BAM!
Mr. Tarrou.
In this video we are going to continue our
look at how derivatives can help us graph
functions... polynomial functions, rational
functions, ultimately we are going to...
We just got done looking at the first derivative
which allows us to find out where functions
are increasing and decreasing which allowed
us to then identify where the relative max
and mins where in a function, and actually
find out where those where as opposed to PreCalculus
where we just had to guesstimate what those
values were.
Or use a calculator to tell us what the relative
max and mins were.
We are going to be looking at the second derivative
test.
With the second derivative we are going to
be able to talk about concavity, we are going
to be able to talk about inflection points.
We are going to be using that Second Derivative
Test if we have the second derivative available
to help us identify relative max and mins,
just like the derivative test does for us.
I actually prefer the First Derivative Test
(sometimes) if I have that available.
So let's talk about concavity right now.
Let f be differentiable on an open interval
I, the graph f is concave up on that interval
if the derivative of f is increasing on the
interval and concave down if the derivative
of f is decreasing on the interval.
So concave... concave up... concave down.
If you look at this graph, there is a portion
of this graph... that if I could pour water
in it, it could hold water.. assuming it was
three dimensional.
And it is kind of.. you know... it is curving
up.
Right.
Well, this function.. we look at it and the
first derivative is going to be...
Actually not equal to zero, but negative because
the graph is falling.
We have the first derivative here which is
going to be equal to zero.
So in other words the graph has a negative
slope, the graph has a slope equal to zero,
and then over here... now over here we have
a graph whose first derivative or slope is
positive.
And the graph continues to increase right.
You can see that it starts to... the rate
in which the slope is increasing starts to
slow and then it starts to fall again right.
Instead of a relative min somewhere else,
a relative max possibly.
Well, where the that rate of increase on the
slope.. or the first derivative stops increasing
and then starts to decrease, at that point
of inflection, then we are going to end up
somewhere else where the function is maybe
like this diagram concave down.
So concave up and then concave down.
You can see here as well, that I basically
have the first derivative test up here.
We have a point, a critical value, where the
first derivative is equal zero.
To the left the first derivative is negative
and to the right it is positive.
That would be your relative min.
We will talking later about how the second
derivative helps us to identify this as well.
And to the left of this critical value where
the first derivative is equal to zero we have
a positive derivative and a negative derivative,
or positive slope and then negative slope,
setting up a relative maximum.
Ok.
So test for concavity.
Let f be a function whose second derivative
exists on an open interval.
If your second derivative is positive for
all x in that I, then the graph of f is going
to be concave up.
If the second derivative is negative, or that
is going to be the same thing except that
is going to be indicating that the function
is concave down.
So in other words in this graph right here
I have.. you know.. just some curve representing
a function and actually only identified the
values or the sign of the values for the first
derivative.
But if I went another step further and took
the derivative of the first derivate and found
the second derivative, then within this interval
my second derivative would be positive.
And if I took again that... look at that second
derivative or take the derivative of my first
derivative, then in this interval my second
derivative value would be negative.
Now we are going to look at some graphs on
the next page and just help us visualize what
this first and second derivative is telling
us, where might we find relative max and mins,
or even relative max and mins where the first
derivative is undefined.
Alrighty!
What we have in this screen is what I forgot
to mention on the bottom right hand corner
of my chalkboard which is how do we set up
the interval to test for concavity.
Well, you need to find that second derivative
of course.
You are going to then basically do the same
thing that you did when you did the first
derivative test.
You are going to find out where the second
derivative is equal to zero, where it is undefined,
and like you can see in my first example here
where is the function.. the original function...
undefined.
Where are the discontinuities.
Within each of those intervals you might have
the function be concave up or concave down.
So find the second derivative, find out where
it is equal to zero or undefined, and find
out where that function... the original function...
is undefined.
Now if the function, the original function,
is undefined at some value like this first
example it is undefined at values of pi and
3pi well then of course the first and second
derivatives are going to be undefined as well.
So why worry about where the original function
is undefined?
Well, when you have the concavity change you
might have.. let's say like this example..
now these all are kind of special case examples...
And well, actually we have concavity that
is both up to the left and the right at the
same time.
But we are going to be talking about inflection
points and relative max and mins.
And actually that is what I was thinking of,
this function here goes from concave to concave
down and back to concave up.
You would not want to try and identify an
inflection point where the function is undefined.
And a function can be continuous where the
first and/or second derivative is undefined.
so you need to check all three things.
So here we have, well what kind of function
do I have graphed up here by the way?
How much do you remember your Trig?
I have got sort of a shadow graph starting
at one, going through pi, dipping to a minimum
value at 2pi, going to 3pi and then finally
4pi.
So I have graphed as a guide a cosine with
a period of 4pi and then when in the opposite
direction.
I kind of blew the graph inside out.
Well, if this is a cosine function then I
must have up here the graph of secant.
So we have secant graphed up here with a period
of 4pi.
It continues of course forever but we are
going to look at this interval of 0 to 4pi.
We have a relative min, a relative max, and
a relative min again.
We have intervals where the function is concave
up, concave down, and then finally again concave
up.
