Using the First Derivative Test for Finding Relative Extrema

Video Series: Calculus 1 (Explore, Discover, Learn) Series

(D4M) — Here is the video transcript for this video.

00:00
[Music] a function is monotonic if it is increasing or decreasing
in this video i discuss finding extrema of functions by
studying monotonicity an essential tool in this regard are the first and second
derivative tests finding extrema becomes much easier when you master
these theorems all right welcome back everyone this is the episode
monotonic functions and derivative tests so in this episode we’re going to talk
about first monotonicity in other words we’re going to study

00:01
where functions are increasing and decreasing
and then we’re going to talk about the first and second derivative tests
and then after those two tests um we’ll have a much better
uh understanding of finding a streamer we’ll be able to find local extrema and
absolute extrema and we’ll have more powerful tools to do that
and then at the end we’re going to study
we’re going to talk about some exercises so i’ll have some good exercises for us
at the end so let’s go ahead and get started [Music]
okay so up first is monotonicity we’re going to
uh look at that first and that will give us a lot of insight into the shape of a
function you know where is the function increasing and where is it decreasing
now in order to determine uh where function is increasing and decreasing
we’re going to begin by finding critical numbers so that’ll be the first step

00:02
and the problems that we work out are finding critical numbers
now if you’re not sure what critical numbers are then in a previous episode
um i talked about critical numbers in the extreme value theorem uh episode
we went over critical numbers in in in with lots of examples
so um in the description below is a link to the playlist the calculus one
explore discover learn series and so you’ll be able to find all the
episodes there in particular if you’re not sure what critical numbers are
look at that episode extreme value theorem okay so but in this video in this
episode we’re going to uh you know start off by finding critical numbers and
the whole point of the critical numbers are is that it’s going to divide the
x-axis into intervals so we’ll be able to test the function over these intervals

00:03
and we’ll be testing the function by looking at the sign of the first derivative
over each of these intervals so the the point is is that
when you try to look at the shape of a function you know there’s
there could be infinitely many points on the function and so it
it can be quite daunting task to look at the shape of a function
and so the idea behind the critical numbers is you break the
x-axis up into chunks and you look at each
chunk and each interval and you see what the function is doing on that interval
and that will give you insight into how the function looks overall
by looking at all of these intervals so this procedure is uh often called the
first derivative test and we’re going to use it to find the local extrema
and find out where the function is increasing
and where it’s decreasing in other words we’re going to find the intervals of
monotonicity um so we’re going to first start off by
talking about what a function is uh when we say increasing so

00:04
of course we’re reading left to right and so that
that means that x1 is less than x2 and if if we have f of x1
is less than f of x2 whenever x1 is less than x2 then we’re going to say it’s
increasing so just to look at that visually
if you have a function and we say this right here is increasing
so this right here is two heights and let’s say here this is x1 and this
is x2 here so this will be f of x1 and this right over here will be
f of x2 so f of x one is less than the f of x two you know provided

00:05
x1 is less than x2 in other words as we’re reading left to right
x1 is less than x2 f of x1 is less than f of x2 the function is increasing
so you know this is the way we see it in words
but you should see a graph you should see an illustration also
similarly we have a function is decreasing
now the inequality is switched so f of x1 is greater than
f x2 so looking back here if the function starts to decrease
and then now i’ll call these over here x1 and x2 now i have these points right
here and these over here now change so this is x1 up here
so now this is f of x1 and over here i have f of x2
and so now what do we have happening now we have

00:06
f of x2 is less than the f of x1 whenever x1 is less than x2
in other words as we lead as we read left to right
the graph is falling down it’s decreasing and what we mean by that is that
f of x2 is less than the f of x1 all right so how do you write that in
words right so a function is f is called decreasing on an interval
so um one thing to point out there is that we always say we’re decreasing or
increasing on an interval we never really say that we’re increasing at a number
can’t really be increasing at a single point you need to compare two values
to be increasing or decreasing so we always say
decreasing on an interval and then we say the condition
now a function is monotonic it’s just a short way of saying it’s either

00:07
increasing or decreasing so if you say it’s monotonic then you
certainly know that it’s not constant notice that the inequalities
there are strict f for increasing its f x one is strictly less than
f of x two so that rules out the function is constant right so
a monotonic function is not a constant function monotonic just means it’s either
increasing or it’s decreasing okay so um here’s our first theorem here and i
would like to mention before we see the rest of the theorem
that that looks a lot like the hypothesis or the mean value theorem
f is continuous on a closed interval and
it’s differentiable on the open interval now if the first derivative is positive

00:08
then we’re going to say that f is increasing so
if we look at a graph something like this so i’ll just make a nice smooth graph
because it’s continuous let’s say we’re right here at
a and we’re right here at b [Music] and we’re continuous and we’re
differentiable in other words there’s no sharp corners like this
right so differentiable um means the derivative exists everywhere on the
interior between a and b so there’s no sharp corners like this or this it’s
differentiable meaning it’s nice and smooth and
if the derivative is positive the slope of the tangent line is positive
for example right here the slope of this tangent line is positive the line

00:09
the tangent line is going up and same same thing right here
um is it still positive right here it doesn’t look like it’s
positive right so that doesn’t look like it’s increasing
so to illustrate number one here let’s keep it increasing
there now the derivative is the slope of the tangent line is always positive
um and so it’s increasing that’s all number one
says if the derivative is positive your function must be increasing number two
if the derivative is negative then the function is decreasing [Music]
and here’s some examples here so if you look at c1 what’s happening at
c1 right there at c1 we have a horizontal tangent line

00:10
because the derivative is zero and then to the right of c1 the derivative
is positive we’re increasing to the left of c1
where the derivative is still positive we’re increasing
so function could go increasing and then have a horizontal tangent line
and then it still can be increasing after that
and then it’s increasing until we get to c2 and then at c2 what are we
what’s happening at c2 the derivative is zero again
we have a horizontal tangent line now that see the difference between c one
and c two though both of c one and c one and c two the derivative is zero
which means we have a horizontal tangent line but look to the left of c
one the derivative is positive it’s increasing
and look to the right of c one the derivative is still positive we’re still

00:11
increasing so that means we have no extrema
because as you pass through c1 it’s just just keeps increasing now on c2 compare
that though to the left of c2 the derivative is positive
so we’re increasing all the way until we get to c1
but then when we get to c1 the derivative now becomes
us when we get to c2 now on the right side of c2 the derivative is negative
that means we’re decreasing so we’re going to see here that if there’s a change
in the derivative one on one side it’s positive and then on the other side the
derivative is negative that gives us relative extrema we have a local max at c2
okay so um that will use that in a in an upcoming uh
in a couple minutes here all right so there’s that theorem there if the
derivative is positive it’s increasing if the derivative is
negative it’s degree decreasing all right so let’s look at our first

00:12
example here so this says um find open intervals where we’re increasing and
where we’re decreasing and so to do that we’re going to look at the
critical numbers so first let’s start off by finding the derivative
so the derivative will be 3x squared and then the two comes down right
so minus three x and then where is this zero
so we can factor out a three x and then say x minus one
so x equals zero and x equals one zero here and then one here and so these
are the only critical numbers now why are these the only critical numbers

00:13
so first off both of these numbers make the first derivative zero
so and they’re in the domain of the function
this is a polynomial so the domain is all real numbers
so certainly zero and one are in the domain of the function
and they make the first derivative zero so it’s
easy to see that these are critical numbers now why are they the only
critical numbers well these are the only two numbers that
make the first derivative zero and the first derivative is defined
everywhere else um so that’s the only possible outcome
now when we look at the critical numbers here zero and one
way i usually like to do that is to put the critical numbers on a number line
and what i’m going to do is i’m going to divide up the number line
the x-axis into different intervals here so i have one two three intervals here
and so i’m going to look at this like i’m going to say

00:14
x is less than zero and then x is equal to zero
and then x can be between zero and one and then x could be equal to one and
then x could be greater than one so that’s the only possibilities for
this number line here so whatever this function looks like x
to the third minus three x minus three over two x squared whatever
that looks like this is gonna help us determine the shape of that here
so i’m gonna be looking at the function and
i’m going to be looking at the the first derivative
and i’m going to be making some conclusions from that
so i’ll put a column here or conclusion here
so what’s happening here i’m going to be checking the sign of the derivative
because if the derivative is positive that means i’m going to be increasing
if the derivative is negative that means i’m going to be decreasing

