Introduction

Welcome to AP Calculus AB and BC's Unit 5! In the first topic of Unit 5, we will be covering Topic 5.1 - Using the Mean Value Theorem. In this article, we will learn what the mean value theorem is and when it can be used. We will also go over common applications of this concept on the AP Exam.

The Mean Value Theorem

We use the mean value theorem (MVT) to justify conclusions about a function over an interval, just like the intermediate value theorem.

Let's start by looking at the technical definition of the MVT. Do not worry if it sounds confusing, with a few example problems, it should become easy!

Technical Definition

If a function $f$ is continuous over the interval $[a,b]$ and differentiable over the interval $(a,b)$, then the mean value theorem guarantees a point $c$ within that open interval where the instantaneous rate of change equals the average rate of change over the interval. 

In Mathematical Terms

If the following conditions are met:

1. $f(x)$ is continuous over the interval $[a,b]$

2. $f(x)$ is differentiable over the interval $(a,b)$

Then there exists some $c$ on $(a,b)$ such that

$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$

I would strongly recommend memorizing the above. You especially need to memorize the conditions.

Understanding the Mean Value Theorem with Practice Problems (Part 1)

Question 1

1. Use the function $f(x)=x^2+10x+4$ to answer the following questions.

(a) On the interval $[1,4]$, what is the average rate of change?

(b) On the interval $(1,4)$, when does the instantaneous rate of change equal the average rate of change? 

Solution

(a)

You should have learned how to do part (a) from Topic 2.1. If you forgot, make sure to review that topic, as Topic 2.1 is guaranteed to be on either the MCQ or FRQ portion of your AP Exam.

Average Rate of Change on $[1,4]$ is given by

$\frac{f(4)-f(1)}{4-1}=\frac{(4{)}^2+10(4)+4-((1{)}^2+10(1)+4)}{3}=15$

$15$ is our answer for part (a).

(b)

We are asked when the instantaneous rate of change equals the average rate of change on the interval $(1,4)$. From part (a), we know that the average rate of change on $[1,4]$ is $15$. We know that the instantaneous rate of change is given by $f^{\prime }(x)$. Once we find $f^{\prime }(x)$, we just set it equal to $15$, and that's our answer!

$f^{\prime }(x)=2x+10$

$2x+10=15$

$2x=5$

$x=\frac{5}{2}$

That's our answer, $x=\frac{5}{2}$. This is the $x$ value on $(1,4)$ where the average rate of change equals the instantaneous rate of change on the given interval.

*You may have noticed that the one of the intervals did not include $1$ and $4$, while the other one did. This is just because the average rate of change requires a closed interval. However, the instantaneous rate of change requires an open interval. This is because when we take the derivative, the derivative cannot be computed at an end point of an interval. This is something you should just memorize.

Now, you may be wondering, how did question 1 have anything to do with the MVT. Well, let's work through the conditions of the MVT, and see how it is applied here.

Conditions for MVT:

1. $f(x)$ is continuous on $[a,b]$

In question 1, $f(x)=x^2+10x+4$ and $a=1$ and $b=4$. $f(x)$ is indeed continuous on $[1,4]$

2. $f(x)$ is differentiable on $(a,b)$

In question 1, $f(x)=x^2+10x+4$ and $a=1$ and $b=4$. $f(x)$ is indeed differentiable on $(1,4)$.

Because the conditions are met, we can apply the MVT. Recall that it states that if conditions are met:

There exists some $c$ on $(a,b)$ such that

$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$

In question 1, there exists some $c$ on $(1,4)$ such that

$f^{\prime }(c)=\frac{f(4)-f(1)}{4-1}=15$

or simply

$f^{\prime }(c)=15$

This statement shows us that there is in fact an $x$ value on the interval such that the instantaneous rate of change equals the average rate of change. That is what the MVT is saying! In question 1, that $x$ value, or $c$, is equal to $\frac{5}{2}$.