Of course this will repeat forever if I let
the domain go off to.. you know.. negative
and positive infinity.
And see what is happening is I have got a
place where the second derivative is undefined
at pi and at 3pi.
To the left of that value, or the interval
between zero and pi I have a graph that is
concave up.
I have between pi and 3pi a graph which is
concave down, and then back to concave up.
So I have identified some areas of concavity
that are different.
But I don't have an inflection point because
at pi and at 3pi the graph is not continuous.
There is no inflection point, and so again
you need to find out where that original function
is undefined.
Ok.
And just again highlighting here your Second
Derivative will be positive, negative, and
positive indicating that concavity of being
up, down, and back up.
ok.
So again, no inflection points.
Here we have a function that has a vertical
tangent line which means that both... really
the first derivative and second derivative
are going to be undefined and that is a critical
value.
Again you need to find out where the second
derivative is zero, undefined, and where the
original function is undefined or has a discontinuity
to set up your intervals.
So we have a second derivative which is undefined,
a second derivative which is equal to zero,
and again that would set up my number line
into some intervals.
I will test just one number, I only need to
do one number, in each of these intervals
plugging into the second derivative and seeing
if the sign is positive, negative, or positive.
Now this one I have drawn is smooth and continuous
everywhere so we have inflection points at
(C1,f(C1)).. so when I find that interval
where the concavity goes from being positive
to negative, then I am going to want to take
that value of c and plug it back into the
original function to identify the inflection
point that I am sure that you are going to
be needing to find and tell your teacher just
like my students are going to need to do.
Ok.
So..
Oh...
Also what I want to mention here is just because
you find a critical value does not necessarily
mean, like we learned in the First Derivative
Test that you are going to find a relative
max or min.
This graph actually takes a couple of breaks.
It has a place where there is a vertical tangent
line, a place where there is a horizontal
tangent line, but it just keeps climbing again
and that first derivative never changes from
being positive (other than the critical values).
So this graph would not have a relative max
or min even though we ended up finding a couple
of critical values.
Here we have a sharp bend which means... and
it is continuous.
And we are looking at a function here that
is not smooth, but continuous everywhere.
But the first and actually second derivative
is going to be undefined at some value of
c.
Maybe I should have written that up here.
So there is some value of c where the function
has a second derivative which is undefined,
so it is a critical value.
And we test maybe the left and right intervals
to the left and right of c.
We find that our second derivative is positive
on both sides.
We can also see that we have a slope which
is positive so our first derivative is going
to be positive.
Our first derivative on the right hand side
is going to be negative.
So we have a relative maximum value that we
could identify with the first derivative test.
Testing to the left and the right of that
critical value to see if we indeed have a
relative max or min.
And if we go from positive to negative (1st
derivative) we have a relative max.
So that is at least an idea with a few examples
here of some special cases where you are going
to be looking for how the concavity changes.
Let's take a look at a smooth curve and identify
what could possibly the first derisive and
then second derivative look like given that
original function before we get onto some
more notes, and finally those three examples.
BAM!
So we have a function that appears to be a
cubic function, or a third degree where there
is two bends here.
So I want to lay over this what could potentially
be a first and second derivative function,
and then fill in some information at the end
for your notes.
So seems to be that the function is bottoming
out here.
So we know that the first derivative at that
point is going to be zero.
It is a critical value so I am going to come
down here and highlight that.
We are going to make sure that we hit that
point on the x axis as something that is quite
important.
We have another place over here where the
first derivative appears to be zero.
I am going to drop that down as a critical
value.
And then since we are talking about inflection
points even though I have not really defined
it completely for you yet, the graph is falling
and it kind of like looks like if I just kept
going it would just sort of swing around like
and then there is a place where the graph
is sort of cupped down if you will.
Well, it is concave down.
So somewhere in the middle there, there is
an inflection point where the graph.. the
rate of change for slope... it is increasing
increasing increasing, and the rate of change...
the slope starts to decrease, the rate of
change for the slope.
So somewhere in here is the inflection point.
I am going to highlight that right here.
And let's get the first derivative up here.
Now don't forget the first derivative of course
describes the slope of the original function.
And along here my slope is negative, so my
first derivative needs to be giving me values
which are negative.
Or in other words the derivative needs to
be below the x axis, at least until we get
to here where the first derivative is equal
to zero.
So, actually let's pick a different color.
Maybe it will look something like that.
Now the graph goes from being zero, and the
slope is increasing and getting greater and
greater until at the inflection point it seems
like the slope is as steep as it is going
to be before it starts to fall again.
So I draw my first derivative that needs to
be where the highest value that my derivative..
you know roughly is because this is only a
sketch and it is a guestiment here.
But maybe it looks something like this.
And then it starts to fall until at this point
where the slope again stops being positive,
goes to zero, and then starts to fall.
Well, from here the slope is negative so my
first derivative needs to be below the x axis.
Whoo!
So pretty.
Alright, so that is f prime or the first derivative
of f.
And um, looks like my highest point is not
really my inflection point.
Let me just cheat and move that over a little
bit.
Ok.
I am sure that was so important to move that
half an inch, right.
