BAM! Mr. Tarrou. In this video we are going
to learn how the first derivative will allow
us to find open intervals in a function where
the function.. we are going to identify where
the function is increasing and where the function
is decreasing, and where it is constant. We
are also going to learn how the first derivative
will allow us to identify those relative max
and those relative mins whether at a smooth
bend or at a sharp bend. We are going to finish
up after we get a lot of note about the definitions,
theorems, and what not... vocabulary, we are
going to go through three algebraic examples
and then a fourth example where we just look
at a function... a graph of a function...
basically a picture... I will draw a picture
of a function. We will come up with a sketch
of what the first derivative could look like
and again try and allow it to help us identify
those increasing, decreasing, and constant
intervals, and relative maximums and minimums.
So increasing and decreasing functions and
the first derivative tests. When a graph is
read left to right, you want to read these
like most languages I think you read left
to right...so you read like a book I like
to say. A graph is increasing when it rises
to the right. Or in more math style language,
a function f is increasing on an interval
if for any two numbers x sub 1 and x sub 2
in the interval, if X1 is less than X2 this
implies that f(X1) is less than f(x2). What
does that say? It says that if you have a
bigger x, you have a bigger y. So it goes
up to the right. A graph is decreasing when
it falls to the right. Again that very formal
math definition, a function f is decreasing
on an interval if for any two numbers x sub
1 and x sub 2 in the interval X1 is less than
X2 implies that f(X1) is greater than f(X2).
So again, what does that say? I wrote all
of these notes and properties and what not
from the textbooks because a lot of us have
trouble reading them. I am trying to decipher
them for you. So what does this say? X sub
2 is greater than X sub 1, so if you have
a bigger x and you put that into the function
you actually get a smaller y value. You know
f(x), it is function notation for basically
y or the dependent variable. A test for increasing
and decreasing functions. Let f be a function
that is continuous on the closed interval
of a to b and differentiable on the open interval
of a to b. If f prime of x, or the derivative
of f(x) is greater than zero for all x's in
that closed interval then f is increasing
on that closed interval. If the first derivative
is less than zero, or negative, then it is
decreasing. The function is decreasing along
that interval. If the first derivative is
equal to zero, then you have a constant. On
the next page here we are going to look at
steps for finding these open intervals and
I am going to give you a graphical display,
just some crazy graph that I have made up
to show you how these intervals look... increasing,
decreasing, constant, and have some sharp
bends as well, and some special cases we will
have to look out for as we do our homework.
Steps for finding open intervals of increasing
and decreasing. Step one, well this is all
about the first derivative so you are going
to have to find the that first. Step two,
solve for... set the derivative of f(x) equal
to zero. Then set it equal to, find out if
there are any values where it can be undefined.
Like maybe you have a rational function and
there is an x in the denominator. If that
denominator can be equal to zero, what are
those values making the function undefined.
Or Trig functions like sine of theta is y
over r, so... Well actually that is a bad
example because sine defined with the domain
of all real numbers. Maybe like tangent, tangent
of theta is y over x. So what rotation would
give you an x value equal to zero thus making
the tangent undefined? So find what makes
your derivative equal to zero or undefined
and those are going to be your critical values
in your open interval. And you are going to
use those critical values to determine or
step up your test intervals. Basically take
your number line from negative infinity to
positive infinity, or whatever domain is set
up in the problem, and breaking it up into
sections based on these critical values. Then
you are going to plug in one value from each
of those intervals into your derivative. If
you get a positive answer then the function
is increasing in that interval, if you get
a negative answer it is decreasing, and if
it ends up being zero... that is kind of...
you set the first derivative equal to zero
to find your critical values. Unless it ends
up just being a constant function like a horizontal
line which is kind of a special case. Strictly
monotonic functions only increase or decrease
on the entire interval. So let's say that
you have a function that has a degree of three,
like a regular polynomial with a degree of
three where it comes up and it does a little
bit of bend in the middle but then just continues
to increase. That would be a monotonic function.
