Mean Value Theorem
Learning goal: students start to learn this cornerstone of calculus
We’ve had some technical theorems about continuous functions. Let’s recall them:
Intermediate Value Theorem (IVT): If f(x) is continuous on the interval [a, b], and y is any
value between f(a) and f(b), then there is a c on the interval [a, b] with f(c) = y.
Extreme Value Theorem (EVT): If f(x) is continuous on the interval [a, b] then it achieves both
a minimum and a maximum on the interval. That is, there is a c on [a, b] so that f(c) ≥ f(x) for
any x in [a, b]. Similarly, there is a d on [a, b] so that f(d) ≤ f(x) for any x in [a, b].
Note that these theorems don’t tell you anything about how to actually find these intermediate or
extreme points, they only tell you such points exist somewhere.
Also note that every theorem has a hypothesis, which is the “if” part. The hypothesis gives the
condition when the theorem applies. When the hypothesis is true, the conclusion of the theorem
is then automatically true—that is the nature of a theorem! The conclusion is what the theorem
says must happen. But only if the hypotheses are true. If the hypotheses are false, the theorem
doesn’t apply. The conclusion might be true or false, we just don’t know.
Example: if there were a theorem that said “If it’s raining, then it is Tuesday” which of the
following statements must also be true?
a) It’s raining, so it must be Tuesday.
—True. It is raining, so by the theorem, it is Tuesday.
b) It only rains on Tuesdays.
—True. The theorem doesn’t allow it to rain any other time (any time it is raining it must be
Tuesday!).
c) It always rains on Tuesdays.
—False. Just because it is Tuesday doesn’t guarantee that it will rain. It just can’t rain any other
day.
d) It’s not raining, so it must not be Tuesday.
—False. There can be Tuesdays when it doesn’t rain; there just can’t be rain on any other day.
e) It’s not raining, so it must not be Tuesday.
—False, again, there can be Tuesdays when it doesn’t rain.
f) It’s not Tuesday, so it isn’t raining.
—True. This is the “contrapositive.” Since it only rains Tuesdays, if it’s not Tuesday it can’t be
raining.
It is important to distinguish what is the hypothesis (a condition that may or may not be true) and
the conclusion (the result that happens when the conclusion is true—it might happen other times,
too!). And always keep in mind that if the hypotheses aren’t met, the theorem isn’t false, it’s
irrelevant.
<<Students should work through the MVT Intro sheet here>>
Now we are going to look at a technical theorem about differentiation. The intro sheet might
have suggested the idea that if a function is differentiable, then somewhere in the interval, its
exact slope equals its overall slope (i.e., its average rate of change).
Example: if you drive 100 miles in two hours, your average rate of change (average speed) is
50mph. If the Mean Value Theorem (MVT) applies, then you are guaranteed to have been going
exactly 50mph at some point in the trip.
(Note: What if you went 20mpg for the first hour, and then 80mph the second? It turns out that
if your position is a differentiable function of time, this can’t happen. How did your speed
suddenly jump up like that? At one point in time, you have infinite acceleration, and this
function won’t be differentiable at that point.)
The hypotheses of the MVT are a little awkward, but are there to make it as applicable as
possible:
Theorem: (Mean Value Theorem): If f(x) is continuous on [a, b] and differentiable on (a, b),
f (b) − f (a)
then there is some point c in (a, b) for which f′(c) =
.
b−a
Example: Let f(x) = x2. Then f(1) = 1 and f(4) = 16. The average rate of change here is 5 (it is
(16 – 1) / (4 – 1) = 15/3). Since f is both continuous and differentiable we can apply the MVT to
know there is someplace between x = 1 and x = 4 where the derivative is exactly 5. Of course,
f′(x) = 2x, and 2x = 5 when x = 5/2.
The weird hypotheses allow the function to be not differentiable at the endpoint, as long as its
continuous. This is important for functions whose domains are intervals, and the derivative
doesn’t exist at the endpoints. For instance, a function like f (x) = 1− x 2 whose graph is a
semicircle, but which is not differentiable at the ends because it is infinitely steep. It does not
apply to function like tangent, where not only does the derivative blow up but so does the
function.
Example: an airplane is cruising at 36,000 feet, and comes in for a landing. It takes 15 minutes
from the beginning of the descent until touchdown. What does MVT tell us?
Solution: since airplanes don’t disappear and reappear at a lower altitude, presumably the
altitude is continuous. And since infinite accelerations are bad, presumably the airplanes altitude
is also differentiable. So the plane descends from 36,000 to zero in 15 minutes. By MVT, there
must be some time during the descent that it was going (0 – 36,000)/15 = –2400 feet/minute, i.e.,
descending at 2400 feet per minute (= 40 f/sec). Sometimes it might be dropping faster, others
maybe slower, but at least one instance it is dropping at exactly the right rate.
Question: Can’t we just apply the IVT to the derivative to learn this same fact? Answer: no—
there are function that are differentiable, but whose derivative is not continuous. There is such a
best on the new Take-home.
<<Students should work through MVT practice sheet here>>
Some special applications of the MVT….
Rolle’s Theorem: if f is continuous on [a, b] and differentiable on (a, b), and f(a) = f(b), then
f′(x) = 0 for some x in (a, b).
This is just the MVT for the special case where the ends are at the same height, so the average
rate of change is zero. It turns out to be easier to prove this first, then prove the MVT from it.
(We’ll get to that later.)
Increasing Function Theorem: if f is continuous on [a, b], differentiable on (a, b), and the
derivative is always positive, then the function is always increasing. (If the derivative is always
non-negative, then the function is non-decreasing.) There is a corresponding decreasing function
theorem.
Proof: Let x1 < x2 be two points in the interval [a, b]. By MVT, we know that there is some
point c between x1 and x2 where f′(c) = (f(x2) – f(x1)) / (x2 – x1). Thus, f(x2) – f(x1) = f′(c)(x2 – x1),
the product of two positive (nonnegative) numbers. Thus f(x2) > (≥) f(x1) so f is increasing (nondecreasing).
Constant Function Theorem: if f is continuous on [a, b] and differentiable on (a, b) and f′(x) is
zero for all x on (a, b) then f is constant.
Proof: same as above, only now f′(c) = 0, so f(x2) = f(x1) for all x1 and x2.
We already “knew” this, but it actually requires the technicality of the MVT to actually prove it!
A very useful, prosaic way to think of the MVT is the racetrack principle. It says this: if two
racers start in the same place, and one is always faster than the other, the faster one will always
be ahead after the beginning. Similarly, if they end in the same place, the slower one was always
ahead. In fancy words, let f and g be continuous on [a, b] and differentiable on (a, b), and let
f′(x) ≥ g′(x) for all x in (a, b). Then:
(a) If f(a) = g(a), we have f(x) ≥ g(x) for all x in [a, b].
(b) If f(b) = g(b), we have f(x) ≤ g(x) for all x in [a, b].
The idea is that we can use the increasing function theorem on the difference f – g, whose
derivative (f – g)′ = f′ – g′ is always positive (non-negative). So f – g keeps getting bigger.
Practice Problems (Hughes-Hallett section 3.10): 1 – 10.