How Do You Find the Extrema Using the Extreme Value Theorem

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

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

00:00
[Music] so [Music] the extreme value theorem is important and helpful
it says that if a function is continuous on a closed bounded interval
it must attain its maximum and minimum values but what are critical numbers
relative extrema and absolute extrema we go through all of these concepts
in detail okay welcome back everyone uh this is the episode
extreme value theorem relative and absolute extrema so in this video i’m

00:01
going to first talk about relative extrema and then i’m going to
talk about critical numbers and what they are
and um then we’re going to talk about absolute extrema
in particular we’re going to cover the extreme value theorem in great detail
and then stick around to the end where we’re going to talk about some exercises
so let’s get started okay so let’s up let’s first talk about relative uh
relative extrema so if a function is defined on an open
interval and if at some point in that interval the function reaches a
maximum or minimum value relative to that interval then we say
the function has a relative extrema on that interval now
let’s see what that looks like intuitively
so i usually like to draw like a roller coaster ride

00:02
so let’s say we have some roller coaster going on here
and so let’s uh say this just goes down here forever
and this goes down here forever actually let’s make it um
let’s make this curve up a little bit more and let’s make it go up here forever
[Music] so there we are and let’s say right here we have some values right here
so i’m looking at these relative extrema right here [Music]
so let’s call this here one a1 a2 a3 those are some interesting points there
why are they interesting this one here is a relative maximum and what does that
mean that means relative to some interval around a1 here
this is a maximum value so if i just look at this part of the curve only

00:03
some open interval around a1 then we can
see that this is a maximum there so it’s not a maximum
everywhere in fact this keeps going up so this has
no global maximum so global maximum is when you consider the entire domain
so this keeps increasing so this has no global maximum but this
is a relative maximum in fact here’s another relative maximum right here because
around a3 we can find some open interval right here
where this is the maximum so it’s a maximum relative
to that interval right there in here similarly we have relative extrema
a relative minimum because around e2 i can find some open interval
where this is a minimum value right there and similarly
similarly for a4 if you can find an open interval
where it’s the minimum value then that’s called a relative minimum right there
so relative minimum and maximum relative maximum

00:04
collectively are called relative extrema and
they help you better understand graphs so for example
if this graph was representing a roller coaster
i like to think of roller coasters but uh this you know the relative maximums is
when you have the most excitement in the ride right because
you have the place where you’re going down right here
and you’re going down you’re falling down and so that’s that’s really exciting
whereas these relative minimums right here so that’s when you’re going the
slowest and so now you have to start climbing
back up right so those are the boringest parts of the ride
well whatever your function is modeling the relative extrema
and the global extrema or the absolute extrema
um are the important places to look at often
and so that’s what that’s what the topic of this
uh video is about relative extrema an absolute extrema and when we start

00:05
looking at absolute extrema we’re going to cover the extreme value theorem
so next we’re going to say f is a function defined on some open interval and
if f of c is greater than or equal to f of x in other words
if f c is the greatest value there um for all x in that interval then f of
c is called the relative maximum so you know back on our graph here
if i if i call this a c right here then this height right here is f c [Music]
and f of x is less than or equal to f of c
for all x in here no matter what x you choose in here if you go up to the graph
it’ll be smaller less than or equal to f of c right here
so this is what we mean by relative maximum uh using an inequality uh

00:06
and an interval and similarly we have relative minimum if f c is less than or
equal to f of x for all x in the interval then f of c is called the
relative minimum of f okay so and if it’s either a relative maximum or
relative minimum then we’re just going to call it relative
uh extrema okay so let’s see here um now this last uh line here says that
uh if it’s if it’s a relative maximum well some people call it a local
maximum so if you want to call it relative maximum or local maximum
same thing with minimum it’s it’s either called a relative minimum or a local
minimum the words are just meaning the same thing
all right so the next thing is to acknowledge vermont vermont looked at
this first uh in terms of being able to use it with analytic geometry

00:07
and so the the following examples will show that
even though the derivative is zero right so so why would i be looking at the
derivative of zero well let’s go back and look at this right here right
so if we look at the relative extrema here
so like if i look right here we have a horizontal tangent line
if we look right here we have a horizontal tangent line remember
horizontal tangent lines are where the derivative is zero
so it seems like wherever i have relative extrema
i have a horizontal tangent line or in other words the derivative
is zero the derivative is zero at c one the derivative is zero at c
two and you can just see that by looking at the graph the derivative
is zero i’m calling these a’s sorry these derivatives are all zero at a one
and a two a three a four we have horizontal tangent lines

00:08
okay but what this is pointing out here is that
even if your derivative is zero there may not be a maximum
or minimum value there so that’ll be something that we need to watch out for
so let’s keep our eye on that and in other words the converse of
vermont’s theorem is false in general so what is from um
what is from what’s there so there may be an extreme value
even if the derivative is not zero or where the derivative doesn’t exist so
we’re going to see some very specific examples of this but here’s the
relative extrema theorem so this is not the
theorem our main theorem of the of the of the episode
but it’s a preliminary theorem and it says that if f has a relative extrema at c

00:09
and if the derivative exists then the derivative has to be zero
so there’s two things you have to check before you would know that you have a
horizontal tangent line in other words you have to
check f has a relative extrema and you have to check that the
derivative exists and then you could say oh the derivative must be zero
so let’s look at some examples so here’s the statement of the theorem
if f has a relative extrema and the derivative exists
then the derivative must be zero so first let’s look at this example here
so when i look at this example here what do we have
so let’s see if we can make a reasonable sketch
and we’re looking less than two-thirds and greater than or equal to two-thirds
so um what’s happening right here this line

00:10
uh 2 2x minus 3 right so that’s coming through here
something like that and we’re chopping it off
at two-thirds so i think two-thirds is going to hit about right here
we’re gonna have some height right there and this is about a minus three let me
make it a little bit more steeper something like that and then we have
this hole right here and this height right here is what uh two
and then a two-thirds in here and then minus three right so that’s uh four
thirds minus nine thirds right so minus five thirds is that height right there
but there’s a hole there because there’s a strict inequality over here
and so this is a minus three right here and then now what about the other side
of two thirds right here so here’s two thirds