It kind of, you know, the slope is not constant
but it never stops say increasing within its
domain or within whatever interval you are
interested in. Over here I have got a visual
representation of what I have put in these
steps. I have got some kind of crazy squiggly
line. I just made that up. I can't imagine
what the function would look like that would
make something like this. Except I can pretty
much guarantee it would be a piecewise function
with multiple pieces. We have a function that
is increasing in this interval and that first
derivative, because the function is rising
to the right, would be positive. Then my graph
levels off. So the first derivative at some
point, at least the way that I have drawn
it, maybe comes out to be zero. So the function
is increasing, it levels off and becomes horizontal
maybe just for one point on that function
if you will, and then it starts to rise again.
So when you find a critical value such as
when your derivative is undefined or especially
at zero, it does not mean that your graph
is going to... you know... stop rising and
start falling. It can level off and then continue
up again. Then we have another interval where
the first derivative as I have drawn it would
be positive. We have again another critical
value where our first derivative is equal
to zero. This time I do have the graph starting
to fall. So this is a preview of what is on
the next screen where I give you the formal
language called the First Derivative Test.
But if your first derivative goes from being
positive to negative that means that the function
stopped rising and started to fall. Well then
at that point somewhere within that rise and
fall you have got a relative maximum. And
we are falling with the negative first derivative
and then it switches from a negative first
derivative to a positive, and that means that
you have bottomed out. There you are going
to find a relative minimum. Now your relative
max and mins don't have to happen at a smooth
continuous point. It can show up at a sharp
bend. That is why we need to find out where
the first derivative is equal to zero and
where it is undefined. It is not necessarily
going to be a vertical asymptote like maybe
you would have in a rational function. Maybe
it is an absolute value function. So at that
sharp bend, if the first derivative goes being
positive to negative, that is a critical value
and would have helped you set up your intervals,
and in this case the way it is drawn that
would be another relative maximum. Then the
graph, maybe the graph of the original function
is a rational function... and it is undefined
at some value. So it just shoots down to negative
infinity and this value here of... whatever...
c makes the original function undefined. So
we don't... hmmm... Sometimes we do need to
back and make sure that the function itself
is not undefined somewhere. Like at this point
my first derivative would be undefined. But
if my function is continuous there, well then
I have a sharp bend like an absolute value
function. But sometimes when the... well and
when the first derivative is also again undefined
there is a possibility that not only the first
derivative undefined but the original function
is undefined as well. That could be in the
form of a hole, or like I have it drawn in
this diagram a vertical asymptote. So that
would be a discontinuity right there. You
are not going to have a relative max or min
at a point of discontinuity. Even if it is
a hole, and maybe the y value of that hole
is 4, you know you could be coming infinitely
close from the bottom and infinitely close
from the top and maybe you are coming up to
3.9, 3.99 and 3.999... but as you soon as
you pick a value, I can get just a little
bit closer. So there is not going to be a
relative maximum coming to a hole. Just like
when we talked about finding max and mins
with open and closed intervals. You can't
have an open dot be an absolute maximum or
min for the matter if it was going in the
other direction. Then of course we have another
positive first derivative because the graph
is rising to the right. Then I just drew an
arrow here where the slope of the original
function is horizontal, that would be a first
derivative value of zero. It is a constant
value. For some reason this becomes horizontal
over there. So that a visual representation
of how you would... the importance of these
critical values helping you set up these intervals
and find those increasing and decreasing intervals
along the function. I gave you the preview
of the First Derivative Test. So let's get
to it right now with some official notes.
