The Mean Value Theorem

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Anything You Want to Talk About?

Extreme values and endpoints?

The key idea is that when you’re looking for extreme values over a closed interval, one or both extremes might occur at one of the ends of the interval rather than at a critical point:

The process for solving extreme value problems on closed intervals is thus a modification of the one for open or unbounded intervals:

1. Find the function’s values at critical points (but you can ignore any critical points outside your interval)
2. Find the function’s values at the endpoints of the interval
3. The extremes are the minimum/maximum of the values from (1) and (2).

SI

The next session is Sunday, 6:00 - 7:30. Emiliana will announce the link shortly before.

The Mean Value Theorem

From section 4.4 in the textbook and this discussion of the Mean Value Theorem.

The Stereotypical Example

Suppose f(x) = x². Find the value of c between 1 and 2 such that f′(c) = ( f(2) - f(1) ) / (2 - 1).

This is asking you to do the Mean Value Theorem calculation, i.e., find a value c between a and b such that f′(c) equals ( f(b) - f(a) ) / (b-a). In this case f(x) = x², a = 1, and b = 2:

Plugging in numbers gives us that ( f(b) - f(a) ) / (b-a) = 3:

Finally, we need to set f′(c) equal to 3 and solve for c. The Mean Value Theorem says that the solution should be somewhere between 1 and 2, and sure enough it is:

Paraphrase

In working with the above example, someone realized that the right-hand side of the Mean Value Theorem formula is really a slope:

In particular, it’s in some sense the average slope, or average rate of change, of function f between x=a and x=b. So an English paraphrase of the Mean Value Theorem that captures why the word “mean” (as in “average”) appears in the theorem’s name would be:

For a continuous function, the instantaneous rate of change must somewhere equal the average rate of change.

Remembered in this form, the Mean Value Theorem isn’t particularly surprising, you would expect that a continuous function (in this case, the derivative) somewhere has a value equal to its average value.

A More Fun Example

Geneseo’s official College Magic Carpet flies from Geneseo to Rochester and back, taking a non-zero amount of time. Assuming that during the trip the distance from Geneseo to the carpet is a continuous function of time, is there ever a time during the flight when the carpet’s speed towards or away from Geneseo is 0?

The Mean Value Theorem says there must be: speed is the derivative of distance, so in asking about the carpet’s speed we’re also asking about the derivative of its distance. And since the distance starts and ends at 0, over a non-zero amount of time, the Mean Value Theorem says that somewhere the derivative of distance (i.e., speed) has to equal 0 - 0 divided by a non-0 value, i.e., 0:

Antiderivatives

What does Theorem 4.7 (that if f′(x) = g′(x) over some interval, then f(x) = g(x) + C over that interval, for some constant C) say about how complete our rules for finding antiderivatives are?

It says that even though we came up with the rules in an ad hoc way (basically saying, “hey, here’s a derivative rule, so if we use it in reverse we can find an antiderivative”), without worrying about whether there might be other or more general antiderivatives for the function in question, we actually got the most general antiderivatives after all (since we’ve been remembering to include the “+ C”) part.

Next

Another corollary of the Mean Value Theorem says that over any interval where f′(x) > 0, f(x) is increasing, and over any interval where f′(x) < 0, f(x) is decreasing. This lets us start to use derivatives to understand and predict the shapes of functions’ graphs.

To explore this idea further, please read “The First Derivative Test” in section 4.5 of the textbook by class time Thursday, and participate in this discussion of derivatives and shapes of graphs.

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