Lesson 20: Mean Value Theorem & Rolle's Theorem

(a) Opening Question

You drive onto a toll road. Your entry ticket is stamped 2:00 p.m. at mile marker 0; your exit ticket is stamped 3:00 p.m. at mile marker 75. The speed limit is 65 mph the whole way.

1. What was your average velocity over that hour?
2. A trooper argues: "I can prove you broke the speed limit at some instant, without ever clocking you." Is the trooper right? On what assumption about your motion does the argument rest?
3. Suppose instead you left and returned to mile marker 0 (a round trip). Must there have been an instant when your velocity was exactly zero? Why?

Spend two minutes before reading on. Question 1 is a slope: (75 − 0)/(3 − 2) = 75 mph. The trooper's claim — that your instantaneous speed must have equaled your average speed at some moment — is exactly the Mean Value Theorem, the centerpiece of this lesson. The round-trip question is Rolle's Theorem, the MVT's most famous special case. Both rest on one assumption the trooper needs: that your position function is continuous and differentiable (you can't teleport, and your velocity never glitches to undefined).

(Answers: 75 mph; yes, by the MVT your instantaneous velocity equals 75 mph > 65 at some instant, assuming position is continuous and differentiable; yes, by Rolle's Theorem, since you start and end at the same place, v(c) = 0 for some interior time c.)

(b) Core Concepts

The Mean Value Theorem

The Mean Value Theorem (MVT) connects the average rate of change of a function over an interval to its instantaneous rate of change somewhere inside that interval. It is the theoretical engine behind almost every analytical result in Unit 5.

Mean Value Theorem. If f is

(1) continuous on the closed interval [a, b], AND

(2) differentiable on the open interval (a, b),

then there exists at least one number c in the open interval (a, b) such that

`

f'(c) = [f(b) − f(a)] / (b − a)

`

Read the conclusion carefully. The right-hand side, [f(b) − f(a)] / (b − a), is the slope of the secant line through the endpoints (a, f(a)) and (b, f(b)) — the average rate of change. The left-hand side, f'(c), is the slope of the tangent line at x = c — an instantaneous rate. So the MVT says:

Geometric meaning. Somewhere strictly between a and b, there is a point where the tangent line is parallel to the secant line through the endpoints.

• the secant line drawn through the endpoints (a, f(a)) and (b, f(b))
• a tangent line drawn at an interior point (c, f(c)) that is parallel to the secant
• vertical dashed lines at x = a, x = c, x = b
• label: secant slope = [f(b) − f(a)]/(b − a); tangent slope at c = f'(c); the two lines are parallel]

Both hypotheses are essential. Notice the asymmetry that the AP exam loves to test: continuity is required on the closed interval [a, b] (endpoints included), but differentiability is required only on the open interval (a, b) (endpoints excluded). This is deliberate — we need the function to connect to its endpoints (continuity there) but we do not need a derivative at the endpoints, since the guaranteed point c lives strictly inside. If either hypothesis fails, the conclusion can fail. For f(x) = |x| on [−1, 1], the secant slope is 0, but f'(x) is never 0 (it is −1 or +1 and undefined at 0); there is no c, because f is not differentiable at x = 0.

Finding c: a worked example with full hypothesis-checking

Worked example. Find all values of c guaranteed by the MVT for f(x) = x² on [1, 5].

Step 1 — Verify the hypotheses (never skip this). f(x) = x² is a polynomial, so it is continuous on the closed interval [1, 5] and differentiable on the open interval (1, 5). The MVT applies.

Step 2 — Compute the average rate of change (secant slope).

`

[f(5) − f(1)] / (5 − 1) = (25 − 1) / 4 = 24 / 4 = 6

`

Step 3 — Set f'(c) equal to that slope and solve for c in (1, 5). Here f'(x) = 2x, so

`

2c = 6 ⇒ c = 3

`

Since c = 3 lies in the open interval (1, 5), the value guaranteed by the MVT is c = 3. (At x = 3 the tangent slope is 6, matching the secant slope — exactly as the theorem promised.)

Rolle's Theorem — the special case f(a) = f(b)

Rolle's Theorem is just the MVT applied when the two endpoints have the same height.

