Today, we will be talking about a theoretical fact known as the mean value
theorem. The mean value theorem relates the notion of “instantaneous” rate of
change with the notion of “average” rate of change.
Before beginning on this topic, we need to ask: what does the phrase “average rate of change” mean? When we talk about an “average” rate of change,
we’re not referring to the process of “adding up all the rates of change and
dividing to find the average.” This is a point that calculus students have a lot
of trouble with, so it bears repeating: the average rate of change is not
computed by adding instantaneous rates of change and then dividing
by something. When we talk about the average rate of change of a function
f (x) on an interval [a, b], we mean the unique number such that, IF the derivative f ′ (x) were constant and equal to this number on [a, b], the total change
f (b) − f (a) would be the same. What we mean is something like this:
Let’s say that a car travels 200km in 2 hours. At some points during the
trip- the car may be travelling faster or slower than 100km/h, but if the car
were travelling at a constant speed, that speed would be 100km/h. This
is what it means to say that the average speed of the car is 100km/h.
Let’s say this car has distance function f (t). We didn’t need to know the
instantaneous speed of the car at any point in the journey to find the average speed, and we didn’t obtain the average speed by “adding and dividing”
anything. We just took the distance travelled (the difference in position) and
divided by the time it took (the difference in t).
So if f (x) is a differentiable function on [a, b], the average rate of change of
f (x) is the value
f (b) − f (a)
.
b−a
Note that this is exactly the “rise over run” formula for calculating the slope
of a line- the average rate of change is the slope of the secant line connecting
(a, f (a)) and (b, f (b)).
So what is the connection between average rates of change and instantaneous rates of change? Well, one way they’re related is by the definition of the
derivative:
f (x) − f (c)
lim
.
x→c
x−c
But the problem here is that this statement doesn’t directly imply any quanti(c)
are, so the definition doesn’t
tative information about what the values f (x)−f
x−c
′
really help us convert information about f into information about the average
rate of change of f . What we need is the following fact:
Theorem 0.1 (Mean Value Theorem). Let [a, b] be an interval and let f (x) be
a differentiable function on (a, b) that is continuous on [a, b]. Then there exists
a c ∈ (a, b) such that
f (b) − f (a)
f ′ (c) =
.
b−a
What this theorem says is that somewhere between a and b, there is some
place where the instantaneous rate of change is exactly equal to the average rate
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of change between a and b. So in our car example, even though we don’t know
much about the behavior of the car, we know that at some point, the car must
have been travelling at exactly 100km/h. It’s possible that this only happened
once the entire trip, but it must have happened at least once.
A particularly important case of the mean value theorem is when f (a) =
f (b). This tells us that, if f is continuous on [a, b] and differentiable on (a, b),
with f (a) = f (b), we have that f ′ (c) = 0 for some c in between a and b. In
particular, f must have a critical point between a and b.
The immediate value of the mean value theorem is the following fact that
you have likely seen and will certainly use again.
Theorem 0.2. Let f (x) be a function such that f ′ (x) is strictly positive on
[a, b]. Then f (b) − f (a) > 0 for any b > a. (We say that f (x) is strictly
increasing if this happens).
The point here is that if f (b) − f (a) is less than or equal to zero and b − a
is positive, then the average rate of change on [a, b] is negative, so the instantaneous rate of change must be negative at at least one x-value on [a, b]. This is
how the mean value theorem lets us turn information about the instantaneous
rates of change of f into information about the average rate of change of f .
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