Concavity and Curve Sketching (Graphing Functions Using Derivatives)

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

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

00:00
[Music] what’s the difference between the functions
taking the square root of a number or squaring a number
both of these functions are increasing but one of them is
increasing much faster than the other comparing the growth
of functions is the idea behind concavity
and is foundational in the subjects like computer science let’s get started
[Music] alright everyone welcome back to the calculus one

00:01
explore discover learn series uh this episode is inflection points and concavity
curve sketching and we are going to uh talk about
uh concavity test in this uh episode and we’re going to talk about inflection
points um and then we’ll talk about uh vertical tangents and cusps
and once we get those two items out of the way then we’ll be able to talk about
curve sketching where we apply and combine everything that we’ve talked about
first derivative test concavity test that we’re learning today
vertical tangents because and also of course horizontal and vertical tangent uh
horizontal and vertical isotopes and so the previous uh episodes are very
important to to watch first so i have a playlist in the description below
where i uh have video an episode over um isotopes and
the first derivative test and the extreme value theorem
so those videos or are of course uh useful

00:02
to watch before this one right here but the main topic of this one is the
concavity test and that’s we’re going to cover first let’s go
okay so up first is the concavity test of course to understand the concavity
test what is concavity and why is it important so first thing is
we’re going to need a number in our domain
so we’re going to say c is in the domain of a function
we have a function given and we’re interested in a number
and we’re going to call that number c a first order critical number
if the derivative is zero or the derivative does not exist
now that should look familiar to you because that’s just what we’ve been
calling a critical number we talked about critical numbers um i
think in the episode extreme value theorem first
and so you know make sure and understand what critical number is

00:03
but the reason why we’re calling them now instead of just critical numbers
now we’re also calling them second order critical numbers
um because we need to have different uh derivative tests one will be a
concavity test and one will be the first derivative test
so for the first derivative test we use the
first order critical numbers and if you look back at that episode we just call
them critical numbers but now we’re going to have a concavity
test in the concavity test is going to use the second order
critical numbers i want to distinguish between
the two types of critical numbers the first order in the second order
the second order is the same as the first order critical numbers
it’s just the only difference is we’re looking at the second derivative now
if the second derivative is zero or if the second derivative doesn’t exist
then they’re called critical number second order critical numbers
of course you always start off with a number in the domain of the function

00:04
all right so um what does concavity mean well suppose we have a function that’s
differentiable and we’re looking at the um first
derivative here if the first derivative is increasing then we call the function
concave upward on that interval if the first derivative is decreasing
then we call the graph of f concave downward on that interval
and so let’s see what that means let’s see what that means
so if the graph of f lies above all of its tangents
on an interval then it is concave upward on i
on the enter on that interval if the graph of f
lies below of its tangent is concave downward on that interval and
if you have a change in concavity if if if at a number c to the left of it

00:05
to the left of c it’s and then to the right of c it’s a different concavity
then we say we have a change in concavity so for example to the left of c
might be concave upward but then to the right of c
the graph might be concave downward and then we would have an inflection point
because it would be a change in concavity
so let’s look at here this graph here so you can see what i mean
you know when we have this part right here from a to b
here so i’m looking on the interval right here a to b
and i’m looking at the shape of the graph right here
so this is an example of a concave downward graph
and the reason why is because if you pick a point any point
and you were to draw the tangent line the tangent line would be above the graph
for example if i pick the point right here pick a point right here along the
x-axis pick a number and they come up here to the graph and
draw the tangent line the tangent line would be above the graph

00:06
now from b to c on this interval the graph is now concave upward looks
like the shape of a bowl so if i pick a point between b and c say one right here
and i go up and i draw the tangent line the tangent line is now below the graph
so what about c to d is it concave downward or is it concave upward
so if i come over here and i can see that’s concave downward
and here it’s concave upward [Music] and here it’s concave upward and here’s
concave downward notice there’s a inflection point at p
because to the left of p we’re concave upward
and to the right of p we’re concave downward so there’s a change in
concavity at p notice that b is also a inflection point because to the left

00:07
where concave down and to the right were concave up and same thing
for point c here we’re concave up and then to the right of point c here
now we’re concave down so there’s a change in concavity at c
so c point c is a an inflection point so you know notice the definition of an
inflection point says the we have an inflection point
provided f has a tangent line at c so notice that at d here there is a
sharp corner and so there’s no derivative at d um so right here also
at this there’s no inflection point at e where concave down we’re concave down
there’s no inflection point at this number here e
okay so if the graph lies above all this tangent on an interval is

00:08
concave upward if the graph lie lies below all this tangent it’s concave
downward and we have the definition of here of an inflection point
so here’s what the concavity test is because the concavity test is going to
allow us to break the number line the horizontal axis
into pieces into a finite number of pieces
and we’ll be able to test each of these pieces each of these intervals
and determine whether it’s concave up or concave down
so to use a concavity test though you need to know that your function is twice
differentiable in other words the derivative has a derivative
now if you have that then what you can say is that
if the second derivative is positive then the graph of f is concave upward
so if you know the second derivative exists on the interval and it’s positive

00:09
then the graph has to be concave upward if the second derivative is negative
on an interval then the graph is concave downward
and so let’s get started with an example so you can see
all this put together so we want to find all the inflection points
all the local extrema and we want to sketch the curve
so here we go with our first example here
so when i’m looking at this example here we need to take the first and second
derivative test now it says to find local extrema right
there right so that’s the first derivative test
we can use the first derivative test on that and to find the inflection points
that’s the concavity test which we just talked about and we’re going to now
understand better so here we go so the first derivative three x square

00:10
and then minus three and we can factor out three and we can factor all this
x minus one x plus one and i’m gonna set it equal to zero and so
x equals plus or minus one um before we would say these are critical
numbers now let’s say these are first order critical numbers
just to make sure i distinguish them between the second order critical
numbers i’m about to find so first order critical numbers
and so we could go apply the first derivative test now
um let’s go ahead and take our second derivative now when i’m taking the
second derivative i’m not looking here i’m going to look right here
so this will be 6x and that’s it and so i’m going to set this equal to 0
and solve and so x equals 0 is the second order critical number [Music]

00:11
second order critical number now make sure that you always keep in mind
the domain of the original function we have a polynomial
it is defined everywhere is continuous everywhere
is differentiable everywhere so we have our first order critical numbers
and he says and this is the second order critical number here
so in order to find the local extrema and inflection points
i’m going to find i’m going to work out the first derivative test and the
concavity test so first i’ll do the first derivative test
so i’m looking at the uh first order critical numbers here
so i’m going to look less than minus one between minus one and one
and then and then let’s go ahead and put equal to minus one
and then let’s put between and then equal to one and then greater than one
and so these are the first order critical numbers minus one and one

00:12
right so i broke up the number line i’m less than i’m equal i’m between
i’m equal or greater than and so here i’m gonna have a column for f
and the derivative and i’m gonna make some conclusions
right so we covered this in detail in the uh episode that derivative test
monotonic and derivative test all right so i’m going to be testing
this this right up here and if i choose a number less than minus
one for example minus two then what’s happening over here we’re
gonna get a negative and a negative what’s a negative times a negative we’re
gonna get a positive [Music] and let’s test the number in here for example zero
we test zero into our first derivative so we’re going to get a negative times a
positive so we’re going to get this whole thing as negative
now i’m going to test the number greater than one for example
two just test the two here positive positive so the whole thing’s positive

00:13
all right so now we know that this is increasing we know this is increasing
we know we’re decreasing right here [Music]
all right so what’s happening at these points here do we have any relative
extrema so we’re changing from increasing to decreasing
so at -1 we have a change in monotonicity we’re changing from
increasing to decreasing so think of it like this we’re
increasing and then we start to decrease right so that’s going to be a relative
max so at minus one we have a relative max
now we’re we have a change also a change in ma
monotonicity so we’re decreasing and then we start to increase that’s going
to be a relative min here so at at one we have a relative min
now because we have to go and sketch the graph here

00:14
we probably should just go ahead and plug in these points here what’s happening
at minus one what’s happening at one so at minus one
we’re having minus one to the third minus three times minus one and then
plus one and so that’s what minus one uh plus three plus one so that is three so
at minus one we have a three and then what’s happening at a one
we have one to the third minus three times one plus one and so that’s one
and one that’s two so minus three plus two so that’s a minus one
all right so we have a relative max the value is three
and occurs at minus one we have a relative min
of minus one and it occurs at x equals one
and so there’s our first derivative test it gives us our real

