Extreme Value Theorem (College Board AP® Calculus AB)

Study Guide

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Written by: Roger B

Reviewed by: Dan Finlay

Extreme value theorem

What is the extreme value theorem?

The extreme value theorem states that:
- If a function $f$ is continuous over the open interval $(a, b)$
  - then $f$ has at least one minimum value and at least one maximum value on the closed interval $[a, b]$
This tells us that a continuous function
- will always have a minimum and a maximum value on a closed interval
  - but that those values might occur at the endpoints of the interval
- If a minimum or maximum value does not occur at an endpoint
  - then it will be a local minimum or local maximum within the open interval
    - In this case if the function is differentiable on $(a, b)$
    - then the derivative of the function will be zero at that point

Examiner Tips and Tricks

When using the extreme value theorem on the exam:

Be sure to justify that the theorem is valid
- I.e. that the function is continuous 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 $7 \leq t \leq 12$ , at which $m^{'} (t)$ , the rate of change of the mass, equals 0 kilograms per day.

Answer:

You need to prove that the derivative has a particular value at an unspecified point

At first this might seem like a job for the mean value theorem

But there are no two points in the table between which the average rate of change is zero

Instead, the stated result may be proved by using the extreme value theorem

Start by showing that $m$ is continuous; then the extreme value theorem will be valid

Remember that differentiability implies continuity

$m$ differentiable $\Rightarrow m$ continuous on $[7, 12]$

The extreme value theorem says that $m$ will have a maximum value on $[7, 12]$

Show that the maximum value does not occur at either of the endpoints of the interval

$m (7) = 70.3 < 89.7 = m (10)$

and

$m (10) = 89.7 > 89.1 = m (12)$

Because $m$ is differentiable on $(7, 12)$ , this means the maximum is a local maximum point somewhere in that open interval, at which point $m^{'}$ is equal to zero

$m (t)$ is twice-differentiable, which means $m (t)$ is differentiable, which means $m (t)$ is continuous on $[7, 12]$

By the extreme value theorem, $m$ has a maximum on $[7, 12]$

That maximum does not occur at $m (7)$ or $m (12)$ , therefore $m$ has a maximum on the interval $(7, 12)$

Because $m$ is differentiable, $m^{'} (t)$ must equal 0 at this maximum

Last updated: 5 August 2024

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