00:15
so on this interval right here we can test any number we want in this interval
because we’re testing the whole interval the whole interval all the real numbers
less than zero so pick a nice number less than zero for example
minus one and what’s happening if we use a minus one
into our derivative right here right this is our derivative
so i like to look right here because this is factored already when i use a minus
one i’m going to get a minus three so it’s negative
when i use a minus one here i get a minus two
now the number is not actually important it’s the sign of the number what’s a
negative times a negative that’s a positive so that means the function is
increasing on that interval right there now let’s test the number between zero
and one between zero and one for example tests say

00:16
one half one half is a number in here so what happens when we test one half
we’re going to get a three over two now three over two is not what’s important
it’s important if it’s positive or negative
so when i test i’m testing a one-half and i’m going to get a positive times a
negative and a positive times a negative is negative
so on this interval right here the function is decreasing
now i’m gonna test greater than one so pick a convenient number greater than one
to test this whole interval right here so for example i can test two
when i use a two i’m gonna get positive and a positive
so a positive times a positive is positive and so i’m going to get increasing
so this means increasing this symbol right here means decreasing

00:17
and this is increasing now what’s happening in between
so i like to put this column here also so this is where the derivative
is zero and this is where the derivative is also zero
we that’s where we found those by setting the derivative equal to zero
so we know we plug in zero of course we get out zero and we plug in one we get
out zero what’s happening at these two points
here though so what happens when we plug
in zero into the original function right here we plug in zero into the original
function we get out zero so we get zero zero what happens when we
substitute in one so i come down here real quick f of one
is one to the third minus three over two times one and so that’s just one minus
three over two which is two over two minus three over two
which is minus one half so we get minus one half here at this one

00:18
just by plugging in one and working that out
so what’s happening here when we change these uh when it goes from increasing to
decreasing so whatever this function looks like
right here in fact let’s come down here and see if we can look at it a little
bit um x to the third minus 3 over 2 x to the square x squared [Music]
so when i go graph this what i’m saying is that something’s happening at 0 0
but it’s increasing to the left of zero so right here is to the left
here’s zero so i’m going to be increasing to zero
and we also have a critical number of one so at one here we know we’re getting a

00:19
minus one half so i’m going to come down here and label minus one half
now what’s happening um less than zero we’re increasing all right we got that
less than zero we’re increasing now what’s happening between
zero and one between zero and one we’re decreasing so i gotta be decreasing now
and i’m going to go through that point right there let me make it a little bit
more clear so we’re increasing and then we’re decreasing
and then now what’s happening to the right of one to the right of one
so greater than one our derivative is positive so now we’re increasing
so now i’m going to be increasing the function looks something like that
this is zero zero right here and because it’s switching here it’s going from
because it’s switching here increasing to decreasing

00:20
so as we can see right here it’s increasing and then decreasing
so this right here is a relative max i’m gonna go label that in my conclusion up
here relative max and then it’s decreasing our our table
our chart right here table says decreasing and then increasing right we
can see that on the graph it’s decreasing and then it’s increasing
so this right here is a relative min i’m going to put that right here
relative min so i like to put my conclusion and i
like to say where it’s increasing and the relative extrema decreasing increasing
and i i like to put my column f and my column derivative here
so this has all the information right here now we did a lot more of the problem
than they asked all they asked was where it was increasing and decreasing
but i went ahead and found the relative extrema just so that you can see how

00:21
that works if there’s a change in the sign of the derivative
if it goes from increasing to decreasing then we’re going to have a relative max
and you can see that intuitively if it’s increasing
and then it’s decreasing that’s going to be a relative max
on the other hand if it’s decreasing and then it starts to increase
at that critical number right there that’s going to be a relative
minimum um and so this is our first example we’re going to do
a lot more examples obviously but this is our first example
right here oops sorry and so let’s just make sure and answer
the question right here right where find the open intervals where it’s
increasing so let’s come down here and say that the function f is increasing
on minus infinity to zero so that’s where it’s increasing all the way to here

00:22
and is decreasing on um zero to one and is increasing on
one to positive infinity so we found the intervals where it’s dec increasing
minus infinity to zero and it’s also increasing on one to positive infinity
and it’s decreasing on zero one so those are the intervals
and what’s doing on each one of them all right let’s move on to the next example
okay so for this next example here we’re looking at three x to the fourth minus

00:23
four x to the third and we’re gonna find out where it’s uh monotonic
where is it increasing and where is it decreasing so let’s take a peek at the
graph there’s x to the fourth now someone gives you a graph
then it’s a lot easier to see where it’s increasing and decreasing
especially if the points are labeled so we can kind of tell there
that is decreasing at the beginning from minus infinity to one there but
you may not have the graph so how do we do this
we’ll start by finding the derivative so what is the derivative here

00:24
the derivative is 12x to the third and then minus 12x squared and then minus 24x
[Music] and we’re going to set this equal to
zero so how many x’s can we factor out x you can also factor out a 12.
so i’m going to factor out a 12x and i’m going to say this is x squared
and then this will be minus x and then this will be minus 2.
so can we factor this more let’s go with
minus two and plus one that will give us
what we need so the critical numbers are zero two and minus one
i like to lay them down on a number line just
organized just to organize my work so when we look at this function here this
function to the fourth power here we’re only we’re gonna we’re going to break

00:25
uh break it down into one two three four
intervals and that will allow us to take a problem a graph a function
that has infinitely many points on it and only check
for intervals here and we’ll be able to tell where it’s increasing where it’s
decreasing by only checking four intervals that’s a
really powerful observation to be honest because if you look at this fourth
degree polynomial up here the function f you know maybe it’s hard to see what
it looks like maybe you have no idea what that function looks like
of course i just showed you in any case we have the derivative here and see we
we can say the function is continuous and differentiable it’s a polynomial
so to test where it’s monotonic we’re going to divide the x-axis up into
well we’re going to look at the 12x we’re going to look at the

00:26
second factor x minus 2 and we’re going to look at x plus 1. so when we um
break this up into intervals we’re going to have
a less than minus 1 and we’re going to have between -1 and zero so these are the
intervals we came up with here one two three and four so between zero
and two and then greater than two and so when we break this up i’m
breaking it up by my rows here so there’s my four rows
now the second column is i’m picking a number in that interval
and i’m testing the 12x so for example for the first row pick a number but
that’s less than minus one for example minus two so what does minus
two look like at 12x well it’s minus 24 but it’s not minus 24 that matters it
matters that it’s negative so i’ll just put a negative there so on
the first row i’m basically testing the minus two i test the minus two into each

00:27
expression so when i do minus two minus two that’s still negative and when i do
minus two plus one that’s still negative now to get the
next column the derivative column i simply multiply because that’s what
the derivative is the derivative is 12x times x minus 2 times x plus 1.
so what is a negative times a negative times a negative it’s negative
so that’s how we get that uh derivative column
so now let’s test a number between minus one and zero
to test a number between minus one zero we can for example test
minus one half so twelve times minus one half
is negative minus one half minus two is negative minus one half plus one
is positive and now i ask what is a negative times a negative
times a positive and that’s a positive so my derivative is positive
and so similarly we check the other two rows

00:28
choose a number between zero and two for example one
twelve times one is positive one minus two is negative
one plus one is positive and so what’s a positive times a negative times a
positive that’s negative and finally for the last row
choose a number greater than two you can
choose any number you want for example a hundred
twelve times a hundred positive a hundred minus two
positive a hundred plus one positive so what’s a positive times a positive
times a positive is positive now if the derivative is negative
then we’re decreasing so we’re decreasing on that interval
and if the derivative is positive then we’re increasing
and so we have decreasing increasing decreasing and then increasing so
that chart and the derivative and factoring the derivative and solving the
derivative equal to zero can do all that without looking at a

00:29
graph the graph is just gravy so if you’re given a graph first
that’s really nice but you don’t need it to find the intervals where we’re
increasing and decreasing we find the first derivative test
and we find the critical numbers assuming we can
in this case we can because we’re given a nice derivative all right next example
all right so let’s see here um determine on where the intervals are increasing
and decreasing so again i’m going to find the first derivative
so we have one minus one over x squared or said differently x squared minus one
over x squared and we’re gonna ask what are the critical numbers

00:30
now plus or minus one so we set the derivative equal to zero
so x is plus or minus one and we also look where the derivative is undefined
so we’re saying x equals zero also so there’s
three potential critical numbers however you must also always check out the
original function 0 is not in the domain of the original
function so this is not a critical number
so rule that one out what about plus or minus one is that in the domain of the
original function and the answer is yes so plus or minus one
are the critical numbers [Music] now what i usually like to do after find
my critical numbers is put them on the number line however
many you have sometimes you’ll have no critical number sometimes you’ll have one