Well now this is my first derivative.
Ok.
Now let's talk about graphing the second derivative.
By the way, this kind of looks like a parabola,
right.
I have a function that appears to be a third
degree and now I have a function that appears
to be a second degree.
And that should be normal, right.
Because when you find the derivative of say
y=x^3, it becomes y=3x^2.
So it seems normal, and actually very correct...
this first derivative looks like a parabola.
Now let's put the second derivative up here.
Now the second derivative is the slope of
the first derivative.
It just keeps going in steps.
And from here up until where my first derivative
ends up sort of maxing out..
Actually I can kind of just, I could just
say that this is the original function, right.
My function has a positive slope, or my derivative
has a value slope until it gets to this critical
value, and then it starts to fall.
So if I want to draw the second derivative
which is the slope of the first derivative,
it had better be above the x axis where the
first derivative has got a positive slope.
So we are going to put that dot right there
on this critical value where the first derivative
goes from being positive... excuse me.
Stops rising and starts to fall.
And let's draw this line right here.
Hit that critical value which ends up actually
being the inflection point.
So that definition is coming up very very
soon.
And that is purple line is the sketch or an
approximation of what our second derivative
is going to look like.
So again, our original function is falling
until we get to this point.
So the first derivative slope represents that
falling, or that negative slope.
The bottom out.
We get a slope which is positive so my first
derivative is above the x axis, where my original
function is rising.
Then it starts to fall so that first derivative
starts to be under the x axis.
Now getting to the second derivative, the
first derivative has a positive slope until
we get to here.
So the second derivative is above the x axis.
Then the first derivative starts to fall with
a negative slope, and thus the second derivative
is below the x axis.
So what we see here is where the second derivative
is equal to zero is a place where we possibly
find an inflection point.
You will find an inflection point if their
is a sign change.
Those inflection points can happen where the
second derivative is equal to zero or where
it is undefined.
Now again, if the second derivative is undefined
you need to make sure that your original function
is not discontinuous at that point.
I am going to pause the video and write some
notes up on this diagram now that we have
it drawn so you can include them in your notes.
[nananananana] Alright.
I just filled in some notes.
i am not going to talk about this too much,
because obviously I just did and it is repetition
of a lot of what I have already pointed out.
Other than, here we have the inflection point
and I will give you a formal definition of
that in the next screen before we get to the
examples.
And these orange dots, the relative min and
the relative max, this is a preview or a visual
representation of your Second Derivative Test
which will also be on the next screen.
If your first derivative exists and is equal
to zero and the second derivative is positive,
then at that point of whatever...
(C1,f(C1)), you have a relative min.
If the first derivative is equal to zero and
the second derivative is negative at that
same point, then you are finding a relative
maximum.
Honestly again, there is less work if you
start from the original function it is less
work to get to the first derivative obviously
than the second derivative, and you will see
in the next screen when we talk about the
Second Derivative Test that it does not always...
it is not always able to tell you if a point
is a relative min or max.
So honest, I kind of really prefer the First
Derivative Test for identifying those relative
max's and min's (sometimes 2nd Der Test much
quicker!)
But the Second Derivative Test is still something
we have to be aware of and occasionally use
depending on how they give us the information
in the problem.
Definition of Point of Inflection.
Let f be a function that is continuous on
an open interval and let c be a point in that
interval.
If the graph of f has a tangent line at this
point (c,f(c)), then this point is a point
of inflection of the graph if the concavity
of the function changes from either being
positive.. or concave up to concave down,
or concave down to concave up.
So basically concavity changes on each side
of this value of c, or to the left and right
of (c,f(c)).
What this is highlighting here is my book
puts this in the definition of a point of
inflection, some books apparently do not,
and what that means is that my book would
only look at this point as a point of inflection.
At this point (c,f(c)) to the left I am concave
up and to the right I am concave down.
And this would just highlight as a relative
minimum.
But, some books would say that this function
is continuous here, it is concave down to
the left and concave up to the right thus
this is also a point of inflection at that
sharp bend.
Remember you cannot find... you don't have
a tangent line a sharp bend.
So, this point would fail in my particular
book's definition of point of inflection.
Now of course if the function is undefined,
then that is still one of the test point intervals
remember...
That where the function has got a discontinuity
or where the second derivative is undefined,
to the left and to the right we have a change
of concavity.
But, you cannot have an inflection point of
there is no point.
By the way, the second derivative would be
undefined at a sharp bend as well.
It is just my book is not going to call that
a point of inflection, in some books it will.
Ok.
Points of inflection occur where the second
derivative is equal to zero or where the second
derivative is undefined.
Remember that first screen with the three
pictures we had that vertical tangent line.
And we had a point of inflection there.
It went from being concave up to concave down.
Second Derivative Test, or let f be a function
such that the first derivative of f... or
f prime of c is equal to zero.
So our first derivative is equal to zero,
or the slope is zero of our original function,
and f double prime exists on an open interval
containing c.
Well, again we are only talking about where
the first derivative is equal to zero.
If f double prime of c is greater than zero,
then f has a relative minimum at (c,f(c)).