00:15
it gives us our local extrema so you can put local max here or local men
it says local extrema here so maybe i should use the word local but
to me they’re interchangeable okay so now let’s go do
the concavity test so um we only have one second order critical number so it’s
going to be a lot easier easier to do the concavity test now when
you have multiple tests maybe you might want to put here in this
box here first derivative test so now when i go make the concavity test
i’ll have a ct here for concavity test and we’re looking less than zero equal
to zero and greater than zero because those are the only choices x
equals zeros are only second order critical number
of the concavity test here i’m gonna look at the function
in the second derivative and i’m gonna make a conclusion here
all right so that says conclusion there all right so

00:16
what’s happening to the second derivative i’m going to test right here
at minus six so i’m going to choose something less
than zero for example minus one and i get a minus six that’s negative so
the second derivative is negative that means concave down so concave down
and some people draw you know this right here showing it’s concave down
like i draw the arrows here some people accept only this some people accept only
that just depends upon who you’re communicating with let’s look greater than one
let’s say a two we tried to the second derivative is positive without concave up
and so do we have a change in concavity and the answer is yes on the left side
of zero or concave down on the right side we’re concave up so

00:17
this right here is an inflection point so just like here we had to change in
monotonicity so we had a relative extrema
here we have a change in concavity so we have an inflection point
now what is that point the inflection point happens at x equals zero but what
is the actual output value so if we go but look back at the
original function it’s at one okay so now if we look at all this
information here we should be able to sketch the graph
so let’s see here i’m going to try to sketch it right here
and see if you guys can see all of this here we need to take into account the
first derivative test so i need to break it up between minus
one and one something’s happening at those points right there
so i’m gonna stay on minus one is here and a one is here
and i’m gonna be increasing and then i’m gonna reach that height of three and

00:18
then i’m gonna be decreasing so i’m not gonna pay attention to this
yet i’m going to just be increasing to minus one
i’m gonna hit a height of three at minus one
so let’s say minus one is here i have a height of three here
so i’m going to be increasing to there and then so here’s minus one
here’s minus one three that’s the point i have right here
so i’m going to be increasing to minus 1 3 that’s where i have my relative max
right here so now i’m going to be decreasing between
-1 and 1 and where do i cross the y-axis
that’s this point right here zero one so let’s put a one here
so i’m going to be increasing make sure you don’t have a sharp corner there
because it’s not because the derivative is
exists so it’s nice and smooth and then i’m going to be

00:19
at one here we have a relative min so that’s at minus one so
let’s say a minus one is here so i need it to be nice and smooth through there
and then after one so nice and smooth through here
we don’t want to make a sharp corner there now at one
greater than one then we start to increase we know this doesn’t come back
down because it’s going to be increasing and we know this only gets here it’s
increasing here and then it’s decreasing here
and it hits somewhere in here we don’t know exactly where yet
alright so this is increasing and then decreasing and then increasing
and i have my relative max and my relative min and that all comes from the
first derivative test now let’s see if i sketched it right
according to the concavity test because this is increasing here but
there’s different ways to increase wait wait a minute what do you mean

00:20
there’s a different way of increasing well just as a quick aside example here
think about these two functions here i was like to have these two functions in
my mind when i’m thinking about concavity and
and the expression there are different ways of increasing right
when you have x squared it’s just a parabola going through here
when you have y equals square root of x it’s just going like this
so both of these functions are forever increasing when you’re greater than zero
it’s just increasing and this right here when you’re greater
than zero you’re just increasing but this one right here increases much
slower this one increases much faster so how do you how do you
like quantify that well concavity remember takes into account
the second derivative this is concave up here concave up is
going to increase much faster this is concave down right here and this
is going to increase much slower in any case back to our problem [Music]

00:21
when i’m less than zero i’m concave down so here’s zero right here did i draw a
concave down yes when i’m greater than zero
oh that says one when i’m greater than zero sorry if that confused anybody
when i’m greater than zero it’s going to be concave up
and so do i have it concave up yes and so right here at 0 1 is an inflection
point and that tells me exactly where i changed concavity
so this is a relative max relative min and that’s a change in concavity right
there so it looks like i’m concave up looks like i’m concave down
and if you don’t see it quite if you’re not happy with it too much
you know make it make it more obvious ah
make it more obvious you know don’t make it so straight it’s nice and smooth
and here this is you almost want to like draw it first

00:22
and then label everything this is one one minus one yeah so there there’s a
nice smooth curve and you can see the relative max and the relative extrema
but what’s new is you can see the inflection point right there at 0 1.
all right so there’s our first example there and we can see it now
so that’s what we came up with right there this is a nice smooth curve going
through those points and we have a minus one three is our relative max
we have one minus one is our relative min and we have our inflection point right
there as zero one all right let’s look at the next example
okay so determine where this curve is concave upward
where it is concave downward and what are the inflection points

00:23
so this one here we just need to do the concavity test let’s do the concavity
test here so to do the concavity test we need the second order critical numbers
so let’s first start off with the first derivative
um so let’s see here we’re going to get 4x to the third
and then minus 12x squared and we don’t really need to set this
equal to zero and solve it because we don’t need the first order critical
numbers if you’re looking for relative extrema
and doing the first derivative test of course but here i’m going to go on to
the second derivative so this will be 12x squared minus 24x
and to find the second order critical numbers i’m gonna look to see where this
is zero and where this is undefined now look at our original function
it’s a polynomial it’s continuous and differentiable everywhere
and the second order uh the second derivative is also continuous
and differentiable everywhere so i just need to find out where this is zero

00:24
so i’m gonna set this equal to zero and let’s solve so we can factor out an
x and get an x minus two and so we get x equals zero and x equals two
and there’s our [Music] second order critical numbers here so
let’s go and do the concavity test so i’m gonna look less than zero [Music]
equal to zero between zero and two equal to two and greater than two
so this will be my concavity test and i’m going to be testing the function f
and to do that i’m going to be using the second derivative and i’m going to make
some conclusions [Music] all right so let’s look less than zero
so when i’m testing my derivative i want to look right here
this is the easy one to test the second derivative is right here [Music]

00:25
so i usually like to underline it just so it’s easier to find but
there’s my second derivative right there
i’m going to test it’s nice and factored for us
i’m testing less than zero for example minus one that’ll be a negative and a
negative so multiply we get a positive so it’s concave up
uh between zero and two i test the number for example one
i get a positive times a negative so that’ll be negative so it’ll be concave
down now is there a change in concavity yes punk gave up concave down
so at zero we have an inflection point now you you can’t just say um
if it was concave up and concave up there would be no inflection point
or if it was concave down and then concave down there would be no
inflection point there has to be a change in concavity
all right so now let’s test greater than two for example a three
so i go to my second derivative in test three so here is positive and positive

00:26
multiplied together we get positive so it’s concave up we have another
change in concavity so we have another inflection point here
so here’s an example where we have two inflection points
we have an inflection point zero what happens when we substitute zero into our
graph we have zero what happens when we substitute in two
so two to the fourth minus four times eight all right so that’s what uh
16 minus 48 and what is that minus 32 okay so we have the inflection points here
in here and um we’ve determined where it’s concave up
it’s concave up less than zero and greater than two
and we determine where it’s concave downward that is between zero and two

00:27
and we determine what the inflection points are so we got everything labeled
here in this table and we found these these outputs here make sure you find
the outputs because you need the inflection point so you need an x and a y and
everything is here and so now we can go look at the graph
all right so there we go so we can see the concavity that is concave
up less than zero and that is concave down between zero and two
and then you can see that it’s concave up when you’re greater than two
so all that matches the the concavity test that we just saw all right very good
let’s go on to the next example uh now let’s look at vertical tangent
some customs all right so now i’m going to cover vertical tangents and cusps

00:28
but not not all calculus courses do that in fact if you look in some calculus
books some of them cover vertical tangents some of them don’t
i’m going to go ahead and put it in here so we have a complete um
understanding of of how to make really good sketches um at least
in terms of concavity and tangents and cusps and i actually find them quite fun
now when we’re looking at the derivative um there’s four possibilities i’m going
to show you so we’re going to assume we have a continuous function
we’re going to assume we have a continuous function and
we are going to talk about the four possibilities so
two of the possibilities we’re going to have a vertical tangent and the other
two we’re going to have a cusps so to find the vertical tangent

