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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 righthand limit of both sides
of this.
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View Full DocumentWe are also assuming that
exists and recall that if a normal limit exists then it
must be equal to both onesided 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 lefthand 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
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 Fall '08
 sc
 Derivative, Formulas

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