Rolle's Theorem. If f is

(1) continuous on [a, b], (2) differentiable on (a, b), AND (3) f(a) = f(b),

then there exists at least one c in (a, b) such that f'(c) = 0.

Why it follows: if f(a) = f(b), the secant slope is [f(b) − f(a)]/(b − a) = 0, so the MVT's guaranteed c satisfies f'(c) = 0 — a horizontal tangent. Geometrically, a differentiable function that returns to the same height must level off somewhere in between (a peak or a valley). This is the round trip from the opening question: same start and end position ⇒ velocity zero at some interior instant.

The Extreme Value Theorem

The MVT's proof leans on another existence theorem you must be able to cite by name.

Extreme Value Theorem (EVT). If f is continuous on a closed interval [a, b], then f attains an absolute maximum value and an absolute minimum value on [a, b] — that is, there exist numbers d and e in [a, b] with f(d) ≤ f(x) ≤ f(e) for all x in [a, b].

Two conditions carry the weight: the interval must be closed and the function must be continuous. Drop either and the guarantee dies. On the open interval (0, 1), f(x) = x attains neither a max nor a min (it gets arbitrarily close to 0 and 1 but never reaches them). And f(x) = 1/x on [1, ∞) (not a closed bounded interval) has no minimum. The EVT is why, when we hunt for absolute extrema on a closed interval later, we know a max and min genuinely exist — we just have to locate them among the critical points and endpoints.

Consequences of the MVT

The MVT is not just a curiosity about tangent lines; it is the logical bridge from the sign of f' to the behavior of f. These consequences justify everything in Lessons 21–24.

Consequence 1 (constant functions). If f'(x) = 0 for every x in an interval, then f is constant on that interval.

Why: take any two points a < b in the interval. By the MVT there is a c between them with f'(c) = [f(b) − f(a)]/(b − a). But f'(c) = 0, so f(b) − f(a) = 0, i.e. f(b) = f(a). Every pair of outputs is equal, so f is constant. (This is also the reason two antiderivatives of the same function differ only by a constant — the foundation of the "+ C" you will meet in Unit 6.)

Consequence 2 (increasing/decreasing). On an interval, if f'(x) > 0 for all x, then f is increasing; if f'(x) < 0 for all x, then f is decreasing.

Why: for any a < b, the MVT gives f(b) − f(a) = f'(c)(b − a). Since b − a > 0, the sign of f(b) − f(a) matches the sign of f'(c). If f' > 0 throughout, then f(b) > f(a) whenever b > a — that is the definition of increasing. This is precisely the First Derivative Test you will formalize in Lesson 21.

MVT vs. IVT — do not confuse them

Both are existence theorems that require continuity, and both are favorite distractor material on the AP exam.

One-line memory hook: IVT is about heights; MVT is about slopes. If a problem asks you to guarantee a particular output value, that is IVT. If it asks you to guarantee a particular rate, velocity, or slope, that is MVT.

(c) Worked Examples

Example 1 — Verify hypotheses and find c (foundation) · [NO CALC]

Problem. Let f(x) = x³ on the closed interval [−1, 2]. Verify that the MVT applies and find all values of c it guarantees.

Strategy. Check continuity and differentiability, compute the secant slope, solve f'(c) = secant slope, keep only the c in (−1, 2).

Solution.

Hypotheses: f(x) = x³ is a polynomial, hence continuous on [−1, 2] and differentiable on (−1, 2). The MVT applies.

Secant slope:

[f(2) − f(−1)] / (2 − (−1)) = (8 − (−1)) / 3 = 9 / 3 = 3

Solve: f'(x) = 3x², so

3c² = 3  ⇒  c² = 1  ⇒  c = ±1

Only c = 1 lies in the open interval (−1, 2); c = −1 is the excluded endpoint. The MVT guarantees c = 1.

Justification (model language). "By the Mean Value Theorem, since f is continuous on [−1, 2] and differentiable on (−1, 2), there exists c in (−1, 2) with f'(c) = [f(2) − f(−1)]/(2 − (−1)) = 3. Solving 3c² = 3 gives c = 1 (rejecting c = −1, which is not in the open interval)."

Example 2 — Rolle's Theorem application (procedural) · [NO CALC]

Problem. Let f(x) = x² − 4x + 1 on [1, 3]. Show that f satisfies the hypotheses of Rolle's Theorem and find the c it guarantees.