Alright then. To our formal introduction to
what I had in that previous diagram, the First
Derivative Test. Let c be a critical number
of a function, again the first derivative
equal to zero or undefined, a function f that
is continuous on an open interval I containing
c. If the function f is differentiable on
the interval except possibly at c like those
sharp bends I was talking about, then f(c)
can be classified as follows: If the first
derivative f changes from negative to positive
at c, then the function f has a relative minimum
at (c,f(c)). If it changes to positive to
negative, rising and then falling, then you
have a relative maximum at (c,f(c)) and then...
like at the beginning of my diagram where
it rose, leveled off, and then rose again...
if the first derivative does not change sign
on both sides of c, then f(c) is not a relative
max nor is it a min. WHOOO! Alright. All of
these fancy notes are done. Let's get to some
examples. Remember three algebraic examples
and one graphical one. [bad singing] Let's
get critical... critical... f(x) is equal
to 1/3x^3-1/2x^2-6x+4. I want to find all
of the critical values so I can use them to
find the intervals where the function is increasing
and decreasing. Ok. I am not going to start
off with a really hard example. So we need
our first derivative. So f prime of x, or
the derivative of f, this is just a very basic
polynomial. It does not look basic, but it
will once I get finished taking the derivative.
So we have got 3 times 1/3 which is 1. So
we have x squared. We have got, bring the
2 down, we get minus 1x. Then negative 6x,
the derivative of -6x is simply -6. That is
a constant so the derivative of a constant
is zero. Our critical numbers are where the
first derivative is equal to zero or undefined.
So I want to set this equal to zero. Where
is again f prime equal to zero? That is going
to be x squared minus x minus six is equal
to zero. This one is very easily factored.
We have (x-3) times (x+2). Easily factored
and I am making a mistake. That was plus 2.
That means that we have critical values of
x equals 3 and -2. When we set up those intervals
make sure that you go from low to high, or
from negative infinity to positive infinity,
or how ever your intervals are defined in
your problem. So again where our first derivative
is equal to zero is not the only possible
critical values. Now I want to find out where
f prime is equal undefined. This is a parabola
opening up. It is a polynomial. It is smooth
and continuous everywhere within its domain,
or everywhere from negative infinity to positive
infinity. So the solution to where is f prime
equal to being undefined, there is no solution.
So our intervals are from negative infinity
to negative two, now we are going to go from
negative two to three, and then finally from
three to positive infinity. And we are looking
of course for the signs of our first derivative.
Our first derivative. I only have to test
one value within each of these intervals.
So let's say like negative three. Well negative
three squared is 9, and 9 minus negative 3
is plus... so 9+3 is 12, and then 12 - 6 is
a positive number. So in the interval from
negative infinity to negative two our function
is increasing. Now I have another interval
from -2 to 3. Ok. Let's try testing a value
of zero. If I plug in zero, I get 0 squared
is zero, 0 minus 6 is negative six. I just
said it, it is negative. So we have a function
which is again increasing, decreasing. Now
this is not automatically be increasing again.
You can... you know... the graph can level
off at that critical value of 3 and then continue
to fall. So let's make sure that we check
it. Four, 4 squared is 16, 16-4 is 12, and
12-6 is 6. I don't really care that I know
that it is 6, all I really care about again
is that I know that it is positive. So the
graph is once again increasing. Alright. Well
use the First Derivative Test to determine
all relative maximums and relative minimums.
So in other words is there a place where the
graph bottoms out setting up a relative min,
and if that is the lowest y value in the entire
function it is actually an absolute minimum.
But relative minimums have less criteria so
we start with calling them that. And then
we are going to look, and we tested that -2
and tested that 3, so let's see what happens.
Our critical values. Negative two and positive
three. Our function stopped increasing and
started to decrease at -2. So we went from
being positive for our first derivative, went
from being positive to being negative. So
it looks like we have a positive... negative...
looks like we have a relative maximum at the
value of -2. We have a relative max at (-2,
f(-2)). Now you are going to need to take
that -2 and remember to plug it back into
the original function and not your first derivative
like you were doing for your testing of intervals.
When you plug that -2 back into this function,
right off the top of my head, you are going
to get a y value of 11.3. So we have a relative
max at (-2,11.3). Now at our other critical
value at 3, our f prime again... I am just
summarizing this again... I don't know why,
I could just point to it... We go from negative
to positive. That means that we were falling
and started to rise again. So we have a relative
minimum at (3,f(3)). And when you take that
value of 3 and plug it back into the original
function, right off the top of my head, we
have a relative minimum at (3,-9.5). And if
you were to put this into your calculator
we have a graph that is increasing from negative
infinity up until you get to a value of -2,
this will be a very rough sketch, (-2,11.3)...
the graph is going to fall. It is going to
fall until you get 
to (3,-9.5). And then since that was our last
critical value the graph is going to start
to increase and go off to positive infinity.