If your second derivative is greater than
zero, then you are concave up and of course
at the bottom of that concaveness... or the
bottom of that curve you are going to have
a relative minimum.
If the second derivative is negative or that
means your function is concave down, then
at (c,f(c)) you have a relative maximum.
Again we are talking about only the points
where f prime of c, the first derivative,
is equal to zero.
Now it is possible in doing the second derivative
test that you find a place where your first
derivative is equal to zero, but when you
go to find f double prime of c it comes out
to be zero.
That means basically you have run a test but
you have not come to a conclusion, thus you
will have to fall back on the first derivative
test.
So if you are giving me an actual original
function and just said find the relative max's
and min's I would probably just find the first
derivative and do the first derivative test
because sometimes the second derivative test
fails.
WHOOO!!!
That is a lot of notes.
So let's get to some examples:)
Man!
Tired of writing all of these notes.
My hand is hurting.
First example.
Determine the open intervals on which the
graph is concave upward or downward.
find the inflection points and any relative
max or min's.
We are going to find those relative max or
min's by looking for critical values of the
first derivative, but I want to first talk
about concavity.
That means that I need the second derivative.
So if this is our polynomial y, and by the
way polynomials have a domain of all real
numbers so we don't have any issues with the
original function being undefined or having
discontinuities.
So our domain is all real numbers and y prime..
the first derivative, is 2/3 times 3 is equal
to 2x^2...
2 times 2 is 4x.. and then minus 6.
We are going to be using that again to find
those relative max and min's in a second.
But, I am going to find the second derivative.
That means just repeat the process.
We have 2 times 2 is 4x... plus 4.
Ok, now remember that those inflection points
and the changes of concavity occur where the
second derivative is equal to zero or undefined.
If that original function has a discontinuity
there is no inflection point.
We already know that does not.
We are going to take a look at the second
derivative.
And we are going to look for where y double
prime of x is equal to zero.
We have 4x+4 is equal to 0.
We are going to subtract both sides by 4.
Divide by 4.
We get x is equal to 1.
Excuse me, negative one.
Well.
I also need to find out where the second derivative
is undefined.
And where the second derivative is equal to
undefined, well that has no solution.
So the only critical values I need to use
to set up my intervals to check for concavity
is the x value of -1.
So my intervals are... my test intervals are
going to go from negative infinity to -1 and
then from -1 to positive infinity.
And my second derivative in those intervals,
I am going to take one value from this interval,
plug it back into the second derivative and
determine if it is positive or negative.
so I want to see.. let's say... y double prime
of -2.
Ok, well 4 times -2 is -8, -8 plus 4 is -4.
So my second derivative in this interval is
negative.
That means it is concave down.
And on my interval from -1 to infinity I want
to test one point in there like 0.
So y double prime... well 4 times 0 is 0...
0 plus 4 is 4, and so it is nice to know it
is four, but really all I care about is whether
that is positive or negative.
Since my second derivative is positive, then
I have a concave up situation on that interval
of -1 to infinity.
Now at the value of -1 my function, which
I know is defined at -1 because it is defined
everywhere, at that value of -1 my function
goes from being concave down to concave up.
That means I have an inflection point.
So my inflection point is going to be (-1,f(-1)),
or -1... now I am going to right off the top
of my head take -1 and plug it into this original
function and find that the y value is 8 and
1/3.
And I am out of space so let me clean this
up a little bit and I will highlight or write
down what information we have found so far
where the function is concave up and down,
and the point of inflection and continue this
problem.
Relative max or min's.
Well, we have already studied the First Derivative
Test and learned all about that first derivative.
So we are going to take that first derivative
and we are going to set it equal to 0.
Find out where it is equal to zero.
So we have got 2x^2+4X-6 is equal to 0.
We are going to factor out a 2, or actually
we could just divide both sides by 2.
And we get x^2+2x-3=0.
That is a quadratic.
Of course we can solve it by factoring, we
can solve it by completing the square, we
can solve it by the quadratic formula.
But I knew I was going to have to do this
example on the video right in front of you
of the top of my head, so it is pretty easily
factorable.
So this is going to be (x+3) times (x-1) is
equal to zero.
So our first set of critical values here are
x is equal to -3 and 1.
Our other critical values would be for where
the first derivative is equal to undefined.
And that we have no solution.
So we just need to check for relative min's
and max at the x value of -3 and 1.
Now how are we going to do that.
I could do the first derivative test and you
know plug in value to the left of -3 the right
of -3, and to the left of 1 and the right
of 1.
But I am supposed to be teaching you the Second
Derivative Test not the First Derivative Test.
So I am going to take that value of -3 and
I am going to decide if it is a location of
a relative min or relative max.
I can plug it into the second derivative which
is 4x+4.
But, -3 is in this interval.
And within the interval of negative infinity
to -1, well the original function is concave
down.
So if the function is concave down and somewhere
in that interval...
-3... the first derivative is equal to zero,
then I kind have already done the Second Derivative
Test.
Where the first derivative is equal to zero
my second derivative is... well... let's see...
Concave down, so that second derivative was
negative.
So we have a relative maximum at (-3,f(-3)).
