Download 4.2 Mean Value Theorem

CHAPTER 4. APPLICATIONS OF DERIVATIVES
4.2
104
Mean Value Theorem
Comments. We “know” that the derivative of a function tells us about the function
itself right? Well, how do we know this? Suppose that f 0 (x) = 0 for all values of
x. Is it obvious what this means about f (x)? Suppose if f 0 (x) > 0 for all x in an
interval. Is it obvious what this means about f (x)? It turns out that all of our
knowledge about how f 0 (x) a↵ects f (x) depends entirely upon the following result.
Theorem 1 (Mean Value Theorem). Given a function f and the interval [a, b]
(subject to the fine print2 ), then there exists an x-value c such that
f 0 (c) =
f (b)
b
f (a)
.
a
f (b) f (a)
is the
b a
average slope of f (x) from x = a to x = b, or the average rate of change. So
this theorem can be imprecisely stated as follows:
• In this theorem “mean” means “average”. The fraction
There is some point where the derivatve equals the average rate of
change.
• Geometrically, the theorem means this
m = f 0 (c)
f
f
f
m=
f (b)
b
f (a)
a
a
a
b
b
Start with f , a and b
Add secant line, calculate
slope.
a
c
Some tangent line has the
same slope.
Example 1. Let f (x) = x3 x2 6x + 2. Solve for c that satisfies the conclusions
of the Mean Value Theorem applied to the interval [0, 2].
Solution. The average slope is
0
2
f (2)
2
f (1)
=
1
4.
f (x) = 3x
2x 6.
2
3x
2xp 6 = 4
1± 7
x=
3
We take p
the positive solution since it’s the one in [0, 2]
1+ 7
x=
.
3
2
b
We need f to be continuous on the interval [a, b] and di↵erentiable on the interval (a, b).
CHAPTER 4. APPLICATIONS OF DERIVATIVES
Example 2. Show that x3 + x
105
1 has exactly one root.
Solution. Let f (x) = x3 + x 1.
f (x) is continuous everywhere.
f (0) = 1 which is < 0.
f (1) = 1 which is > 0.
Therefore, by the Intermediate Value Theorem, there exists a number a, between
0 and 1 such that f (a) = 0.
Now we show that f (x) has only one root. Suppose b is a second number such
that f (b) = 0. Then
f (b) f (a)
0 0
=
=0
a b
a b
If this were true, then Mean Value Theorem would show that there is a number c
such that
f 0 (c) = 0
But, f 0 (x) = 3x2 + 1, which can never equal 0.
Therefore, there is no such c, and therefore there is no such b, and therefore a
is the only root.
Comments. Now we mention one important application of the MVT that will later
be crucial to studying anti-derivatives.
Corollary. If f 0 (x) = 0 for all x in some interval, then f (x) is a constant function
on that interval.
Proof. Let C = f (x0 ). Then if x is any other x-value, we have
f (x)
x
f (x0 )
= f 0 (c)
x0
For some c. Since f 0 (x) is always 0, we have f 0 (c) = 0, and so
f (x)
x
f (x0 )
= 0 ) f (x)
x0
f (x0 ) = 0 ) f (x) = C
Corollary. If f (x) and g(x) have the same derivative, then f (x) = g(x) + C for
some constant C.
Proof. Let F (x) = f (x) g(x). Then F 0 (x) = f 0 (x) g 0 (x) and F 0 (x) = 0 since we
have assume that f 0 (x) = g 0 (x) for all x. Then by the first part we have F (x) ⌘ C
(i.e. F (x) equals C for all x). Then f (x) g(x) = C and so f (x) = g(x) + C.
Optional results
Comments. One of uses of the MVT is to provide all kinds of error estimates. In
Calculus II you will (probably) learn about the Taylor Series and Taylor’s Theorem.
The latter tells you how you can approximate any function by polynomials, and it
also tells you how big the error in the approximation can be. The error result is
proven using the MVT. The following example gives a very small taste of this type
of application.
CHAPTER 4. APPLICATIONS OF DERIVATIVES
106
Example 3. Suppose you know that f (3) = 7 and that f 0 (x) is always between 1
and 2. Find upper and lower bounds for f (4.5) and give a geometric interpretation
of your reasoning.
Solution. Graphically, we have the following picture, where the straight lines come
from the slopes of 1 and 2, and give upper and lower bounds for f (x) in general,
and f (4.5) in particular.
Although the picture is pretty convincing, we want an algebraic approach to finding
these upper and lower bounds, and show that indeed f (4.5) lies between them.
We apply the MVT with the interval [3, 4.5],
f 0 (x) =
f (4.5)
4.5
f (3)
.
3
We know something about f (3), we know something about f 0 (x) and we want to
find something out about f (4.5). Let’s start with the upper bound
2 > f 0 (x) =
f (4.5)
1.5
7
which we can easily solve for f (4.5)
2>
f (4.5)
1.5
7
) f (4.5) < 10
Now we get the lower bound
1 < f 0 (x) =
f (4.5)
1.5
7
) f (4.5) > 8.5
Comments. We all know that if f 0 (x) > 0 then the graph of f (x) is increasing
right? In fact, we could just take this as the definition of increasing couldn’t we? We
could say that by definition, f 0 (x) > 0 means a function is increasing. But before
we ever knew calculus, we probably knew what increasing meant. It meant that a
second value of f was bigger than a first value. To put it a little more precisely, it
meant this: if s < t then f (s) < f (t). If this follows so obviously from f 0 (x) > 0
then we should be able to prove it right? Right?!
CHAPTER 4. APPLICATIONS OF DERIVATIVES
107
Corollary. Suppose that f 0 (x) > 0 for all x in the interval [a, b]. Then f (x) is
increasing everywhere in [a, b].
Proof. Suppose that f 0 (x) > 0 for all in the interval [a, b].
Let s, and t be two x-values in the interval [a, b] with s < t. We want to show
that f (s) < f (t).
Apply the Mean Value Theorem to f on the interval [s, t]. Then there is some
number c such that
f (t) f (s)
f 0 (c) =
.
t s
But, since f 0 (x) is always positive, this means that f 0 (c) > 0 and so
f (t)
t
f (t)
f (t)
f (s)
= pos#
s
f (s) = pos# · (t
f (s) > 0
s) = pos# · pos#
f (t) > f (s)
This is where we ended on Wednesday, November 6.