00:31
sometimes you may have four critical numbers but i usually like to put them
in order because that helps me identify the intervals here
so i’m going to have the intervals less than -1
and i’m going to have between -1 and 1 and i’m going to have greater than one
so those are the possibilities here and we’re going to go look at the sign
of the derivative so this first derivative here
so 0 didn’t give us a critical number there so i’ll just get rid of that
all right so we need to check the derivative so here’s the derivative here
now another way you might want to look at the derivative is [Music]

00:32
is x minus 1 times x plus 1 over x squared reason why i like to i write it like
that is because it’s a little bit easier to test
but i just factored the numerator right difference of two squares
so now if i test a number less than minus one
for example minus two and i check it up in here to see if the derivative is
positive or negative now look at the bottom the bottom is
always positive i wouldn’t even worry about the bottom it’s just positive
because it’s a square so when i’m checking say a minus two
something less than minus one i’m going to check minus two here and
minus two here so minus two here is a negative and a negative here
so that would be negative times a negative over positive so that’ll be positive
now when i check between zero and minus one here i can check something like

00:33
well we can’t check 0 actually right so because 0 is not a number
in the domain of f in fact if we look at what f
is let’s come down here and take a look at f a little bit more
so f is x plus one over x right so another way of saying that is just um
x squared plus one over x so x square plus one over x has a
vertical isotope at zero so whatever this graph looks like
at zero here we have a vertical isotope so the graph is the graph is gonna
be going something like this and something like this around zero

00:34
so how do i know that it’s doing this right here i need to check zero here so
you know if is not differentiable at zero because it’s not defined there
so i need to check zero also so i’m going to put zero here and
let’s add zero here to our intervals here to check
so i’m gonna go with between minus one and zero
and then between zero and one and then greater than one [Music]
and the reason why again is because if you look at the
function f it’s not differentiable at zero it’s not even continuous at zero
so we have to check to the left and to the right of it
how do we know it doesn’t do this how do we know it doesn’t do this
so we have to check if it’s uh if it’s increasing or if it’s decreasing

00:35
to the left of its isotope all right so between minus one and zero
so check in here we can check something like minus one half
so let’s go to the derivative and check minus one half
so minus one half here that will give us a negative
minus one half here that will give us a positive so what’s a negative times a
positive times a positive over a positive so that would be negative
that’s why i said it was decreasing down here now we know this one is ruled out
and what about between zero and one so what’s the number between zero and one
to test so between zero and one say a one-half let’s test the one-half into the
derivative so one-half we get a negative and then a positive
and then a negative sorry positive so that would be a negative times a
positive times a positive so that’ll be positive [Music]
sorry i don’t think i checked that right i think it’s decreasing

00:36
let’s try a number between zero and one again like half
so when i check a half it’s negative yeah okay so negative times positive
over positive so that’s negative so it’s decreasing
it’s increasing it’s decreasing and then we’ll check
greater than one let’s check greater one for example two
if i check it two everything is positive so this is positive here
and so it’s increasing and so it’s going to start increasing and
this is just going to be increasing and then it starts to
decrease right here so it’s increasing and then at minus one right here
and at one right here so let’s make sure we get this chart understood
positive it’s increasing the derivative is positive it’s increasing

00:37
so we can see that right here less than minus one now between minus one and zero
between minus one zero we come out with negative so we’re decreasing
so we’re decreasing right here now between zero and one we’re decreasing
the derivative is zero so we’re decreasing so we’re decreasing right here
then we hit one critical number and then to the right of one we find the
derivative is positive so now it’s increasing so the graph must
look something like that this is just shaped like that some
people don’t like to use those arrows only arrows on the axis anyways there’s a
the kind of the shape of the graph right there
in the next video we’re going to study isotopes in much deep much more detail
so this vertical isotope actually this graph has a slant isotope

00:38
which we’ll learn about next time so we’ll we’ll understand these uh
curved sketching a great deal in the next video the next episode
so for this example here though where is
it increasing and where is it decreasing right so we got that here’s where it’s
increasing and it’s also increasing on this interval here
and it’s decreasing on these open intervals right here all right so next example
um actually time for a quick break [Music]
all right so now it’s time for the first derivative test
so the first derivative this is our first derivative test
and it’s called the first derivative test so not very original there but

00:39
traditionally it’s called the first derivative test so i did this
uh first derivative test actually on the very first example we did in this
episode but i just didn’t call it the first derivative test so
we’re going to start off with the statement of the first derivative test and
we’re going to see here what all this means but basically if you have a critical
number [Music] and you need continuity for the function to happen
then if your derivative is positive and then if your derivative changes sign
positive to negative then f is the relative maximum so this
is what we said this is what illustrated in the first example
but we’ll do this again real quick here if your first derivative is positive
so this is a relative max here as we can see

00:40
and here the derivative is positive the slope of the tangent line is positive
and over here on this side the slope of the tangent line is negative
make that a little bit more tangent so the slope of the tangent line there
is negative the line is going down the slope is negative so that’s what this
this theorem is saying is that under certain conditions
if the derivative is positive um and so we got the a and the c and the b
let’s put that down here so let’s call this here um a and a b and a c here’s a c
so we have a relative max at c [Music] and so if the x is in between here
so actually let’s call that here an x right there where i put that

00:41
let’s call that x and let’s call this right here an
a and it’s the same thing over here let’s call this here x
and let’s call this right here b so if i pick an x here in between a
and c and the slope of the tangent line that that x is positive
but over here at this x between c and b over here on the right side
then the slope of the tangent line here is negative [Music]
so for x between a and c the derivative is positive
for the x between c and b the derivative is negative
so that’s what this right here says the derivative is positive for an x between
a and c and the derivative is negative for an x between c and b
then f has a local maximum so let’s draw the second sentence there

00:42
and see if we can make a illustration of that so now you can see
we have a relative minimum so relative minimum
so something like this let’s say we have a relative minimum right there
and we’re right there at some c and so right here let’s say we have an
a and we have a b right so we’re stuck right there on a and b
and we’re we’re looking between a and b right there
now for this x right here if i picks it x between a and c
what is the derivative doing it’s negative
the slope of the tangent line is going down the the tangent line is going down
so the slope is negative and if i pick an x between
c and b and i come up here and i look at the slope of the tangent line
now it’s positive so here we have the derivative

00:43
is positive and here we have the derivative is negative so
if that happens then this has to be a relative minimum
so that’s what we’re saying here if the derivative is negative
to the left of c if the derivative is right
positive to the right of c then f has a relative minimum at c now
if one and two don’t hold then f has no relative extrema at c
so let’s look at our first example here using the first derivative test
so let’s see what our derivative is and we’re going to get 3x squared
and minus 12 and that’s it so we can factor out a 3 and get x squared minus 4
and we can factor that into three x minus two

00:44
and then x plus two and we’re looking where that is zero
so again this function right here f is the polynomial so it is continuous
everywhere and differentiable everywhere and so the only critical numbers
that can happen and even the derivative is the polynomial so it’s continuous and
differentiable everywhere so the only place where we could have critical
numbers are at two and minus two so let’s put that down only critical numbers
[Music] x equals minus two and two so when we um go and make our table here
i have interval i’m going to do less than less than 2
and i’m going to say equal to -2 and then um between minus two and two [Music]
and two and greater than two so that’s the only possibilities that

00:45
can happen along the x-axis you can either be less than minus two equal to -2
between -2 and 2 equal to 2 or greater than 2. that’s the only possibility
so i’m going to have a column for my function i’m going to have a column for my
derivative and you can put the x’s here if you like
and then i’m going to have a column for my conclusion
so here we go we need to test and what am i testing
this is my derivative right here that i’m testing
i found it equal to zero that’s how i found these numbers so i already know
that this is a zero right here and a zero right here
so we already found those zeros the question is though we don’t know what’s
happening at those numbers yet until we test for increasing and decreasing

00:46
so this says conclusion [Music] conclusion all right so let’s test
a number less than -2 for example -3 and all we need is the sign to know if
it’s increasing or decreasing so let’s check minus three so minus
three in here will give me a negative a minus three and here would give me a
negative so that’ll all be positive so that means it’s increasing now a
number between minus two and two is zero so what’s happening at zero so we’re
going to get a negative times a positive so the whole thing is negative
so we’re decreasing and we’re greater than two so pick a number like three
so this will be positive positive everything is positive
so we’re going to be increasing again [Music]
so let’s go ahead and write it out in case some people needed to have it