Before we take a look at another problem, let's review the difference between the mean value theorem (MVT) and intermediate value theorem (IVT).

Mean Value Theorem vs Intermediate Value Theorem

Understanding the Mean Value Theorem with Practice Problems (Part 2)

Question 2

2. A plane takes off from the ground. The twice-differentiable function $h$ models the plane's height, measured in meters, at time $t$, measured in minutes. The table below gives values of $h(t)$ of the plane at selected times $t$.

(a) For $0\le t\le 2$, must there be a time $t$ when the plane is $34$ meters in the air? Justify your answer.

(b) For $2<t<3$, must there be a time $t$ when the plane's velocity is $125$ $meters/min$. Justify your answer.

Solution

(a)

The first thing we need to realize is we must use the IVT. This is because we want to guarantee the existence of a $y$-value (see IVT vs MVT above). Let's check the conditions:

1. $h(t)$ is continuous on $[0,2]$

Yes! We are told $h(t)$ is twice-differentiable, so it must be continuous.

2. $h(0)\ne h(2)$

Yes! $h(0)=0$ and $h(2)=75$

3. $34$ is between $h(0)$ and $h(2)$

Yes!

Therefore there exists some time $t$ on $0\le t\le 2$ when the plane is $34$ meters in the air according the IVT.

(b)

The first thing we need to realize is that velocity is given by $h^{\prime }(t)$. We then need to realize we can use the MVT here to check if there is some $t$ on $(2,3)$ such that

$h^{\prime }(t)=\frac{h(3)-h(2)}{3-2}=125meters/min$.

Let's check conditions:

1. $h(t)$ is continuous on $[2,3]$

Yes! We are told $h(t)$ is twice-differentiable, so it must be continuous.

2. $h(t)$ is differentiable on $(2,3)$

Yes! We are told $h(t)$ is twice-differentiable, so $h^{\prime }(t)$ must be differentiable.

Therefore there exists some $t$ on $(2,3)$ such that

$h^{\prime }(t)=\frac{h(3)-h(2)}{3-2}=125meters/min$

Using the MVT and IVT on FRQs

In order to receive full credit, you must show that all the conditions are met. I would recommend memorizing the table under the MVT vs IVT portion of the article. 

Practice Problems

Free-Response Practice Problems

Questions

1. A particle moves along the $y$-axis so its velocity at any time $t\ge 0$ is given by
$v(t)=t^2+2t-8$. Must there be a time $t$ for $1<t<3$ such that the particle's acceleration equals $6$. Justify your answer.

2. (Calculator Active) Let $f$ be a function with $f(0)={\pi }^2$, $f(4)=\frac{1}{\pi }$, and
$f^{\prime }(x)=4\sin \left(\frac{\pi }{x}\right)-4$. For how many values of $x$, for $x>1$, does the instantaneous rate of change of $f(x)$ equal the average rate of change of $f(x)$ on $[0,4]$.

Solutions

1.

There are two ways of solving this problem. You can either use the MVT or the IVT.

Mean Value Theorem:

Because $v(t)$ is continuous on $[1,3]$ and differentiable on $(1,3)$, there must be some time $t$ on $(1,3)$ such that $v^{\prime }(t)=a(t)=\frac{v(3)-v(1)}{3-1}=6$.

Intermediate Value Theorem:

$v^{\prime }(t)=a(t)=2t+2$

$a(1)=4$

$a(3)=8$

Because $a(t)$ is continuous on $[1,3]$, $a(1)\ne a(3)$, and $6$ is between $a(1)$ and $a(3)$, there exists some time $t$ on $(1,3)$ such that $a(t)=6$.

2. 

Use Desmos to plot the average rate of change of $f(x)$ on $[0,4]$ and $f^{\prime }(x)$ to see how many times they intersect for $x>1$.

From the graph, we can see that there are two values of $x$, for $x>1$, such that the instantaneous rate of change of $f(x)$ equals the average rate of change of $f(x)$ on $[0,4]$

Multiple-Choice Practice Problems