Today, we will resume our discussion of the mean value theorem as we did
on Monday. Let’s restate the mean value theorem, with a greater focus on the
conditions required for the mean value theorem to hold.
Of course, what the mean value theorem says is that, under the right conditions, a function f defined on an interval [a, b] will have the property that its
instantaneous rate of change at some point c ∈ [a, b] will equal the average rate
of change of f over the interval [a, b].
Let’s go over the conditions again:
Theorem 0.1 (Mean Value Theorem). Let f (x) be a continuous function on
[a, b] that is differentiable on (a, b). Then there exists a c ∈ (a, b) such that
(a)
.
f ′ (c) = f (b)−f
b−a
This theorem says nothing about where this point c is, and in practice, it
may be hard to find.
Let’s just be clear about why we need the continuity at a and b and the
differentiability on (a, b).
Suppose that f (x) is discontinuous at a. Then f (a) could be any value we
want without changing f ′ (x) at any point in (a, b). For example, consider
n
f (x) = x2 if 0 < x ≤ 2 5 if x = 0 .
This function is defined on [0, 2] and differentiable on (0, 2). The derivative is
the positive-valued function 2x. However, we have
4−5
−1
f (2) − f (0)
=
=
,
2−0
2
2
which is negative. There is no c such that f ′ (c) = 2c = − 12 .
In our car example, we should think of this as the situation where the car
teleports at the beginning or the end of the journey.
Note that we could have found such a c if f (0) were, say, 3. It’s still possible for the conclusion of the mean value theorem to hold even though the
assumptions don’t.
Now, let’s consider an example of a function that’s continuous but has a
corner.
n
f (x) = 10x if −1 ≤ x < 0 x2 if 0 ≤ x ≤ 1 .
This function is clearly continuous on [−1, 0) and (0, 1] and differentiable on
(−1, 0) and (0, 1). Therefore, it is only necessary to check what happens at 0.
We can see that f (x) is continuous at 0:
lim f (x) = lim+ x2
x→0+
x→0
=0
lim f (x) = lim− 10x
x→0−
x→0
=0
f (0) = 0
1
So f (0) = limx→0 f (x) and therefore f (x) is continuous at zero.
However, we can verify that f (x) is not differentiable at 0:
f ′ (0) = lim
x→0
which is equal to limx→0
f (x)
x
f (x) − f (0)
x−0
because f (0) = 0. But we have that
lim+
x→0
f (x)
= lim+ x
x
x→0
=0
but
lim
x→0−
f (x)
= lim 10
x
x→0−
=1
so f (x) is not differentiable at x = 0. Therefore, we have no reason to expect
the mean value theorem to work:
11
f (1) − f (−1)
=
1 − (−1)
2
But the derivative of f is equal to 10 on (−1, 0) and equal to 2x on (0, 1). This
is never equal to 11
2 .
In our car example, this corresponds to an immediate change in the speed of
the car. This isn’t physically realistic for this example, but it is mathematically
important.
As we’ve stated, the place where the average velocity and instantaneous
velocity are equal may be difficult to find. However, it often is not hard to find
at all:
Example 0.2. Let f (x) = x3 − 10x2 + 5. Find the point c in the interval [1, 3]
for which the instantaneous rate of change at x = c is equal to the average rate
of change on the interval [1, 3].
We can solve this problem by differentiating: the derivative of x3 − 10x2 + 5
(1)
is 3x2 − 20x, and the average rate of change over the interval is f (3)−f
= −27.
2
So we need to solve
3x2 − 20x + 27 = 0
which has the solutions
x=
which comes out to
20 ±
√
400 − 324
6
√
20 ± 76
x=
6
2
Note that
20
6
is a little bit bigger than 3, so the value we want is
√
20 − 76
.
c=
6
For this example, there was only one such c, but there could be many more!
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