And again right off the top of my head, that
comes out to be (-3,19).
Now remember I am plugging that negative three
back into the original function.
Ok.
My other critical value is at 1.
Well, again the second derivative test says
where the first derivative is equal to zero,
if the second derivative is positive then
we have a relative minimum and if the second
derivative is negative... so it is concave
down... we have a relative maximum.
Well at 1 that is in this interval where the
function is concave up.
So my first derivative is zero, my second
derivative is positive, so I am going to take
this value and say.. ok.. well my relative
min is equal to (1,f(1)) which turns out to
be 1... comma... take that 1 and plug it back
into here.
Right off the top of my head we have a relative
minimum at (1,-2 1/3).
Ok.
On to the next example.
It is going to be obviously...
I am going to go harder as we get through
our examples.
This was a basic polynomial.
The next function is going to involve finding
the derivative using the Product Rule and
dealing with some square roots.
Second example.
f(x) is equal to x times the square root of
x+6.
So we are going to rewrite that as f(x) is
equal to x times (x+6) to the 1/2 power.
Walk through the product rule.
f prime of x is going to be x times... now
welcome to your chain rule... we have got
1/2 times x+6 to the -1/2.
And if you want to just make sure you are
practicing the chain rule, times the derivative
of the inside which is going to be simply
equal to 1.
Which of course means that is really not necessary
unless you are just making sure you are reinforcing
your skills of the Chain Rule.
Now plus...
Ok.
So that was the first times the derivative
of the second factor.. plus the second factor
which is going to be (x+6) raised up to the
1/2 power... times the derivative of the first
factor which is going to be equal to 1.
Ok, so usually when you have radical, or especially
when you are working the quotient rule, there
is going to be a little bit factoring going
on.
Both of these terms have a factor of (x+6).
You are going to take out the lowest power
which is the negative 1/2.
And I like to, when I have a coefficient which
is a fraction because there is going to be
some stuff floating.. as you can see with
the negative exponent to the denominator..
take that fraction out as well.
So I am going to factor out a 1/2 and the
(x+6) to the -1/2 power out of both of these
terms.
And 1/2 divided by 1/2 is 1.
Anything divided by itself again with that
(x+6)^-1/2 is going to be equal to 1.
So we are just left with x... plus... and
then when you divide like bases of course
you subtract the exponents.
I am forgetting that I am dividing by that
1/2, so 1 divided by 1/2 is 2... times (x+6)^1/2
divided by (x+6)^-1/2.
When you divide like bases you are supposed
to subtract the exponents.
Well, 1/2 minus -1/2 is 1/2 plus 1/2 which
1.
So we end up with (x+6) to the first.
Now we are going to distribute this 2 through
and we are going to show all of our steps
as we do these problems because they are very
meticulous and if you make a small mistake
you can carry it through very easily and not
notice it until you are done and you check
your answer.
The more work you show the easier it is going
to be to find your mistake and then fix it.
So this two is already in the denominator,
that 1/2 that we factored out and the x+6
has that negative 1/2 power.
So we are going to write the first derivative
of f, or f'(x), is equal to...
Alright so we have x+2x, just showing all
of the steps, 2 times 6 is 12.
On the bottom we have 2 times... well..
I would write the square root of x+6 if knew
I was done.
But, I want to take one more step and find
the second derivative for the concavity and
the inflection points.
I am going to leave that as x+6 to the 1/2
power in the denominator.
Of course we add like terms and we get the
first derivative is 3x+12 over 2 times (x+6)^1/2.
Now I have to go through another derivative
process but this time using the Quotient Rule.
It is going to be a lot of steps and I don't
want to talk and write something incorrect
and not catch it in the video.
So I am going to actually pause it and just
sort of reveal the answer.
It would be a good time to make sure that..
well... if you are really good with your derivatives
maybe you could just see the answer.
If you sometimes make some small mistakes
or just want to challenge yourself for a second
to review your quotient rule for finding derivatives,
give it a shot.
Pause the video.
Ok.
So we have got.. well just the quotient rule.
I have the...
What did we have here.
The denominator times the derivative of the
numerator minus the numerator times the derivative
of the denominator over... just showing every
single step and doing the times one again
because I just showing the continuation of
the chain rule until I am done even though
i am just multiplying by one.
I might not show that in my own work.
These again both have a common factor of (x+6).
You take out that lowest exponent.
And what else did I want to show you?
Oh!
When I bring this x+6 to the negative 1/2
down to the denominator, where the 3/2 came
from was again I am bring this down.
When I bring this down it is going to be (x+6)
to the 1/2 power, and when you multiply like
bases you add the exponents.
So 1 plus 1/2 of course is 3/2.
And well, now we have our second and our first
derivative and I am out of space.
So I am going to clean this up, put them over
here, we will find those critical values for
the second derivative and find the intervals
for concavity, the inflection point, go back
to the first derivative and... excuse... we
are only doing the second derivative test,
and find those relative max and min's with
the Second Derivative Test after we find out
where the first derivative is equal to zero.
It was a dark and stormy night when I was
learning about the Second Derivative Test!
