MISSION · Calculus
The Mean Value Theorem
Somewhere along the way, your speedometer read exactly your average speed
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The idea in one sentence
If you drove 120 miles in two hours, then at some instant your speedometer read exactly 60 mph — not approximately, not on average, exactly.
What it actually says
If \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there is at least one \(c\) strictly between \(a\) and \(b\) with
\[f'(c) = \frac{f(b) - f(a)}{b - a}\]The right-hand side is the average rate of change over the whole interval — the gradient of the straight line joining the endpoints, the secant. The left-hand side is the instantaneous rate at one particular moment, the gradient of the tangent there.
So the theorem says: somewhere inside, the tangent runs parallel to the secant.
Finding the c
The question usually comes as a calculation. Take \(f(x) = x^2\) on \([0, 4]\).
Average rate: \(\dfrac{f(4) - f(0)}{4 - 0} = \dfrac{16 - 0}{4} = 4\).
Set the derivative equal to it: \(f'(x) = 2x = 4\), so \(c = 2\).
And \(2\) does lie strictly inside \((0, 4)\), as promised. Three steps every time — average rate, set \(f'\) equal to it, solve and check the answer is in the open interval.
Both conditions do real work
Neither hypothesis is decoration; drop either and the conclusion fails.
Continuity on the closed interval. A function that jumps can get from \(f(a)\) to \(f(b)\) without any gradient matching the average, because part of the change happened at the jump where no gradient exists.
Differentiability on the open interval. \(f(x) = |x|\) on \([-1, 1]\) has an average rate of 0, but its derivative is \(-1\) or \(+1\) everywhere it exists, and never 0. The corner at \(x = 0\) is exactly where the guarantee would have had to live.
Note the asymmetry: continuity is required at the endpoints, differentiability is not. \(f(x) = \sqrt{1 - x^2}\) on \([-1, 1]\) has vertical tangents at both ends and the theorem still applies.
Rolle’s theorem is the special case
If additionally \(f(a) = f(b)\), the secant is horizontal, so the theorem hands you a \(c\) with \(f'(c) = 0\) — a stationary point somewhere between.
That is Rolle’s theorem, and it is not weaker than the MVT so much as the same statement after tilting the picture. Prove Rolle’s, subtract the secant line from \(f\) to flatten the endpoints, and the MVT falls out.
Why anyone cares
The theorem is almost never the answer to a question. It is the reason other answers are true.
- ”\(f' > 0\) means increasing” is proved with it. Take any \(x_1 < x_2\); the MVT gives \(f(x_2) - f(x_1) = f'(c)(x_2 - x_1)\), which is positive. Every sign chart you have ever drawn rests on this.
- ”\(f' = 0\) everywhere means constant”, and so two antiderivatives of the same function differ by a constant — the \(+C\) in every integral.
- Error bounds for Taylor polynomials and for Newton’s method come from it.
- Speed cameras that time you between two points and issue a ticket for the average are, legally speaking, relying on the MVT.
Common mix-ups
- Expecting one c. There can be several. The theorem promises at least one.
- Accepting a c at an endpoint. It must be strictly inside \((a, b)\).
- Confusing it with the Intermediate Value Theorem. The IVT is about \(f\) hitting every value in between; the MVT is about \(f'\) hitting the average gradient. Continuity alone gives the first; you need differentiability for the second.
- Averaging the endpoints instead. The average rate is a difference quotient, not \(\frac{f(a) + f(b)}{2}\).
Check yourself
For \(f(x) = x^2\) on \([0, 4]\), find the average rate of change and the \(c\) the theorem guarantees. Then say which extra condition turns the MVT into Rolle’s theorem.
The average rate is \(\frac{16 - 0}{4} = 4\). Solving \(2c = 4\) gives \(c = 2\), safely inside. Rolle’s is the case \(f(a) = f(b)\), where the secant is flat.
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