00:29
i’m going to look where the derivatives um on the left and on the right
are both matching and they’re both positive infinity
if that happens you have a vertical tangent at c
if the both the derivatives the limits as approaching left and right
if they both match again but this time they’re both minus infinity
then we also have a vertical tangent and so that’s two
of the possibilities the other possibility is
if they don’t match so we’re taking the limit of the derivative
from the right and if that’s positive infinity and then from the left is
minus infinity then we have something called a cusp
at c and then the last final possibility is
if the derivative the limit from the uh right of the derivative
is minus infinity and then from the other side
the derivative approaches positive infinity then we also have a cusp so notice
in all of those cases we are looking at the

00:30
limit of the derivative so in order to find out if you have a vertical tangent
or vertical or cusp you have to look at the limit of the derivative
and so let’s look at some examples so we can get a
good feeling for what this means here so here’s our first example here now
when i’m looking here at the uh curve here we have to ask ourselves
where do we start what do we what do we need so let’s go back for a second here
how do we find like like like let’s look at number one for a second
how do we know what the seas are how do we know where to look
how do we know when the derivative is going to be infinite
so we covered infinite limits um i think last episode we talked about
horizontal and vertical isotopes but we also talked about limits
involving infinity but in a way a way to get an infinite

00:31
limit whether it’s positive or negative infinity
is when you have a fixed number in the numerator over something that’s going to
zero and so we’ll look for that case we’ll look for those cases
but first of course we need to find our derivative alright so here we go
so our derivative is going to be all right so we’re going to use the
product rule here so i’m going to take the derivative of
the first piece so it’ll be nine fifths x to the three-fifths
minus one and then times the second one which is five minus x minus four x
squared plus now the first one so three x to the three fifths
times the derivative of the second one which is minus one minus eight x

00:32
and then times the derivative of the uh yeah that’s good all right there we go
yeah so um the first piece times the derivative of the second
and that’s it all right so now um in order to see
where what the seas are where i’m going to be taking my limits of the derivative
i need to get a common denominator so let’s put all this together
so this has a negative exponent here so and we have a 9 up here so let’s write
this as 9 times five minus x minus four x squared
all over five and then we have x to the two fifths power plus
and now all of this is in the numerator um so let’s just write it down again
and right so i just i didn’t change anything just write it down again
think of this as being over one but we need a common denominator

00:33
so i need to multiply by five x to the two fifths over five x to the two fifths
and so now we have a common denominator which is five x to the two fifths
and let’s see what’s happening in the numerator so here we’re getting 45
minus nine x minus 36 x squared 36 x squared
all right and what’s happening over here here we’re getting a
three times a five so let’s call this um 15 x um
so i’m just taking this times this part here
and i’m going to say 15 and then an x right because that’s three fifths
plus the two fifths that’s five fifths which is one
and i’m just going to leave this out here for right now [Music]

00:34
all right so i’ll distribute that next and let’s see if we get anything
interesting [Music] so let’s distribute that we’re going to get minus 15x
and then we’re going to get a minus a 15 times a which is obviously 120
x squared all over five x to the two fifths [Music] all right and that’s a 15.
[Music] okay so so far so good and the that can be simplified more but
if the x is zero here remember this is ours um first derivative here [Music]

00:35
and right so that’s the function is f so there’s our first derivative and
um we can simplify that but what c are we going to be looking for so remember
when we are looking at the um when we’re looking for vertical tangents
and cuts we need those the limit of the derivative to go to blow up to
positive or negative infinity so i need to find those c’s to look at
it has to be some c where the function is continuous so
when i’m looking here i’m looking at the c
equals zero here because the derivative here at at x equals zero is not defined
so it and when i plug in zero i’m gonna get all those
zeros there i’m gonna get 45 over zero so i know that if i take the limit as we
approach zero we’re going to get um some infinity so now i need to look at

00:36
the limit as we approach zero from the right
of the derivative and i need to look at the limit
as we approach zero from the left of the derivative
so i’m looking for vertical tangents or vertical cuts
and i look where the derivative is going to blow up we approach zero of this
derivative right here we’re approaching from the right so
what is this what is this limit going to be if i try to plug in 0
i’m going to get 45 over 0 right so i know it’s going to be some infinity
same same for both of these so here it’s very crucial that if you
don’t understand this you go back and look at the episode over
uh limits involving infinity the horizontal and vertical isotope
uh episode but as we approach zero from the right we’re going to get 45 over
zero the question is is it positive or is it negative infinity

00:37
so as you can see all these other terms over here don’t really matter
what matters is as you’re approaching zero from the right
this is getting extremely small these are getting really small
think of x for example is like point zero zero zero one
so that’s really close to zero these are all really close to zero
the 45 is dominating so it’s going to be positive
and if i choose a uh any any number right here whether it’s pot
to the right of zero to the left to zero that x and that’s going to be a squared
and then i take the fifth root this is always going to be positive out
here so both of these limits are positive infinity right here
and so what that means is that we’re gonna have a vertical tangent [Music]
at x equals zero we have a vertical tangent there [Music]
so that’s what these two limits give us right here and now if i try to sketch

00:38
the graph it’s gonna there’s gonna be a vertical tangent here
so when i’m approaching zero the function
the derivative the slope of the tangent line is increasing without bound so
something like this right here [Music] so if i draw a tangent line for example
right here the slope is positive but if i draw
another tangent line say right here the slope is still positive but it’s
it’s greater than the slope that it was right here the closer we get
to zero the tangent lines increase the tangent lines keep
increasing and so we have a vertical tangent there
the slopes of the tangents are just growing
unbounded as you get closer and closer zero in fact let me try to make it even

00:39
a little bit more obvious let’s see if i can do that
i mean it’s almost like a vertical line there i’m going to come in
and look really close [Music] so it’s still a function
but there’s a vertical tangent right there so if i start taking um slopes of
tangent lines that you see the slopes keep increasing
when you’re really close to zero the tangent line is almost vertical
here the tangent line is positive when i get a little closer the tangent line is
still positive alright so here’s what graph would look like
right there uh and you can see the vertical tangent at x equals zero
so there’s an example of a vertical tangent and we found it by looking where the
first derivative was going to blow up [Music]

00:40
so let’s look at another example now so an easy way to generate vertical
tangents and cusp is to be looking at fractional powers
and so here’s another one let’s look to see if we have a vertical tangent or
cusp on this one here so here we go let’s find the derivative and see where
the derivative blows up so here we go the derivative is all right so
derivative of the first one is two thirds x to the two thirds minus one
and times the second one [Music] plus now leave the first one alone
times the derivative of the second one which is two x
plus five all right so now because i have a negative exponent here

00:41
i know i need to get a common denominator here
so i’m gonna write this as a two x squared and then it
plus 10x and then a minus 40 and then in the denominator i have three
x to the one third and over here i just going to write down the same thing
it’s a positive exponent there so i’m thinking about this as being over
one and i want to multiply and get a common denominator
so i’m gonna do three x to the one third and then three x to the one third
all right so now we have a common denominator
our denominator is 3x to the one-third and now let’s see what we get on top now
i see a 2x and a 10x let me go double check that
i think that should be 2x squared i forgot my squared here
so this will be 2x squared plus 10x minus 40.

