Mean Value Theorem (College Board AP® Calculus AB)

Revision Note

Author

Roger B

Expertise

Maths

Mean value theorem

What is the mean value theorem?

The mean value theorem is an important result in calculus
It 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 $f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
- I.e. 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**

Exam Tip

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

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 $m^{'} (t)$ , the rate of change of the mass, equals -0.3 kilograms per day.

Answer:

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 hour

Therefore by the mean value theorem there must be at least one time, $t$ , for $10 \leq t \leq 12$ , at which $m^{'} (t)$ equals -0.3 kilograms per hour

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 $f^{'} (c) = 0$
- I.e. 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**

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