Strategy. Confirm continuity, differentiability, and the equal-endpoints condition f(a) = f(b); then solve f'(c) = 0.

Solution.

Hypotheses: f is a polynomial, so continuous on [1, 3] and differentiable on (1, 3). Check the third condition:

f(1) = 1 − 4 + 1 = −2
f(3) = 9 − 12 + 1 = −2

Since f(1) = f(3) = −2, all three hypotheses of Rolle's Theorem hold.

Conclusion: f'(x) = 2x − 4, so

2c − 4 = 0  ⇒  c = 2

c = 2 lies in (1, 3). By Rolle's Theorem there is a c = 2 in (1, 3) with f'(2) = 0 — the vertex of the parabola, where the tangent is horizontal.

Example 3 — EVT and closed-interval reasoning (AP level) · [NO CALC]

Problem. A function g is continuous on [0, 4] with g(0) = 3 and g(4) = 3. A student claims, "Because g(0) = g(4), the function must attain an absolute maximum on [0, 4]." (i) Is the conclusion (a max exists) correct? Which theorem actually justifies it? (ii) Does g necessarily have an absolute max at an interior point?

Strategy. Separate the true conclusion from the wrong reasoning; cite the correct theorem; test the interior claim with the definition.

Solution.

(i) The conclusion is correct, but the cited reason is wrong. A maximum exists not because the endpoints are equal but because g is continuous on the closed interval [0, 4] — that is the Extreme Value Theorem. The EVT guarantees both an absolute maximum and an absolute minimum.

(ii) No. The EVT guarantees a max somewhere on [0, 4], but it could occur at an endpoint. For example, if g is the constant function g(x) = 3, every point — including the endpoints — is an absolute max. Or if g dips below 3 in the middle and rises back, the maximum value 3 is attained at the endpoints. The EVT does not promise an interior extremum.

Justification (model language). "By the Extreme Value Theorem, since g is continuous on the closed interval [0, 4], g attains an absolute maximum value on [0, 4]. The theorem does not specify where; the maximum may occur at an endpoint."

Example 4 — Forensic "average velocity ⇒ some instant" application (AP level) · [CALC]

Problem. A car's position (in miles) along a straight road is s(t), a differentiable function of time t (in hours). The car is at mile s(0) = 0 and at mile s(2.5) = 165. The posted speed limit is 65 mph.

(i) Use a theorem to prove the car exceeded the speed limit at some instant in (0, 2.5). Name the theorem and verify its hypotheses. (ii) If additionally s(t) = 30t + 12t² − 1.6t³, find a specific time c in (0, 2.5) where the instantaneous velocity equals the average velocity.

Strategy. This is a textbook MVT "speeding ticket." Establish hypotheses, compute the average velocity, invoke the MVT, then for (ii) solve s'(c) = average numerically.

Solution.

(i) Hypotheses: s is given as differentiable on its domain, so it is differentiable on (0, 2.5) and (since differentiable ⇒ continuous) continuous on [0, 2.5]. The MVT applies.

Average velocity:

[s(2.5) − s(0)] / (2.5 − 0) = (165 − 0) / 2.5 = 66 mph

By the MVT there exists c in (0, 2.5) with s'(c) = 66. Since 66 > 65, the car's instantaneous velocity was 66 mph at some time c, exceeding the speed limit.

(ii) s'(t) = 30 + 24t − 4.8t². Set s'(c) = 66:

30 + 24c − 4.8c² = 66  ⇒  −4.8c² + 24c − 36 = 0  ⇒  4.8c² − 24c + 36 = 0  ⇒  c² − 5c + 7.5 = 0

By the quadratic formula, c = [5 ± √(25 − 30)] / 2. The discriminant is 25 − 30 = −5 < 0, so there is no real solution — meaning this particular s(t) does not actually reach s(2.5) = 165. (Quick check: with this formula s(2.5) = 75 + 75 − 25 = 125 ≠ 165, so the formula and the endpoint datum are inconsistent.)

