We want to determine any relative extrema
using the second derivative test of the given function.
Let's do a quick review of the second derivative test.
The first step is to determine the critical numbers
of the function,
which is where the first derivative equals zero
or where it doesn't exist.
Remember, these are the possible locations
of any relative extrema.
And then what we'll do is determine the second derivative
and then evaluate the second derivative
at these critical numbers.
If the second derivative is positive at a critical number,
that means the function will be concave up,
and therefore at that critical number,
we would have a low point or relative minimum.
And if we evaluate the second derivative at its negative
at the critical number,
that means the function is concave down,
and therefore at the critical number,
we have a relative maximum.
And then if it happens the second derivative is zero
with the critical number,
the test fails, and we have to go back and use
the first derivative test.
So let's take a look at our example.
So the first step is to determine the critical numbers,
so we'll determine when the first derivative is equal to zero
or undefined.
So our derivative will be 12X squared minus 30X minus 18,
and the derivative of 12 would be zero.
This will never be undefined,
so let's go ahead and set it equal to zero
to determine our critical numbers.
The greatest common factor of these three terms would be six,
so let's factor out six.
We'd be left with two X squared minus 5X minus 3.
This is going to factor into two binomials.
Two factors of 2X squared would be 2X and X.
And the second position must come from the factors of -3,
so the sum of the inner and outer product equals -5X.
So if we put -3 here, that’ll give us -6X,
and +1 here, that’ll give us +1X, which does give us the -5X.
So the derivative equals zero when 2X + 1 equals zero
or when X - 3 equals zero.
So solving this equation for X, we'd have X equals -1/2,
and here we'd have X equals 3.
And these are the critical numbers,
which means these are the possible locations
of any relative extrema.
So now what we'll do is determine the second derivative,
and then determine the sign of the second derivative
at these two critical numbers.
So the second derivative
would be equal to the derivative of the derivative.
So here's our first derivative,
so the second derivative would be 24X minus 30.
So let's go to the next slide
and organize this information in a table.
So here are our critical numbers,
and then we'll determine the sign of the second derivative
which will tell us the concavity
and therefore tell us whether we have a relative max or min
at these two locations.
So we need to determine the sign of F double prime of -1/2
and F double prime of 3.
And here's our second derivative from the previous slide.
So we have 24 times -1/2 minus 30.
Well, this is going to be -12 minus 30, that's -42,
so that means the second derivative is negative
at negative 1/2.
And then at F double prime of 3, we're going to have 24 times 3 minus 30.
Well, 24 times 3 equals 72 minus 30 equals +42,
so the second derivative is positive at 3.
So now we can summarize our results.
Since the second derivative is negative at negative 1/2,
that means we have a relative maximum,
and since the second derivative is positive at 3,
that means we have a relative minimum.
And that wraps it up for our second derivative test.