00:11
now on the other side of two thirds we look like the line
three minus seven x so where does that intersect so let’s say two-thirds in here
and so this will be three minus fourteen thirds
let’s say that’s nine thirds so that’s again is minus five thirds
so that’s what that is right there so this one is filling it in
and then it has a slope of that line has slope of negative seven right so we’ll
just make a line there so this is the branch
y equals so this is the branch coming from the lower part when we’re greater
than or equal to two thirds and this is the line here two x minus three
so there’s some kind of reasonable sketch for it
and we notice right what’s happening right here is that
um there may be a sharp corner right so how do we determine
if we have a sharp corner if the derivative exists

00:12
at two-thirds so let’s remember what the derivative at two-thirds
is the derivative at two-thirds that is the limit as h approaches zero
of f of two thirds plus h minus f of two thirds all over h
um but the problem with trying to evaluate this limit
on both sides of zero is that when h is less than zero this will be less
than two thirds which means we’ll use a different branch
for that if you look at how the function over here
is defined so we’re going to have to compute this from
this limit from the left and from the right so the question is does this exist
right here so let’s look at the limit from the left here
or as we call this in a previous video the derivative from the left

00:13
and then f of two thirds [Music] so we calculated this right here as
minus five thirds that’ll be a negative negative that’ll be a positive
so this will be um which piece will i choose here
two-thirds plus h but h is a negative number
so this would be less than two-thirds so we’ll use the first piece
we’re going to use it to and then this is my x here two-thirds plus
eight so the function says two x minus three and then minus and then a
negative five-thirds all over h and then if you calculate this limit
um the numbers are going to cancel and we’re going to get a 2h
and the h’s are going to cancel and we’re going to get a 2 out of all this
so the limit from the left is 2 and this is the limit of the
difference quotient so this is the derivative from the left is 2.

00:14
now if we look separately the limit from the right of f of two thirds plus h
minus f of two thirds all over h this limit here now when h is
approaching zero from the right that’s going to be two-thirds plus a positive
number that’s going to be greater than two-thirds so when i’m greater than
two-thirds i use the bottom piece here so here we go limit h approaches zero
and now i’m going to use the three minus seven
and then the input x and our input is two thirds
plus h and then minus and then this will be a negative five thirds
and then all over h and then if we add up all the numbers they add up to zero
and here we’re going to get a minus seven h and the h’s are going to cancel
so if you do all that you get minus seven
so the derivative from the left is 2 and the derivative

00:15
from the right is minus 7 so the derivative does not exist so f
prime f prime at 2 3 does not exist so the example wanted us to determine
if the relative extrema theorem applies and it doesn’t because the derivative
has to exist so it says there if f has a relative extrema and
if f prime exists so we just show that f prime at two thirds does not exist
so therefore the relative extrema does not apply
so that’s a very nice powerful theorem but you cannot just use it wherever you
want you have to check the two things you have to check that f has a relative
extrema and you have to check that your derivative
exists if you check both of those then you know the derivative would have to be

00:16
zero that didn’t work for this example there’s a sharp corner
so we have to keep keep in mind that this theorem right here doesn’t always hold
okay so next example um so i just wrote up what we just worked out
the function has a maximum value at two thirds
and we saw that from looking at the sketch here it has the maximum value
right there and this is a relative extrema so we can find a
open interval around two-thirds where this is the maximum value
so that is a relative extrema the relative extrema
does exist here but the derivative does doesn’t exist so we cannot say we have a
horizontal line here um there’s all kinds of tangent lines
no there’s no tangent line here right so you don’t have to have a horizontal
tangent line to have a relative extrema you could have a relative extrema and no

00:17
horizontal tangent line it just so happened that the first
example that i draw drew with the roller coaster all of the relative extrema had
horizontal tangent lines but that doesn’t have to happen is what
this example is showing us okay so the derivative doesn’t exist at two
so the relative extrema theorem does not apply okay so let’s look at this
another case so when we look at the relative extrema theorem
there’s two things we have to check f has a relative extrema
and the derivative exists in our first example we saw one where the derivative
didn’t exist and this is an example where we see that the first
hypothesis f has a relative extrema doesn’t hold
so let’s look at a you know a sketch of x to the third plus four

00:18
right that’s our function now x to the third plus four
what does that look like right so that’s just a cubic shifted up by four
so let’s just keep it going right through the origin but it’s been shifted
up for right so it looks like something like
that and as you can see here there’s no relative extrema
um this is if you look at this right here it does not do this
that would be relative extrema here and here but this graph does not do that
this graph is increasing everywhere so this graph is increasing and then it’s
decreasing in here right so these graphs are not the same
this is increasing everywhere so this is x to the third plus four it’s
increasing everywhere so there’s no relative extrema on this right here
so we would not expect that to have a horizontal tangent line anywhere
okay so the derivative is three x squared and the derivative

00:19
is zero at zero but we do not have any relative extrema
f does not have a relative extrema relative extrema theorem does not hold
all right next part [Music] okay so now we’re going to talk about
critical numbers what they are and why we need them and how to find them
so let’s first get our understanding of why we need them
um so in general critical numbers divide the domain of a function into
intervals um which which the sign of the derivative remains the same
so you know we were looking at horizontal tangent line with it
with the derivative of zero but they’re going to be intervals where the
derivative is negative there may be intervals where the derivative is positive

00:20
and so we’ll be able to use critical numbers to find those intervals
so if a function is defined on an interval
it is either increasing or decreasing on that interval
um where we’ve broken the domain up using the critical numbers
so a particular graph cannot change directions on that interval
so let’s go back to that example we were looking at a minute ago where we had
you know x to the third plus four and we see the derivative is three x squared
and except for at zero the derivative is always positive
um you know it’s greater than or equal to zero
it was only at four or s at x equals zero and the y was four
but everywhere else the derivative is positive now if the

