Real Analysis 41 | Mean Value Theorem [dark version]

Q: Why does the mean value theorem require differentiability on (a,b) and continuity on [a,b]?

The theorem’s conclusion compares endpoint values f(a), f(b) with an interior derivative f′(x̂). That comparison uses the secant slope (f(b)−f(a))/(b−a), which depends on f being well-defined at both endpoints. Differentiability on (a,b) is what makes the tangent slope f′(x̂) meaningful and allows the Rolle’s theorem reduction to work after constructing G(x) = f(x) − (secant line) − f(a).

Q: How does subtracting the secant line turn the mean value theorem into a Rolle’s theorem problem?

When f(a)≠f(b), the secant line has slope (f(b)−f(a))/(b−a). Defining G(x) = f(x) − [(f(b)−f(a))/(b−a)](x−a) − f(a) forces G(a)=0 and G(b)=0, so the new function matches at both ends. That endpoint equality is exactly the setup for Rolle’s theorem, which then guarantees an interior point where G′(x̂)=0.

Q: What does G′(x̂)=0 mean in terms of f′(x̂)?

Differentiating G gives G′(x) = f′(x) − (f(b)−f(a))/(b−a). Setting G′(x̂)=0 yields f′(x̂) = (f(b)−f(a))/(b−a). So the “zero derivative” condition for G becomes the “matching slopes” condition for f.

Q: Why is the mean value theorem an existence statement rather than a uniqueness statement?

The theorem guarantees at least one interior point x̂ with the required slope, but it doesn’t claim only one such point exists. For a constant function f, the secant slope is 0 and f′(x)=0 for every interior x, so infinitely many x̂ satisfy the condition.

TL;DR

If f is differentiable on (a,b) and defined on the compact interval [a,b], there exists x̂ in (a,b) with f′(x̂) equal to the secant slope (f(b)−f(a))/(b−a).

Briefing Cornell Notes

Briefing

The mean value theorem guarantees that a differentiable function on a closed interval has at least one point where its instantaneous slope matches the average slope across the interval. Concretely, if a function f is defined on a compact interval [a,b] with a<b and is differentiable on (a,b), then there exists some x̂ in (a,b) such that

f′(x̂) = (f(b) − f(a)) / (b − a).

Geometrically, the right-hand side is the slope of the secant line through (a,f(a)) and (b,f(b)). The theorem asserts that as the secant line “slides” into the graph, there is a tangent line somewhere in the interior whose slope equals that secant slope. Importantly, the theorem is an existence claim, not a uniqueness claim—multiple interior points x̂ can satisfy the condition. A constant function illustrates this: its secant slope is 0, and every interior point has derivative 0, so many x̂ work.

The proof strategy uses a previously established Rolle’s theorem as a reduction step. First, handle the special case where f(a)=f(b). Then the average slope becomes 0, and Rolle’s theorem directly produces an x̂ in (a,b) with f′(x̂)=0, which already matches the mean slope.

For the general case f(a)≠f(b), the argument “shifts” the problem into the Rolle’s theorem setup by subtracting the secant line from the function. Define a new function G by

G(x) = f(x) − [ (f(b) − f(a)) / (b − a) ]·(x − a) − f(a).

This construction forces G(a)=0 and G(b)=0, so the endpoints of G match—exactly the condition needed to apply Rolle’s theorem. Since G is built from differentiable pieces (a difference of differentiable functions), it remains differentiable on (a,b). Rolle’s theorem then yields an x̂ in (a,b) where G′(x̂)=0. Differentiating G shows

G′(x) = f′(x) − (f(b) − f(a)) / (b − a),

so G′(x̂)=0 rearranges to the mean value theorem identity f′(x̂) = (f(b) − f(a)) / (b − a).

A key payoff comes immediately: monotonicity criteria based on the sign of the derivative. If f′(x) is positive everywhere on an interval, then for any two points x1<x2, applying the mean value theorem on [x1,x2] gives a point x̂ where f′(x̂) equals the average rate of change. Because both f′(x̂)>0 and x2−x1>0, the average rate of change is positive, forcing f(x2)>f(x1). Since x1 and x2 were arbitrary, f is strictly monotonically increasing. The same logic yields strictly decreasing behavior when f′(x)<0 everywhere, and if the derivative is only nonnegative or nonpositive, the conclusion weakens to monotone (not necessarily strict) increase or decrease.