00:42
and what’s happening over here we have x to the two thirds
times an x to the one-third so that’ll be x to the three-thirds which is just
an x so i’m going to get a 3x multiplying that part in that part
and then i’ll just leave this alone for for now
in fact there’s really no point in expanding this out because
you know this is our first derivative here
right that’s the first derivative and we need to look where this blows up
right we don’t need to go simplify this the the goal of this exercise isn’t to
find the first derivative it’s just to find the vertical tangents and cusps
oh in fact they actually tell us uh sorry just looked at the problem
the i’m asking why there is a cusp at zero all right so
pretend you didn’t know that where is the derivative blowing up

00:43
so if i look at zero makes the bottom zero the denominator zero
what about the numerator that’s going to zero that’s going to zero
minus forty and then because of that zero right there that’s all going to zero
so this looks like it’s going to minus 40 over 0
so that’s going to blow up so i’m going to look again at zero so i’m going to
look to the left of zero of the derivative make sure don’t put
the function there we’re looking at the behavior of the derivative [Music]
and then as we approach zero from the right [Music]
so because when we’re approaching zero from either side these are going to zero
and that’s going to zero so the top is going to
a fixed number and the denominator is going to zero
so from those two things i know that this has to be some infinity here it has
has to be some kind of blow up the question is is it
positive or negative infinity so i look at where i’m approaching i’m approaching

00:44
from the left that means the x’s are negative so
the top is going to be dominated by -40 that’s
really close to zero that’s really close to zero and that’s all really close to
zero and then minus forty right so the top is going to be negative
and then now what about the bottom approaching zero from the left means
we’re like negative point one so when i do that fractional exponent a negative
will still be a negative so i’m looking at negative over negative
and so i’m looking at positive infinity here and what about this one i’m still
looking at negative but now i’ll be looking at positive down
here so negative over positive will be negative
now because there’s a change here this satisfies the condition of a cusp
at x equals zero if there’s if there’s if one of those positive in one of those
negative infinity then there’s a cusp if they both match then that’s a

00:45
vertical tangent so now we see why we have a cusp at zero
and if we look at the graph we can see that so there’s a cusp
at x equals zero there um as you approach zero from the left the
slope of the tangent lines are going to infinity and if we approach
zero from the left the slope of the tangent lines are approaching minus infinity
so i’ll try to illustrate what i mean by that looking back here
so if we sketch the graph here it’s coming in like this and it’s going
back out like this and so what happens what is this right here mean
right so again the derivative is the slope of the tangent line
we’re approaching zero so we’re close to zero so let’s say we’re right here

00:46
right what’s the x value right there not too close to zero but it’s a little
bit the tangent line is right here so what’s that slope
now let’s look a little bit closer here now the slope of the tangent line is
is greater so as we move closer and closer the slopes of the tangent line
are getting greater and greater and greater so the slope of the tangent
lines are going to positive infinity now as we approach zero from the right
like take this number here it’s a little bit close to zero now what
does the slope of the tangent line look like it’s going down
it’s negative and if i move a little bit closer it’s still negative it’s a
greater negative if i keep moving closer and closer the negative
becomes more and more negative right there so you can see that one right there
is also holding so those are the tangent lines there
and that’s the behavior there and you can see that we have a cusp right here
[Music] all right very good so let’s look at one more

00:47
and this time i’m not telling you which one it is
we’ll have to we’ll have to work it out and find out together
so here we go we’re going to look at the first derivative
and see where the first derivative blows up if if any places at all all right so
our first derivative is so derivative of the first one so two thirds
minus one and then x minus one to the one third plus
now x to the two thirds and times the derivative of the second one
so derivative of the second one is one-third x to the minus one and then now my
one-third minus one and then times the derivative of x minus one

00:48
which is just one all right so on this one right here [Music]
we’re going to write this as two and then x minus one to the one third
over three and then we have x to the one third [Music]
plus now here we have x to the two thirds on top [Music]
three’s in the denominator and then x minus one to the two thirds
okay so here we need to get a common denominator we cannot look at these
individually and determine where the derivative blows up we have to
look at one numerator one denominator to get that kind of behavior
so what’s going to be our common denominator
i’ll show this step out i’ve seen this um not be the friendliest thing for
students they both have three so i’ll see this is x minus one

00:49
um and i’m gonna multiply x minus one over x minus one to the two thirds here
[Music] all right so multiply numerator and
denominator by x minus one to the two thirds because that’s what’s missing
here this one has it but this one doesn’t plus x to the two thirds
over three and then leave this one alone and then what are
we missing over here this one doesn’t have
the x to the one third so to multiply by x to the one third
over x to the one third all right so that could be your next step
if that’s not your next step that you’re
written then you’re probably visualizing that
whatever case may be what do we get um we get the common denominator right
three and then what’s our common denominator x to the one third

00:50
and then x minus one to the two thirds there’s our common denominator sorry
there’s our common denominator there and so what we end up with here x
x minus one to the three thirds so x minus one so we’re gonna get two times
x minus one plus and here we’re just gonna get an x and so
our derivative simplifies quite nicely it’s a two x plus x so that’s a three x
minus three x minus two over three and then x to the one third and
then x minus one to the two thirds all right so there’s our derivative
so now where are we going to look now we have two places to look
where x is zero because why x equals zero well it makes the
denominator zero and makes the numerator non-zero

00:51
so the derivative is going to blow up at zero same thing at one
when i try to use x equals one in here i get zero down here and i get a
non-zero up here so we’re going to have unbounded behavior
so let’s go look at these four limits we’re going to approach versus 0 right
here we’re going to look at the left and we’re gonna look at the right [Music]
make sure we have derivative here so what are these limits
so again we’re approaching zero so the numerator is going to go to minus 2
and the denominator is going to go to 0 so we know that these are some type of
infinity here so when i’m looking less than 0 say a negative 0.1
well i’m going to get a negative and then a negative
and then this will be positive because of that square right here this term

00:52
right here will always be a positive and so i’m going to use a negative 0.1
so i’ll get a negative over a negative so that’ll be positive [Music]
and when i look at 0 from the right of this derivative right here so say
like 0.1 so that negative 2 will still dominate so i’ll get negative
and then i’ll get a positive and this will be positive
so that’ll be negative over positive so this will be negative infinity right
there so we have a cusp at x equals zero now what about at the minus one also
so i need to look at these two limits also limit as we approach uh one from the
left of the derivative and the limit as we approach
one from the right of the derivative so what are these limits
derivative all right so one from the left so again this is always gonna be

00:53
positive because that’s square so when i’m looking at one from the left
like .9 well that’s positive and if i’m looking at one from the right like 1.1
that’s still positive so for either one of these
the denominator is positive and for either one of these
well let’s look at it let’s look at this one we’re approaching one from the left
like point nine that is going to be three that that’s really close to three
minus two that’s positive so i’m getting positive over positive
so i’m getting positive infinities for these
for both of these for example one point one
that’s a little bit greater than three but that’ll be
minus two that’ll still be positive so we’re gonna have a vertical tangent here

00:54
vertical tangent at x equals one [Music] so this is what we found for the
derivative and by looking at the derivative right here
we’re able to find these limits right here where there’s blow ups
and we’re able to find these uh limits here and we can determine if they’re cuts
because these limits don’t agree and these two limits agrees
is a vertical tangent here are very good let’s look at the graph
okay so then you can see we have two different
uh types of unbounded growth for the derivative
at zero you can see that we have a cusp and at one you can see that we have a
vertical tangent all right so let’s go on now next part

00:55
okay so now we’re going to talk about curve sketching
curve sketching so now we’re going to put everything together
that we’ve talked about and i’m going to put it all like in a checklist format
for us so we can see what we have accomplished so far [Music]
and putting it all together putting all together might seem boring
but at the same time if you’re new to all this
when you work it all out in one problem then you can kind of see how
they’re different from each other and how they
how it all works together to give you a really good
sketch so i think it’s really fun to do at least a couple of them all together
but after a while it gets you know kind of a long problem to do
but anyways if you’re given a function you know one of the first things you
want to look at is domain and range and of course you want to simplify the

00:56
function if possible um now i say algebraically simplify the
function if possible but you know also trigonometrically
so for example if you if you were given um hey go graph the function for us
sine x squared plus cosine squared x right graph that function for us
well hopefully you would know the trig identity that no matter what x is
that this is always one right so you’re just going to be graphing the horizontal
line y equals one and there’s really nothing to do there
so not only algebraically simplify but also
trigonometrically simplify or just flat out simplify the function um
but make sure that you don’t alter the function so for example
you don’t want to go overboard in your simplification

00:57
in the sense that you’re actually changing your function for example
look at these two functions here so these are two different functions this
is a function right here that looks like the line y equals one
and this is a function right here that looks like y equals one but in fact
they’re not both the same graph this one has a hole at one
uh sorry this one has a hole right here at zero
right so if someone says here’s your function right here
well i would not simplify it to this one because these functions have different
domains this function the domain is all real numbers this function is all real
numbers except 0. so you want to simplify your functions
but not alter your function all right in any case
step three determine where the function is continuous
so we have a large collection of continuous functions that we know
about for example polynomial and rational functions are