If you remember graphing polynomials, and
we are getting back to this, in PreCalculus
you graphed polynomial functions of higher
order degrees. We talked about end behavior
and we talked about x and y intercepts. What
we missed in PreCalculus was the ability to
find those relative max and mins and actually
know how high the graph rose before it started
to fall again. So we are actually going to
go back to that topic from Precalculus of
graphing higher order polynomials. That being
said, we have a third degree polynomial with
a positive leading coefficient. We did talk
about, last year, end behavior of these functions
going down to the left and up to the right.
Indeed we can see that with this function
being verifies by our First Derivative Test.
Next example! Y equals 2x^2 over x^2-4. A
quotient. So a little bit more work to find
the derivative. Same thing though, find the
critical values, find the intervals of increasing
and decreasing, is the First Derivative Test
to find the relative max and mins. I will
finish with the sketch of this graph as well.
I will do a little better job of it compared
to the last one. We need to find the derivative
of this quotient. That means that we need
the quotient rule, and that means that we
are going to take the denominator which is
x^2-4 and multiply it by the derivative of
the numerator which is going to be 4x... I
am using lot of parenthesis because I have
multiple terms here and I want to make sure
that all of my distributive properties work
and I have everything correct. Now minus the
numerator which is 2x^2 times the derivative
of the denominator which that is going to
be zero so we just have 2x. That is only one
term so I really did not need the parenthesis
there... All over the denominator squared.
Ok, let's clean this up a little bit. We can
apply... OH! Well we are finding the derivative.
It would be nice if my notation is correct.
So my derivative, the derivative of y... We
are going to take 4x and multiply it through
the parenthesis and get 4x^3-16x. Negative
2x^2 times 2x is going to be negative 4 x
cubed. Again over that denominator of (x^2-4)
squared. Of course 4x^3 minus 4x^3 is going
to cancel out. And our derivative again is
equal to, come down here so we can see it
more clearly, -16x over (x^2-4)^2. Just want
to make sure that I did not do something silly.
Good, it checks out. Our critical values.
That is where the first derivative is equal
to zero or undefined. Now we have a rational
function. If the numerator is equal to zero,
the function is equal to zero. If the denominator
is equal to zero, well then it is undefined.
Set our first derivative is equal to zero.
Where is that at? Well -16x is equal to zero.
That means that our solution after dividing
both sides by -16 is x equals zero. So that
is definitely a critical number we will use
to set up our intervals. Another thing that
we need to solve for is where is our derivative
undefined. Well f prime is equal to undefined,
or is undefined, where the denominator is
equal to zero. That is x^2-4 is equal to zero.
Add 4 to both sides. We get x squared is equal
to four. That means that the first derivative
is undefined at x values of positive and negative
2. Now these might be sharp bends in the graph
or it might be where the graph itself, the
original function, is undefined. If the original
function is defined at 2 and -2 and it is
just the first derivative that is undefined
then we are looking at probably sharp bends
here and we will have possibly some relative
max or mins at these points. We will not know
that until we finish the problem. But we are
not going to have relative mins and max if
this is where the original function is discontinuous.
Indeed, I hope you did not think that I forgot
this denominator has the function... or is
the polynomial of x^2-4... just like in the
original function. So the original function
is undefined at 2 and -2. So these are actually
discontinuities and we are not going to have
a relative min or max there. But they are
still going to help set up our test intervals.
So the domain of this function is basically
all real numbers except that x cannot equal
2 and -2. All of this Precalculus, and actually
Algebra 2 coming back, of graphing these rational
functions in Calculus. Ok, our test intervals
are going to be from negative infinity to
negative two, from negative two to zero putting
these in sequence, from zero to positive two,
and then from two to positive infinity. We
are looking for the signs of our first derivative.