Cornell Notes

For a differentiable function f on a closed interval [a,b] (with a<b), the mean value theorem guarantees an interior point x̂ where the tangent slope equals the secant’s average slope: f′(x̂) = (f(b)−f(a))/(b−a). The proof reduces the general case to Rolle’s theorem by subtracting the secant line from f, creating a new function G with G(a)=G(b). Rolle’s theorem then supplies x̂ with G′(x̂)=0, which translates directly into the mean value theorem equation. A major consequence is monotonicity: if f′(x)>0 everywhere, then f is strictly increasing; if f′(x)<0 everywhere, f is strictly decreasing. If the derivative is only ≥0 or ≤0, the function is monotone rather than strictly monotone.

Why does the mean value theorem require differentiability on (a,b) and continuity on [a,b]?

The theorem’s conclusion compares endpoint values f(a), f(b) with an interior derivative f′(x̂). That comparison uses the secant slope (f(b)−f(a))/(b−a), which depends on f being well-defined at both endpoints. Differentiability on (a,b) is what makes the tangent slope f′(x̂) meaningful and allows the Rolle’s theorem reduction to work after constructing G(x) = f(x) − (secant line) − f(a).

How does subtracting the secant line turn the mean value theorem into a Rolle’s theorem problem?

When f(a)≠f(b), the secant line has slope (f(b)−f(a))/(b−a). Defining G(x) = f(x) − [(f(b)−f(a))/(b−a)](x−a) − f(a) forces G(a)=0 and G(b)=0, so the new function matches at both ends. That endpoint equality is exactly the setup for Rolle’s theorem, which then guarantees an interior point where G′(x̂)=0.

What does G′(x̂)=0 mean in terms of f′(x̂)?

Differentiating G gives G′(x) = f′(x) − (f(b)−f(a))/(b−a). Setting G′(x̂)=0 yields f′(x̂) = (f(b)−f(a))/(b−a). So the “zero derivative” condition for G becomes the “matching slopes” condition for f.

Why is the mean value theorem an existence statement rather than a uniqueness statement?

The theorem guarantees at least one interior point x̂ with the required slope, but it doesn’t claim only one such point exists. For a constant function f, the secant slope is 0 and f′(x)=0 for every interior x, so infinitely many x̂ satisfy the condition.

How does the mean value theorem prove that f′(x)>0 everywhere implies f is strictly increasing?

Pick any x1<x2 and apply the mean value theorem on [x1,x2]. It produces x̂ in (x1,x2) with f′(x̂) = (f(x2)−f(x1))/(x2−x1). If f′(x)>0 everywhere, then f′(x̂)>0, and since x2−x1>0, the fraction (f(x2)−f(x1))/(x2−x1) is positive. That forces f(x2)>f(x1). Because x1 and x2 were arbitrary, f is strictly monotonically increasing.

What changes when f′(x) is nonnegative instead of strictly positive?

The same mean value theorem step still gives f(x2)−f(x1) = f′(x̂)(x2−x1). If f′(x̂)≥0, then f(x2)−f(x1)≥0, so f is monotone increasing but may fail to be strict (flat segments can occur where f′=0).

Review Questions

State the mean value theorem precisely, including where x̂ must lie.
Outline the construction of G(x) used to reduce the general mean value theorem to Rolle’s theorem.
Explain how the sign of f′(x) on an interval determines whether f is strictly increasing, strictly decreasing, or merely monotone.

Key Points

1
If f is differentiable on (a,b) and defined on the compact interval [a,b], there exists x̂ in (a,b) with f′(x̂) equal to the secant slope (f(b)−f(a))/(b−a).
2
The theorem is about existence, not uniqueness; multiple interior points can share the same matching-slope property.
3
Rolle’s theorem handles the special case f(a)=f(b) immediately by producing an interior point where the derivative is 0.
4
For the general case, subtracting the secant line from f creates a new function G with G(a)=G(b), enabling Rolle’s theorem.
5
Differentiating the constructed G shows that G′(x̂)=0 is equivalent to the mean value theorem identity for f′(x̂).
6
If f′(x)>0 for all x in an interval, then f is strictly monotonically increasing; if f′(x)<0, then f is strictly monotonically decreasing.
7
If f′(x) is only nonnegative or nonpositive, the conclusion becomes monotone (not necessarily strict).

Highlights

The mean value theorem matches an interior tangent slope to the average secant slope: f′(x̂) = (f(b)−f(a))/(b−a).

The proof trick is to subtract the secant line, forcing endpoint equality for a new function G so Rolle’s theorem can be applied.

Monotonicity follows quickly: a everywhere-positive derivative forces f(x2)>f(x1) for any x2>x1, making f strictly increasing.

Constant functions show why the theorem guarantees existence but not uniqueness: every interior point works when the secant slope is 0.

Topics

Mean Value Theorem
Rolle’s Theorem
Differentiability
Secant and Tangent
Monotonicity