(a,b)$ with $f'(\xi)=\frac{f(b)-f(a)}{b-a}.$

Proof

The secant through the endpoints is $\ell(x)=f(a)+\frac{f(b)-f(a)}{b-a}(x-a),$ whose derivative is the constant average slope $m=\frac{f(b)-f(a)}{b-a}$ . Put $g(x)=f(x)-\ell(x)$ . Then $g$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and since the secant passes through both endpoints, $g(a)=g(b)=0$ . Rolle gives $\xi$ with $g'(\xi)=0$ , i.e. $f'(\xi)-m=0$ .

The chain: build the secant → subtract it → equal endpoints → Rolle → conclusion. The entire trick is "subtract the secant".

Which hypothesis cannot be dropped

Example 3.1Not differentiable: f(x) = |x|

On $[-1,1]$ the secant slope is $0$ , but $f'$ takes only the values $-1$ and $1$ , and $f$ is not differentiable at $0$ . No $\xi$ satisfies the conclusion. Differentiability on the open interval cannot be dropped.

Example 3.2Discontinuous at an endpoint

Let $f(x)=x$ for $0\le x<1$ and $f(1)=2$ . The secant slope is $2$ ; the interior derivative is identically $1$ . Continuity at the endpoints cannot be dropped, even at a single point.

Common mistake

$\xi$ exists, but the theorem says nothing about where it is, how many there are, or how to find it. Never treat $\xi$ as a quantity you can solve for; the theorem's value lies in estimates: $|f(b)-f(a)|\le\max|f'|\cdot|b-a|$ .

Applications

ApplicationZero derivative means constant

If $f'\equiv0$ on an interval $I$ , then for any $a<b$ in $I$ , $f(b)-f(a)=f'(\xi)(b-a)=0$ . This corollary is used in part II of the fundamental theorem in The integral and in the uniqueness proof for $y'=ky$ in What is an ODE. Without the mean value theorem both proofs are empty.

ApplicationMonotonicity test

$f'>0$ on $(a,b)$ implies $f$ is strictly increasing: for $x_1<x_2$ , $f(x_2)-f(x_1)=f'(\xi)(x_2-x_1)>0$ .

ApplicationTaylor's theorem

The Lagrange-remainder proof in Taylor expansion is Rolle's theorem applied $n+1$ times to a carefully built function.

Remark

The mean value theorem is the bridge from local information to global behaviour: knowing the derivative at every point controls the difference of function values between any two points. Nearly every statement of the form "a property of $f'$ implies a property of $f$ " rests on it.

Exercises

For

f(x)=x^3

[0,2]

, find

\xi\in(0,2)

with

f'(\xi)=\dfrac{f(2)-f(0)}{2-0}

(three decimals).

For

f(x)=|x|

[-1,1]

the secant slope is

0

, yet no point has derivative

0

. Which hypothesis of the mean value theorem fails?