00:58
continuous on their domain and trigonometric functions are
continuous on their domain and exponential and logarithmic so and
then we know how to combine them together using addition subtraction
multiplication and composition so we know a lot about
continuous functions and and we have building blocks we know how to create
more continuous functions so having a good idea where your function
is continuous and not continuous is always the important part when you’re
sketching of course you know as you did in
precalculus spend some time looking at your intercepts
they can help your sketch your graph sometimes they’re hard to find but in
many cases you’ll be able to do that and make sure
that you’ve watched the episode over asymptotes make sure you know how to
find your vertical asymptotes and your horizontal asymptotes
make sure you understand how to find those using limits
because when you’re sketching the graph it’s the limits that are actually

00:59
uh very very important all right and so then the heavier uh techniques are
finding the critical numbers and using the first derivative test
finding the second order critical numbers and using the concavity test
and then what we just talked about a little bit ago vertical tangents any cusps
a lot import plot important points for example
any points that you might find that are extrema those are obviously important
any points that are tangent vertical tangents or cuz
intercepts you know plot those important points and then
finally sketch in between all that information
and make sure your sketch is reasonable in the sense that don’t put any sharp
corners where there shouldn’t be if you put a sharp corner that means
you’re telling the reader that the function is not differentiable
there so if you know the function is differentiable

01:00
make sure you make it nice and smooth so
if i was to sketch the graph of y equals x squared versus the graph of y equals
absolute value of x and i was just lazy this graph right here has a sharp corner
it’s not differentiable at zero and if i was to get lazy and just sketch that
graph and you know can you tell which one that is right so
that’s all i’m saying is make sure that you don’t put any unnecessary sharp
corners where there are not any all right so let’s begin that procedure
now and let’s look at this function right here
and so we’re going to sketch the graph here and i’m going to um
look at we’re going to look at this graph together here
and try to figure out what’s going on so i’m going to start

01:01
as is my habit of finding the first derivative so here we go so we have low
derivative high which is 4x minus low all over 4x wait
so this will be 2x squared and then this will be 2x
all over x squared minus 1 squared so there’s our first derivative let’s
see if we can simplify this so this will be 4x to the third
minus 4x and this will be minus 4x to the third also okay so

01:02
in other words our derivative is just the minus 4x the 4x to the thirds
add up to zero okay so i’m going to erase this part right here
because we’re going to need all the space we can get
and just write the first derivative here is minus 4x
over x squared minus 1 squared okay so there’s our first derivative
now we can go find the local extrema if there are any we can look where it’s
increasing we can look where it’s decreasing
um what about any kind of concavity what about
inflection points and stuff like that so maybe we’ll take a look at the second
derivative so here we go um low times derivative high

01:03
minus high times derivative low which will be two x squared minus one
so 2x squared minus 1 and then times another 2x
and then all over x squared minus 1 to the fourth
all right and let’s see if this is going to be something we can work with here
[Music] i see i have x squared minus 1 squared
and over here we have an x squared minus 1 but only to the first power so
what can we factor out from see this minus right here is separating
things so out of this we can factor the minus four and one of the x squared
minus ones i’m gonna write that here minus four and then x squared minus one
and then what do we have left we still have one of these x squared

01:04
minus ones left we factored out the minus four now we have a minus
now i factored out that minus four so we still have an x
left and it’s going to be times and i factored out that x squared minus
one so it’ll be times four x and then all over in fact let’s just put
it over here all over x squared minus one to the fourth
okay so let’s see here what we’re going to get
we’re going to get minus 4 x squared minus 1.
and then what are we getting here we already have an x squared and we’re
going to have a minus 4x squared so i’m going to say minus 3x squared

01:05
and then minus 1 and then all over x squared minus 1 to the fourth okay so
now i’m going to look at the minus 4 those are minuses and i’m going to
cancel here and so what i’m going to end up with here is a 4
and then the 3x squared plus 1 and then that’s going to cancel with one
of those so x squared minus 1 to the third okay so
what are the second order critical numbers now when i’m looking at my second
derivative here i’m looking at hey it’s undefined at plus or minus one
but the original was undefined at plus or minus one also
right so we’re gonna need to be careful at plus or minus one because
um you know what’s happening there is we have vertical isotopes there

01:06
so let’s write our second derivative over here um yeah so let’s do that
right here our second derivative is four times three x squared plus one all over
the x squared minus one to the third all right so this numerator is always
positive um and so we’re going to be looking here
at this third here there may be a change in concavity there
all right so now that we’ve done that grunt work we found the first and second
derivative now we can start trying to find the uh relative extrema
and the concavity the points of inflection and all that stuff here so

01:07
let’s do the first derivative test first um so
yeah okay let’s just do that all right so i’m looking at the
critical numbers what are the critical numbers
um actually before we do the first derivative test let’s just go and take
care of the isotopes right so we need some limits remember to
do the vertical isotopes we need some limits
so i’m looking at the plus or minus one here
so what’s happening is we take the limit as we approach the
um one of the function here and the limit as we approach one
from the left and from the right and what about these limits here when we
approach minus one from the left and minus one from the right
so we need to know these limits here because we’re getting

01:08
unbounded growth there we need to be able to have that on our sketch
so when we look at one from the left of this function up here so notice when
we look at that function up there the top is always positive
but when we’re approaching one we’re going to get two over zero so we’re gonna
so all of these are going to be under bounded growth
because all of these limits the numerator is approaching a non-zero number
and the denominator is approaching zero so we’re getting unbounded growth
so the question is is it positive or negative so when i’m
approaching one from the left like for example like 0.9 now if i do
0.9 squared minus 1 then that’s going to be
negative so when i’m looking at 1 from the left this will be negative infinity

01:09
when i look at 1 from the right for example like 1.1 now 1.1 squared
minus 1 that makes that positive and a positive to the third power will still
be positive the numerator is always positive i’m
sorry yeah the numerator is always positive 2x squared right
all right so we got those now what happens if we approach
minus one from the left you know something like minus one from the left
would be like minus 1.1 and when i square that
i’m going to get something greater than 1 and then minus 1. so that will be
positive so when i approach -1 from the left this is positive
and when i approach minus one from the right like
um you know something like negative point nine
and then but if i square that and then subtract one
that’ll be negative so when i’m approaching minus one
from the right i get minus infinity so clearly we have a vertical isotope at one

01:10
and vertical isotope at minus one but it’s not just that we have isotopes
we actually need to understand all this behavior
because we need to put this on the graph [Music]
all right so that’s the behavior we needed to understand that and now when i
break up and look at the uh where it’s increasing and decreasing and all that
i’m going to take into account um the zero right here
and you know when when x is zero that’s in the domain of our original and
it’s a critical number here because it makes the derivative zero when x is zero
we’re going to get zero over one so that’s so we’re going to look at
so this will be less than minus one um between minus one and zero equal to zero

01:11
and then between zero and one [Music] and then greater than one
and so i’m going to be looking at the function the derivative
and i’m going to be making a conclusion all right so let’s do our first
derivative test first derivative test so when i’m looking
less than -1 what’s happening to our derivative
the derivative is right here now notice on our derivative the denominator is
always squared so it’s always positive so when i’m looking less than minus one
i am looking at say like minus two what’s happening at minus two it’s
because it’s positive so it’s increasing and when i look between
minus one and zero for example negative one half
well that’s that’s positive so negative one half so that’s positive still

01:12
that’s positive so still increasing and so there’s you know no
at -1 there’s no relative extrema which we knew already because there’s a
vertical isotope there but what about between 0 and 1 say a half
so now let’s test a half so now we’re going to get negative over a positive
so that’ll be negative and then now what about greater than one
so greater than one for example say two and that’ll still be negative
and so this will be decreasing so we know it’s decreasing on these
intervals here increasing here and increasing here
all right what’s happening at zero it was increasing and then it was decreasing
right so it was increasing and then decreasing that’s a relative max

01:13
we have a relative max at zero now what is the function
at zero when we plug in zero into the original function we get out zero
zero over minus one which is just zero so that’s our relative max
and now we can go start trying to look for concavity um
before we do that though where is the derivative blowing up
is the derivative ever blowing up remember we need to find those
values that make the denominator 0 in the numerator
non-zero so the numbers that make the denominator zero which are plus or minus
one um right so so you know we’re not going to have any vertical
isotopes or cuts at 1 or -1 because we already know we
have vertical isotopes there so we don’t need to worry about vertical

