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[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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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.

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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i hope they’re helping you if they are and you want to follow along

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and i just want to say one last time thank you for watching

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