Lagrange's Mean Value Theorem

Basis
Last updated: Tags: Calculus, Mean Value Theorems

Prerequisites

Drive 120 km in two hours. Your average speed was 60 km/h. At some instant during the trip, the speedometer must have read exactly 60 km/h. Lagrange’s Mean Value Theorem is this fact stated for any differentiable function.

Statement

Theorem (Mean Value Theorem / Lagrange’s Finite Increment Theorem). Let f:[a,b]Rf : [a, b] \to \mathbb{R}. If

  1. ff is continuous on [a,b][a, b], and
  2. ff is differentiable on (a,b)(a, b),

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

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

Proof

The idea is to subtract from ff the linear function whose graph is the secant line through (a,f(a))(a, f(a)) and (b,f(b))(b, f(b)), turning the problem into one where Rolle’s theorem applies.

Define the auxiliary function

g(x)    f(x)f(a)f(b)f(a)ba(xa).g(x) \;\coloneqq\; f(x) - f(a) - \frac{f(b)-f(a)}{b-a}(x - a).

Then:

  • gg is continuous on [a,b][a, b] and differentiable on (a,b)(a, b) (inherited from ff).
  • g(a)=0g(a) = 0 and g(b)=f(b)f(a)(f(b)f(a))=0g(b) = f(b) - f(a) - (f(b)-f(a)) = 0.

By Rolle’s Theorem, there exists c(a,b)c \in (a, b) with g(c)=0g'(c) = 0. Computing:

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

so g(c)=0g'(c) = 0 gives f(c)=f(b)f(a)baf'(c) = \dfrac{f(b)-f(a)}{b-a}, which is exactly (1)(1). \square

Geometric meaning

The right-hand side of (1)(1) is the slope of the secant line through (a,f(a))(a, f(a)) and (b,f(b))(b, f(b)). The theorem says the tangent line at some interior point is parallel to that secant. In other words, the instantaneous rate of change equals the average rate of change at some point.

Corollaries

Constant functions

Corollary. If f(x)=0f'(x) = 0 for all x(a,b)x \in (a, b), then ff is constant on [a,b][a, b].

Proof. For any x[a,b]x \in [a, b], apply the MVT to [a,x][a, x]: f(x)f(a)=f(c)(xa)=0f(x) - f(a) = f'(c)(x - a) = 0, so f(x)=f(a)f(x) = f(a). \square

Monotone functions

Corollary. If f(x)>0f'(x) > 0 for all x(a,b)x \in (a, b), then ff is strictly increasing on [a,b][a, b].

Proof. For ax1<x2ba \leq x_1 < x_2 \leq b, apply the MVT to [x1,x2][x_1, x_2]: f(x2)f(x1)=f(c)(x2x1)>0f(x_2) - f(x_1) = f'(c)(x_2 - x_1) > 0. \square

The analogue with f<0f' < 0 gives strict decrease; f0f' \geq 0 (or 0\leq 0) gives non-strict monotonicity.

Lipschitz bound

Corollary. If f(x)M|f'(x)| \leq M for all x(a,b)x \in (a, b), then f(x2)f(x1)Mx2x1|f(x_2) - f(x_1)| \leq M|x_2 - x_1| for all x1,x2[a,b]x_1, x_2 \in [a, b].

This is the Lipschitz condition with constant MM, widely used to control approximation errors.

Summary

  • Mean Value Theorem: if ff is continuous on [a,b][a,b] and differentiable on (a,b)(a,b), then f(b)f(a)=f(c)(ba)f(b) - f(a) = f'(c)(b-a) for some c(a,b)c \in (a,b).
  • Proof: apply Rolle’s theorem to g(x)=f(x)(secant line)g(x) = f(x) - \text{(secant line)}.
  • Corollaries: zero derivative → constant; positive derivative → strictly increasing; bounded derivative → Lipschitz.
  • The theorem links global change f(b)f(a)f(b)-f(a) to a local quantity f(c)f'(c), making it the workhorse for proving properties of differentiable functions.