00:47
written out decreasing and increasing and now i have a change in monotonicity
there’s a change if it was increasing and increasing
this would not be any relative extreme at all but there’s a change here so we
have something here what is it relative min or relative max
so the derivative is positive and then the derivative is negative which one is
that so what i usually like to do is to sketch it just make a little
a little sketch over here it’s increasing and then it’s decreasing
so that’s a relative max right here so this is a relative max
and then i’ll just erase that now it’s decreasing
and then it’s increasing so what does that look like what’s happening at two
well it’s decreasing to the left of two and then to the right of two it starts
to increase so this looks like a relative min so this is the relative min here
so on one hand i think it’s important that you

00:48
be able to read the sentences read the math
as it’s written up in a book which is what we did a couple slides ago
right we read all that and we made sure we understood all that
because you need to both be able to read it and do it
so that’s the reading of it right there but then when we get to this part right
here we’re i’m not like memorizing what i i just did a little sketch out here so
that i can intuitively understand what this what
this is going on what’s happening here in any case this is a relative max here
when the derivative is 0 and that’s when x is 2
so when x is 2 we need to go plug in into our function here and so let’s do that
our function here is x to the third minus 12x
minus five i need to go plug in the minus two
into the function so go plug in minus two

00:49
and then we’re going to get minus eight and then a 24
minus five out of all that you’ll get 11. and then now we need to go plug in 2
into our function when we plug in 2 we’re going to get 8
and then minus 24 and then minus 5. so out of all that you’ll get -21
and so this will be the relative max here at minus 2
and 11 and this will be a relative min here
at 2 and -21 and this will tell us what the graph looks like
it’s increasing it’s decreasing and then it’s increasing
so if we were to go sketch all that information now it doesn’t ask
us to do that this is the first derivative test we identified intervals
on where it’s increasing and decreasing and we also determine extrema
relative max remote to men it didn’t specifically ask us to do that
but i wanted to just do that and take that opportunity to
to understand that there now even one more opportunity is to go and

00:50
interpret all this information maybe we can do it over here
i’ll just make a quick little sketch over here interpreting this information
right here so what do we have here we’re looking at minus two
and two let’s put a two here and so at two i know i’m at minus 21
and then at minus two i know i’m at 11. and i’m going to be increasing to to
minus two so i’m going to be increasing up to here and this is our relative max
and then between minus that’s a minus two and two
we’re decreasing so between these two right here we’re decreasing
now what’s happening when it hits the y-axis
when it hits the y-axis when x is zero we can look at the original function and
see we get a minus five there so i’m going to come down here and say
that’s some well that’s a minus 21 maybe make
five a little higher let’s say minus five is right there
so i’m going to come down here and be decreasing here

00:51
and i’m still decreasing to here so all this is decreasing and it goes through
minus five and then to the right of two it starts to increase
make sure don’t put any sharp corners because remember this is differentiable
everywhere so this is a nice smooth curve through here
nice and smooth when i get down here don’t make it look like it’s a sharp
corner right because that’s not what’s happening should be a nice smooth curve
i usually like to label the points afterwards shape it and then label it
all right so there’s what the function looks like
just by interpreting this information here so if you look back at our original
function x to the third minus 12x minus five
the power in all this is the fact that there’s infinitely many points on that
graph but yet we just looked at a couple of intervals here

00:52
and we tested the derivative and knowing what’s happening to the derivative
we’re able to say something about the function and how this function is shaped
it’s increasing it’s decreasing and it’s increasing
now in the next episode we’ll actually get even more information out of the
derivatives to be able to shape this even more precisely so that was a 11 here
all right so let’s look at the next example
so apply the first derivative test so by the way this table
right here is what i’m calling the first derivative test
when i apply the first derivative test that’s the information that i want to
give back to somebody there’s a whole lot of words here in this table
that we don’t need to write simply because it’s organized so well
so let’s look at this new function now here we go this function

00:53
has a fractional exponent so it looks more fun
let’s see what’s happening um we have our function is um let’s go
up here actually can we find the derivative of this [Music]
so product rule right so derivative of the x is one times the second
plus now leave the x alone and derivative of the second
which is two-fifths and then one minus x and then two-fifths minus one
and then times the derivative of the inside part here which is minus one
now our goal is to find the critical numbers we need the critical numbers
to do the first derivative test now um when i’m looking at this right here i
need to look where this is zero and where this is undefined by the

00:54
way the original function which i should have mentioned first is
so tempting to just start taking derivatives
but the original function notice its continuous um for
all real numbers and it’s differentiable
well let’s see where it’s differentiable
so let’s look at this right here this is a minus here
so i’m going to write this as a minus and a two and an x that goes in the
numerator right there so let’s write this as one minus x to the two fifths
plus and then i’m gonna say i have a minus two x
over and then one minus x to the three fifths
okay so that that could be a way that you write it right there
so we have a minus one and a two and an x so oh and we have a five here sorry
all right so this is still not ready uh to start testing with it let’s

00:55
get a common denominator so let’s put this over one
for example and let’s multiply top and bottom here
by this denominator right here so here we go i’m going to do that over here one
minus x to the two fifths and then times five
to the one minus x to the three fifths [Music]
over five times one minus x to the three fifths
and then we still have this little piece over here which i’m going to try to
squeeze in over here minus two x over five times one minus x to the three fifths
so that says three-fifths right there so basically just copy this over here
but what i’m doing to this one is i’m multiplying top and bottom let’s multiply
i guess we don’t need anything for multiply all right so
i just multiply top and bottom by the same thing so now we have common
denominator so what is this going to be equal to common denominator is 5

00:56
1 minus x to the three fifths and then now what are we getting up here
we’re getting a five times and this is one minus x to the two
fifths and this is one minus x to the three-fifths
so two-fifths plus three-fifths is five-fifths which is just one
so this is one minus x to the one power and then i still have a minus two
x so long story short what do we have here five and then minus five x and minus
another 2x so minus 7x [Music] so this will be our first derivative here [Music]
so this is the derivative that we want to do our testing with
this is nice and simplified so there’s all the steps right there to find the
first derivative and we want to get space here

00:57
and so i’m going to erase all this work and just write this is our derivative
so let’s just write it here [Music] this will be 5 minus 7 x
over 5 times one minus x to the three fifths all right so there’s our first
derivative that’s a five make funny facts all right so
there’s the first step is what’s the derivative
now notice this derivative is not defined at one
so but the original function is defined at one
so in particular one is going to be a critical number so
x equals one that’s where the derivative is undefined
but the function is defined when you plug in one you’re going to get out zero
so one is in the domain of f but one is not in the domain of the derivative
in other words f is not differentiable at one
another critical number is taking the numerator equal to zero

00:58
so that’s just a quick solve so that’s just
minus seven x minus five x equals five sevenths
right so five sevenths if i put that under
that’s a little bit less than one right so maybe i’ll write it as
my critical numbers are five over seven and one five over seven and one
all right so there’s our critical numbers
so when i make up my table for my first derivative test
i need to have my critical numbers in there [Music]
all right so let’s see if we can go to that now
all right so i’m taking a peek what the function looks like
it looks very different than say a polynomial
in the sense that we have that sharp corner there because with a fractional
exponent it’s not differentiable one there’s a sharp corner there
now suppose we didn’t know any of that let’s find our first derivative test so

00:59
first step is to find the derivative set it equals
zero and also look where the derivative is undefined
and so there’s our first derivative that’s exactly what we found a second ago
five minus seven x over over that expression there
so the derivative is zero by setting the numerator equal to zero
notice for that five sevenths x equals five sevenths the derivative is zero um
yeah so the the numerator is zero and the denominator is not zero so the
derivative is zero now f prime at one does not exist but f of one does exist
therefore the critical numbers are zero and five sevenths [Music]
so i’m going to put this in a chart i’m going to set up my intervals here
and i’m going to be looking less than five-sevenths
and i’m going to be looking between five-sevenths and one
what’s the number between five-sevenths and one think of one is seven over seven

01:00
so a number in between would be like six over seven and then i’ll be looking at
greater than one so looking at those intervals there
i’m going to look at the numerator first only
so if i choose something less than five sevenths for example
zero then we get five minus seven times zero and we get one minus zero
so positive over a positive is positive so for the first row there the
derivative is positive now let’s look at the second row x is
i’m going to test 6 over 7 because i think it’ll be pretty easy
so if i test 6 over seven right so if we have
five minus seven and we have six over seven
the sevens cancel we get five minus six in other words we get a negative number