Replacing the endpoint datum with the value this formula actually produces, s(2.5) = 125, the average velocity is 125/2.5 = 50 mph, and we solve s'(c) = 50:

30 + 24c − 4.8c² = 50  ⇒  −4.8c² + 24c − 20 = 0  ⇒  4.8c² − 24c + 20 = 0  ⇒  c² − 5c + 25/6 = 0
c = [5 ± √(25 − 50/3)] / 2 = [5 ± √(25/3)] / 2 ≈ [5 ± 2.887] / 2

giving c ≈ 1.057 or c ≈ 3.943. Only c ≈ 1.057 lies in (0, 2.5). (Calculator: graph Y1 = nDeriv(30X+12X²−1.6X³, X, X) and Y2 = 50, then 2nd → CALC → 5:intersect to confirm c ≈ 1.057.) The instantaneous velocity equals the average velocity of 50 mph at c ≈ 1.057 hours.

Teaching note: part (i) is the genuine AP skill — the existence argument holds for any differentiable s with those endpoint values. Part (ii) is a deliberate consistency check that rewards verifying data before trusting it.

(d) Common Mistakes

Applying the MVT without checking continuity and differentiability.

What students do: Jump straight to computing [f(b) − f(a)]/(b − a) and solving for c, never mentioning the hypotheses.

Why it's wrong: If f has a corner, jump, vertical asymptote, or other failure inside [a, b], the guaranteed c may not exist — and the AP rubric awards a separate point for verifying the hypotheses.

Fix: Always write one sentence first: "f is continuous on [a, b] and differentiable on (a, b) because …" Polynomials, sin/cos, and eˣ qualify everywhere; rational functions, tan, √, and |x| need a check.

Confusing the MVT with the IVT.

What students do: Use the MVT to conclude a function value exists (e.g., "by the MVT f(c) = 7"), or use the IVT to conclude something about a slope.

Why it's wrong: The MVT's conclusion is about f'(c) (a derivative/slope); the IVT's conclusion is about f(c) (a value). They answer different questions.

Fix: Ask "Am I guaranteeing a height or a slope/rate?" Height ⇒ IVT. Slope, velocity, or average-rate match ⇒ MVT.

Using the closed interval where the open one belongs (and vice versa).

What students do: Accept an endpoint as a valid c, or demand differentiability at the endpoints.

Why it's wrong: The guaranteed c must lie in the open interval (a, b); continuity (not differentiability) is what is required on the closed [a, b]. A c equal to a or b is invalid.

Fix: Solve for c, then discard any solution equal to a or b. Require continuity on [a, b], differentiability only on (a, b).

Forgetting Rolle's third condition, f(a) = f(b).

What students do: Invoke Rolle's Theorem to claim f'(c) = 0 without checking that the endpoints are equal.

Why it's wrong: Rolle's f'(c) = 0 conclusion needs f(a) = f(b). Without it you get the general MVT (f'(c) = secant slope), which need not be zero.

Fix: Before citing Rolle's, compute f(a) and f(b) and confirm they are equal. If they are not, use the full MVT instead.

Treating the EVT as if it located the extremum.

What students do: Say "by the EVT the maximum is at the critical point" or assume the extremum is interior.

Why it's wrong: The EVT only guarantees that a max and min exist; it never says where. The extremum can sit at an endpoint.

Fix: Use the EVT to assert existence, then find the location separately by comparing critical-point values and endpoint values (the Closed Interval / Candidates Test, Lesson 21).

(e) Practice Problems

For multiple choice, choose the single best answer. Calculator labels match AP rules.

(Justification)Let f(x) = x² − 2x on [0, 4]. Verify that f satisfies the hypotheses of the MVT on [0, 4], then find all values of c guaranteed by the theorem. Write your verification in complete, AP-style sentences.

(Justification)A function f is differentiable everywhere with f(1) = 4 and f(6) = 4. Use an appropriate theorem to justify that f'(c) = 0 for some c in (1, 6). Name the theorem and verify its hypotheses.

(Short answer)At noon a hot-air balloon is at altitude s(0) = 200 ft; at t = 8 minutes it is at s(8) = 200 ft, and s is differentiable for 0 ≤ t ≤ 8. Use a theorem to justify that there is a time in (0, 8) at which the balloon's vertical velocity is exactly 0. Name the theorem.

(Short answer)Explain in one or two sentences why the MVT and the IVT are different theorems, referencing what each one's conclusion is about.

