A Question That Sounds Too Simple to Need a Proof
If you drive 60 miles in one hour, at some instant your speedometer must have read exactly 60 miles per hour. Not "close to" 60 — exactly 60. It feels obvious: you can't average 60 without ever hitting 60, the way you can't average two grades of 70 and 90 without some grade in between landing at, say, 80 along the way. But calculus asks a sharper question: why must this be true, for any differentiable function whatsoever, not just for cars? The answer is the Mean Value Theorem, and Spivak calls it, without exaggeration, "probably the deepest result about derivatives." Getting there takes one clever warm-up theorem first.
Rolle's Theorem: The Easy Case
Suppose a function f is continuous on [a, b] and differentiable on (a, b), and suppose it starts and ends at the same height: f(a) = f(b). Rolle's Theorem says there must be some point x strictly between a and b where f′(x) = 0 — the graph must level off somewhere in between.
The proof leans on a theorem about maxima that's genuinely easy: if f has a maximum or minimum at an interior point x and is differentiable there, then f′(x) = 0. (Picture it: secant lines from the left have slope ≥ 0, secant lines from the right have slope ≤ 0, and the only way both limits can agree is if the derivative is exactly 0.) Since f is continuous on a closed interval, it's guaranteed to attain a maximum and a minimum somewhere on [a, b]. Spivak's proof then splits into cases: if either the max or the min happens at an interior point, that point is your x, and the derivative vanishes there by the theorem above. The only remaining possibility is that both the max and the min sit at the endpoints — but since f(a) = f(b), that would force the maximum value to equal the minimum value, which means f is constant on the whole interval. And for a constant function, the derivative is 0 everywhere, so literally any interior point works.
Every case is covered, and none of them required a hard argument. That's the whole proof.
From Rolle's Theorem to the Mean Value Theorem
Rolle's Theorem is really just the Mean Value Theorem in the special case where the two endpoints happen to have equal height. The trick for the general case is to manufacture that special case out of thin air. Given any continuous, differentiable f on [a, b], draw the straight line L connecting (a, f(a)) to (b, f(b)), and define a new function h measuring the vertical gap between f and that line:
h(x) = f(x) − [ (f(b) − f(a))/(b − a) ] (x − a) − f(a)
This h inherits continuity and differentiability from f, and a quick check shows h(a) = 0 and h(b) = 0 — exactly the equal-endpoints setup Rolle's Theorem needs. So there's some x in (a, b) with h′(x) = 0. Differentiating h and setting the result to zero at that x leaves you with:
f′(x) = [ f(b) − f(a) ] / (b − a)
That's the Mean Value Theorem: the instantaneous rate of change at some point in the interval exactly equals the average rate of change over the whole interval. Geometrically, the tangent line at that mystery point x is parallel to the secant line joining the two endpoints. Back to the driving example: if f(t) is your position at time t, then [f(b) − f(a)] / (b − a) is your average speed over the trip, and f′(x) is your instantaneous speed at some moment x. The theorem guarantees they coincide at least once. Your speedometer has no choice in the matter.
The Payoff: Why Zero Velocity Everywhere Means You Never Moved
Here's where the theorem stops being a curiosity and starts doing real work. Consider a question Spivak poses as deceptively hard: if f′(x) = 0 for every x, must f be a constant function? Physically this seems unavoidable — if a particle's velocity is always zero, surely the particle never moves. But "obvious" physical intuition and "proved" are different things, and the definition of the derivative as a limit of difference quotients gives no obvious foothold for concluding anything about the function's actual values.
The Mean Value Theorem supplies the missing bridge, in one line. Take any two points a and b in the domain with a ≠ b. By the theorem, there's some x between them with:
f′(x) = [ f(b) − f(a) ] / (b − a)
But f′(x) = 0 everywhere, by hypothesis. So the right-hand side must equal 0 as well, which forces f(b) = f(a). Since a and b were arbitrary, every two values of f agree — that is, f is constant. What looked like an intractable question about limits collapses into a two-line argument once the Mean Value Theorem is in hand.
The same idea, run with an inequality instead of an equality, proves something you've probably used without ever seeing it justified: if f′(x) > 0 for every x in an interval, then f is increasing on that interval. Take a < b in the interval; the theorem gives some x with f′(x) = [f(b) − f(a)]/(b − a), and since the left side is positive and b − a > 0, it follows that f(b) − f(a) > 0, i.e. f(b) > f(a). This single fact is the engine behind almost every "sign of the derivative tells you where the function rises and falls" argument you'll ever run in a calculus course — and it's not an assumption. It's a direct consequence of the Mean Value Theorem.
A Theorem That Earns Its Name
It's worth pausing on how unusual this is. Most of the derivative rules you learn early on — the power rule, the product rule — tell you how to compute f′ once you know f. The Mean Value Theorem runs in the opposite direction: it extracts hard information about f itself from nothing but information about f′. That reversal is exactly why Spivak calls it the deepest result about derivatives in the whole book, and why graph-sketching, optimization, and the entire theory of increasing and decreasing functions all quietly depend on it.
Spivak's Calculus, 4th Edition builds to this theorem across all of Chapter 11, "Significance of the Derivative," working from the maximum-point lemma, through Rolle's Theorem, to the Mean Value Theorem and its corollaries — each step small enough to verify by hand, the whole chain strong enough to justify everything you already believed about derivatives on intuition alone.
Further reading: How to self-study Spivak's Calculus · Table of contents · Index · Calculus, 4th Edition · Answer Book
Next time you're on a long drive, can you convince yourself — using nothing but the definition of average speed — that some instant matched it exactly?