01:01
so that five minus seven x column is negative now let’s put six over seven
into one minus x right so that’ll be one
minus six over seven that’ll be positive and a positive number to the three
fifths power will still be positive so the next question is the derivative
so that’ll be a negative over positive so that’s negative
all right so then the last row will be to choose something greater than one
for example ten so five minus seven times ten
negative one minus ten to the three fifths power is also negative
the derivative is the fraction so negative over negative is positive
and so we’re increasing and so we can see the increasing and then the decreasing
and then the increasing again we can see all that behavior
in the graph so that’s the first derivative and we can see the local extrema

01:02
right actually i didn’t specifically write down the local extrema did i
so um if you look there is a change in monotonicity between
at five sevenths at five sevenths to the left of five sevens we’re increasing
and then to the right of five sevenths we’re decreasing
so at five sevens we’re going to have a relative maximum and
so we have to actually um find that number um so let’s say here [Music]
f has a local max at x equals five sevenths and then looking back here
is there a change in monotonicity at one to the left at one we’re decreasing and
to the right of one we’re increasing so yeah we are

01:03
decreasing so that’s going to be a relative minimum so [Music]
this is a relative min here at x equals 1. so a local min at x equals one
[Music] so we could put that down there f has a local min at one
so i usually like to put all that on my table
but when i type this up i didn’t put it in my table some people don’t put it in
their table all right so next example [Music] so find the relative extrema
[Music] i’m going to use the first derivative test we have another fractional

01:04
power here so let’s start off by finding our derivative [Music]
so let’s see here we have two thirds and then x squared minus four and then
we have two thirds minus one so minus one third
and then times the derivative of this inside part here so times two x
so let’s write this as what four x over three and then x
squared minus four to the one third [Music]
so that’s what the derivative looks like there x squared minus four
and so that’s the 4x [Music] so what are the critical numbers right
so at 0 x equals 0 is a critical number because the derivative is 0 and the

01:05
function is defined at zero so f of zero is just what something and then
the derivative is zero at zero so if we set four x equal to zero we get x equals
zero so what about uh plus or minus two that makes the denominator zero
that makes the derivative undefined but can we use plus or minus two into the
original function yes because plus or minus two into the original function
will just output zero so critical numbers are going to be
minus two zero and two so these are the critical numbers [Music]
all right so once we identify the critical numbers now we can go
and do our first derivative test so we can find our
relative extrema so here we go we’re going to look at

01:06
less than minus two equal to minus two so that’s what i’m saying i usually like
to put all that information in here and then between -2 and 0
and then equal to 0 and then between 0 and 2 between 0 and 2.
equal to 2 and then greater than 2. and so for these three critical numbers here
that breaks the number line up into these possibilities here
less than equal between equals between equals and then greater than and i’m
going to be looking at the function and the first derivative and i’m going
to have me a conclusion [Music] and so let’s do this that’s the derivative here

01:07
and that’s the function here so what’s happening less and i’m going
to be testing this right here right so what’s happening say less than 2
we’re going to get a negative and so something like less than negative 2 is
like negative 3 right so when i do negative three down here
that’ll be a nine minus four so that’ll be positive and then negative
three up here so that’d be negative so negative over a positive is a [Music]
negative [Music] now choose a number between minus two and zero
for example minus one and so then that’ll give me a negative
and if i choose a minus one here that’ll be a negative so negative over negative
will be positive now choose a one between zero and two right
so positive over negative so that’ll be a negative

01:08
and then now here greater than two so choose something like a three so we
have positive positive so that’ll be positive
so this will be where it’s decreasing [Music]
this will be where it’s increasing and then decreasing
and that’s positive so that’ll be increasing again so we’re going to see
we’re going to have relative extrema here if it’s decreasing and then increasing
it looks like this decreasing and then increasing so it’s a relative min here
[Music] if it’s increasing and then decreasing it’s going to be a relative max
if it’s decreasing and then increasing it’s going to be a relative min
and then we can go plug in those numbers and fill those values in right there
so at -2 if we substitute -2 into this function right here you see
we’re going to get out 0. same thing when we plug in 2 we’re going

01:09
to square it get a 4 minus 4. so this is another 0. what’s happening at 0
so at zero we’re gonna get um you know minus four to the two thirds [Music]
so we can call this here is cube root of 16. i’ll just put that here
cube root actually here cube root of 16. all right that’s good
so we found the relative extrema right here minus 2 0 and 0 cube root of 16
and at 2 0 so those are the relative mins and the relative maximums
and we found all of them and if we were to go like try to give a

01:10
shape of it the graph we’re decreasing less than -2 so let’s put a minus two
about here and then a two about here so we’re decreasing
and then we hit this place where the derivative is not defined
here so there’s a sharp turn but we have a relative minimum
and so it’s going to come up until we hit this value at 0
cube root of 16 let’s say it’s about right there and then we’re going to
start to decrease again and then we’re going to start to
increase again and again it’s not going to be differentiable here so it’s going
to be a sharp curve and then i’m going to start to increase again [Music]
uh some people don’t like those arrows there [Music]
all right so that’s something how this function looks x squared minus four to
the two thirds power so it’s not jagged like that it’s nice and smooth

01:11
anyways so there we go next example [Music] next example
there we go let’s look at this one here x to the fourth plus one over x squared
so what is the derivative first off so we have
low x squared times derivative four x to the third
minus x to the fourth plus one times two x all over
x to the fourth now before we start looking at critical numbers we should

01:12
simplify this so let’s see here we’re going to get 4 x to the fifth
and here we’re going to get minus um two x to the fifth
and then a minus two x then all over x to the fourth
okay so looks like we’re getting two x to the fifth minus the two x
all over x to the fourth and let’s factor out a two x in the numerator
and so we’re gonna get x to the fourth minus one
over x to the fourth and so let’s say that’s
two and then x to the fourth minus one and then x to the third so [Music]

01:13
that looks good enough there [Music] so the critical numbers are going to be
-1 and zero sorry minus one and one critical numbers are minus one and one
that’s what makes the numerator zero now we also look at where the
denominator is zero to figure out where the derivative is undefined
but the derivative is not defined at zero
but neither is the original function in fact the original function
has a vertical isotope at zero so we need to be aware of that because
it’s not differentiable on an open interval containing zero
so let’s back up here a little bit and take this derivative right here and
write it over here so 2 x to the fourth minus 1

01:14
all over x to the third and we have our critical numbers
so i’ll just put them on a number line minus one to one
but we also need to look at zero because the function has a vertical isotope at
zero so i’m going to break this up into less than minus 1
equal to minus 1 between minus 1 and 0 not equal to 0
because that’s not in the domain of the original function
and then between zero and one equal to one
and then greater than one so these are the possibilities here
given this function and this derivative here these are the possibilities here
negative between here less than equal between here not equal to here because
the original function between zero and one and then greater than one

01:15
and so we can check the function and the derivative and write a conclusion
[Music] so let’s test uh less than negative one
into our derivative right here so let’s test something like minus two
that’ll be negative that’ll be positive positive over negative is negative
between minus one and zero i can test something like a half
a minus a half so that’ll be negative minus a half that’ll be to the fourth so
that’ll be positive but then it’ll be minus one
so that’ll be negative over negative so that’ll be positive
now check a one half so that’ll be positive that’ll be negative
so that’ll be negative now check greater than ones for example something

01:16
like a two so test a two so positive over positive it’s all positive
so it’s positive here and so now we can try to fill in some of this information
here we’ve tested all of our intervals here so we see we’re decreasing [Music]
decreasing we’re increasing and then where um [Music]
sorry i put that in the wrong place between zero and one like a one-half
and then when we test the one-half into our derivative here
we’re going to get positive negative over positive so that’s a negative
so i put that one in the wrong place there um decreasing here
and then positive so we’re increasing [Music]
i love those symbols i don’t know if they’re used everywhere for everyone but

01:17
to me they’re just so intuitive all right so we can ask now for relative
extrema which is what it’s asking for if we’re decreasing and then we’re
increasing this will be a relative min [Music] and if we’re increasing and then
decreasing but the point is not on the function
there’s no point at x equals zero don’t forget that because that’s the
original function there so we don’t have anything right there in
fact at x equals zero we can see we have a vertical isotope
um and then if we’re decreasing to increasing
then we’re going to have a relative [Music]
min again so this graph right here has two relative mins
and no relative maxes let’s see if we can try to get a shape of what’s going on
here so we have the numbers right here minus one and one

