Proofs of Derivative Applications Facts/Formulas
In this section we�ll be proving some of the facts and/or
theorems from the
Applications of Derivatives chapter. Not all of the facts and/or theorems
will be proved here.
Fermat�s Theorem
|
If
has
a relative extrema at
and
exists
then
is
a critical point of
.
In fact, it will be a critical point such that
. |
Proof
This is a fairly simple proof. We�ll assume that
has
a relative maximum to do the proof. The proof for a relative minimum is nearly
identical. So, if we assume that we have a relative maximum at
then
we know that
for
all x that are sufficiently close to
.
In particular for all h that are sufficiently close to zero (positive or
negative) we must have,
or, with a little rewrite we must have,
(1)
Now, at this point assume that
and
divide both sides of
(1) by h. This gives,
Because we�re assuming that
we
can now take the right-hand limit of both sides of this.
We are also assuming that
exists
and recall that if a normal limit exists then it must be equal to both one-sided
limits. We can then say that,
If we put this together we have now shown that
.
Okay, now let�s turn things around and assume that
and
divide both sides of
(1) by h. This gives,
Remember that because we�re assuming
we�ll
need to switch the inequality when we divide by a negative number. We can now
do a similar argument as above to get that,
The difference here is that this time we�re going to be
looking at the left-hand limit since we�re assuming that
 .
This argument shows that
.
We�ve now shown that
and
.
Then only way both of these can be true at the same time is to have
and
this in turn means that
must
be a critical point.
As noted above, if we assume that
has
a relative minimum then the proof is nearly identical and so isn�t shown here.
The main differences are simply some inequalities need to be switched.
Fact, The Shape of a Graph, Part I
|
1.
If
for
every x on some interval I, then
is
increasing on the interval.
2.
If
for
every x on some interval I, then
is
decreasing on the interval.
3.
If
for
every x on some interval I, then
is
constant on the interval. |
The proof of this fact uses the
Mean Value Theorem which, if you�re following along in my notes has actually
not been covered yet. The Mean Value Theorem can be covered at any time and for
whatever the reason I decided to put where it is. Before reading through the
proof of this fact you should take a quick look at the Mean Value Theorem
section. You really just need the conclusion of the Mean Value Theorem for this
proof however.
Fact, The Shape of a Graph, Part II
|
Given the function
then,
- If for
all x in some interval I then is
concave up on I.
- If for
all x in some interval I then is
concave down on I.
|
The proof of this fact uses the
Mean Value Theorem which, if you�re following along in my notes has actually
not been covered yet. The Mean Value Theorem can be covered at any time and for
whatever the reason I decided to put it after the section this fact is in.
Before reading through the proof of this fact you should take a quick look at
the Mean Value Theorem section. You really just need the conclusion of the Mean
Value Theorem for this proof however.
|
has
a relative extrema at
and
exists
then
is
a critical point of
.
In fact, it will be a critical point such that
.
has
a relative maximum to do the proof. The proof for a relative minimum is nearly
identical. So, if we assume that we have a relative maximum at
then
we know that
for
all x that are sufficiently close to
.
In particular for all h that are sufficiently close to zero (positive or
negative) we must have,
and
divide both sides of
we
can now take the right-hand limit of both sides of this.
exists
and recall that if a normal limit exists then it must be equal to both one-sided
limits. We can then say that,
.
and
divide both sides of
we�ll
need to switch the inequality when we divide by a negative number. We can now
do a similar argument as above to get that,
.
and
.
Then only way both of these can be true at the same time is to have
and
this in turn means that
must
be a critical point.
has
a relative minimum then the proof is nearly identical and so isn�t shown here.
The main differences are simply some inequalities need to be switched.
for
every x on some interval I, then
is
increasing on the interval.
for
every x on some interval I, then
is
decreasing on the interval.
for
every x on some interval I, then
is
constant on the interval.
then,
for
all x in some interval I then 
is
concave up on I.
for
all x in some interval I then 
is
concave down on I.