I don't know if you are going to hear the
thundering, but it just started to downpour
as I do this video.
Alright.
So I want to test for concavity.
I need to find the critical values for the
second derivative.
Now I kind of have a bad habit of this, where
I just start into the problems and I keep
forgetting to check the domain of the original
function.
We cannot square root a number, or even root
excuse me... it is true for square root too,
we cannot even root a number which is negative.
So before we try to find our critical values
and identify where the function is concave
up and concave down and where the inflection
points are, don't do any work where the function
is undefined, right.
So this function's domain, I was going to
talk about that earlier in the video, the
function's domain is.. well..
x+6 has to remain greater than or equal to
zero.
So our domain is x is greater than or equal
to negative six.
So any values that we find, a critical values
outside that domain... you know... is not
going to be usable.
And I could write that from -6 to infinity
of course.
So ok...
So for our second derivative our intervals
of interest we need to find out..
Well we are not going to set up our intervals
if we don't know our critical values.
Where is the second derivative equal to zero?
Where is f double prime equal to zero?
Well, for a rational function to equal the
numerator has to be zero.
So we have got 3x+24 is equal to zero.
That is 3x is equal to negative 24.
And x is equal to negative 8.
So that should be one of our critical values,
but -8 is not within our domain so that is
not a usable number.
Our second derivative, where is it equal to
being undefined?
The original function, other than this domain
of -6 to infinity does not have any other
issues with domain, but now our second derivative
has a variable in the denominator.
So what is going to make that equation undefined?
Well, that is going to be where x+6 is equal
to zero.
And x therefore is equal to negative 6.
That is where our second derivative is undefined.
That is just at the very edge, as I stand
in front of it, of our domain.
Our domain actually starts at negative 6 and
goes to the right.
So we only have one variable of interest as
far as checking concavity.
That is from -6 to infinity.
So..
I can put that here.
Our interval, intervals that we care about
for the second derivative is from -6 to infinity.
If I want to find out where the function...
if the function is concave up or concave down
in that interval I need to take just one value
within this interval.. like...let's pick the
easiest one and use a test value of 0.
Well, the second derivative in this interval,
if I take 0 and I plug it into my second derivative
I get 3 times 0 is 0... plus 24 so the top
is positive.
In the bottom I have 0 plus 6 which is positive.
So I have a positive number in the numerator
and a positive in the denominator, so my second
derivative from -6 to infinity is positive.
And if my function has got a positive second
derivative, then that means that we are concave
up.
Ok.
So we need to define concavity, we need to
find inflection points.
We only have one test interval so there is
no place for the function to change from being
concave up to concave down, so we have no
inflection points.
Therefore the only thing left is to identify
any relative max or min's.
Alright, so let's see.
Let me get out of the way and get this board
cleaned up.
I keep writing so big and we will come back
with that First Derivative and find any relative
max or min's.
Ok to help us finish up the last piece of
this puzzle which is where are the relative
max and min's.
I went ahead and cheated a little bit because
you could very be doing these problems without
the aid of a calculator in some schools.
But, I drew us a function just to remind of
a couple of things here.
We can see the concave up that we find between
-6 and infinity.
Secondly we are going to find a relative min
in here somewhere.
It should be obvious because we only have
a concave up situation.
But I also wanted to just highlight with this
particular question that it is defined at
-6.
I can take -6, remember that is where our
domain started, and I can take that -6 and
plug it.
-6 plus 6 is zero, the square root of zero
is zero, and thus the answer is 0.
So we do have....
RELATIVE MAX AND MIN'S CANNOT BE AT THE ENDPOINTS
That is a closed dot, it is not an open dot.
If it was open at -6, we would not be able
to find a relative maximum there.
Ok, so relative min's and max occur where
the first derivative is equal to... well has
a critical value whether it is equal to zero
or being undefined.
So let's find those values.
f prime, where is that equal to zero?
Well, it is again a rational function so let's
just set the numerator equal to zero.
We have got 3x+12 is equal to zero.
That is 3x is equal to negative 12.
Divide by 3 and x is equal to -4.
Ok.
Well, again at the x value of -4 the first
derivative is equal to zero and the second
derivative, if I come back here to my interval
where I determined that the function was concave
up because my second derivative was positive..
If where your first derivative is equal to
zero and your second derivative is positive,
you have a relative minimum.
So the relative minimum 
is at (4,f(4)).
Which is.. coming back to my original function,
4 plus 6 is 10.
The square of 10, I am just going to leave
it like that with a 4 out front.
Excuse me.
Don't forget those negatives.
(-4,-4 SquareRoot 10) Well that is going to
change that answer, isn't it.
Let's try that again.
-4 plus 6 is 2.
And we have -4 square root of 2.
Alright.
Just make sure.
Excellent.
Ok. Bam!
Last example.
Now like before I am going to walk out and
we are going to find that first and and especially
second derivative so we can talk about concavity
and the inflection points.
We are also going to be highlighting the domain
of this original function after I reveal the
derivatives to you.
If you want to pause the video and practice
your derivative skills with your trig functions.
Make sure you remember, and my students will
hopefully remember these already.. have them
memorized, the derivative of secant of x is
equal to secxtanx and the derivative of tanx
is equal to secant squared.