01:18
coming from right here and we know we have a vertical isotope at x equals zero
so i’ll just label that x equals zero here
now what’s happening at the one when we look at the original function
at one we’re gonna get two over one so we get a height of two
let’s say it’s right here at two what happens at minus one
well the fourth power and the square take care of all that we get the one
back out again so to the left of minus one we’re decreasing
so we’re going to be decreasing to here and then between minus one
and zero we’re increasing so we’re going to hit this relative min right here and
be increasing again this is a vertical isotope so it’s not going to cross
here and then over here it’s going to be decreasing
and then it’s going to hit this relative min and it’s going to be increasing

01:19
again so graph is shaped something like that
it’s just kind of rough sketch of the graph right there [Music]
with this isotope here so that’s just a rough sketch of that
graph there from this information right here rough sketch right there right next
example [Music] how about this one right here fourth degree
let’s see what’s happening [Music] all right [Music]
so we have our first derivative here is 4x to the third
and then a minus 12x squared and let’s shape this up a little bit here

01:20
let’s factor out a 4x squared so i’m going to have an x left and i’m
going to have a minus 3 left and so we’re looking at 0 and 3 [Music]
and so the original function f is a polynomial
the derivative is a polynomial they’re all continuous
and differentiable everywhere and so our only critical numbers is where the
derivative is zero and so we have zero and then three
so let’s go and do our first derivative test so i’m looking less than zero
equal to zero between zero and three [Music]
equal to three and greater than three and i have my function my derivative and

01:21
my conclusion [Music] and so let’s make our let’s make our table here now
so i’m gonna test here so when we’re less than zero for example
minus one what’s happening now notice the four x squared is always
positive we don’t need to worry about what that is that’s positive
so when i test uh a minus 1 i’m going to go here and say that’s a negative
negative times positive right so it’s negative between 0 and 3 i tested 2
that’s negative so it’ll be negative when i test greater than 3
like a 4 then that’s positive and that’s always positive so it’s positive
so decreasing decreasing increasing so i’ll just write those out real quick
decreasing decreasing again and then increasing now remember if there’s no
change in monotonicity there’s no relative extrema

01:22
there’s no relative extrema here it’s decreasing and then it’s still decreasing
now here it’s decreasing and increasing so this will be a relative min
and we can go find out what that is by plugging in three
into the function so if we plug in three into the original function there
we’re gonna get three to the fourth minus four times three times to the third
and then a plus 20 plus a 12. so we’re gonna get minus 15 here
let me calculate all that up now even though i don’t have a relative
extrema there i want to plug in zero um when i plug in 0 everywhere i
get a 12 out and so now we can kind of make a
shape over here interpretation of what all this looks like
all this data in terms of a shape and so let’s put that right here

01:23
at zero i’m right here at 12 and i’m going to be decreasing to there
and then when i get there i’m on the other side i’m still decreasing
i’m going to go right through there so i’m decreasing and then decreasing
so notice that’s not a relative extrema and then at 3 we’re going to be at -15
so let’s say 3 is about right there so i’m decreasing to the 3 and then at
minus 15 then on the other side i change direction i start to increase
so we’re just going up like that and this is the point here three
and then that is minus 15 and then that’s a 12. and so that’s kind
of roughly the shape there it’s decreasing all the way to three
and then on the other side of three it’s increasing all right so very good
so let’s look at another example let’s look at some fractional power some more

01:24
[Music] so here we go so let’s find our first derivative [Music]
so we have 15 times two thirds and then x to the two thirds
minus one so minus one third and then now we have minus three times
the five thirds [Music] and then x to the five thirds
minus one right so five thirds minus one is just two thirds [Music]
okay so there’s our derivative well let’s simplify it
so this one will be what 10 over x to the one third

01:25
and then minus and then this will be -5 over or yeah just five
x to the two thirds [Music] okay so let’s get a common denominator
let’s put that over one so 10 over x to the one third
minus five x to the two thirds and then multiply this denominator so x
to the one third over x to the one third so this will be ten
minus x to the two thirds x to the one third that’s three thirds
add them up that’s three thirds which is one so this is minus five
x those together give us an x and then x to the one third so we got a
common denominator there so we got 10 minus 5x over x to the one-third

01:26
now this zero right here when x is zero the original function is defined because
zero to the two thirds is zero zero to the five thirds is zero so we’re
going to get zero minus zero so we’re gonna get zero
so we plug in zero to the original function it’s defined we get zero
but we plug in zero into the derivative we get undefined right here so zero is a
critical number and two is a critical number
because two makes the derivative zero right plug in a two here we get zero
so we got two critical numbers here zero and two and so let’s go and make our uh
chart here so let me erase this right here and write the derivative
10 minus 5x over x to the one-third that’s a nice looking derivative

01:27
and zero and two are the critical numbers so now let’s make our first derivative
test table so i’m looking less than zero let’s write f f prime and conclusion
first this time and so we’re looking less than zero
equal to zero between zero and two equal to two um equal to zero yes because
the the function is defined at zero so equal to two and then greater than two
so we divided the number line up into all the possibilities [Music]
so let’s test the intervals now when we’re less than zero say a minus one
what happens we use minus one so when i use a minus one here

01:28
i’m going to get a positive over a negative so that’ll be negative when i use a
number between zero and two for example a one
when i use a one in here the ten is going to dominate it’ll be ten minus
five so to be positive over positive so it’s positive and when it’s greater than
two say for example a three i test a three so now this will determine the sign
that’ll be minus and then positive so minus over
positive it’ll be negative so this will be saying it’s decreasing
and this is positive so this is increasing
and this is negative so it’s decreasing and so so this is negative so it’s
decreasing so do we have a change in monotonicity yes

01:29
decreasing to increasing at zero here we have to have a relative min
and increasing and then decreasing we have to have a relative max at two
so now we know that something is happening here
what’s happening at zero well we already plugged in zero we got zero out
relative max what happens when we plug in two
when we plug in two we’re going to get some value out so what would that be um
nothing very pretty it’ll be so f of 2 will be 15 to the two to the two thirds
which isn’t anything pretty and then minus three times two to the five thirds
[Music] which also isn’t very pretty i’ll just call this number
here m so i’ll just put an m here whatever that number is you may

01:30
be tempted to approximate it on the calculator
and that might be fine for some uses but whatever that number is i’ll call it m
i don’t really rewrite all that over here [Music]
now let’s look at how this is shaped we found the relative extrema already 0 0
and 2m whatever that m was and then now what is the shape of it look like
because we have all this information here it’s just a shame to not
to not take a peek at what it looks like right so what does it look like
so we’re decreasing to zero and then we have a relative min
so i’m decreasing to zero now at zero the derivative is not defined
remember at zero the derivative is not defined
so i’m going to have a sharp corner right here at 0
and then i’m going to be increasing and then to this 2 right here

01:31
so we’re increasing up to the 2 and then after that we’re decreasing
so we’re decreasing like that so this two here this height
that’s the value that we found there m and that’s the relative max
and here’s the relative men [Music] so that’s the shape there make sure you
make it a sharp corner there because it’s not differentiable there [Music]
okay so next example okay so this one let’s change it up a tiny bit let’s say
find the local and absolute extrema um values for this function right here
on this interval so now i’m giving us a close bounded
interval right here minus one and three and i’m saying let’s find the local
and the absolute extrema now we dealt with a problem of finding the absolute

01:32
extrema for this type of problem right here in a previous episode
where we use the extreme value theorem so
we can use the extreme value theorem and the first derivative test
to answer both of these questions here so there’s what a peak of what it looks
like but you don’t need to know what it looks like to
to determine all this information out so first off i’m going to notice is that f
is a polynomial you know if you don’t look at the
restriction you just say x to the third x y is 2 squared that’s the polynomial
is continuous and differentiable so it will satisfy the conditions
that we need for our theorems the extreme value theorem and the
first derivative test on that on that closed interval there
so we’re going to find the critical numbers and we’re going to look at the
first derivative and see where it’s zero and for that there’s the first

01:33
derivative now let’s check that first derivative real quick [Music]
so let’s go over here and check that first derivative real quick
so we’re gonna get here f of x is um so i mean you can expand it out but we
don’t really need to we can just you know take the derivative real quick
derivative of the first [Music] times the second
plus the first times the derivative of the second
to the first power and the derivative of the x minus two is one
so we’re going to get here two things here and let’s factor some
things out of these terms so they both have x minus twos in them
and they both have so that’s an x to the third
and that’s an x squared so they both have x squares in them