00:21
derivative is positive that means our slope of our tangent line is positive
and that means the function is increasing so for example right here
our slope of our tangent line is positive and the function is increasing so the
critical numbers are going to divide the domain up into sections where the graph
is either increasing or decreasing [Music]
so the crucial idea is behind using the derivative
to analyze graphs of functions and so these critical numbers
are going to break the function you know because
if you have a function how many points do you have on a function
you have infinitely many points and you cannot just sit there and check at
every single point so what you want to do is you want to break up
the domain into intervals where you can check infinitely many values at once
to see if the function is increasing or decreasing

00:22
so that’s kind of the main idea there so let’s go ahead and define what a
critical number is [Music] so looking at our previous examples in particular the
relative extreme theorem is very important to know where the derivative
was defined and where the derivative was zero and
that’s what we’re going to base our critical numbers off of
if we have a number in the domain of the function and the derivative is zero
or if we have a number that’s in the domain of the function
but it’s not in the domain of the derivative in other words
f prime at c doesn’t exist if either of those two cases happen then
we have a critical number and what we’re going to do with these
critical numbers again is divide the domain up into sections
and we’re going to create derivative tests so that we can test those intervals
and that’s much more efficient than testing at every single point
on the function so in in many of these examples that we work on we’re going to

00:23
have a finite number of critical numbers which will which
means we’ll have a finite number of intervals to check
and that’s you know advantageous to checking every number or every point
on the graph so here’s our theorem finally if a function is continuous and has a
relative extrema then c is a critical number of f
in other words if you know if you have a relative extrema it
has to happen at a critical number and so we’ve looked at those examples before
which helped us understand this theorem a little bit better
but let’s look at it so here’s our first example
find the critical numbers if there are any for this function right here
so let’s look at this function right here
so what does this function look like um well it looks like uh absolute value

00:24
of x something like that um but it’s been shifted over two
so one two and now it looks something like that and then it’s been shifted
down three so then i’ll just shift it down three
and then it looks something like that right so let’s make that over here [Music]
so it’s been shifted over and then down and let’s see where does it cross the
x-axis so we have something like x plus two equals three
and x plus two equals minus three so we’re looking at
one and minus five so let’s say here this is one here and then this is -5 here
so the graph looks something like that so the question is does it have a corner

00:25
there so what are the critical numbers if any are there for this function
so we can check right here and this was at the point here minus two
um we can check right here and this was minus three
if there is a derivative there or not because
two um and so another way to look at this function here is
as a piecewise function right here right so we’re gonna be broken up at
-2 and 2 right so we can say this is going to be x minus 1
and minus x minus 5 and this is when x is greater than or equal to -2
and when x is less than minus 2 right here
and so you know that just comes from the basic absolute value function right the
absolute value of x is just x and then minus x when we’re broken up at

00:26
at zero so let’s say greater equal to zero and less than zero so
if we have a plus two in there and then a minus three
so that’s how we get this um you know adding a two and then a minus three
um and here we have to um you know add the two and minus three um
we’re gonna get what here um x and then x plus two
minus three right so that’s how we get the minus x minus two minus three minus
five there so here’s a sketch of the graph and here’s what it looks like as a
piecewise function so this is the same thing as
absolute value of x plus 2 minus 3. so when i look at it like this
i can say here that the derivative is one and minus one when x is

00:27
um greater than two and when we’re less than minus two
and notice i didn’t put equals here because
at minus two we have a sharp corner and the way we can show we have a sharp
corner is by finding the derivative at minus two does not exist
and the way that you would see that the derivative at -2 does not exist
would be by finding these uh limits here the limit as h approaches 0 from the
left of minus 2 plus h minus minus 2 all over h and by finding the
derivative from the right which is this one-sided limit here
minus 2 plus h minus f of minus 2 all over h and so for these limits here

00:28
we’ll get so from the left this is going to be -1 and this will be one
um so you can provide those steps there and check them
um i did an example earlier try to look back at that example and see
if you can fill in those details there so anyways f prime at minus 2 does not
exist and the derivative exists everywhere else
so -2 is a critical critical number [Music]
are there any other critical numbers so we need to check where the derivative is
0 or where the derivative doesn’t exist so the derivative is nowhere else is
zero it’s either gonna be one or minus one
or at minus two it doesn’t exist so that looks like our only critical number
there okay so here we go so we can write this

00:29
um absolute value function that’s been shifted around a little bit
as a piecewise function and then we can analyze the derivative
we can say there’s a sharp corner by by looking at the left derivative
and the right derivative and so it’s not defined at minus two
and so that tells us that we have a critical number at minus two
so the derivative is not defined uh it’s not zero anywhere
so that’s not giving us the critical number so you know if we look back at the
definition of a critical number it is c is in the domain of f we have to check
that and the derivative has to be zero that didn’t work anywhere
the derivative was never zero for our example
so then we have to check that it’s in the domain and that c is not
in the domain so we’re looking at minus two is minus two in the domain of f
and minus 2 is not in the domain of f prime

00:30
and that’s we have for this example here minus 2 is not in the domain
of the derivative the derivative doesn’t exist but minus 2
is in the domain here we get out minus 3. so that’s why we say -2 is a critical
number there all right so next example find the critical numbers
of x to the third plus two so let’s look at that here i’m going to
um take the derivative so this is a polynomial it’s
it’s um continuous everywhere the derivative exists everywhere the derivative is
just three x squared [Music] derivative alright so um where’s the
derivative zero so x equals zero is a critical number

00:31
why well if zero is plugged into the original function
we get out two so zeros in the domain of g and the derivative is zero at zero
so those two things combined together to say x equals zero is a critical number
are there any other critical numbers and the answer is no because g the
derivative is zero only at zero and the derivative is defined
everywhere else so that says x equals zero is the only critical number [Music]
so notice that when x is 0 the derivative is 0 and so 0 is a critical number
now some people are very careful they’ll say so zero is a critical number
but what you’re are trying to do is to find
all of the critical numbers so you can’t just say oh i found one and stop

00:32
right you have to argue that there are no other ones
that you have found all of them so we will check
all the conditions notice that zero is not a relative extrema or an absolute
extrema but as i argued a minute ago um you know this is a derivative
3x squared is only 0 at 0 and the derivative is defined
you know 3x squared the domain is all real numbers
so there are no places where the derivative is undefined
so that tells us that 0 is a critical number and is the only critical number
okay let’s look at another example here about uh sine of x right so remember
what sine of x looks like it’s just going to go um something like this and so on