01:14
tangents and cuts because we already have vertical isotope there
now what about horizontal isotopes let’s look at horizontal isotopes before
we go do the concavity test so let’s look at horizontal isotopes are
there any so i need to find the limit as we approach positive infinity
of the function two x squared over x squared minus one
and in order to find this limit here i’m going to divide by the highest power
so i’m going to divide by x squared on top and x squared on the bottom [Music]
so what is this limit here so we have 2x squared over x squared x
squared over x squared minus 1 over x squared i divided everything by x squared

01:15
and that limit is after simplifying it 2 over 1 minus 1 over x squared
so this limit is 2 right because that’s going to go to 0
so we have 2 over 1 that’s 2. so we know that y equals 2 is a horizontal isotope
is it the only horizontal isotope what about if we put a minus here
we go from here to here we’re just dividing everything by x squared
so that’ll still work even if you have a minus there
and now i’m simplifying x squared over x squared is 1
and then x squared over x squared is another one and then we have 1 over x
squared now even if this is a minus infinity
this still goes to zero we still have two so y equals two is the only
horizontal asymptote all right so good we have a we have the
two vertical asymptotes we have the horizontal isotope we know

01:16
where it’s increasing decreasing in the relative extrema and so
we need now the concavity test so when we’re looking here at this second
derivative here um all we have to look at are the plus or minus ones
and we know that there’s no inflection point
at those but we need to figure out the concavity so let’s look at [Music]
less than -1 between -1 and one and greater than one
and so we’re looking at the function the second derivative
we’re going to make our conclusion so this will be the concavity test
so when we’re looking less than -1 for example -2
what’s happening to the derivative the derivative is all positive up here

01:17
but when we’re looking at -2 here that’ll be a positive to the third
so that’ll be positive so it’s all positive so this is concave up here on
this interval here so to the left at -1 remember there’s a
vertical isotope at -1 but to the left of it we know it’s concave up
and in between minus one and one what’s happening there so
between minus one and one i can choose zero so when i choose zero
everywhere now i get minus one to the third so that’s negative
so i’m gonna get concave down here so between minus one and one is concave
down now greater than one it’s going to be like a two
so that’ll be positive here and it’s always positive here
so that’ll be positive it’s concave up again so you know you have a change in

01:18
concavity so you you might be tempted to say oh
there’s an inflection point there but remember that x equals negative one
is not a second order critical number um because at x equals minus one
the function is not defined that’s technically why it’s not a
critical second order critical number what’s happening at minus one we already
know we have a vertical isotope there so of course there cannot be an
inflection point because there is simply no point
when x is -1 it’s not on the function okay but now we know where the function
is concave up and now we know where it’s concave down
okay so putting all that together the first derivative test the concavity
test the horizontal and vertical isotopes and knowing that this function doesn’t
have any vertical tangents or vertical cusps
we can put all that together and look at a sketch

01:19
and so that’s what it basically looks like um
except for the fact that those vertical those vertical asymptotes should be
um dashed in okay but less than minus one we can see that we’re concave up
between minus one and one we can see where concave down
and greater than one we can see that we’re concave up
and less than minus one we can see that we’re increasing
and between minus one and zero we can see that we’re still increasing
and between zero and one we can see we’re decreasing
and we and greater than one we can see that we’re decreasing
and since there’s a change in monotonicity
at zero in other words to the left we’re increasing into the right of zero we’re
decreasing at zero zero there has to be a relative max and we can see that
and then also i don’t have on their dash the horizontal isotope

01:20
of y equals two um so we could see that at y equals two
there’s going to be a horizontal isotope
and we should dash that in to finish off the sketch
okay so let’s go on to the next one [Music] and let’s get started on this one
so on this one right here we’re going to um are there going to be any vertical
isotopes on this one right here that might be what we’re
looking at first here is this denominator every 0
the answer is no the x squared plus 1 would never be 0. on the previous one we
had x squared minus one but if you have x squared plus one that’s never going to
be zero so there are no vertical isotopes um
what about horizontal isotopes we’ll just put here no

01:21
vertical isotopes so what about horizontal isotopes
so let’s take the limit as we go to infinity and i’m looking at the
x square plus one so that’s sometimes known as x squared plus two x plus one
all over x squared plus one and i’m going to divide by the highest power
so here i’m going to divide by x squared and 2x by x squared and the 1 by an x
squared the next square by an x squared and the 1 by an x squared
and i’m going to find this limit now all these are going to zero
as we saw in the episode uh horizontal vertical isotopes
um but this is x squared and that’s an x squared

01:22
so this is going to one over one which is just one
and it doesn’t matter if we have plus or minus infinity here
so from this step to this step all i did was divide everything by x squared
now when i find the limit here this is still going to zero and zero and zero
and this is still a one and a one so it doesn’t really matter if you’re going to
plus or minus infinity all of these squares on here they’re the
same this is a one and that’s a one so this is all going to one so y equals
one is the horizontal isotope and there are no vertical isotopes
all right so now let’s look at the first derivative so here we go we have low
derivative high and i’m going to look at this and i’m taking the derivative of
that numerator you could just use the chain rule if you
want but actually x plus one squared is just sitting right here

01:23
and so what is the derivative of it it’s just two x plus two
and then all right so low times derivative high minus high so x plus 1 squared
times derivative of the denominator which is 2x
and then all 1 plus x squared over squared okay so what can we factor out here
i’m going to think of this as 2 times x plus one
so i have an x plus one squared here [Music] um
so probably it’s just easier to just multiply it all out one times the two x
plus two x squared so that would be two x to the third

01:24
and then plus two x squared and then minus there’s a squared there
so that’ll be negative and then this right here is x squared plus 2x plus 1
all times 2x right so i’m working on this part right here
so this will be minus and then 2x to the third and then minus the 4x
and then minus 2x now this will be third and this will be a squared here
and then minus the 2x so by just by multiplying that out i get those four
and then we’re going to get these three right here minus 2x to the third
minus 4x squared minus 2x all over 1 plus x squared squared
and so then simplifying that okay so i just multiplied all that out and all

01:25
that out and what do we get here these add up to zero
and here we’re going to get 2x squared and a minus 4x squared so i’m going to
get a minus 2x squared and then these add up to zero and then i get a plus two
[Music] and then all over one plus x squared squared so that’ll be minus 2
and then x squared minus 1 1 plus x squared squared
so minus 2 and then x minus one x plus one [Music]
all right so that’s what i like my first derivative to look like right there
so we just had to expand all that out and then combine it together and

01:26
simplify it factor it and now this one is easy to test it looks like
something’s happening at minus one and one where we have the
first derivative here um on the other hand this one’s kind of handy here
because we could take the second derivative here or actually even here so
let’s go up here and erase this and put our first derivative here so
it is minus two now let’s write this one minus 2x squared plus 2
over 1 plus x squared squared and then that’s also equal to this one x minus one
x plus one all over one plus x squared the reason why i like two is one’s easy

01:27
to test but still we need our second derivative so
there we go now we can erase all that let me do it here all right so we got
our first derivative accomplished we can look at this right here for
testing we can look at this right here for the second derivative now
so here we go low [Music] times derivative high and that’ll be minus 4x
and then minus and then minus 2x squared times the derivative here which will be
2 1 plus x squared to the first power times another 2x
and then now this is 1 plus x squared to the squared to the squared

01:28
so to the fourth okay so what can we factor out all this
how can we simplify this we could just go expand it out again
but i want to can’t i want to keep one plus x square and see if i can
cancel with one of these this minus right here is separating all
that that minus and then we have all of that
so think in terms of this one and this one
and what do they have in common one plus x squares
and then they have some twos in common where we are fours we have a four and a
four and this one has a minus and this one has a minus so let’s factor that out
let’s factor out the um minus 4x and then the 1 plus x squared
only to the first power though all right so what are we going to have left here
we factored out that we factored out one of these
so we have a 1 plus x squared left we’re factoring out that minus

01:29
and that 2 and that 2x and we get that and so what do we have left here we have
a plus and then we have and we factored that out too
so we really just have left a minus 2x squared plus 2.
and we’re not touching the bottom yet so let’s double check this
so i’m thinking about we have that and that two different things and i
factored this out and one of these out so i still have one of these left
now i factored out the 2x the 4 and the minus sign so i have all that out
and i factored that out so i have this left right here with a positive in front
of it so i have a positive and then it’s a minus two x squared plus two

