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

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 $\left[a,b\right]$

  • and differentiable on $\left(a,b\right)$

• 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

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 $10\le t\le 12$, 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 $m\text{'}\left(t\right)=-0.3$

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 $m\left(t\right)$ is continuous; along with being differentiable, that will make the mean value theorem valid

Remember that a differentiable function is automatically also continuous

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

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

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

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

$m\left(t\right)$ is twice-differentiable, which means $m\left(t\right)$ is differentiable, which means $m\left(t\right)$ 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 $10\le t\le 12$, at which $m^{\text{'}}\left(t\right)=-0.3$, which means that the total mass of the garden gnomes is decreasing at a rate of 0.3 kilograms per day at this time

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.

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

Must there exist a value of $c$, for $10<c<50$, such that $v\text{'}\left(t\right)=0$? 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 $\left[10,50\right]$

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

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

Apply the theorem

$v\left(t\right)$ is differentiable, which means $v\left(t\right)$ is continuous

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