If I plug -3 in here, well negative three
plugged in for x, -3 times -16 is going to
be positive 48, and the denominator I know
is going to be positive because the entire
expression is being squared. So even if x^2-4
comes out to be negative, it is going to become
positive because of that power of two. I plug
in a negative, it becomes positive, it is
over positive, so the first derivative is
going to be positive. The function is going
to be increasing on that interval. Now we
skip over this critical value and go to our
next interval from -2 to 0. Let's try -1.
-16 times -1 is positive again in the numerator.
The denominator is going to be positive guaranteed.
So positive over positive means that our first
derivative is going to be positive which again
means that our function is going to be increasing.
Then moving over to our interval of 0 to 2,
if I plug in 1, 16 times 1 is positive....
Excuse me... let's try that again. -16 times
1 is negative over positive value so our first
derivative is going to be negative. That means
the function is decreasing. And it is going
to, if you plug in a value of 3, it is going
to be negative again. The function is decreasing.
So we have here it is... um... again.. we
have three critical values but the first derivative
shows that the function is increasing almost
everywhere. There is one place where the function
actually changes direction and creates what
is going to be a relative maximum because
we are going from positive first derivative
to a negative first derivative where it is
going to stop rising and start to fall. And
the function is defined at the x value of
zero. If it was undefined, if the original
function was undefined at zero, again it would
not be a relative max or min. You cannot have
a relative max or min when you are approaching
a hole a vertical asymptote. So our First
Derivative Test says that we have a relative,
let's see again positive to negative... rise
to fall, so a relative maximum at (0,f(0)).
That is zero, go to my original function I
plug in zero for x and we are going to have
zero over zero minus four. Zero over anything
other than another zero is equal to zero.
So we have a relative maximum at zero. I am
going to pause the video, erase a little piece
of this board, and just see what this function
looks like to verify everything that we observed
in our First Derivative Test. nanananana...
Here we go. And we have a vertical asymptotes
at x equals negative two and positive two.
We our relative maximum at (0,0) that our
First Derivative Test showed us. And just
to point out a couple of things again. Rational
function... where a rational function is equal
to zero in the denominator, it is undefined.
Now if that zero in the denominator can be
factored, excuse me not factored, but cancelled
with the numerator.... like maybe we have
an x-2 and an x-2 up here... if that zero
in the denominator can be cancelled out with
the numerator remember that is where the holes
are. This x+2 and x-2 does not factor away
so we have those vertical asymptotes because
of that. And we are going to be talking about
infinite limits, or the limits as x approaches
negative infinity and positive infinity, which
is very reminiscent of looking for your horizontal
asymptotes in PreCalculus. Our numerator and
denominator both have a degree, they are polynomials
and they both have a degree of two, and when
the degrees of your numerator and denominator
polynomials are equal you have a horizontal
asymptote where the leading coefficients are.
So it is y equals 2/1, and so we have a horizontal
asymptote at y=2. And here we are increasing
as we approach that first critical value of
-2, and we are increasing again. We are coming
from the opposite ends of positive infinity
and negative infinity, but the graph is still
rising. Again that (0,0), so we have a first
derivative that goes from being positive to
being negative. We get to the asymptote and
it continues to fall. So where is my remote?
[laughing] Let's get to that next example.
Y is equal to x minus 5 cosine of x. We have
got a little trig there. Same steps of find
the derivative, find the critical values,
set up the intervals, find out where the function
is increasing and decreasing. This time I
have an interval of zero to 2pi. Then use
that information to find any relative max
or mins. Ok. So that means that we need the
derivative. Y prime, the derivative of x is
equal to one minus five times the derivative
of cosine. Now cosine, the derivative of cosine
is negative sine. Negative sine so we are
going to come back here and change that to
a positive. Excellent. So we have got y prime
is equal to 1+5sin(x). Where are the critical
values? Well they are where the derivative
is equal to zero. Let's set this up. We have
got 1+5sin(x) is equal to 0. We are going
to subtract 1 from both sides and get 5sin(x)
is equal to -1. We are going to divide both
sides by 5 and get sin(x)=-1/5. Now I need
to get the sine function away from the x.
