Extreme Value Theorem (College Board AP® Calculus AB): Revision Note

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 $\left[a,b\right]$

• Remember that if a function is differentiable on an interval

  • then it is also continuous on that interval

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 $m\left(t\right)$ measured in kilograms and $t$ measured in days.

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

Justify why there must be at least one time, $t$, for $7\le t\le 12$, at which $m^{\text{'}}\left(t\right)$, 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 $\left[7,12\right]$

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

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

$m\left(7\right)=70.3<89.7=m\left(10\right)$

and

$m\left(10\right)=89.7>89.1=m\left(12\right)$

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

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

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

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

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

In the worked example above, there are multiple other methods to show the same result.

You could use the intermediate value theorem to show that there is a value $c$, for $7<c<10$, such that $m\left(c\right)=89.1$.

You could then use the mean value theorem or Rolle's theorem to show that there is a value $d$, for $c<d<12$, such that $m\text{'}\left(d\right)=\frac{m\left(12\right)-m\left(c\right)}{12-c}=\frac{89.1-89.1}{12-c}=0$.