Subsections

• Purpose
• Background
• Exercises

A Really Mean Theorem

Purpose

The purpose of this lab is to use Maple to help you visualize and understand the mean value theorem (MVT) and its applications.

Background

The mean value theorem is one of the most important and useful theorems from calculus. Before stating it, here is a (rather silly) application of it.

Just after you get on the Pike by Auburn, you pass a car pulled over by the highway patrol. Feeling glad that it isn't you, and relatively safe because you have a radar detector, you blast on by. One hour and eighty miles down the road, however, another patrolman pulls you over. You're not worried, because your detector had warned you to slow down. Imagine your shock, however, when he proceeds to write you a ticket for going 80 miles per hour! When you protest, he tells you that the patrolman at Auburn had radioed your position ahead. He then claims that since your average speed was eighty, the MVT says you must have been going eighty at least once in your journey. Then you see the ``MathNet'' patch on his uniform, and you really start to get that sinking feeling.

Mathematically, the MVT can be stated as follows.

Theorem 333 (MVT)

Suppose that f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Then for some c between a and b,

$$f'(c) = \frac{f(b) - f(a)}{b - a}.$$

Geometrically, the MVT says that for at least one point strictly between a and b, the slope of the tangent line to the graph of f(x) is equal to the slope of the straight line between the two points (a,f(a)) and (b,f(b)). (Note: This line is often called the secant line between a and b.) Obtaining a formula for the secant line isn't hard, but to cut down on the amount of Maple drudgery, a procedure, called secantline has been written that does this for you. The secantline command takes three arguments. The first is a function or expression, the next one is the base point, and the third is the increment, b-a. Before you can use this procedure, you must first load it using the with(CalcP) command, as shown in the following example session.

  > with(CalcP):

  > f := x -> x^3;

  > g := secantline(f(x),x=-1,2);

  > plot(f(x),g,x=-1..1);

Looking at the graph displayed in the previous example, you should be able to see that there are two values of x between x=-1 and x=1 where the slope of f(x) = x³ is the same as the slope of the secant line between the two points (-1,f(-1)) and (1,f(1)).

To find these values, we need to solve for the values of x in the interval (-1,1) where the derivative, f'(x), is equal to the slope of the secant line, which is 1 in this example. Maple commands to do this, and plot the two tangent lines are shown below. In this case the solve command finds both solutions. Note the use of the label sol and the notation sol[1] and sol[2] to access the two roots.

  > sol := solve(diff(f(x),x)=1,x);

  > plot(f(x),g,tangentline(f(x),x=sol[1]),
tangentline(f(x),x=sol[2]),x=-1..1);

The solve command can't always do the job if the function f(x) is complicated. When the solve command fails, there is no output from the command, as shown in the example below. In this case, use the fsolve command instead, as shown below.

  > f := x -> sin(x^2);

  > g := secantline(f(x),x=4,1);

  > plot(f(x),g,x=4..5);

  > solve(diff(f(x),x)=diff(g,x),x);

  > fsolve(diff(f(x),x)=diff(g,x),x=4..5);

It is important to restrict the x interval in the fsolve command to (4,5) in this case, because the equation

$$2x \cos(x^2) = \sin(25) - \sin(16)$$

has an infinite number of solutions. In fact, there are three solutions to this equation in the interval (4,5). The fsolve command given above found one of them. The other two can be found by restricting the range in the fsolve command to an interval that includes only the desired root. To do this, you need to determine approximately where the roots are. Looking at the graph of the function and the secant line is probably the best way to do this - just look for points on the curve where the slope of the tangent line is the same as the slope of the secant line.

If you do this for the previous example, $f(x) = \sin(x^2)$, you should be able to recognize that one of the roots is in the interval (4,4.3). Using this interval in the fsolve command produces the following results.

  > fsolve(diff(f(x),x)=diff(g,x),x=4..4.3);

There is nothing special about the interval (4,4.3); any other interval that included the root at $x = 4.159\ldots$, but didn't include any other roots would have worked. That is, the interval (4.1,4.2) would have worked equally as well, but the interval (4,4.6) is not a good choice because it includes two roots.

Exercises

1.

For each of the following functions, verify graphically that the MVT holds and find all values of c, a < c <b, that satisfy the condition

$$f'(c) = \frac{f(b) - f(a)}{b - a}.$$

(a)

f(x) = x³-3x, $a = - \frac{1}{2}$, $b= 2\frac{1}{2}$.

(b)

$f(x) = \cos(x)+\cos(2x)$, a=1, b=4.

(c)

f(x) = x + sin(x), $a=\frac{\pi}{4}$, $b = \frac{5 \pi}{4}$.

(d)

$f(x) = x^2 \exp(x)$, a=0, b=2.

2.

Explain how the MVT applies to support the patrolman's claim.

3.

For which values of a and b would the MVT apply to $f(x) = \mid x \mid$?

4.

In some cases, the slope of the secant line can be interpreted as an average value. For example consider the falling body problem from the text. If s(t) is the position of a body, then the quantity

$$\frac{s(b)-s(a)}{b-a}$$

is change in position over the change in time, or the average velocity of the body over the time interval. Because the derivative of s(t) is the velocity v(t), the mean value theorem says that there is a time between a and b at which the instantaneous velocity is equal to the average velocity.

Suppose that a body starts at a height of 100 feet and that air resistance can be neglected. At time zero, the body is released and subsequently falls to the ground.

(a)

Find the time $t_{\mathrm{hit}}$ at which the body hits the ground.

(b)

Find the average velocity of the body from when it was released to when it hits the ground.

(c)

Find the time at which the instantaneous velocity is equal to the average velocity.

5.

In the previous problem, you should have found that the time at which the instantaneous velocity is equal to the average velocity is exactly half the time it takes for the body to hit the ground. Is this always true? Go through the following steps to show that it is.

(a)

Use s(t) = -16 t²+ H₀ as the position function, where H₀ is the initial height of the body, and find the time at which the body hits the ground.

(b)

Using the value of t you just computed, find the average velocity of the body.

(c)

Use v(t) = -32 t to compute the value of t at which the instantaneous velocity is equal to the average velocity. Then show that this value is half the time it takes for the body to hit the ground.