The mean value theorem, explained with examples
The mean value theorem links the average rate of change over an interval to the instantaneous rate of change at a point. It is a favourite on AP free-response questions, where the justification matters as much as the answer.
On this page
The statement
If f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one number c in (a, b) such that
f'(c) = (f(b) − f(a))/(b − a).
The right-hand side is the gradient of the secant line joining the endpoints, the average rate of change. The left-hand side is the gradient of the tangent at c, the instantaneous rate of change. The theorem says that somewhere between a and b, the tangent is parallel to the secant.
What it means in real life
Suppose you drive 120 miles in 2 hours. Your average speed is 60 miles per hour. The mean value theorem says that at some moment during the trip, your speedometer read exactly 60 mph. You might have gone faster and slower at different times, but you cannot average 60 without passing through 60 at some instant, provided your speed changed smoothly. This is the idea behind average speed cameras on motorways.
Why the conditions matter
Both conditions are needed. Take f(x) = |x| on [−1, 1]. It is continuous, and the average rate of change is (1 − 1)/2 = 0. But f'(x) is −1 or 1 everywhere except x = 0, where it does not exist, so no tangent has gradient 0. The theorem fails because f is not differentiable at 0. Similarly, a function with a jump can have an average rate of change that no tangent matches. On AP questions, you must state that the conditions hold before using the theorem.
Rolle's theorem
Rolle's theorem is the special case where f(a) = f(b). Then the average rate of change is zero, so there is a c in (a, b) with f'(c) = 0, a horizontal tangent. If a smooth curve starts and ends at the same height, it must turn around somewhere in between.
Worked examples
Example 1: finding c
f(x) = x² on [1, 3]. Find the value of c guaranteed by the mean value theorem.
- f is a polynomial, so it is continuous on [1, 3] and differentiable on (1, 3).
- Average rate = (9 − 1)/(3 − 1) = 4. f'(c) = 2c = 4, so c = 2, which lies in (1, 3).
Example 2: a cubic
f(x) = x³ − x on [0, 2]. Find c.
- Average rate = (6 − 0)/2 = 3. f'(c) = 3c² − 1 = 3, so c² = 4/3.
- c = ±2/√3 ≈ ±1.155. Only c = 2/√3 ≈ 1.155 lies in (0, 2).
Example 3: Rolle's theorem
f(x) = x² − 4x + 1 on [0, 4]. Show Rolle's theorem applies and find c.
- f(0) = 1 and f(4) = 1, and f is a polynomial, so the conditions hold.
- f'(c) = 2c − 4 = 0, so c = 2.
Example 4: AP free-response style, from a table
A car's position s(t), in metres, is differentiable. s(2) = 10 and s(6) = 50. Must there be a time in (2, 6) when the velocity is 10 m/s? Justify.
- s is differentiable, so it is also continuous on [2, 6].
- Average rate = (50 − 10)/(6 − 2) = 10 m/s.
- By the mean value theorem, yes: there is a time c in (2, 6) with s'(c) = v(c) = 10 m/s.
Example 5: when the theorem does not apply
f(x) = 1/x on [−1, 1]. Does the mean value theorem guarantee a c with f'(c) = (f(1) − f(−1))/2?
- f is not defined at x = 0, so it is not continuous on [−1, 1].
- The theorem does not apply. Indeed, the average rate is (1 − (−1))/2 = 1, but f'(x) = −1/x² is always negative, so no such c exists.
Example 6: bounding a value
f is differentiable with f(0) = 2 and |f'(x)| ≤ 3 for all x. What is the largest possible value of f(4)?
- By the mean value theorem, (f(4) − f(0))/4 = f'(c) for some c, and f'(c) ≤ 3.
- So f(4) ≤ 2 + 12 = 14.
The three existence theorems compared
AP Calculus has three theorems that guarantee something exists. Students often mix them up, so it helps to see them side by side.
- Intermediate value theorem (IVT): if f is continuous on [a, b], then f takes every value between f(a) and f(b). Use it to show f(c) equals a given number.
- Extreme value theorem (EVT): if f is continuous on [a, b], then f has an absolute maximum and minimum on [a, b].
- Mean value theorem (MVT): if f is continuous on [a, b] and differentiable on (a, b), then f'(c) equals the average rate of change for some c. Use it to show f'(c), a rate, equals a given number.
A quick test: if the question asks about a function value, think IVT. If it asks about a derivative or rate, think MVT.
How to write the justification
A full-credit answer has three parts: state the conditions (“f is differentiable, so it is continuous”), compute the average rate of change from the given values, and conclude with the theorem by name (“by the mean value theorem, there is a c in (a, b) with f'(c) = …”). When the function is given only in a table, “differentiable” in the question stem is the condition you quote.
Common mistakes
- Forgetting to state that the conditions hold.
- Using the theorem on a closed interval where the function has a corner, a jump or an asymptote.
- Giving a value of c outside the open interval (a, b).
- Using the mean value theorem when the question is about a function value, which needs the intermediate value theorem.
Practice questions
Try each one before opening the answer.
Q1. f(x) = √x on [0, 4]. Find c.
Show answer
Q2. f(x) = sin x on [0, π]. Find c from Rolle's theorem.
Show answer
Q3. g is differentiable, g(1) = 4 and g(5) = −4. What does the MVT guarantee?
Show answer
Q4. Does the MVT apply to f(x) = x2/3 on [−1, 1]? Why?
Show answer
Frequently asked questions
Is the mean value theorem on the AP Calculus exam?
Yes, in both AB and BC. It is in Unit 5 and appears often in free-response questions, usually with a table of values.
Do I need to find c exactly?
Sometimes. If f is given as a formula, you may be asked to find c. If f is given in a table, you only need to show that such a c exists.
Key takeaways
- Continuous on [a, b] and differentiable on (a, b) are both required.
- Some tangent is parallel to the secant.
- Rolle's theorem is the case f(a) = f(b).
- IVT is about values; MVT is about rates.
- State conditions, compute the average rate, name the theorem.