01:34
so i’m gonna factor out an x squared and an x minus two
and what does this one have left it still has a 3 left
we took out the x squares we took out one of the x minus 2 so we still have an
x minus 2 left let’s change that to a bracket now all right so
when i factor out the x squared i still have a three left
and i’m only factoring out x minus two to the first power so i have an x minus
two left and now what about this second part here
we have a two left i never factored out the two
i factored out the x minus two x to the third well i only factored out two of
those so i still have an x left [Music] so this is x squared x minus two and
then this right here is what three x plus a 2x so that’s a 5x
and then we’re going to have a minus 6 here [Music]

01:35
all right so that’s our first derivative there
so there’s so the first derivative is x squared x minus two times five x minus
six so the critical numbers are zero and two and then 6 over 5
from setting each of those factors to 0. all right so now we can look at the
table here and so i’m going to be looking between -1 and 0. so notice our
critical numbers here are um zero two and from this one right here if you set
this equal to zero you’re going to get you know six over
five right so when i go to a number line i say okay here’s zero here’s six over
five and here’s two but always remember the domain of the function

01:36
the domain starts at minus one and goes to three so we don’t need to check
less than zero we need to check between minus one and zero
and we need to check this one and this one and this one
that’s why my table here it starts at that minus one
x is between minus one and zero and that’s because of the given restriction
of the function so our next interval would be between zero and six fifths
and then between zero uh six fifths and two and in between two and three and
the x square column is always positive because it’s the squared
and so then we test the x minus two column pick a number
x in the interval and test x minus two same thing with the five five x minus
six test that column so choose a test value for example the first row
i would choose x is negative one half and

01:37
plug that into a five x minus six right you get out a negative
multiply all three columns together positive times a negative times a positive
that’s how you get the positive for the derivative
so some people like putting all those columns in there
but as you see in the previous example i didn’t even bother to put all those
columns in there it just depends upon what you want to do
or what your instructor is asking you to do but in any case the derivative is
positive or increasing the derivative is negative we’re decreasing [Music]
so it’s left to find the relative extrema and the absolute extrema
so as you can see from the graph at the minus 1
we’re going to have a absolute minimum and at the 3 we’re going to have an
absolute maximum so to find those values we plug in the minus 1
into the original function and we get out the number

01:38
we plug in 3 into the original function and for that one we get 27
so those are the absolute extreme values the absolute maximum is 27 and happens
at x equals three and the absolute minimum value is
what happens when you plug in minus one there so you’re going to get minus one
times um minus three squared right so you get minus nine so
the absolute minimum value is minus nine and it occurs at minus one
x equals minus one now we can also find the relative extrema
is there a relative extrema at six uh sorry
is there a relative extrema at x equals zero
and the answer is no to the left of zero
it’s increasing and then to the right of zero
it’s still increasing and you can see that from the graph it’s just

01:39
increasing right through zero is there a relative extrema at six fifths
the answer is yes because there’s a change in monotonicity
to the left of six-fifths it’s increasing and to the right of
six-fifths it’s decreasing so that means at x equals six-fifths i have a
relative maximum and then you can see we have a relative minimum at x equals two
okay so there’s a lot of examples that we just did
and i hope that helps and let’s see what’s next
okay so what’s next is the second derivative test
now the second derivative test allows you to

01:40
find relative extrema also in some cases but it does so a lot faster so you
notice for the first derivative test we had to build this whole table so the
second derivative test tries to allow you to get to the relative extrema
a little bit quicker without having to go make a whole table
but to do so you have to um come up with a new test so a slightly different test
it involves the second derivative that’s why it’s called the second
derivative test not because it’s the second one in the order of derivative tests
but because it involves the second derivative
so let’s read the second derivative test
first thing we notice is that the second derivative has to be continuous
so in other words you look at your function you take the derivative
first derivative and then you take the second derivative and then you look to
see if that’s continuous if that’s continuous then the second

01:41
condition you check is where is it zero where is the first derivative
zero so we have those values c where the first derivative is zero
and then we have two conditions if the second derivative is
positive where the first derivative is zero then we have a relative minimum the
second sentence says if the second derivative is negative
where the first derivative is zero then f is the relative maximum
so let’s look at some examples of that but first i just want to make clear that
if one of those two cases doesn’t hold then the test is inconclusive in other
words if the first derivative is zero and the second derivative is zero
the test is inconclusive and the second derivative test doesn’t tell you

01:42
anything in that case so if we look at two examples here the x to the third
and x to the fourth [Music] so let’s look at these two functions
here and see what can happen so let’s look at that here
when i look at the first derivative here is 3x squared and the first derivative
over here is 4x to the third and i look at the second derivative here
and this is 6x and i look at the second derivative of g and that’s 12x squared
now notice the very first condition so look at the theorem second derivative
test suppose the second derivative is continuous
are these second derivatives continuous absolutely

01:43
they are continuous so check the second condition up here
is the first derivative zero well let’s see
where is the first derivative zero when x is zero
the first derivative is zero so that means x is zero
same thing over here if i set this equal to zero
x equals zero if i set this equal to zero
that tells us that x must be zero so in both of these cases
the first derivative is zero at zero the first derivative is zero whenever x
is zero but the second derivative is also zero
here right because if we plug in zero here the second derivative tells us
we’re also zero same thing here if we plug in zero here we get
zero so we do not have number one holding number one does not hold does

01:44
not hold in order for number one to hold
the second derivative has to be positive but here the second derivative is
zero and two doesn’t hold either does not hold same thing over here both
one and two do not hold does not hold [Music]
so for these two functions here x to the third
now what does x to the third look like it just looks like this
there’s no relative extrema at zero even though the first derivative is zero
at zero there’s still no relative extrema
it’s just increasing right through there so the second derivative
is zero and that means the second derivative test
does not hold so in other words for some functions you cannot use the second

01:45
derivative test in other words if i wanted to look at the function x equal
y equals x to the third i cannot use the second derivative test
the second derivative test will be inconclusive the second derivative test
is inconclusive when both the first derivative
is zero and the second derivative is zero at that c
if they’re both derivatives of zero then the derivative test is inconclusive
here’s why here’s a here’s an example where the first derivative is zero
and the second derivative zero and there’s no relative extrema
on this hand the first derivative is zero the second derivative is zero at zero
but what is x to the fourth look like well when it’s x to the fourth it’s
actually going to look something like that it’s not a parabola because it’s a
fourth power but the point is is that it has a relative minimum right here

01:46
so if your first derivative is if your first and second derivative
are both zero who knows what’s happening
it’s inconclusive sometimes there may be no relative extrema
sometimes there might be a relative extrema so you have to really be careful
if you find out a c in this case zero if you find out a number where the first
derivative and the second derivative are both zero it’s inconclusive now
having said all that there’s a lot of cases where the second derivative test
does work and it is really faster and we’ll see some examples like that
right now so for some functions i just want to
reiterate for some functions the second derivative test
might be easier to find local extrema because remember first derivative test
we had to go make out a whole table and you know sometimes you want to know
just really quick is this a relative max or relative min
you know let me know right away really fast

01:47
in some cases the second derivative will work really fast
in particular when we go solve some applications
you may not be interested in sketching a graph looking where it’s increasing
decreasing you may not be interested in all that you may just be interested in
and finding optimization and solving that problem
and so sometimes the second derivative test will be much quicker
so let’s look at some examples like that now
so use the second derivative test to determine
if each first order critical number is a relative max or relative min or neither
so now we’re going to use the second derivative test
so here we go first thing is i need first derivative
here we go so i need first derivative what is it

01:48
so we’re going to go with a 15 x to the fourth and then minus 15 x squared
and we’re going to set that equal to zero now to solve that i’m going to factor
out a 15 x squared and go with x squared minus one so x equals zero
and plus or minus one so these are called the
first order critical numbers first order critical numbers because they’re the
critical numbers coming from the first derivative
now in the next episode we’re going to study concavity
and we’ll have a concavity test and then you’ll have the first derivative test
the second derivative test and the concavity test
and using those three tests together you can really
shape the curve and you really have a lot of flexibility in terms of finding