So x is equal to the inverse sine of -1/5.
Now hopefully you remember from last year
in Precalculus, we have done a little bit
of trig already this year in Calculus, that
you cannot do the inverse sine... Like if
you look at the original sine function and
think about what it looked like. It's range
was from negative one to positive one. What
do you put into a trig function? You put in
an angle measure. What do you get out? The
sides of a right triangle or the ratio of
y/r. The inverse trig function therefore you
are putting in the y/r and you are looking
for the angle measure to come out. Well if
this value is outside the range of -1 to 1,
then your calculator is going to give you
an error because there is no solution. But
of course -1/5 is within the standard range
of sine so we are going to get an answer for
this. Right off the top of my head, because
you know I can do this... I know my entire
sine tables in my head, the inverse sine of
-1/5 is equal to -.2014. Now remember our
calculator is a tool. Our calculator is not
the end all... be all. This angle measure
is not within my domain, not within the restricted
domain or interval that I am looking to work
this problem in. I have to think a little
bit. This is a negative ratio of y/r. Remember
you put the sides of a triangle into your
inverse trig function, so where is the sine
function negative. Well, sine is y/r right.
So when you start with the initial rotation
of zero, whether it be radians or degrees,
your rotate pi/2 and that is quadrant, pi...
quadrant 2. If you are in quadrant 1 to 2,
your y values are positive. Your above the
x axis. So I need a couple of angle measures
that land me in the quadrants where the sine
is negative. So this is basically my reference
angle. Don't forget inverse sine from your
calculator is going to give you answers between
-pi/2 to pi/2 or -90 to 90. So this is in
quadrant four, but it is not within our restricted
interval. I will have to erase this board
to finish the example, but for our quadrant
3 angle.... let's see. This is basically my
reference angle (.2014), so I want to rotate
over to pi and then keep rotating a little
bit more, rotate .2014 radians into quadrant
3. So I am going to have pi plus .2014. I
will write that answer down in a second off
my paper. And our angle in quadrant 4 that
we are going to be using.... quadrant 4. I
am going to rotate through 1, 2, through quadrant
3, and don't forget about reference angles
which is what this basically is... the number
of degrees between the x axis and the terminal
side, not the y axis. We are going to go to
2pi and then back up to get me in quadrant
4. I don't want to be in quadrant 1. So ... I
am thinking of a number. 2pi, or approximately
6.28, minus .2014 means that the x's that
I am actually going to work with in the problem,
my critical values, are 3.343 and 6.082. But
I am already taking up my entire chalk board
here, so I am going to erase this scratch
work and just put my critical values over
there so I have more room to work. So here
are my critical values again. And I have used
them to set up our test intervals. And if
I take a value of 1, 2, 3, 3.1 and I plug
them into our first derivative I am going
to get a positive value. Make sure that your
calculator is in radian mode if you are doing
your problem in radians which you really should
be doing. Then when I plug in a value anywhere
between 3.343 and 6.082, like maybe 4 or 5
just to make it easy, this my first derivative
is going to come out to be negative. Finally
if I plug in maybe something like 6.1 again
into my first derivative, I am going to again
get a positive answer. Yes. Ok. So actually
this should be 2pi just to be exact with my
interval that was defined at the beginning
of the problem. And again these functions
are continuous everywhere so I don't have
any issues with any discontinuities. So when
this first derivative switches between positive
to negative, that means that i am going to
have a relative maximum at ... let's see.
I am going to take the value 
of 3.343 and plug that into my original function.
I get (3.343,8.24). We switch between our
first derivative being negative to positive,
falling and rising again, so we have a relative
minimum at.... drum roll please... (6.082,1.18)
That is the end of our last algebraic example.
I have just got one more that is going to
be purely visual looking at the comparison
between the original function and its first
derivative and identify again the idea of
critical values, increasing and decreasing,
and so on. Thank you very much for watching.