01:30
so i think that looks good there and then so obviously we still have the
same denominator all right so now we look like we have a minus x squared here
and then a three and so let’s go with that minus four x one plus x square
we’re gonna cancel these one plus x squares
we’re gonna get minus four 4x here and then we’re going to get a 3
minus x squared all over 1 plus x squared and then now to the third power
all right so there’s our second derivative there all right so
after we do all that work now let’s erase all that and move our second

01:31
derivative up here so here we go and let’s move this up so minus 4x
over uh 3 minus x squared and so we can see what the second order
and first order critical numbers are now so we can start making our derivative
test here in our concavity test so let’s do the first derivative test
first so we’re looking at 1 and -1 as first order critical numbers if we
substitute them into our function they’re both defined there and they make
the derivative zero is there anywhere where the derivative is going to blow up
and the answer is no this is never going to be zero down here
so there’s not going to be any vertical tangents there’s not going to be any
cusps on this problem and so we can just go do the first
derivative test so i’m looking less than minus 1 equal to minus 1

01:32
between -1 and 1 equal to 1 and greater than one and i’m going to
look at the function the derivative and i’m gonna make a conclusion [Music]
and so here we go when we’re less than minus one
for example a minus two now keep in mind the denominator is always positive
and this is always negative so if i look at minus two
that’ll be a negative a negative so i have three negatives and so that’ll be a
negative it’s gonna be decreasing now between
minus one and one for example is zero then this is going to be negative
positive so what’s a negative times a negative
it’ll be positive so now i’m increasing so decreasing increasing
and greater than 1 say for example 10 right positive positive that’s that

01:33
negative there is still there so it’s negative so it’s decreasing so decreasing
all right so is there a change between decreasing and increasing
yeah so this is going to be a relative min at -1 and at one we’re going from
increasing to decreasing so this is going to be a relative max [Music]
now what happens when we plug in minus one
um when we plug in minus one we’re going to get zero we plug in one
substitute in one into our original function and we get out two oops
two and here we put a zero here in the wrong place
all right so at the point -1 0 we have a relative min
and at the point 1 2 we have a relative max

01:34
and so we know where it’s increasing and where it’s decreasing
and so this was the first derivative test and so now let’s look at our second
derivative here and let’s make our concavity test so when we say it’s
inc when we say it’s increasing we’ll know exactly how
it’s increasing and when it’s decreasing we’ll know exactly how it’s decreasing
so um i’m looking here at zero and i’m looking here
at plus or minus um well what makes that zero right
so plus or minus square root of three so we need to um
look at those places there at zero so i’ll just put it on a number line
mine like that so we need to look at one two three four intervals there
and we need to start doing some testing so we’ll look at less than square root

01:35
of three equal to negative square root of three between the
negative square root of three and zero equal to zero and between zero
and square root of three equal to square root of three
and greater than square root of 3. and so those are the places that we need
to look and we’re going to look at the function the second derivative
and we’re going to make our conclusion all right so here’s our derivative we’re
testing right here now keep in mind square root of three it’s
about 1.7 ish so think of that as negative 1.7 but when i’m less than it
like say like negative 10 when i square negative 10 that’ll be
like like that’ll be like a really big number but it’ll be 3 minus that number

01:36
so that negative will dominate and when i use the negative right here
that’ll be negative times the negative so that’ll be positive so this will all
be positive here so um actually if this is a negative
here and that would be a negative times a negative times a negative
that’ll be three negatives sorry i missed that two negatives there
right so think of that as negative 10 so negative 4 times negative 10
and then this is negative because it’s 3 minus negative 10 would be like 100. so
what’s 3 minus a hundred right so that’s negative so negative times
negative times negative you can’t look at that x and assume it’s
positive the x is less than a negative square root of three so you know that x
is negative all right so we know it’s concave down here so concave down

01:37
all right and then let’s see what’s happening between negative
square root of three and zero so that’s like negative
one 1.7 so think of like negative one so what’s happening when we look at the
derivative second derivative at negative one
so negative four times negative one so that’s positive
now when i use a negative one squared that’s a one so what’s three minus one
that’s positive so i’m going to get a negative times a negative times a positive
so i end up with a positive right here so this is concave up [Music]
and then between zero and three so square root of three is about one
point seven so think of the x as being a one
so when x is a one here so that would be negative four times a one so that’ll be
negative and when i use x is one here that’ll be three minus one

01:38
so that’s a positive so we’re going to get a negative times a positive times a
positive so now it’s negative so now it’s concave down again and then um
greater than square root of 3 for example a 10.
so if i look at 10 up here now if i look at 3 minus
10 square right that’s 3 minus 100 right so that’s a negative
and a positive and a negative so that’ll be negative times a positive times a
negative so that’s that’s positive so let’s concave up again
so is there a change in concavity square root of three
and is square root of three in the domain of the function
yes so this right here is an inflection point inflection point

01:39
and that minus square root of three there’s a change in concavity
this is also an inflection point all right so um and what’s happening at zero
um concave up concave down is x equals zero on the graph yes so we
still have another inflection point so we have three inflection points
when x is zero we’re going to get out of one when x is square root of three and
negative square root of three we get out some numbers
um for example we’re going to get out 4 in the denominator so
we can write those numbers over here f of square root of three is

01:40
square three plus one over one plus square root of three square so w four
so let’s square that and then f of minus square root of three is minus
square root of three plus one over four again and this will be uh
squared so i’ll just call this one m1 and this one m2
and so we have m2 right here and m1 right here and that way we can say
yeah we found those numbers they’re over there they’re not exactly
we could go find those as decimals but all right so we
found the horizontal isotopes we found the no vertical isotopes
we did the first derivative test we did the concavity test
and we noticed that this first derivative is never going to blow up
so there’s no vertical tangent so there’s no there’s no cusp

01:41
all right so let’s see the graph all right so there we go and let’s make
sure everything makes sense let’s look at the first derivative test
that we found so if we’re less than -1 we should be decreasing
at -1 we should have a relative minimum between -1 and 1
it should be increasing and then at 1 2 we should have a relative max and then
greater than 1 we should be decreasing now what about the inflection and the
concavity when we’re at square root of three and zero
and minus square root of three we should
have some inflection points there and we can see those on the graph
when we’re less than minus square root of three we’re concave
down um between minus square root of three and zero
we’re concave up and you can see that on the graph

01:42
between zero and square root of three we’re concave down
and you can see that and then greater than square root of three
we are concave up all right so yeah everything checks out
oh yeah and then the last thing is the horizontal isotope is y equals one that
we found and you can see that on the graph right there and then
of course there is no vertical isotopes and we plotted the x-intercept and we
plotted the y-intercept and so that looks like a pretty good
shape shape of the graph there okay [Music] so let’s look at one more example

01:43
here we go so this time it has fractional exponents so i’m anticipating some
vertical tangents and or cusps happening but let’s see what we get so here’s the
first derivative and so here we go we have um two thirds x to the two thirds
minus one times the second one and then plus x to the two thirds
times um one third six minus x to the two thirds minus one
and then with the minus one out here okay so there’s the product rule there
right and now we have some negative exponents so
let’s get a common denominator we have a negative exponent here

01:44
meaning we have a denominator and we have a denominator here
and we need to put them together so first i’ll write this as
two and then six minus x to the one third and then this will be over three x to
the one third and then i’m gonna save some space here and then
what do we have going on over here we have a minus
one so we have a minus x to the two thirds
and we have a three and then we have six minus x to the two thirds
what’s going to be our common denominator they both have threes
so i’m going to use six minus x to the two thirds here
over six minus x to the two thirds now we have the and then over here i’m
going to use a x to the one-third and then x to the one-third

01:45
okay so now we have a common denominator everywhere
so our common denominator is x to the one third
three times x to the one third and then six times six minus x to the two thirds
okay so far so good right there’s our common denominator
now what are we going to get in the numerator this is 6 minus x and 6 minus
x we’re going to add the exponents one-third plus two-thirds three-thirds
so we’re gonna get two times six minus x
to the first power and then we have plus and then here we have x and x so we add
the exponents we get three thirds we get an x
so we’re just going to get minus x out of that
okay and let’s just simplify this a little bit this is going to be
12 and then minus 2x minus another x so minus 3x and this will be over

01:46
three we have a three here and we’re the arrows already have three all right so
three and then x to the one third and then six minus x to the two thirds
so simplifying some threes here we have four minus x over x to the one third
six minus x to the two thirds so that looks like the best first derivative there
so right we can simplify some freeze there so i like that so
let’s see if we can rewrite this first derivative up here and save us some space
so let’s see here it’s 4 minus x over x to the one third