00:33
and so where are the critical numbers right so the critical numbers are found by
looking at the derivative which is just cosine and so
you know we’re going to have relative extrema for sine there’s going to be
infinitely many relative extrema but where’s the derivative 0 where’s cosine 0
first of all the derivative is defined everywhere
so i’m not going to have any critical numbers coming from the derivative being
undefined i’m only going to have critical numbers where the derivative is 0.
and where is cosine 0 is it ever 0 right so cosine is like this
right so it hits zero at pi over two and then at three pi
over two and then five pi over two and seven pi over two and it hits at
minus pi over two [Music] and so on and so i’m going to say pi over 2

00:34
plus there’s a pi there’s always a pi in between so i’m going to say pi k
so pi over so pi over 2 plus another pi plus another pi
plus another pi so i’ll say k pi where k is like one two three four and so on
but also minus one because if i have a minus one then i hit right there
and case could be minus two and i hit right there
um so there’s actually going to be infinitely many places where we have
critical numbers infinitely many critical numbers
so the derivative is cosine which is defined for all real numbers
so the only thing to check is where the derivative is zero
so you know that’s sort of the the from a practical point of view is
you want to see what is the derivative and then you ask
two questions where’s the derivative zero and where is the derivative undefined
and for all those numbers that you get get to answer those two questions you go

00:35
back and check if those numbers are in the domain of the original function
so all of these numbers that we’re about to find are in the domain of sign
because the domain of sign is all real numbers so certainly for those
x values pi over 2 plus pi k any of those pi over 2 or 3 pi or 5 pi over 2
all of those are in the domain of sign so those are the um critical numbers
and if you look at the graph this is just a little tidbit there but
if you look at the graph you can actually see that those are the absolute
extrema okay so let’s look at another one here maybe you don’t have a
idea at all how this function looks and that’s the powerful part of the
critical numbers as we’re going to see later in an
upcoming episode is finding the critical numbers
allows us to help us shape the graph so anyways find the critical numbers so

00:36
let’s practice the strategy that i just gave find out where the derivative
is zero find out where it’s undefined and then for those numbers that we get
for those two steps go back in and make sure that you can
plug them into the function so here we go what’s the derivative [Music]
so the derivative is so this is a product function so i’m
going to take the derivative of the first
so two thirds x to the two thirds minus one and then times the second function
plus now x to the two thirds and then times the derivative of the
second one which is what minus one so there’s the product rule there now
we need to look where the derivative is zero and where the derivative
is undefined so the way i wrote it right now
it’s not very friendly for looking at that so let’s rewrite it i’m going to say

00:37
this is 2 1 minus x over 3 and then this is cube root of x
that’s cube root of x minus because i have a minus here
and then this is x to the two thirds so i’ll say cube root of x squared
and i want a common denominator so this is cube root so this
is three cube roots of x and then times three cube roots of x
so that would give me a common denominator there so i’m multiplying top
and bottom by three times the cube root of x so that gives me a
common denominator of three times the cube root of x and let’s see what we
get on top we’re going to get a 2 times a 1 2 times minus x and then here we’re

00:38
going to get a 3 with a minus and then what’s the cube root of x
squared times the cube root of x that’s just x
so in the end we end up with is two minus five x
over three times the cube root of x so there’s our derivative there nice and
simplified and now we’re ready to ask the question where is the derivative 0
and where is the derivative undefined right so f prime is 0
at x equals right so set that equal to zero
set the numerator equal to zero right so that’s just simply
you know two equals five x you know two fifths right so two fifths [Music]
so f prime is zero at x equals two fifths
that’s where i set the numerator to zero that’s where the derivative is zero

00:39
and f prime is undefined at where’s the denominator zero
so x equals zero so f prime is defined is undefined at x equals zero [Music]
if i try to use zero into the derivative here i’m going
to get two over zero right so we cannot do that um f prime is zero at two-fifths
if we substitute in two-fifths we’re gonna get zero up here
and we’re gonna get non-zero down here so so these two statements are correct so
these are two possible critical numbers there’s just one thing left to check
can we substitute both of these back into our function so if we look over
here what the function is can we substitute two fifths in and the
answer is yes you can go and do two-fifths to the two-thirds power and get some
decimal and then times one minus two-fifths so all that’s defined can we do zero
so again we can substitute in zero into all of that so

00:40
these are the two critical numbers right here two fifths and zero
and they’re the only two critical numbers these this is the only one that
makes the derivative zero and this is the only one that makes the
derivative undefined so these are the only two possible critical numbers
okay so there’s the derivative and we need to check where the derivative is 0
and where the derivative is undefined so the derivative is undefined when the
denominator is zero so that’s at zero and the derivative is zero
only when x is two-fifths so we got it and so those are the critical numbers
zero and two-fifths so the idea here um you know just to make sure that we have
a good understanding of of why we’re looking at the critical numbers

00:41
is if i were to try to sketch that graph [Music]
or understand where the graph is increasing and decreasing
i would do something like this i would say here’s zero and here’s two fifths
so i would have three intervals and the function whatever it looks like
is either going to be increasing or decreasing on this interval
you see they’re going to be increasing and decreasing on this interval
and it’s either going to be increasing or decreasing on this interval
and we’re going to find we’re going to come up with derivative tests
and and and some episodes coming up so these critical numbers will be um
you know critical anyways next example let’s look at
square root of 9 minus x squared what are the critical numbers if there are
any so what’s our derivative of course we’re looking at the function

00:42
like this 9 minus x squared with a one half so the derivative is simply one half
nine minus x squared to the minus one half and then times two x minus two x from
the chain rule so the derivative is these cancel we end up with a minus x on top
we have a square root of nine minus x squared down here
so that’s good there’s our derivative now we need to know where the derivative
is zero and where the derivative is undefined so clearly the numerator is zero
um sorry clearly the derivative is zero when the numerator’s zero and the
denominator is not so that gives us x equals zero
um but then we also have to worry about the denominator here
um so you know we need this to be positive and we need it to

00:43
not be zero so we need some strict inequalities so let’s see that so there’s our
derivative minus x over square root of nine minus x squared
so we need to check where the derivative is zero and where the derivative is
undefined so the derivative is undefined
when you use plus or minus three because you’re going to square those and you’re
going to get nine minus nine which is zero
square root of zero is zero don’t divide by 0. so we’re certainly going to
consider those as critical numbers however be careful
you must always check the original function can you plug in 3
into the original function you’re going to get square root of 0
which is of course 0 so yeah you can use
plus or minus 3 in the original function
but you cannot use it in the denominator in the derivative
so we’re going to have some critical numbers there 0 and plus or minus 3.