01:49
extrema but any case there’s the first derivative we set it equal to zero
we found our critical numbers you don’t have to use that terminology right now
but we’ll certainly use it in the next episode these are the critical numbers
and these are the only critical numbers because
f is a polynomial it’s continuous and differentiable everywhere
the derivative is defined everywhere it’s continuous everywhere and the
second derivative so now i need a second derivative so
we’re using the second derivative test so you definitely need the second
derivative so i’m going to go here with a 60x to the third and then a minus 30x
this is 60. all right so i’m going to factor this

01:50
and say 30x and then say 2x squared minus 1. okay so we have here now um [Music]
where the first derivative is zero at these three places
and we have the second derivative here now so
i’m going to look at the zero first and i know that the derivative is zero here
that’s how we found it and so what is f what is the second derivative at zero
it’s zero if we plug in a zero zero to our derivative we get zero
and what is the second derivative at the minus one
and what is the second derivative at the one
so i look at my critical numbers and i plug them into my second derivative
when i plugged in zero i got out zero what happens we plug in
one here we’re going to get zero oh no we’re going to get here uh 1
we’re going to get now we only need to worry about the sign

01:51
well i’ll go ahead and put equals but what is it for equals
it’s going to be here um 1 here so it’ll be -30
and here at one it’ll be another one here so it’ll be 30. so what’s important
though is that it’s less than zero and this one is greater than zero so
let’s refresh our memory what’s happening here
these are the numbers where the first derivative is zero
and then the second derivative says if the second derivative
is negative what’s happening so this will tell us we have a relative max
relative max at x equals minus one and the second derivative says if your
first derivative is zero as we found if your first
derivative is zero and your second derivative is

01:52
positive then that says we have a relative min [Music] at x equals one
and the second derivative is inconclusive here inconclusive here so the
at zero at x equals zero the first derivative
is zero and the second derivative is zero so it’s inconclusive
but for them for the minus one the first derivative
is zero and the second derivative is negative
so that means i have a relative max at one so the first derivative is zero
and the second derivative is greater than zero so that means i
have a relative minimum so whatever this graph looks like
well we don’t know but we know where the relative min and whether
where the relative max are now because the second derivative is zero here it’s
inconclusive there may be a relative min there may be a relative max
or there might not be either so in that case

01:53
the second derivative is inconclusive so in order to finish this problem here
the second derivative isn’t enough because it says or neither
so we found the we used the second derivative test to find the relative max
the relative min but we didn’t do it for each first order critical number yet
so for -1 we did it and for one we did it but not for zero yet
so to finish this problem off we’re going to have to resort to the first
derivative test to finish it off so let’s look at what’s happening
less than zero or between -1 and 0 those are the critical numbers there
right and then also look between 0 and 1 and see what’s happening

01:54
so let’s look at the derivative here and so let’s choose a number between
minus one and zero and let’s see what’s happening to our first derivative
so if i choose the number between minus one and zero so like minus one half
so if i go to my first derivative here and look at minus one half
that’s gonna that’s gonna be a negative right here
negative times a positive so this will be negative so that means i’m decreasing
and now choose so i tested minus one half i got negative so it’s decreasing
if i look at a number between zero and one for example a one-half
so now i test a one-half in here and so that will still be a minus there
minus times a positive so that’ll still be a minus and so this
will be a minus again so so it’s still decreasing so we’re asking
is there a relative max or relative min at zero

01:55
and your answer is no because to the left of zero
the function is decreasing to the right of zero
the function is still decreasing so we don’t have either a relative max or
relative min so the second derivative test is inconclusive
at zero but it was really quick to find these two
but then since it’s inconclusive then you have to go back to the first
derivative test for that one right there and then we can see that
there’s neither a relative max so it’s inconclusive by second derivative test
but by the first derivative test x equals 0 is neither so f has no
extrema at x equals zero that’s what the first derivative tells
us so sometimes the second derivative test is inclusive and

01:56
when that happens you have to use some other method
and in this case we use the first derivative test so we had to go
and do the table but at least we didn’t have to go do the full table
we just needed to check to the left and to the right of zero
and so then we could determine if there is a extrema there or not
okay next example in fact there’s the graph right there if
you wanted to see that you can see there’s a relative max at minus one
and there’s a relative min at one and then at x equals zero
well to the left it’s decreasing and then to the right it’s still decreasing
okay so let’s look at this right here so let’s go here and [Music]
what’s our derivative so the derivative is one minus

01:57
four over x squared or said differently x squared minus four over x squared
and even better x minus two times x plus 2 over x squared
so this is what our first derivative says so we’re going to look
at the first or we’re going to look at the critical numbers
so it says use second derivative test at each critical number
remember the critical numbers come from the first derivative
so we’re looking here at -2 and 2 and what about 0 here
is 0 a critical number normally you say where the derivative is undefined it’s a
critical number but you also have to check the original
function zero is not part of the original function
in fact this has a vertical isotope at zero so zero is not in the domain of the

01:58
original so zero is not a critical number
but there is a iso a vertical isotope at zero
so we just need to check the minus two and two that’s it
for relative extrema so i need the second derivative now
these are the values that make the first derivative zero
but we need to test the second derivative to know if it has a relative
extrema or not so you know if you wanted to use the first derivative test
we could just go start making our table but if we want to use the second
derivative test well we need the second derivative
now where would you want to look to find the second derivative
i don’t want to find the second derivative here i’d rather find the
second derivative here derivative of the one is zero and think of this
as -4 x to the minus two so what’s the derivative of that it’s

01:59
just eight to the x minus three or said differently eight over x to the third
so that’s the second derivative now this is sweet because
that’s a very easy second derivative to work with
what happens when we plug in two we’re going to get eight over two to the
third is eight so we get a one which is greater than zero
and when we look at the second derivative at minus two
now we’re going to get eight over what’s minus two to the third that’s minus
eight so it’s minus one which is less than zero
so very quickly we can find out if we have relative or max or min
because if the first derivative let’s say at minus two if the first
derivative is zero right and at -2 the second derivative is negative
what happens if the second derivative is negative right so this tells us we have

02:00
a relative max [Music] and then at two at two we know we have
the first derivative is zero but at two the second derivative is
greater than zero so this tells us we have a relative min
so this is an example of how to use the second derivative test
i find my first derivative and i look at the critical numbers
and then i use those critical numbers knowing they make the first derivative
zero and knowing that the sign of the second derivative
will tell us if i have a relative max or min
so i know that whatever this curve looks like
at minus two and two i have a relative extrema
in fact we can kind of see what this looks like in fact actually i think we
looked at one similar to this before there’s an isotope here
and it went like this and there’s an isotope there
and it went like this and there’s an isotope there and then right here

02:01
and right here we have the relative min at two we have the relative min right
here and at negative two we have the relative max right here
so the graph looks something like that where there’s a vertical isotope here
okay so there’s an example there where the second derivative test
does quickly help us without having to make the full table
of all the critical numbers and look where it’s increasing and decreasing
sometimes just looking at the second derivative is very quick
so that gives you an option there to use that
okay so there’s the graph there and now for the exercises [Music]
okay so for the exercises here i just wanted to give us some
opportunity here for us to have some interaction

02:02
here are some exercises for you to try and when you try these um take good notes
and you know try to see which ones you have problems on
so there’s exercise one and there’s uh three more exercises
so on the first set of exercises here i’m giving you the function and you’re
asked to do the first derivative test and it’s increasing and decreasing and
find all extrema um you know also feels feel free to try the second derivative
test on all of these as well then for this next batch of problems
here two three and four so you can see number four here um
here we’re giving the opposite i’m giving you
all of the information so you don’t need to go find the
derivatives i’m just telling you the behavior of the function
and all you need to do is give me a sketch of it so those problems are fun to do

02:03
and try to try to sketch those graphs by looking at that
information that kind of tests if you really understand what’s happening
is by kind of reversing the problem so these problems
should be easier but i find that actually in practice that sometimes they’re
they’re hard as well so in any case try try some of those problems
and let me know how it goes in the comments below
okay so i want to say thank you for watching
i really appreciate you watching and taking the time
and if you like what this is if you like the series
hit that subscribe button below um now the next um episode as i mentioned a
couple times it’s about isotopes we’re going to look at
horizontal isotopes and vertical isotopes and we’re going to be taking the limit
as x goes to infinity and some of the limits we get are going
to be have infinite um so that’ll be a lot of fun um and so

02:04
just say thank you for watching and uh i’ll see you next time
if you like this video please press this button
and subscribe to my channel now i want to turn it over to you math can be
difficult because it requires time and energy to become skills
i want you to tell everyone what you do to succeed in your studies
either way let us know what you think in the comments

About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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