I am going to reveal this in small steps incase
you get stuck.
You can kind of see all of that in the next
line and just keep going, or you can just
watch the answers get revealed.
I just wanted to start off by finding the
domain.
So I wrote secant in terms of cosine just
to make sure you realize that when the denominator
is equal to zero the function of course is
undefined.
And cosine of x/2, if you set that equal to
zero and solve it then you are going to...
you know...
You might look at this half angle and go,
"I remember when I worked with multi-angles
in PreCalculus/Trigonometry that the half
angle told me how many times I had to go around
the unit circle."
If was a half angle, I only had to go around
the unit circle half way.
But that is because the answers they were
looking for were between 0 and 2pi, and this
restricted domain for this problem is 0 to
4pi.
So I actually have to go around the circle
basically twice as far as I would normally
do.
Just keep going through and ok...
Inverse cosine of zero.
That is going to be where your pi/2 or 90
degrees and 3pi/2, and maybe you might carry
over.... and let's see here... 1, 2, 3...
maybe you carry on to 5pi/2.
But um... you know... you would stop those
angles where cosine is equal to zero.
you have a half angle, so your step to solve
for x is going to be to multiply both sides
by 2.
And if you happen to make an extra angle that
is beyond this domain, then you would just
forget about it, or just not include it.
So this function's domain is all real numbers
except for where x is equal to pi and 3pi....
excuse me... 3pi.
Hope you got that.
We just of course used the chain rule.
The derivative of secant is secant tangent
of the x/2.
So I have to, because this is not just a simple
x, I have to do the chain rule and do the
derivative of the inside function of x/2.
The derivative of 1/2 x, or x/2 is equal to
1/2.
Now to find the second derivative I have two
factors being multiplied together.
I am going to let that constant sit with the
secant.
So I have 1/2 secant x/2 times tangent x/2.
That means that we have to go through the
product rule.
So it is the first times the derivative of
the second plus the second times the derivative
of the first.
We have secant x/2 times secant squared x/2
times 1/2... doing that derivative of that
inside function of x/2 again... that 1/2 and
1/2 is what is going to give us the 1/4.
And then we have...
What was it?
The second times the derivative of the first.
So I am again just letting the 1/2 sit there.
Pardon that.
I am going to just let that 1/2 sit out front.
so we have the derivative of secant which
is secant x/2 tangent x/2, and again completing
that chain rule with the inside function getting
the 1/2.
And then 1/2 times 1/2 is again 1/4.
To make it easier to find the critical values
of the second derivative I find it easier
to put it into terms of sine and cosine.
So I wrote secant as 1 over cosine, tangent
of course is sine over cosine, secant is 1
over cosine.
Combine all of that and we have a second derivative
which is equal to 1 plus sine squared x/2
over 4 cosine cubed of x/2.
So I am going to erase all of this, get the
first and second derivatives listed over here.
And in this rational form as opposed to leaving
secant and tangent, it will be easier to find
out where the second derivative is equal to
zero and where it is undefined, the two conditions
we need to check for finding our critical
values.
Ok.
So let's find those critical values to find
the intervals in which we are going to test
for concavity.
So second derivative.
Where is the second derivative equal to zero?
Well, that...
I have a rational function again, so the numerator.
Where is the numerator equal to zero?
Well it is going to be one plus sine squared
of x/2 is equal to 0.
And at some point you might, maybe even immediately,
realize there is not going to be a solution
to this equation.
Because we are going to subtract 1 from both
sides and get sine squared x/2 is equal to
-1.
And when I go to...
Maybe now you realize, you try square root
both sides and you cannot square root a negative
number.
So there is no solution to where the second
derivative is equal to zero.
So let's turn our attention to where the second
derivative is undefined... the denominator.
So f double prime is equal to undefined.
Well, we are going to set 4 cosine cubed of
x/2 equal to zero.
Divide both sides by 4.
You get cosine cubed of x/2 is equal to 0.
This is starting to look the same as when
I checked for the domain.
I wonder why?
Secant is 1 over cosine, and our denominator
is just cosine.
So yeah, we are going to get the same answers
again.
We are going to cube root both sides and get
the cosine of x/2 is equal to 0.
And we already know from the previous work,
like I said, that our critical values here
when we take the inverse cosine of both sides....
I was going to just stop, but that was again
x/2 is equal to pi/2 and 3pi/2.
Multiply both sides by 2 and we get pi and
3pi.
That means that our test intervals that we
are going to work with go from negative...
Excuse me.
I was going to write negative infinity, but
our domain starts at zero.
So 0 to pi, and from pi to 3pi, and of course
I would be including my critical values from
where the second derivative was equal to zero
if any existed, now from 3pi to 4pi.
Ok.
So again we are looking for where the second
derivative is equal to either positive or
negative for concave up and down.
And we are going to have to plug these values
into this function.
So let's see.
Do I dare to try and do this on camera?
Let's try pi/2.
Pi/2...
If I take pi/2, because that is between there
divide by 2, I get pi/4.
The sine of pi/4 is square root of 2 over
2 which is a positive number.