00:44
all right so one more example [Music] we’re going to find the derivative
let’s check that out ourselves so we have f prime is
found by looking at the quotient rule so we have low
times derivative high minus high times derivative low all over
x squared minus 3 and then squared [Music]
and so it looks like we’re going to get x squared times
minus 2x but then we’re also going to get a
sorry x squared times positive 2x but then we’re going to get a minus
that so it looks like we’re going to get a minus 6x here
so that’s how we get that minus 6x there all right so again to find critical

00:45
numbers is we find the derivative and we look where the derivative
is 0 for horizontal tangent lines and we look where the derivative
is undefined so for those [Music] we find that we have the square root of
3 because we have to square that and that will give us 0. so in other
words you know you you solve where the denominator is zero to find those
right so we’re going to get um minus three plus x squared equals zero where does
that happen right that’s x squared equals three you know so x
equals plus or minus square root of three any case so those are the
um critical numbers there where the derivative is undefined we get two of them
and then because we have minus six x the numerator is going to be zero when x
is zero so the derivative will be zero when x is zero

00:46
so those are the um three critical numbers um and then we go back and check the
original function so we cannot use plus or minus square
root of three in the original function because that will give us
three minus three in the denominator so x equals zero is the only critical
number x equals zero is the only critical number for that problem there
so there’s an example where it’s important that you don’t just look where
the derivative is zero you don’t just look at where the derivative is undefined
you always have to go back into the original function
and find out if those are actually truly critical numbers or not
all right next example [Music] find any critical numbers
so we have here natural log of square root of x minus 2 and so

00:47
let’s find the derivative so there’s two ways to find this derivative um so
should we bring that half power down [Music]
so we can raise that to the half power and then bring that out as a constant
and that could greatly simplify our derivative
and so it’ll be something like this don’t take derivative yet let’s say this
is natural log of x minus 2 to the one-half
and the one-half comes out so here’s our function
one-half natural log of x minus two and so now the derivative will be one half
times one over times the derivative of the x
minus two which is just one so there’s our derivative right there

00:48
so we’re going to ask where’s derivative zero or where is it undefined so
never is the derivative zero the numerator is just a fixed one
um anyways the derivative right so so looks like 2 will be a critical number
right here because that’s where the derivative will be undefined
however if we look back at our original function can we substitute 2 into our
original function well we’re going to get natural log of 0 natural log of 0 is
undefined so it doesn’t look like we’re going to get any real
critical numbers on this problem so there’s our derivative
we need to check for any x in the domain of f
where the derivative is 0 or where the derivative is undefined
we see that the derivative is undefined at 2
however and the derivative is never zero so the only temptation is x equals two
but that’s not a critical number because it’s not in the domain of the function

00:49
all right so break okay so now we’re going to turn our attention to
absolute extrema so now we’re going to take the whole domain into consideration
in terms of extrema and so we’re going to think more than just relative extrema
um so absolute extreme are defined um i’m going to define that force right now
and the procedure for finding the absolute extrema
on a given closed bounded interval is given and that’s going to be our big
theorem the extreme value theorem so first off let’s define what i mean by
absolute extrema so it’s easy to make a quick example of what they are
if you just look at something like sine of x or cosine of x

00:50
or perhaps something like you know just y equals x squared what are the
absolute extrema well what’s the absolute maximum
none this function just keeps increasing
so this function has no absolute maximum it has an absolute minimum though
because if you take into account the whole domain
of the function then you’ll see that this is the um smallest value the
the height ever attains is zero so zero would be an absolute minimum here
now if we change this domain and just say something like one and minus one here
and we look at this right here on minus one to one
in other words we’re changing the domain if we look at this function right here
then now we can see that it has an absolute maximum value of

00:51
one and it has an absolute minimum value of zero so you really have to take in
consideration the domain and not just the rule that your function is given by
okay so here we go so we’re going to say f is a function
and we need to say what the domain is because when we’re talking about
absolute extrema we need to consider the whole domain so i’m going to call my
domain d so for example right here my domain
is -1 to 1 that would be my d that’s just the domain
of the of the function if i restrict this function to this domain that’s the
domain all right so if if we’re greater than if f of c is greater than f of x
and now we’re not going to say for all x in an interval like we did for relative
extrema now f of c is greater than or equal to f of x
for all x in the domain so that’s what i mean by we’re taking into account the
whole domain so that gives us an absolute maximum and then similarly

00:52
we switch the the ordering f of c is less than or equal to f of x
for all x in the domain and that’s the absolute minimum value
and similarly if if it has either an absolute maximum or an absolute minimum
then we just say oh it has an absolute extrema
so if someone says find all the absolute extrema then they mean find all the
absolute maximum and also find all absolute minimum
and then we also say ffc is an extreme value if you want to be even less precise
because then you could say absolute extrema or relative extrema
okay so absolute extrema are also called
absolute or global extrema so global max sometimes we use that or sometimes we
use global minimum okay so um i thought we would look at

00:53
some examples first without looking at any particular type of rules just some
functions that you’re familiar with just to make sure that we’re all on the
same page here so do linear functions have any absolute extrema
so you know if i go and sketch some linear functions like a line
you know the line keeps increasing so there’s no absolute maximum
the line keeps falling there’s no absolute minimum
on the other hand maybe you have a horizontal line if you have a horizontal line
then you have an absolute maximum and an absolute minimum it’s whatever
the height of the line is right so if i put some coordinate
system down and i say this is zero and this is three this is y equals three
or f of x equals three if you want but this function right here has an absolute

