Second Derivative Test
Suppose that
is a critical point of
such that
and that
is continuous in a region around
. Then,
1.
If
then
is a relative maximum.
2.
If
then
is a relative minimum.
3.
If
then
can be a relative maximum,
relative minimum or neither.
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.
Proof of 1
First since we are assuming that
is continuous in a region around
then we can assume that in fact
is also true in some open
region, say
around
,
i.e.
.
Now let
x
be any number such that
, we’re going to use the Mean
Value Theorem on
. However, instead of using it on the function itself we’re
going to use it on the first derivative. So, the Mean Value Theorem tells us that there is a
number
such that,
Now, because
we know that
and we also know that
so we then get that,
However, we also assumed that
and so we have,
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Or, in other words to the left of
the function is increasing.
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 Fall '08
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 Calculus, Critical Point, Derivative, Mean Value Theorem, Continuous function

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