BAM! Last example. So we have this crazy function
f drawn. I want to come up with what seems
like could be a good plausible sketch of what
the first derivative would maybe look like.
Well, I am going to look through here and
I am going to try to find all the places where
the original function has a slope of zero,
or a slope which is undefined. Because if
the slope of the original function, that is
going to be a critical value right for the
first derivative. So it looks like we have
a critical value here 
where the original function has a slope of
zero. And it levels off here at x=2. And it
stops decreasing and bottoms out, hopefully
my drawing is ok for you guys here, it seems
like we have a slope here for our original
function of zero at 5. It is not really 5,
I am going kind of cheat and say it is!. It
starts to rise and then fall at the x value
of 6. Ok. Let's try and draw what could be
our first derivative. Now our first derivative,
our original function is rising, and if our
function f is rising that means that the slope
of that function is positive. The first derivative
gives us answers for y which is slope. The
first derivative describes the slope of the
original function f. So if our slope is positive,
the first derivative is giving positive answers,
thus it is going to be above the x axis. Then
it is going to cross through the x axis here
at the value of negative one. So very very
positive, and very steep positive slope for
the original function so maybe it is coming
through here like this and our first derivative
is crossing the x axis at this value of -1
where the original function has a slope of
zero. Now the original function has a slope
which is starting to fall so it is negative.
It seems that about right here the original
function has a slope that is going to get
about as negative as it is going to get, or
as small as it is going to get. I am going
to let this sketch of what could be the first
derivative get to its lowest value and reflect
that lowest value of slope right there. Then
the original function is still decreasing,
still has a negative slope, but it is not
as negative. So the value of the slope is
actually starting to increase even though
it is still negative until it gets back to
a value of zero. So this orange graph is going
to come up here and... it is going to touch
here at x=2. But then after the slope starts
to increase, it is still negative, but increasing
to zero and then drops again. So from here
to here this original function is falling
which means all of my slopes are negative
except for maybe here it is equal to zero.
So I need to keep this first derivative below
the x axis. I am going to come up here and
touch the value of 2, and again try and get
that deepest part of the bend somewhere underneath
where it looks like the original function
has the most negative slope. Then it starts
to fall off again. It seems to have the most
negative slope somewhere around in here. Then
it reaches zero again as it starts to rise.
I am going to go ahead and let my first derivative
show its most negative slope value around
that interval and start to approach zero again.
Maybe it looks something like that. Now this
time, at this critical value, the graph actually
starts to rise. So the first derivative describing
that slope, positive slope therefore the first
derivative has to give positive values. That
means that it is above the x axis until we
get to the next critical value of 0 at 6.
We have a first derivative that is zero here,
a first derivative that is zero here. But
that slope between 5 and 6 is positive, so
that first derivative is about the x axis.
Then it starts to fall again. So I think that
this seems like a pretty reasonable sketch
of what that f prime looks like. Again our
critical values here, here,... at -1, 2, 5,
and 6 means that... let's see. From negative
infinity up until -1 our first derivative
is positive. So again that is increasing for
one last summary. Between -1 and 2 f prime
is... let's see... I should have wrote that
as positive. Then between -1 and 2 f prime
is negative, so that function is... Excuse
me the first derivative is positive so the
function is increasing... then it is decreasing.
From the values of 2 to 5 our first derivative
is still below the x axis, so we are again
decreasing. And since our derivative when
from negative and stayed negative, we do not
have a relative min or max at the x value
of 2. Then from 5 to 6, our first derivative
is positive indicating that our original function,
and we can see it in our drawing, was increasing.
Then from 6 to infinity we went back to the
first derivative being negative, and again
we can see that function was decreasing. So
man that was a lot. I hope it helped you to
completely understand the First Derivative.
How it shows a function to be increasing,
decreasing, or constant on intervals. And
also the First Derivative Test helping you
to identify relative max and mins in these
functions. I am Mr. Tarrou. BAM! WOOOO! I
am ready for a break. Maybe you could take
one too. Then, Go Do Your Homework:)