00:54
maximum absolute minimum and it’s a linear function
it’s just a horizontal line um vertical lines of course is
not a linear function anyways what about quadratic functions what can
happen if you have a quadratic function so something like you may have something
like this where you have an absolute minimum or you may have an upside down
quadratic function where you have an absolute maximum what else
what about the six trigonometric function do they have any absolute maximum
and absolute minimum values so we talked about sine and cosine before
you know if you if you don’t put any coordinate axis down
or well even if you do put that down but you know this could be a sine or cosine
graph it’s going to have an absolute maximum of one

00:55
and an absolute minimum value of minus one um what about the tangent graph
or or actually since i just had that here
um you know what if you have coordinate axes through here or or not but
you know what if you’re looking at the um
secant and cosecant graphs so i’ll look at something like comes like this right
here and something that looks like it’s coming through right here where we have
isotopes where it’s hitting that zero we have another isotope right so if
we’re looking at graphs like secant and cosecant
this is going to keep increasing so there’s not going to be any absolute maximum
and this is going to keep falling so there’s not going to be an absolute minimum
and that’s true whether if you’re talking about secant or cosecant
graphs what about tangent or cotangent so tangent cotangent graphs look either
like this or like this and so in either case

00:56
you’re not going to have an absolute max
you’re not going to have an absolute min and the same for something like that
which is supposed to be curvier you’re not going to have an absolute max
or an absolute min so for tangent and cotangent you’re not going
to have any absolute extrema so what about
exponential and logarithmic what do they look like right so if you just look at
say an exponential looks something like that where it’s going to increase
so it’s not going to have an absolute maximum um
does it have an absolute minimum so for example if we look at something like
2 to the x or e to the x or something like that
so this is a horizontal isotope here and so this doesn’t have an absolute
minimum because it’s just it’s bounded by zero by the horizontal axis
but it’s not going to have an absolute minimum and it’s not going to have an

00:57
absolute maximum either and then if you were to sketch the logarithmic graph
again this keeps increasing it’s increasing at a slower and slower rate but
it’s still increasing and here we have a vertical isotope
it’s going to keep falling right here so neither one of these have absolute
maximums or absolute minimums so there’s just some you know common
graphs that you should be familiar with um all right so now we’re ready for
our big theorem the main theorem of this talk
so again there’s two points to this episode here was
to introduce you to relative extrema and absolute extrema
and critical numbers and then the second
is to really make sure we understand the extreme
value theorem so here’s what it is and you know it has two hypotheses there

00:58
oh sorry it has one hypothesis there um continuous function and it has to be
continuous on a closed bounded interval so that’s not an open interval
it’s a closed interval there and then if you have that
then you know it must have a maximum value and it must have a minimum value
at some numbers in that interval s and t could be the same numbers
um but it has to have a maximum it has to have a minimum value
all right so let’s look at um well i just want to point out before we
look at an example that this is an existence theorem
but i am going to cover a procedure that you can follow
so you can use this existence theorem to your advantage
extreme value theorem all right so first of all let’s see here if this has any

00:59
absolute extrema on its domain so this function right here is just a
parabola and it looks like this minus x squared but at zero someone took away
that point there um and what did they replace it with they replaced it with
the minus one so they took this away and they put a minus one here
so let’s put a coordinate axis there and i kind of missed my mark there
so this is open right here and here’s minus one so it’s open there
so it’s not continuous right there so does this have an absolute maximum no
it does not because we’re getting really really close to that zero
and we’re getting really really close to that zero but because someone took that
point away it doesn’t have an absolute maximum so

01:00
what that’s saying is that the um extreme value theorem doesn’t apply when
the function is not continuous another said differently you’re assuming
that f is continuous if it’s not continuous all bets are off
it may not have an absolute max or absolute min
and so here’s an example where it didn’t have a maximum
and the reason why we cannot use the extreme value theorem
is because it’s not continuous even if we were to put a interval right
here say one and minus one so now we could say oh we got our
interval right here minus one to one but it’s not continuous there it it doesn’t
have to have an absolute max so extreme value theorem you have to you
have to make sure that you’re continuous on a close about on a closed interval
there all right so f does not have a global maximum since it takes on values

01:01
less than but arbitrary close to zero or as i was saying a second ago you know
this is approaching zero but doesn’t actually get there someone took that
point away all right so never reaches the value of zero
so this function is not it’s not continuous
and so the extreme value theorem does not apply so here’s another example here
what does this function look like so we are looking like um
x g of x is just equal to x so we’re just going to go through here at x
but that’s only greater than zero and it’s strict so i’m going to open it up
here and then when we’re less than zero we’re just at a three
so let’s say that height is three right there
so this is what the graph looks like and and when i look at this graph right

01:02
here for g of x it is not continuous right here at zero
so this is having the absolute extrema and the answer is no three is not
an absolute height because this function keeps going up and zero is not an
absolute min because someone took the point away it
doesn’t reach it in other words i can get close and close to zero but it
doesn’t reach it so here’s another example where the extreme value theorem
doesn’t apply even if you restrict it to a closed
bounded interval right here like say one um say one
well one one would be down here because it would be one one
but even if i restrict it to a one and a minus one
here closed interval for g it would still the extreme value theorem would
not apply because it’s not continuous on minus one to one

01:03
okay so it does not have a global minimum and it does not have a global maximum
the extreme value theorem does not apply because it’s not defined it’s not
restricted to a closed interval but even if you were to restrict it to a closed
interval something like this it would still not apply
because it’s not continuous alright so [Music]
now we’re ready for our next theorem we saw two exceptions so that tells us that
the hypothesis and the extreme value theorem
is important cannot overlook it so now let’s assume
that the extreme value theorem is holding and the question
is how do we find the absolute extrema how do we find them
so that’s what this theorem tells us so if we’re continuous
on a closed interval a b then here’s the steps we can take

