okay in this video I just want to do

another example of factoring by grouping

so again I usually start thinking about

factoring by grouping when I see four

terms which we have in this case the

first thing I always ask myself is is

what's the greatest common factor of all

the terms that are present in this case

notice there's no variables in common

between all four terms the first one has

a u the second one has a U and a V the

third one has U and V s and the fourth

one only has V so we can't factor out

any variables but I can think about the

greatest common factor of 12 15 24 and

30 and I think the biggest number that

would go into all of those would be

three so I'm just going to factor that

out so I guess we would need a four u

squared when we distribute to get our

first term we would need a 5 u v to get

back the second term we would need 8u v

squared to get our third term and then

it looks like we would need a 10 v cubed

to get back the fourth term and now I'm

just looking inside the brackets and I'm

going to do the same thing as before I'm

going to look at the first two terms try

to factor out the greatest common factor

I'll look at the third and fourth terms

try to factor out the greatest common

factor

so for the first two terms it looks like

there's nothing in term if you look at

the four and the five the greatest

common factor of that is just one so we

won't factor that out unless there's

nothing else to pull out

they both do have used and again we

factor out the smaller exponent so we

have au to the first so I'm simply going

to factor out a you and I think that's

all we can factor out of the first two

terms and then it looks like we would

need for you so when we distribute to

get the 4u squared and it looks like we

would need a 5v to get the second term

back when we distribute so let's do the

same thing with the third and fourth

terms inside the brackets so I think

what's the biggest number that goes into

eight and four the greatest common

factor I think that would be a positive

two notice they both have V's present so

we can factor out the smaller of the

exponents which would be to the second

power and then again I think well what

would I need back inside so 2 times 4

would give me the 8 and it looks like we

would need a you to get 8 UV squared and

then it looks like 2 times 5 would give

me back with the 10 and I think we would

need an extra 5 or excuse me an extra V

as well so at this case at this point

we've got this this case where we've got

the same thing in parentheses and

remember the trick on these is so 3

still hanging out front we just take the

stuff out front bu and the positive 2 v

squared we'll stick that in one set of

parentheses and then we put our common

factor the 4 u + 5 V in the other set of

parentheses and now we're done we have

completely factored

this original this original expression

Factoring By Grouping - Ex 2

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Factoring By Grouping - Ex 2.  Another example of factoring by grouping!


all right in this video I want to look

at trying to prove some identities so

suppose we have uh these couple of

identities and we would try to try to

figure out if they're true or false so

um let's look at part A here first uh 1

minus sin T over cosine t plus cosine T

over 1 - sin t = 2ant t so I typically

you know in general start with the more

complicated side uh and try to work on

that side you know and to me when I look

at this sometimes getting started is the

hard part um you know I see two terms

here uh and I only see a single thing on

the right side so to me it says well you

know I'm going to have to have at least

only one thing on the left side so I'm

going to try to combine these which

tells me let's get common denominators

and uh so we can combine the

fractions okay so for our first fraction

it looks like we would have to multiply

the numerator and the denominator by

this uh one minus sin t

so plus and then our second

term it looks like we just have to

multiply the numerator and denominator

by cosine

T and again I'm kind of wondering does

this equal to secant T at the

end so okay

um so let's simplify the numerators so

we've got 1 * 1 which will be one we

have 1 * sin T which will be negative

sin T then we have negative sin T * 1

then we have a negative sin T and A

negative uh sin T which will give us

positive sin SAR

t uh looks like over here we'll get

cosine time cosine so we'll get

cosine squar of t

in the denominator we still have our uh

we would have our our single denominator

our common denominator of cosine T * 1

- sin

T all right so we can do a couple things

here uh notice we've got sin squar plus

cosine squar and sin squ uh plus cosine

squ remember there's an identity that

simply says that's equal to one so

really we have uh a negative sign T and

A negative sin T that's going to give us

-2 sin T we have a positive one and by

our identity we have another positive

one so we would have

positive2 over cine T * 1 - sin

T well looks like we could factor a two

out of the numerator then we would have

two uh we' have 1 minus sin t over cine

T * 1 - sin T and hey now we can cancel

out our factors the 1us sin T's we're

left with 2 over cosine t uh you can

think about this as being 2 * 1/

cosine of T and 1/ cosine is secant so

this is in fact an identity um it is

correct to say that that original

expression does equal 2 secant t so this

I would say is true

Proving an Identity, Example 1

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okay here we're going to find an