(f) AP Exam Focus

FRQ — Mean Value Theorem from a table (Section II style). Calculator NOT permitted. Total: 9 points.

The differentiable function f gives the temperature, in degrees Celsius, of a metal rod at a distance x centimeters from one end, for 0 ≤ x ≤ 10. Selected values are given in the table. Assume f is differentiable on [0, 10].

| x (cm) | 0 | 3 | 6 | 10 |

|---|---|---|---|---|

| f(x) (°C) | 40 | 52 | 58 | 80 |

(a) Using the data in the table, approximate f'(4.5). Show the computation and give units. (2 points)

(b) Explain why there must be a value c in the open interval (0, 10) such that f'(c) = 4. (3 points)

(c) Must there be a value x in (0, 10) at which f(x) = 70? Justify your answer, and name the theorem you use. (2 points)

(d) Find the average rate of change of f over [3, 6], and interpret its meaning in the context of the problem, using correct units. (2 points)

Model Solution

(a) Approximate the derivative by the average rate of change over the interval [3, 6] containing x = 4.5:

f'(4.5) ≈ [f(6) − f(3)] / (6 − 3) = (58 − 52) / 3 = 6/3 = 2 °C/cm

f'(4.5) ≈ 2 degrees Celsius per centimeter.

(b) f is differentiable on [0, 10], so f is differentiable on the open interval (0, 10) and (since differentiability implies continuity) continuous on the closed interval [0, 10]. The hypotheses of the Mean Value Theorem are satisfied. The average rate of change over [0, 10] is

[f(10) − f(0)] / (10 − 0) = (80 − 40) / 10 = 40/10 = 4

By the Mean Value Theorem, since f is continuous on [0, 10] and differentiable on (0, 10), there exists a c in (0, 10) such that f'(c) = [f(10) − f(0)]/(10 − 0) = 4.

(c) f is differentiable, hence continuous, on the closed interval [0, 10]. We have f(0) = 40 and f(10) = 80, and 70 is a value between 40 and 80. By the Intermediate Value Theorem, since f is continuous on [0, 10] and 40 < 70 < 80, there exists a value x in (0, 10) with f(x) = 70. So yes, such an x must exist.

(d) Average rate of change over [3, 6]:

[f(6) − f(3)] / (6 − 3) = (58 − 52) / 3 = 6/3 = 2 °C/cm

Interpretation: "Over the segment from x = 3 cm to x = 6 cm, the temperature of the rod increases at an average rate of 2 degrees Celsius per centimeter." (Units: °C per cm.)

Scoring Commentary — where students lose points

• Part (a) [2 pts]: 1 pt for the correct difference quotient (58 − 52)/(6 − 3), 1 pt for the value 2 with units (°C/cm). Choosing an interval that does not contain 4.5 (e.g., [0, 3]) loses the first point. Omitting units is the most common single-point loss on the whole problem.
• Part (b) [3 pts]: 1 pt for verifying the hypotheses (continuous on [0, 10], differentiable on (0, 10)); 1 pt for computing the average rate = 4; 1 pt for the MVT conclusion stated with "there exists c in (0, 10)." The classic lost point: students compute 4 and assert the conclusion but never state the hypotheses — the rubric explicitly requires naming continuity on the closed interval and differentiability on the open interval. A bare "by the MVT, f'(c) = 4" with no hypothesis check earns at most 2 of 3.
• Part (c) [2 pts]: 1 pt for naming the Intermediate Value Theorem and citing continuity on [0, 10]; 1 pt for noting 40 < 70 < 80 and concluding such an x exists. The signature trap: using the MVT here. The MVT is about the derivative; this question asks about a function value, which is the IVT's job. Citing the wrong theorem forfeits both points even if the conclusion ("yes") is correct.
• Part (d) [2 pts]: 1 pt for the value 2; 1 pt for an interpretation that includes units (°C per cm) and the meaning (average rate of temperature change with respect to position). Writing "the temperature is 2" (confusing the rate with a value) or dropping units forfeits the interpretation point.

Justification language that earns full credit: "Because f is continuous on [0, 10] and differentiable on (0, 10), by the Mean Value Theorem there exists c in (0, 10) such that f'(c) = 4." Vague phrasing like "by MVT the slope is 4 somewhere" earns no hypothesis credit.