01:04
to find the absolute extrema and and the and the key is is that we know the
absolute extrema exists because of the extreme value theorem so
the next question is okay now that we know they exist i know i’m
continuous on a closed interval so i know the extreme absolute extrema exists so
now the question is how do we find them following steps
number one find the critical numbers and plug those into the function f
so we practice finding critical numbers so we’re good on number one
we’ll practice some more but but we can do that step two is to evaluate f on the
boundary so you know you can actually put all this in a chart you can say
here’s all the critical numbers here’s the boundary a and b
and i’m going to plug in all that information into f
and then i’m going to see what are the smallest and what are the largest

01:05
values and those will be the absolute extreme so let’s see this in practice
the largest is the absolute max and the smallest is the absolute minimum
all right so here’s our first example here we go find the absolute extrema
so first we notice that is that we have a polynomial
and we have a closed interval so i’m going to actually write that down
f is polynomial and so f is continuous on a closed interval
so i’m going to abbreviate that extreme value theorem holds
extreme value theorem holds so in other words this procedure that i said here
the probably the most important step here is the very first sentence

01:06
you need to be continuous on a closed interval
you can’t proceed with one two and three if you don’t have that
so that’s what i want to check first is you know we got a polynomial we know
polynomials are continuous on their domain that’s continuous on the entire real
line but we’re actually looking at the restriction minus three to two
we know it’s continuous there and we know since we have a closed interval
that the extreme value theorem holds so now i can follow through
with that strategy i can go through steps one two three and four
and so that’s what we’re going to do now so
first thing is i notice it’s continuous and then i’m going to go find its
derivative because remember we got to find the
critical numbers how do we find the critical numbers
we look where the derivative is zero and where the derivative
is undefined so if that’s our derivative uh well we have to see where that

01:07
derivative is zero right where is that derivative zero four x to the third
minus eight x where is that zero so i can factor out four x i get x squared
minus two right so we can see where that derivative is zero
so this is going to give us square root of two plus or minus
and that’s going to give us zero e for that for the zero
so there’s three places where we might have critical numbers
and because the derivative is also a polynomial
the derivative is defined for all x and so we only have
three critical numbers where the derivative is zero so we will take
a peek at the sketch of the graph this is the graph of
x to the fourth minus four x squared plus two
and you can see there there we have some
uh and this is on the interval minus two to two

01:08
so we can see there why we need to check the boundary that’s why i want to show
us this graph here so when we’re looking back at this um
steps here we’re looking at steps one two and three
step two is sometimes forgotten is you have to check the function
at the boundary for our problem the boundary is what
minus three and minus two and you can see by looking at the graph
minus three is going to give us the absolute maximum isn’t it
at two we’re not going to get an absolute maximum
all right so anyways we do steps one two one two and three we find the critical
numbers and then i’m gonna make my table now notice in my table here
i have my x column and the x column contains the critical numbers and the

01:09
boundary so the boundary is minus three and two so i have those in my x column
and then i have my critical numbers now for each of those x values i go plug it
into my function so i have to go calculate here what is
minus three to the fourth and then minus four times minus three to the squared
and then plus two right so to get that first r uh you know that 47
i got to go calculate all that up so i’m gonna go put
minus three into my function and calculate this up
and uh you know once we do that that’s where we get that number from
so we get a 47 for all that and then we plug in minus 2 into the function
oh sorry then we plug in minus square root of 2 into the function
and we get out minus two and so on now once we get that second column there
once we plug in all those numbers and get get the outputs
then we can look to see um the conclusion

01:10
now i also put a derivative column just to make sure that i
cover all the cases so the derivative column is there is just for
why are those numbers in that column like y is minus three that comes from
the boundary and why is two there that just comes to
the boundary so i just put dashes for those
but the 3 points plus or minus square root of 2 and 0
i put 0 for that column because that’s how i found them i found it where the
derivative was 0. all right the important column of course
is the conclusion column so the conclusion column says what our conclusion is
what’s happening at -3 well i get a 47 that’s the absolute maximum because if i
look at that second column there 47 is the largest number
what’s the smallest number smallest numbers minus 2
that happens in two different places those are the absolute minimum values
so by looking at the critical numbers i make that table

01:11
and we found all of our absolute minimum
and we found all of our absolute maximum in other words the problem of finding
the absolute extrema we did it we found all the absolute extrema
so that’s our first example let’s look at another
so this time let’s look at x to the four-fifth
and we’re going to be on the interval minus 32 to 1.
so what’s our first step is to find the derivative so let’s do that
so our derivative is going to be four-fifths and x to the four-fifths
minus one so minus one-fifths so this is four over five and then we
have fifth root of x so where is the derivative zero and where is the derivative

01:12
undefined so the derivative is never zero and the derivative right because you
know there’s no solution to this there’s no x’s where this will be zero
in order to have zero here you must have zero up here
um and that’s a four so that’s never going to be zero
now if you had a four x then you could say oh yeah x will be zero
no this is a four so this will never be zero
now where is the derivative undefined the derivative is undefined
at x equals 0 but is that even a number in our domain so we’re looking at minus
32 to 1. so 0 is in there isn’t it so 0 is in there
can we substitute 0 into that rule into that
function and yeah 0 to the 4 5 power which is just 0.
so x equals zero is the critical number and that’s the only one we have that’s
the only place where the derivative is undefined

01:13
and the derivative is never zero so we just have x equals zero
so when we make our table here we just need to look at -32 0 and 1.
and i always put them in order for myself [Music]
now i have to plug those numbers into the function so i need to do 32 to the
four fifths power right our function is four fifths
so if i had to do that that would be what 16.
and when i plug in 0 in here i get 0 when i plug in 1 i get 1.
so this is the largest value so that’s going to be the absolute maximum
and this is going to be the absolute minimum all right so let’s write that up so
that’s what a sketch looks like if you want to take a peek at that
but you can see right there in the absolute maximum and at
absolute minimum all right so here we go we have the function

01:14
and we notice in order to do this procedure that we’re doing
that is only looking at the boundary and only looking at the critical numbers
we have to first check that the absolute extrema
must exist we do that using the extreme value theorem so i’m going to notice
that the function is continuous on that closed interval so that’s the first step
then the second step is to find the derivative we found the derivative
and then i look where the derivative is equal to zero and i look where the
derivative is undefined where it is equal to zero
well it’s never equal to zero and where is the derivative
undefined only at x so that’s my only critical number
so i’m going to check minus 32 0 and 1. when i plug those into the function then
i can make my conclusions [Music] all right so there we go another example