expression for the perimeter of our

rectangle so

suppose we don't have exact numerical

values but we do have these algebraic

expressions so two of the sides are

going to be x plus 4y x plus 4y

and then to the the other two sides will

be 2x plus 3y and 2x plus 3y

so remember to get the perimeter we just

add up each side okay

so that's what we're going to have to do

in this case we'll simply have

to add up the lengths of these four

sides

and of course we'll try to simplify this

expression down

so the perimeter p

that'll be x plus 4y

plus 2x plus 3y

plus x plus 2y

plus

2x plus 3y

again you could have also said this a

little differently

we could say we have two times the side

x plus four y

likewise we have two of the sides two x

plus three y so either one of those

expressions are absolutely correct

probably the second one here is even a

little bit easier to uh well maybe not i

guess it's all it's all the same

difference really

but i'm going to go ahead and do do the

work on this expression

so we have 2 times x

2 times 4y which will be 8y

2 times 2x

which will be 4x

2 times 3y which will be 6y

and now it's just a matter of combining

like terms i see something involving x

to the first power and x to the first

power

so we can combine

those two terms

2 plus 4 will give us 6. again we keep

the x

and then again we have some more like

terms we have a positive 8y

and a positive 6y

well positive 8y plus 6y will give us a

positive 14y

and again at this point x and y are not

like terms so we can't combine those any

further and that would be our simplified

expression

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding an Algebraic Expression for the Perimeter of a Given Rectangle.   Here we find a simplified algebraic expression for the perimeter of a rectangle


all right so one more example here

involving multiplication of decimals so

suppose in this town Lake View uh

there's

23.15 in of rain that uh falls on

average each year and we want to know

how much rainfall we could expect in 4.5

years well we could start you know

adding this up multiple times and then

trying to figure out what half of it is

or equivalently you know kind of the the

shortcut way uh we'll just take 23

.15 we're going to have to multiply that

by

4.5 all right so now just the

multiplication 5 * 4 is 20 so we'll keep

the zero carry the two 5 * 1 is 5 + 2 is

7 uh 5 * 3 is going to be 15 so we'll

keep the five carry the one 5 * 2 is 10+

1 will give us

11 drop down my Zero 4 * 4 is 16 so

we'll keep the six and we'll carry a 1 4

* 1 is 4 + 1 is 5 4 * 3 is 12 so we'll

keep the two and again carry a one 4 * 2

is 8 + 1 is 9 now we'll add these up so

0 + 0 is 0 7 + 6 is 13 so I'll keep the

three carry the one 1 + 5 + 5 is going

to be 11 so we keep the one carry the

one 1 + 1 + 2 is 4 and then 9 + 1 will

be

10 so again I count uh the number of

digits to the right of the decimal sign

we've got three digits total to the

right of the decimal signs so I'm going

to move the decimal place one two three

places over so it says we could expect

uh roughly you know says we could expect

in of rain to fall over the course of

4.5 years and again you know uh counting

decimals is good but also you know think

about it you know this is pretty close

to 20 right it's close to 20 inch you're

doing it for you know if you did it for

four years 20 in Time 4 years is you

know would be 80 in 20 in * 5 years

would be 100 in well this is a little

bit bigger than 20 so it should be a

little bit you know it should be around

you know 100 or so give or take um and

that's exactly what we're getting so not

only does counting you know clearly work

but you know just check the answer with

your intuition and make sure that uh it

seems like a reasonable a reasonable

solution

Multiplying Decimals, Ex 3

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okay here we're going to convert the

polar point negative 5 comma 315 degrees

to rectangular coordinates so we use the

formula x equals R times cosine of theta

and that y equals R times sine of theta

so again in this case our R value will

be negative 5 our theta about theta

value will be 315 degrees so that will

give us x equals negative 5 times cosine

of 315 degrees and Y will be negative 5

times sine of 315 degrees again I always

forget degrees I'm so used to using

radians so I'm gonna convert well let's

see if we're at the negative y-axis

that's 270 degrees which is 3 PI over 2

I guess we would need to add another 45

degrees which would be adding another PI

over 4 radians and if we get common

denominators we would have 6 PI over 4

plus PI over 4 so 7 PI over 4 will be

our our angle in radians so equivalently

we can write this as negative 5 times

cosine of 7 PI over 4 and we can write

the Y value as negative 5 times sine of

7 PI over 4 and to evaluate each of

these recall that at the angle 7 PI over

4 if you're on the unit circle at 7 PI

over 4 on the unit circle you're hitting

the ordered pair what would that be I

guess that would be square root of 2

over 2 comma negative square root of 2

over 2 so just reminding myself of the

values here

so let's see here we'll have you get

another piece of paper so for cosine or

for the x-coordinate we'll have negative

five x cosine of 7 PI over 4 7 PI over 4

will be the square root of 2 over 2 it's

really not a lot of simplification you

can do there so I'm just gonna leave it

negative 5 times root 2 over 2 and for

the Y value we'll have negative 5 times

sine of 7 PI over 4 but 7 PI over 4 is

negative square root of 2 over 2 well

there are two negatives we'll make it

positive so I'll have 5 times root 2

over 2 so in rectangular coordinates we

would use the point negative 5 times the

square root of 2 over 2 comma 5 times

the square root of 2 over 2

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Polar Coordinates to Rectangular Coordinates ; A Quick Example #2


