AP®MathsCollege BoardCalculus BCStudy GuidesUnit 5: Analytical Applications of DifferentiationGraphs of Functions & Their DerivativesMean Value Theorem

Mean Value Theorem (College Board AP® Calculus BC): Study Guide

Written by: Roger B

Reviewed by: Dan Finlay

Updated on 7 May 2026

Mean value theorem

What is the mean value theorem?

The mean value theorem states that:
- If a function $f$ is continuous over the closed interval $[a, b]$
  - and differentiable over the open interval $(a, b)$
- Then there exists a value $x = c$ in the interval $(a, b)$
  - such that $f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
In plain language, it means that there will be a point within that open interval $(a, b)$
- where the instantaneous rate of change $f^{'} (c)$
- is equal to the average rate of change over the interval

A graph of a function with tangent and average rate of change lines illustrating the mean value theorem — **An illustration of the mean value theorem**

For example, consider the differentiable function $f$ such that $f (1) = 5$ and $f (3) = 7$
- The MVT tells you that there is a value $c$ in the interval $1 < c < 3$ such that $f' (c) = \frac{7 - 5}{3 - 1} = 1$
- This means the equation $f' (x) = 1$ has at least one solution

What does the mean value theorem not tell me?

The MVT does not tell you where the derivative takes the value $\frac{f (b) - f (a)}{b - a}$
- It tells you there is a value in the interval $a < x < b$
- But it does not tell you the actual value
The MVT does not tell you how many times the derivative takes the value
- It tells you there is at least one value that satisfies $f' (x) = \frac{f (b) - f (a)}{b - a}$
- But there could be multiple values
The MVT says nothing about whether the derivative takes other particular values
- It guarantees a point where the slope is equal to $\frac{f (b) - f (a)}{b - a}$
- But it does not tell you anything about the slope at other points
For example, consider the continuous function $f$ such that $f (1) = 5$ and $f (3) = 7$
- The MVT does not help you find the actual value of a solution to $f' (x) = 1$
- The MVT does not tell you the number of solutions to the equation $f' (x) = 1$
- The MVT does not tell you whether there is a point that has a slope of -1

Examiner Tips and Tricks

When using the mean value theorem on the exam

Be sure to justify that the theorem is valid
- I.e. that the function is continuous on $[a, b]$
- and differentiable on $(a, b)$
Remember that if a function is differentiable on an interval
- then it is also continuous on that interval

Do not confuse the intermediate value theorem with the mean value theorem:

The IVT is about a function taking a value
The MVT is about the derivative taking a value

Worked Example

A social sciences researcher is using a function $m$ to model the total mass of all the garden gnomes appearing on lawns in a particular neighborhood at time $t$ . The function $m$ is twice-differentiable, with $m (t)$ measured in kilograms and $t$ measured in days.

The table below gives selected values of $m (t)$ over the time interval $0 \leq t \leq 12$ .

$t$

(days)

$m (t)$

(kilograms)

24.9

36.0

70.3

89.7

89.1

Justify why there must be at least one time, $t$ , for $10 \leq t \leq 12$ , at which the total mass of the garden gnomes is decreasing at a rate of 0.3 kilograms per day.

Answer:

You need to show that there is at least one value $t$ for $10 < t < 12$ such that $m' (t) = - 0.3$

Showing that a function's derivative has a particular value at an unspecified point is a job for the mean value theorem

But first you have to justify why $m (t)$ is continuous; along with being differentiable, that will make the mean value theorem valid

Remember that a differentiable function is automatically also continuous

$m (t)$ differentiable $\Rightarrow m (t)$ continuous

Now calculate the average rate of change of $m$ between $x = 10$ and $x = 12$ using $\frac{f (b) - f (a)}{b - a}$

$\frac{m (12) - m (10)}{12 - 10} = \frac{89.1 - 89.7}{2} = \frac{- 0.6}{2} = - 0.3$

Now everything is in place to justify the result using the mean value theorem

$m (t)$ is twice-differentiable, which means $m (t)$ is differentiable, which means $m (t)$ is continuous

The average rate of change of $m$ between $t = 10$ and $t = 12$ is -0.3 kilograms per day

Therefore by the mean value theorem there must be at least one time, $t$ , for $10 \leq t \leq 12$ , at which $m^{'} (t) = - 0.3$ , which means that the total mass of the garden gnomes is decreasing at a rate of 0.3 kilograms per day at this time

Rolle's theorem

What is Rolle's theorem?

Rolle's theorem is a special case of the mean value theorem
- It occurs when $f (a) = f (b)$ in the mean value theorem,
  - Which means that $f (b) - f (a) = 0$
Rolle's theorem states that:
- If a function $f$ is continuous over the closed interval $[a, b]$
  - and differentiable over the open interval $(a, b)$
- And if $f (a) = f (b)$
- Then there exists a value $x = c$ in the interval $(a, b)$
  - such that $f^{'} (c) = 0$
In plain language, this means that there will be a point within that open interval $(a, b)$
- where the instantaneous rate of change $f^{'} (c)$ is equal to zero
This means there will be a horizontal tangent at that point
- and hence a local minimum or maximum point somewhere between $x = a$ and $x = b$

A graph of a function with tangent and average rate of change lines illustrating Rolle's theorem — **An illustration of Rolle's theorem**

Examiner Tips and Tricks

In your answer to an FRQ on this topic, you can refer to the mean value theorem or Rolle's theorem. Just remember to check that the conditions for the theorems are satisfied.

Worked Example

$t$ (minutes)	0	10	25	50
$v (t)$ (inches per minute)	0	3	-2	3

A snail is moving in a horizontal line. The velocity of the snail is modeled by the differentiable function $v$ , where $v (t)$ is measured in inches per minute and $t$ is measured in minutes. Selected values of $v (t)$ are given in the table.

Must there exist a value of $c$ , for $10 < c < 50$ , such that $v' (t) = 0$ ? Justify your answer.

Answer:

This is asking about the derivative, so use the mean value theorem

Check that the function is continuous

$v$ is differentiable ⇒ $v$ is continuous on $[10, 50]$

Find the average rate of change over the interval $[10, 50]$

$\frac{v (50) - v (10)}{50 - 10} = \frac{3 - 3}{40} = 0$

Apply the theorem

$v (t)$ is differentiable, which means $v (t)$ is continuous

By the mean value theorem, there must exist a $c$ , for $10 < c < 50$ , such that $v' (c) = 0$

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Mean Value Theorem (College Board AP® Calculus BC): Study Guide

Mean value theorem

What is the mean value theorem?

What does the mean value theorem not tell me?

Examiner Tips and Tricks

Worked Example

Rolle's theorem

What is Rolle's theorem?

Examiner Tips and Tricks

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Dan Finlay