Lesson 11: The Mean Value TheoremAP

The Mean Value Theorem (CED topic 5.1) is the rigorous bridge between average and instantaneous rates of change: it says that on any reasonable trip, your instantaneous speed at some moment exactly equalled your average speed for the whole trip. It looks obvious; it is actually one of the deepest results in the course, because it is the theorem that makes derivative information *prove* things about functions — including, eventually, the Fundamental Theorem of Calculus.

The statements

Rolle's Theorem. If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there is at least one c in (a, b) with f'(c) = 0.

Mean Value Theorem. If f is continuous on [a, b] and differentiable on (a, b), then there is at least one c in (a, b) with

f'(c) = (f(b) − f(a))/(b − a).

Geometrically: some tangent line is parallel to the secant line through the endpoints. Rolle's theorem is the special case where the secant is horizontal. Note the hypotheses carefully — continuity on the closed interval, differentiability on the open interval — and note the conclusion is *existence*: the theorem guarantees a c but does not find it, does not promise uniqueness, and says nothing if the hypotheses fail.

Worked Example 1: finding c exactly

Problem. Let f(x) = x^2 − 4x + 5 on [1, 4]. Verify the MVT hypotheses and find all values of c the theorem guarantees.

f is a polynomial: continuous and differentiable everywhere. ✓
Average rate: (f(4) − f(1))/(4 − 1) = ((16 − 16 + 5) − (1 − 4 + 5))/3 = (5 − 2)/3 = 1
f'(x) = 2x − 4;  set 2c − 4 = 1 → c = 2.5, which lies in (1, 4). ✓

Answer: c = 5/2. For quadratics the guaranteed point is always the interval's midpoint (the vertex axis symmetry) — a useful check on your arithmetic.

Worked Example 2: the MVT from a table (numerical)

Problem. A differentiable function f has table values f(0) = 2, f(2) = 5, f(5) = 5, f(8) = 11. What conclusions can the MVT draw?

On [2, 5]: average rate = (5 − 5)/3 = 0 → some c in (2, 5) has f'(c) = 0. (Rolle's case.)
On [0, 8]: average rate = (11 − 2)/8 = 9/8 → some c in (0, 8) has f'(c) = 9/8.
On [5, 8]: average rate = (11 − 5)/3 = 2 → some c in (5, 8) has f'(c) = 2.

Solution. Three separate subintervals give three separate guarantees. This is the standard exam format: a bare table, no formula, and the only available tool is the MVT applied interval by interval. The word "differentiable" in the problem statement is doing heavy lifting — it certifies both hypotheses at once.

Worked Example 3: when the theorem refuses to apply

Problem. Let f(x) = |x| on [−1, 2]. The average rate of change is (f(2) − f(−1))/(2 − (−1)) = (2 − 1)/3 = 1/3. But f'(x) is −1 for x < 0 and 1 for x > 0 — never 1/3. Why does this not contradict the MVT?

Solution. The MVT's hypotheses fail: f is continuous on [−1, 2] but not differentiable at x = 0, which lies inside the interval. With a hypothesis violated, the conclusion is simply not promised — and here it genuinely fails. This example is the permanent antidote to the commonest MVT error: computing a secant slope and asserting a matching tangent point without ever checking differentiability. Piecewise functions with corners, and functions with vertical tangents or cusps inside the interval, are the exam's standard traps.

Real-world application: the speed-camera argument

Two toll gantries on a motorway are 100 km apart. A car's transponder is logged passing the first at 14:00 and the second at 14:45 — 45 minutes for 100 km, an average speed of 133 km/h on a road limited to 120. The driver protests: "I never exceeded 120." The MVT settles it: position is continuous and (barring teleportation) differentiable, so at some instant the car's *instantaneous* speed equalled its average, 133 km/h. Average-speed enforcement systems (used on real motorways worldwide) are the Mean Value Theorem deployed as traffic law. The same argument structure recurs in physics (some instant had the average acceleration), economics, and medicine.

Common mistakes

• Applying the MVT to a function with a corner, cusp, or break inside the interval. Verify hypotheses *in writing* before concluding; rubrics award the point for the verification.
• Confusing the IVT with the MVT. The IVT (values, needs continuity) promises a *height* is attained; the MVT (slopes, needs continuity + differentiability) promises a *derivative value* is attained. Exam questions deliberately place them side by side.
• Claiming c is unique or that the theorem finds it. "There exists c in (a, b)" is the whole conclusion.
• Computing the average rate over the wrong interval — reading table rows carelessly. The theorem applies to the interval you actually averaged over, and the c lives in *that* interval.