It gets squared and that is a positive number,
and 1 plus a positive number is a positive
number.
And the cosine of x/2, again pi/4 is going
to be positive.
So this is a positive over a positive, therefore
my second derivative is positive is positive
here.
Now if I try... let's see here between 0 and
3pi.
Let's try 2pi.
That should be pretty straight forward.
If I try to 2pi... just hesitate because I
was working ahead in my head and started to
think I was going to have a sign error.
Ok.
So we are going to try 2pi.
Sorry about that.
So 2pi divided by 2 is pi.
The sine of pi is... zero.
Zero squared is zero, so we have 1 plus 0.
It is positive in the numerator.
Then when we take 2pi and plug it into the
denominator, 2pi divided by 2 is pi.
The cosine of pi is -1.
-1 cubed is still negative one..
times 4 is negative.
So we have a positive divided by a negative,
so the second derivative here is negative.
And then when you plug in.. let's see here...
How about...
I paused just to make sure I gave you a number
between 3pi and 4pi.
So 7pi/2.
If you take that and plug it in, you have
7pi/4.
The sine of 7pi/4...
1,2 , 3, 4, 5, 6, 7pi/4... is negative square
root of 2 over 2.
That squared is going to be positive.
Positive plus one is going to be a positive
answer.
In the denominator at 7pi/4 your cosine is
going to be positive.
Right, because in quadrant 4 your x's are
positive.
Again 7pi/4 is going to be again the square
root of 2 over 2 only this time positive.
That cubed is a positive number.. times 4
is a positive number.
So the second derivative in this last interval
of 3pi to 4pi is positive.
Ok.
So it would seem like I would now to have
to start talking about...
You have the concavity in these intervals...
up, down, and up.
So it would seem like I would start checking
for inflection points because my function
alternates between concave up... concave up
and concave down.
But we know that is not what the secant graph
looks like.
It has a bunch of asymptotes and we have already
checked for our domain.
We know that the function is undefined at
pi and 3pi, which is now where our function
is changing its concavity.
So you cannot have an inflection point where
there is an asymptote.
So we have our concavity, we don't have any
inflection points.
But I need some room to find those relative
max and mins.
For those relative max and mins we are going
to be looking for the first derivative is
equal to zero.
I am not going to check for where it is undefined
because I keep having that same denominator
of cosine of a half angle, so I know the function
is going to be undefined.
No relative max and mins.
So where is that first derivative equal to
zero?
Well, that means again the numerator has to
be zero.
We have 1 plus...
Excuse me, I am looking at the second derivative.
We have sine of x/2 is equal to zero.
We are going to take the inverse sine of both
sides.
Sine is equal to zero where y is equal to
zero.
We have zero, and pi, and 2pi, and do I need
to keep going?
Well, again our restricted domain is zero
to 4pi.
And if I go beyond 2pi to 3pi, I am going
to multiply by 2 to get 6p which is too big.
So that is it.
We are going to multiply both sides by 2.
We get x is equal to 0, 2pi, and 4pi.
Making a few small mental mistakes tonight.
So those are our three critical values where
the first derivative is equal to zero.
Now, the first derivative is equal to zero
here and in that interval the second derivative
is positive.
So this is going to give me a relative minimum.
And at 2pi I am in my interval where my second
derivative is negative, so I am looking at
a relative maximum.
Finally at 4pi I am in my last interval where
the second derivative is positive so this
is going to be a relative minimum.
So let's list those out.
My relative min is going to be zero.
That means that I need to take this zero and
plug it into here.
Well, let's see.
0 divided by 2 is 0, the secant of zero...
Well, cosine of zero is equal to 1.
The reciprocal of 1, because that is secant,
is still 1.
And at 2pi, actually let's do both min values.
The other one is at 4pi.
So 4pi divided by 2 is 2pi, and if I rotate
2pi.. yeah.. right back to where I was in
the first place.
So that is again going to be equal to one.
Then the relative max, well that is at 2pi.
So 2pi divided by 2 is 1pi.
So over here, and the secant is 1 over cosine.
The cosine of pi is -1.
If you take -1 and you flip it, you have negative
one.
So it seems a little bit weird to have a minimum
values that have higher y values then the
relative max but let's not forget what secant
looks like.
There is sort of a... it is not a parabola...
but it is sort of a parabola shape up here.
Then it drops down and so they are relative
min and max's.
When you have a sort of a bowl going up then
that is a relative min.
When you have a another bowl going down, then
you are going to have a relative max.
So that is the end of this example.
I am going to step out and give you sketch
of what this graph looks like just to tie
it all together if you need it.
There you go...
But you cannot make ends of a closed interval
as a relative minimum.
There is our graph.
Before we leave, I just want to make a note.
If this was an open interval and not a closed
interval, then we would not been able to identify
any relative min's these points because we
would be approaching that y value of 1 but
not actually becoming the y value of 1.
So um... it was closed.
We do NOT have those relative min's and max's
and the end of any interval.
That you for watching.
I know this is a long video, but hopefully
with these three examples now 
it is time.. and you can...
Go Do Your Homework.
BAM!
