The following situation illustrates the main result of this chapter. You travel a total distance of 111 miles between toll booths in an elapsed time of one and one half hours. All the toll booths have synchronized clocks accurate to within a few seconds. When you arrive at the exit booth, the officer hands you a speeding ticket. He cites you for 74 mph even though, while you were in his sight you were obeying the 65 mph limit and even slowed cautiously before the toll booth. Were you really speeding? Of course, but how can he prove this?
The officer computes your overall (mean or average) speed as
mph and feels he can prove you had to be going that fast at at least two times.
You actually had to drive over 74 mph part of the time, didn't you? Do you pay the ticket or will you try to argue in court that you averaged 74 mph without ever going that fast?
20.1
We want to formulate this speed problem in a general way for a function y=f[x] on an interval [a,b]. You may think of x as the time variable with x=a at the start of the trip and x=b at the end.
The elapsed time traveled is b-a, or 3/2 hours in the example. (Perhaps you start at 2 and end at 3:30.) You may think of y=f[x] as a distance from a reference point: your odometer reading.
We start at f[a], end at f[b], and travel a total of f[b]-f[a]. The elapsed time is b-a, so the overall "average" speed is (f[b]-f[a])/(b-a). Instantaneous speed is the time rate of change of distance or derivative f'[x].
Sketch the line connecting the end points of the graph (a,f[a]) and (b,f[b]). What is the slope of this line?
Find a point on your sketch where the speed is 74 mph and sketch the tangent line at that point.
Call the point c. Why does this satisfy
The following is what the officer needs in court, since he needs to show that there is a time c when your speed f'[c]=74. We already know that (f[b]-f[a])=111 and (b-a)=3/2, so (f[b]-f[a])/(b-a)=74.
There may be more than one point where f'[c] equals the mean speed or slope.
PROOF
The average speed over a subinterval of length
Suppose we let
The officer wants to preempt an argument on your part that you were not speeding near the toll booths, so he does not want to argue that g[xhi] is either the first or last half-hour.
He also has only cited you for 74 mph, so he would like to argue that there was a half-hour interval inside your trip interval where you averaged exactly 74 mph.
He knows
We will come back to your speed once we establish the fact that there is an x1 where
20.2
Your mean speed g[x] was below 74 at time xlo and above 74 at time xhi, one half hour later.
Was g[x0] ever equal to 74? Most people would say, "Yes." They implicitly reason that speed "moves continuously" through values and hence hits all intermediate values.
This idea is a precise mathematical theorem, and its most difficult part is in correctly formulating what we mean by "continuous" function.
Our definition was given in Chapters 4 and 5 of the main text.
In court, the officer cites the following result to prove that there was a half-hour interval - away from the toll booths, between xlo and xhi - when your average speed was exactly 74. We'll come back to the officer after we see why he's gotcha on this one.
A complete proof is in the Mathematical Background Section 4.3. The related Theorem of Darboux for derivatives is in the Math Background Section 7.2.
Since g[x] is continuous, Bolzano's Intermediate Value Theorem says that there is an x1 between xlo and xhi with
But the officer has had calculus and his argument isn't finished.
He has an interval
The officer continues, "Your honor, I will now show that the defendant's speed was 74 mph to within the accuracy of any known measuring device." He is about to show that there is a time xm and a tiny change in time
Once you know that the average of the small subinterval speeds,
Why is
One way or the other, complete the officer's argument that there was a time when you were going 74. This is the main task of the project.
20.3
In the Mathematical Background, we state the Mean Value Theorem for Derivatives in its ultimate generality, only assuming weakly approximating pointwise derivatives and those only at interior points.
This complicates the proof but is the key in the Math Background to seeing why regular derivatives and pointwise derivatives are the same when the pointwise derivative is continuous.
20.4
Why do we call (f[b]-f[a])/(b-a) an average of f'[x]?
First, if h[x] is a continuous function on a interval [a,b], we show that
Break the interval [a,b] into n tiny subintervals, a,
Why is the overall speed
?
Theorem 20.1
Figure 20.1: Mean Slope and Tangents
is
and this new function is defined and continuous on
.
compute the average of 3 averages, the speeds on
, 
, and
. This ought to be the same as the overall average and the telescoping sum below shows that it is: 
This implies that there is an adjacent pair of subintervals with
because the average of the three subinterval speeds equals the overall average and so either all three also equal the overall average, or one is below and another is above the mean slope. (We know that xlo and xhi differ by
, but we do not care in which order they occur xlo<xhi or xhi<xlo.)
and wants to reason that g[x] attains the intermediate value (of the mean speed, 74) for some x between xlo and xhi. (If your middle half-hour speed is 74, we could use that one, but the officer knows you may have stopped at the Midway Rest Area and have witnesses.)
.
Theorem 20.2
, then g[x] attains every value intermediate between the values
and
. In particular, if
and
, then there is an x0, 
, such that g[x0]=74. 
equals -1 when x=-2, equals +1 when x=+3, but never takes the value
for any value of x. Why doesn't j[x] violate Bolzano's Theorem?
(= 74 in your case). The subinterval
lies inside (a,b), has length (b-a)/3 and f[x] has the same mean slope over the subinterval as over the whole interval. (So far we have only used continuity of f[x].) You see your opening argument forming.
All the officer has done is to show that there is a half-hour interval with the average speed of 74. You will introduce your witness from the Midway Rest Area to show you weren't speeding when you stopped there and argue on the subinterval that you somehow averaged 74 without ever going that instantaneous speed, just the agrument you were planning for the whole trip, but now modified to the interval one third as long.
with
where
He has two ways to proceed.
One way would be to break this interval into three more and apply the same reasoning to get an interval of one ninth the original time period where your speed was again 74. Another is to break this interval into tiny subintervals and average the speeds over the tiny subintervals.
so that
Why does this show that f'[c]=74 for some c?
How does the successive intervals argument of length one third the previous interval lead to the conclusion that there is a c with
?
How could the officer prove that the average of the speeds over many tiny subintervals is the overall speed? Write the average as a sum and show that it collapses into
.
for x running from a to
in steps of
is
, how do you know that there are adjacent x's with
with
?
?
, 
, 
, 
of length
. Since h[x] is continuous,
for any
, and the average of any one value chosen from each of the subintervals is approximately
?
?
the average of the instantaneous speed?