01:15
and let’s look at another function so here we go x over
x plus one and we’re looking on one to two so
first off is where is this function continuous where is that function continuous
it’s continuous on all real numbers except -1 but we’re restricting it
so now we have to ask just the question is it continuous on one to two and the
answer is yes i mean that’s a rational function it’s
continuous for all real numbers except minus one but minus one isn’t
even in there so when i look at this function
on this interval when i look at that all together that’s a function
and it’s continuous on that close bounded interval so the procedure that
we’ve been working on we can do it again so we go find the derivative

01:16
so we have low times derivative high minus high times derivative low
all over denominator squared and so what is that derivative there the
x’s add up to zero we just get one over x plus one
when is this ever equal to zero never there’s no x’s where that will be zero
when is this ever undefined when x is negative one
we’re going to get zero down here so x equals negative one
you’re tempted to say is a critical number but
always look back at the interval and the function
so i cannot use -1 in there it’s not in the domain
if i just look here you can see that but of course also
-1 is not between 1 and 2. so this has no
critical numbers so all we have to do is check the boundary
excellent so that’s what the graph looks like right there

01:17
and you can see nothing is happening between 1 and 2.
now on this sketch that i i i made by computer
i’m only looking between one and two so that’s not the origin there that’s
between one and two but i think this sketch shows you what’s happening here and
what’s happening is basically the graph is just increasing
between one and two and nothing else is happening
so you know that would be the boringest rollercoaster ride
you know thinking as before this is just going up
and up and up and up and then the right is over
anyways the function is continuous on one two we check that so then we check the
derivative and we find that there’s no critical numbers
so now we just check the boundary and we see which one is the absolute maximum
and we see which one is the absolute minimum we plug in one we get out one

01:18
half we plug in two we get out two thirds all right very good
let’s look at one more example here so this is sine plus cosine
which is continuous everywhere if you just look at sine
plus cosine that’s continuous everywhere
so that function is certainly continuous on its domain
0 to pi over 3. certainly continuous there so we have a closed interval
and a function continuous on it so we can use the extreme value theorem we
can use our procedure steps one two and three
first step is to go find the derivative so let’s find the derivative
actually we’re going to take a peek real quick what that looks like
so between zero now pi over three is a little bit bigger than one
so it looks like it goes a little bit past one and there the

01:19
function it you know when it goes up like that and then it starts to come
back down looks like it’s going to have an absolute maximum there
somewhere in the interior you know we’re looking from zero to pi over three so
somewhere in between zero and pi over three it looks like it’s going to have a
critical number there let’s find it all right first thing we notice is that we
have a continuous function on a closed interval
so the extreme value theorem holds so let’s look at the derivative the
derivative of sine is cosine and the derivative of cosine is minus sine
so we look where the derivative is zero now if we
set that equal to zero then all we’re getting is sine equals cosine right so
just to make sure everyone is okay with that where’s the derivative

01:20
that’s just cosine minus sine where is that zero
so just move sine x over we get cosine equals sine
and now if we divide both sides by cosine we’re going to get one
equals tangent so either way you want to look at it
you’re going to get x is pi over four and actually there are infinitely many
values where tangent is one or there’s infinitely many values where cosine
equals sine but for this interval that we’re looking at
zero to pi over three the only one is pi over four
so pi over 4 there is our critical number so pi over 4 is our critical number
there and we can see that on the sketch there at pi over 4 which is
you know less than one all right so now we’re ready for our table
we look at the boundary zero and pi over three
and we look at pi over four also we take

01:21
those three numbers those three x values we plug them into our function we plug
in zero we get out one right sine of zero is zero and cosine of
zero is one so we get out one and then we plug in pi over four
and then um you know you know when you plug in pi over four
you’re gonna get square root of two over two
and cosine pi over four you’re gonna get another square root of two over two
in other words you get square root of two all right and then you can plug in
pi over three and you get that number and then you can see where the absolute
maximum is where the absolute minimum is and if you
need to you can go get some decimal approximations
you know that way you can know like you know more clearly to you
which one is the maximum all right so there’s um several examples on absolute

01:22
extrema and hope that’s been fun [Music] okay so now it’s time to look at some
examples uh some exercises sorry and um now i’m gonna have these
exercises here for you to go and try on your own and then get back to me
and tell me which ones you liked and which ones you didn’t like which ones
you’d like me to see in a video and so let’s go through some exercises
and look at that so the first exercises are find some critical numbers
and then two and three is to find some absolute maximum and minimum values
make sure that you can use the extreme value theorem by checking that you can
use that first we saw some examples in this video where
we used the extreme value theorem but we also saw some examples where you
couldn’t use the extreme value theorem so make sure and watch out for those
four five and six are also and seven are find the

01:23
absolute extrema we we have some equations there or some functions
and we got close close intervals there and then you know look at the graph too
and see what you can find from looking at the graph and then
number nine is several more examples where you can practice and
and hone your skill there all right and so then now find extreme values so
we’re going to be looking at some examples there where we have cosine inverse
where we have fractional powers where we have natural exponents um [Music]
and then we have some function some exercises here where
we’re really going to test to see if you’re paying attention to which
theorems you can use and when um and then some examples here

01:24
where they get more challenging 23 is a good fun interesting problem
and so there you have it there’s 23 exercises there
that i hope you give a good look to and so that’s it for this extra uh
episode and so i wanted to say thank you for watching um
the next episode is over the mean value theorem
um and we’ll also cover rolls theorem and we’ll look at some examples that
we’re going to use the mean value theorem for um
and so that’s a really important theorem and so
i hope you watch that video also i hope you’re getting value out of these videos
i hope they’re helping you if they are and you want to follow along
and you know keep up with the videos i have the social media links below
and i just want to say one last time thank you for watching
and i hope to see you in the next episode

01:25
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math can be difficult because it requires time and energy to become
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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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