AP Exam Experience

1. (No calculator.) For f(x) = x^3 on [0, 2], the value of c guaranteed by the Mean Value Theorem is

• (A) 1
• (B) 2/sqrt(3)
• (C) sqrt(2)
• (D) 4/3

Answer: (B). Average rate: (8 − 0)/2 = 4. f'(c) = 3c^2 = 4 gives c^2 = 4/3, c = 2/sqrt(3) ≈ 1.155, which lies in (0, 2) as required. (The negative root is outside the interval and is discarded.)

2. To which function and interval can the Mean Value Theorem not be applied?

• (A) f(x) = x^2 on [−1, 3]
• (B) f(x) = sin x on [0, π]
• (C) f(x) = 1/x on [−1, 1]
• (D) f(x) = e^x on [0, 1]

Answer: (C). 1/x is not continuous on [−1, 1] (it is undefined and unbounded at 0), so the first hypothesis fails. The others are continuous and differentiable on their closed intervals.

3. A function f is continuous on [0, 3], differentiable on (0, 3), and f(0) = f(3) = 5. Which statement must be true?

• (A) f has a relative maximum in (0, 3).
• (B) There is a c in (0, 3) with f'(c) = 0.
• (C) f is constant on [0, 3].
• (D) f'(x) = 0 for all x in (0, 3).

Answer: (B). This is Rolle's theorem verbatim. (A) does not follow (the guaranteed point could be a minimum or a flat spot); (C) and (D) wildly overclaim — one zero of f' is all that is promised.

Free-response question. A differentiable function f has values given in the table: f(0) = 10, f(3) = 4, f(6) = 4, f(10) = 20.

(a) Explain why there must be a value c in (0, 10) with f'(c) = 1. (b) Explain why there must be a value c in (3, 6) with f'(c) = 0, and name the theorem that applies. (c) A student claims there must be a value c in (0, 3) with f'(c) = −2. Is the claim correct? Justify.

Scoring guideline.

• (a) 2 points. 1 point for the average rate (f(10) − f(0))/10 = (20 − 10)/10 = 1. 1 point for the conclusion with hypotheses: f is differentiable on [0, 10] (given), hence continuous, so by the Mean Value Theorem some c in (0, 10) has f'(c) = 1.
• (b) 2 points. 1 point for noting f(3) = f(6) = 4, so the average rate on [3, 6] is 0. 1 point for the conclusion f'(c) = 0 for some c in (3, 6), naming Rolle's Theorem (or the MVT with equal endpoint values — either name earns the point).
• (c) 1 point. Correct: the average rate on [0, 3] is (4 − 10)/3 = −2, and the MVT (f differentiable) guarantees c in (0, 3) with f'(c) = −2. A "yes" without the computed average rate and the theorem earns no point.

Additional worked practice

Problem. Prove that the equation x^3 + x − 1 = 0 has exactly one real solution.

Existence (IVT): f(x) = x^3 + x − 1 is continuous everywhere; f(0) = −1 < 0 and f(1) = 1 > 0,
  so a root exists in (0, 1).
Uniqueness (MVT): f'(x) = 3x^2 + 1 ≥ 1 > 0 for all x.
  If two roots a < b existed, Rolle's theorem would give c in (a, b) with f'(c) = 0 —
  impossible, since f' is never zero. So there is at most one root.
Conclusion: exactly one real root (it is ≈ 0.682).

Answer: exactly one root. This two-theorem combination — IVT for *at least one*, Rolle/MVT (via a never-zero derivative) for *at most one* — is the standard template for "prove there is exactly one solution" problems, which appear in some form on most AP exams. Notice that neither theorem alone suffices: existence without uniqueness, or uniqueness without existence, leaves the claim unproved.

Extension: what the MVT is *for* — three theorems it proves

The MVT looks like a curiosity until you see it work as the engine of other results.

1. Positive derivative means increasing — proved. Suppose f'(x) > 0 on an interval and take any a < b in it. By the MVT, f(b) − f(a) = f'(c)(b − a) for some c; both factors on the right are positive, so f(b) > f(a). That is the *proof* of the Unit 5 dictionary entry "f' > 0 ⇒ f increasing" — the sign tests of Lesson 12 stand on the MVT.

2. A Lipschitz bound for sine. For any a, b: by the MVT, sin b − sin a = (cos c)(b − a) for some c, and |cos c| ≤ 1, so |sin b − sin a| ≤ |b − a| — sine can never change faster than its input. Error bounds of this shape run all through numerical analysis.

3. The two-runners argument. Two runners start a race together and finish in a dead heat. Consider d(t) = (position of A) − (position of B): d(start) = d(finish) = 0, so by Rolle's theorem there is an instant when d'(t) = 0 — their *velocities were exactly equal* at some moment mid-race, however different their strategies. Any "two processes that start and end equal must somewhere change at the same rate" claim is Rolle's theorem in a costume.