 
Hi, I'm Cheryl Leech.
In this problem,
we're going to be
using the second derivative
test to help us find
relative minimums and maximums.
What does the second
derivative test tell us?
Well, in the second
derivative test,
we're going to be looking
at when the first derivative
at a given point is equal to 0.
That means we have a
horizontal tangent.
Then we want to
examine what happens
to the second derivative.
And in this first picture
that I have drawn,
the second derivative
is greater than 0.
In other words, it's concave up.
And it turns out that if we
have a horizontal tangent
and it's concave up, the
only way that can happen
is if we have a
relative minimum.
 
On the other hand,
what if we again
have a horizontal tangent but
this time our second derivative
at that point is
negative, which means
our graph is concave down?
If we have a horizontal
tangent and it's concave down,
the only way that can happen is
if we have a relative maximum.
 
This is a nice, quick, easy
way to find relative minimums
and maximums, but you
have to be careful.
It doesn't always work.
One thing is that
it doesn't help
when we have a minimum or
maximum because our derivative
is undefined.
For instance, at a cusp.
And the other place you
have to be careful of
is-- what happens if the
second derivative equals 0?
We know what happens if it's
greater than 0 or less than 0,
but what if it
actually equals 0?
In this case, the test fails.
You don't know what happens.
It could be a relative minimum,
relative maximum, or maybe
neither.
So let's go ahead and take a
look at a particular example.
I have the function f of x
equals x to the fourth minus
4x cubed plus 2.
I'd like to find the relative
minimums and maximums.
So I'm going to try to use
the second derivative test.
So the first thing we
need to do is find out
when those first
derivatives are equal to 0.
So the first derivative is
4x cubed minus 12x squared.
We'll set this equal to
0 and then solve for x.
Notice that I can factor a
4x squared out of both terms,
and I'm left with x minus 3.
Two things multiplied
together that equals
0 means that one of
that needs to be 0,
so either 4x squared
is 0 or x minus 3 is 0.
And in this case, x squared
is 0, and so x is 0.
And in my second situation,
x is going to equal 3.
So there are two times
when the first derivative
is going to equal 0.
Now we have to look
at the concavity.
So to look at the concavity,
we need the second derivative,
so let's go ahead and calculate
the second derivative.
 
My second derivative is going
to be 12x squared minus 24x.
So now let's see what
happens at 0 and at 3.
 
The second derivative
at 0 is going to be 0.
Well, now, wait a minute.
That means my test fails.
 
I'm going to hold onto
that for a minute,
and I'll come back to it.
Let's go ahead and
take a look at what
happens when x equals 3.
So I'm going to have 12 times
3 squared minus 24 times 3.
Now this is definitely going to
be a positive number because I
have 12 times 9 minus
72, or 108 minus 72,
which again is definitely
a positive number, which
means my graph is concave up.
And if it's concave up and
slope of the tangent line is 0,
we must be talking about a
relative minimum at x equals 3.
 
So let's go back to the case
where my second derivative test
failed.
When this happens,
we can always go back
to the first derivative test.
So we'll come back to
our first derivative
and make our number
line, marking down
all of the critical numbers.
I want to make sure that I
include that 3 because when I'm
doing test points, I
don't want to choose
anything larger than 3.
I'm just interested in
what's happening at 0.
So pick any number that
you'd like less than 0--
for instance, a negative
1-- and let's plug it
into the first derivative.
If I do that, I'm going to end
up with a negative 4 minus 12,
which is a negative number.
Now I'm going to choose
something between 0 and 3.
Let's try 1.
When I plug a 1 into the first
derivative, I get 4 minus 12,
which is also a negative number.
Notice that there was no change
from increasing to decreasing.
It was strictly decreasing
the whole way through,
which means that there was
no relative extrema occurring
when x equals 0.
Our only relative minimum
was when x equaled 3.
I hope this helps.
Thanks.