01:47
times six minus x to the two thirds all right so there’s our first
derivative there and let’s get the reddish through the rest of this work here
all right we’re doing good so far now um you know when we look at this
uh original function up here the domain is
all real numbers um because we’re going to square an x we can always square an x
we can always cube root so um you know it’s continuous everywhere it’s
not differentiable everywhere it’s not differentiable at six
it’s not differentiable zero so we got some things happening there
um in fact we have some blow ups don’t we when x is zero
right we’re going to get four over zero right that’s going to be a blow up
when x is six we’re going to get another blow up
when x is six we’re gonna get four minus six so we’re gonna get
a minus two we’re gonna get a non-zero over zero

01:48
so we’re gonna have two potential places where we have vertical tangent or
cusp do we have any vertical isotopes or horizontal isotopes
the answer is no if you take the limit as x goes to infinity
you’re just going to get minus infinity um when you take the limit as x goes to
positive infinity and negative infinity they’re just going to
blow up to infinity so we’re not going to get any
vertical or horizontal isotopes so but we are going to get a vertical
tangent or cusp one of the two maybe one of each so let’s look at the limit
as we approach zero from the left of the first derivative
and the limit as we approach zero from the right of the first derivative
right because that’s going to make that zero four over zero so it’s going to

01:49
blow up there and let’s also look at the limit as we approach 6 from the left
and the limit as we approach 6 from the right
of the first derivative those are also blow ups because at six we’re going to
get -2 over zero so non-zero over zero that’s going to be a blow up
all right so now we need to find the sine of these infinities
now if i approach zero from the left that means i’m negative i’m a little bit
negative so this will dominate the four the four will dominate four minus
something really really close to zero that’s going to be really close to
positive 4 so that’s going to be positive there but
if i’m approaching 0 from the left this will be negative
and this will be positive so this whole thing right here will be negative
so we’re approaching 0 from the right now

01:50
something like 0.1 again the 4 is going to dominate so positive
0.1 this will be positive this will be positive so
positive over positive times positive all that’s positive there
okay so now what about as we approach six from the left now we approach six from
the left this is going to be getting really close to six
so this will be four minus something really close to six so this is
all really close to negative two so for both of these the numerator is
going to be negative now if we’re approaching six either from
the left or from the right if you’re 5.9 or 6.1
if you cube root those you’re still positive
so this will be negative and this will be positive
what about this one well this one has a square on it
so if you square something it’s always positive so i’m getting negative

01:51
over a positive times a positive so both of these are negatives
so here’s where we’re going to have a vertical tangent
vertical tangent at x equals six and we’re going to have a cusp
at x equals zero so whatever the sketch that we come up with
it needs to have that cusp and it needs to have this vertical tangent
all right so let’s remember those now let’s go and do our first derivative test
um four is a first order critical number right because four’s in the domain it
makes the derivative zero um what about x equals zero it’s in the domain
it makes the derivative undefined but we already know what’s happening at x
equals zero right but we still need to check to the left and to the right of it

01:52
six is also a first order critical number the function is defined at six
the output is zero at six the derivative is undefined so we have um zero
four and six is our critical numbers so here we go less than zero
equal to zero between zero and four equal to four between four and six
equal to six and greater than six and so we’re going to have the function
that we’re testing and the first derivative and a conclusion
so first derivative test there we go [Music] and i’m just gonna

01:53
write out the derivative over here so that’s the first derivative that
we’re testing there all right so let’s test it less than
zero so something like -1 so we’re going to get a positive and
we’re going to get a negative and this expression is always positive
because we’re squaring that right there so that’s always positive
so when we’re looking at say -1 i’m going to get a positive over a
negative so that’s all negative so it’s going to be decreasing right there
now between 0 and 4 for example a 3 what’s happening we’re gonna get a
positive positive a positive so it’s all positive so we’re gonna be increasing
so between zero and four we’re increasing and then what’s happening at
between four and six so between four and six
say say a five five so now i’m getting negative

01:54
five so we’re getting positive positive so negative
over a positive is negative so i’m gonna be decreasing
and what about greater than six or like for example like a ten so i’m looking at
a negative positive positive so it’s negative
so at six here we’re not going to have a relative max or min
but in fact we already know what’s happening at six we have a vertical
tangent at six and at zero we already know we have a cusp at
here so you know we can see it’s not changing monotonicity there’s
no relative extrema here in fact we have a cusp there now what
about it four though at four we’re increasing to the left and then
increasing to the right so this is a relative max
and we plug in 4 to get out that number into the function

01:55
so we can look at the function right here we plug in a four we get um four
to the two thirds and then six minus four so that’ll be minus two
to the one third and that may simplify i’m just going to call it an m
so at 4 we get this number m here and i would definitely get a decimal
approximation of that so that you can kind of see where where it is
this simplifies to two to the five thirds so maybe you didn’t know that
um so we can say this is two to the four thirds

01:56
and we can write this as minus one and then two to the one third
so actually where am i getting that minus from if i use four
it’s six minus four that’s a positive two
so this will be two to the one third all right so it’s just two to the five
thirds that’s our m i guess we’ll just write that two to the five thirds
all right so we plug in four and we got our crazy
two to the five thirds out that’s just the cube root of 32. [Music]
in any case that’s our relative max and now we can go find our um
second derivative of this and look for the concavity um

01:57
however the video is running long and i don’t think i’m going to take the
second derivative of that here’s the graph we can see what we’re going to get
if we do take the derivative of that and get our second derivative it’ll be gone
it’ll be concave down less than zero it’ll be concave down between zero and six
and concave up when we’re greater than six so
we see that the first derivative that we got it’s in
decreasing left left left of zero between 0 and 4 it’s increasing
and then between 4 and 6 is decreasing and then greater than 6 is still
decreasing and you can see the vertical tangent vertical tangent right here
and you can see the cusp right here at zero that we found

01:58
um and so definitely want to check that concavity out where it’s
where it’s concave down here and it’s concave down here
and then it’s concave up right here so at six here we’re going to have a point
of inflection and if we get the second derivative of that
we’re still going to get these expressions right here
where to go take the derivative of that we’d still get these expressions right
here in the denominator with some powers on them
so it’s not too hard to see that six would be an inflection point
all right so um that was our last example so let’s look at some exercises

01:59
[Music] so as you can see graphing using that curved sketching procedure
can be quite long um because there’s just so many
things that we’ve learned and we’ve done and you know when your functions get
more complicated to show a variety of of uh
features you know can you come up with a function that has
a vertical tangent has some isotopes right so then your derivatives start
getting more complex and the problems keep getting longer and
longer but in any case here i have a good selection of problems for you right
here some of them are straightforward for example one’s a polynomial
number two only has one fractional exponent
but anyways i definitely want to check out these problems here
and see if you can solve them and see how that goes
and then i got some more from you here and number two all the information’s
been figured out all you gotta do is sketch the graph

02:00
and then on number three here um give you a function right here
if you looked at that episode on horizontal isotopes
that function right there you might watch out for how many horizontal
isotopes it has on number four here we have another fractional exponent
so watch out for vertical tangents and tests
uh number five here five years kind of interesting because
the the functions not fully given to you you got some parameters
you want to find the parameters so that you have a vertical isotope
and you have a horizontal isotope so those type of functions
are kind of fun those type of problems are kind of fun
and then we got six seven and eight where we’re concentrating on
vertical tangents and vertical isotopes and then um nine and ten here the last

02:01
two um more curse curve sketching and finding all the information out that
they want and so yeah if you uh get stuck on any of those ten problems
or you have any questions for me i look forward to hearing
from you in the comments and hope you enjoyed this video
so i just want to say thank you for watching and if you like this video please
subscribe and like below and on the next video we’re going to talk about applied
optimization problems so we’ll be solving a lot of word problems
and we’ll be using all the tools that we’ve developed so far like the first
derivative test the rel the extreme value theorem and the second derivative test
will be big so i look forward to uh seeing you there
and 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

02:02
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
Dave White Background Blue Shirt Squiggles Smile

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.

Let's do some math together, today!

Contact us today.

Account

Affiliates

About Us

Blog

© 2022 DAVE4MATH.com

Privacy Policy | Terms of Service