okay here we're going to look at a

couple examples related to the direct

comparison

test so in part A here we're going to

figure out whether the series 1 over n

cubed + n + 4 whether or not that series

converges or diverges again and we'll

compare it to uh a series that we know

converges or diverges so in this case uh

I usually kind of pick out dominant

terms and to me you know again just the

powers that are getting the the biggest

Powers well in the top there's just a

one in the denominator though there's n

cubed and if you think about the series

one over n cubed that would be a

convergent um P Series so let's see if

we think about it you know 1 over n

cubed well if we add n to it I claim

that that's going to be less than or

equal to 1/ n cub and again this is for

n values greater than or equal to one

this would be true and again this makes

sense because because you know um since

we're adding in the

denominator of the left side um must be

bigger than the denominator on the right

side well a bigger denominator makes for

an overall smaller fraction well okay I

think this is true likewise 1 over n Cub

+ n well plus 4 by the same reasoning

would be smaller than uh 1/ n cubed

notice these are certainly greater than

or equal equ Al to uh Z but we know

since you know the series 1 n Cub from

Nal 1 to Infinity that's a convergent P

series that's a convergent P series well

since that's convergent and we know now

that the series associated with one over

n Cub + n + 4 is smaller than that one

and again greater than or equal to zero

since the uh the uh the series on the

right converges and the values uh that

we get out you know the numerical values

we get out are going to be even smaller

uh that would imply by the direct

comparison theorem that our original

series Nal 1 to Infinity of 1 over n Cub

+ n + 4 that also converges

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Direct Comparison Test - Another Example 1.  In this video I show that another series converges or diverges using the direct comparison theorem.  I do not run over the formula/theory in this video (but do in another video, so look around if that is what you need!).


okay so here we're gonna look at some

more examples involving integration of

some trig functions where we first have

to do some substitution so in Part A

here we have secant of cosine times

tangent of cosine times sine X so the

first thing you know again kind of the

things that I see here the first thing

that kind of makes me nervous is I see

the secant of cosine tangent of cosine

and that's kind of a quote unquote

I guess strange function but I know if

it was just secant of something and

tangent of something just a single

variable I know the antiderivative of

secant x times tangent X for example so

that tells me hey maybe if I make this

into a single variable by doing a u

substitution as long as it's D U is in

the problem and well the derivative of

cosine is sine it looks like everything

would work so if we let u equal cosine X

D u would be negative sine X DX and to

get a positive sine X we can just

multiply both sides by negative one and

now we can just fill everything in it

says we would have secant of U times

tangent of U and then our sine X DX we

can replace all of that with our

negative one D U and well the

antiderivative of secant times tangent

is just secant of U plus C and now again

we can just plug our substitution back

in so we would get secant of cosine X

all of that plus C

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Substitution Rule / U-sub Basic Example 3


all right here we're going to look at a

quick little example of finding distance

between three points where the points

are uh located in three dimensions so

we've got the point 1 35 and then4 07 so

if we label these as x sub 1 y sub 1 Z

sub 1 X sub2 Y sub 2 and Z sub 2 uh the

distance formula is going to be uh the

formula at the bottom again it's just uh

what we normally do I think a lot of

people uh you know hopefully you've seen

the distance form fora and algebra um

and there you typically just deal with

two Dimensions so you see just the

formula X2 - X1 2 + Y2 - y12 but now we

have to add on the Z components as

well all right so we can do this I think

without too much

trouble so we've got the square root

Let's see we would do -4 - 1^ 2ar Y 2 -

y1 would be 0 - 3^ 2 and then we would

have 7

-5 or 7 + 5^

2 so this is going to be -5^ 2 which is

25 uh -3 s is going to be posi 9 and

then we'll have 12 SAR which is going to

be

144 so let's see if we add these

together I guess we get 34 and

144 and let's see that's going to give

us the square root of

178 so I guess you could always see if

this uh simplifies I guess guess we can

write 78 as 2 * um well 2 * 90 would be

180 so 2 * 89 should give us uh

178 and 89 I do believe is a prime

number um three certainly doesn't go

into it what else might go into it um

seven definitely

doesn't um let's see is there anything

much bigger that could potentially work

um so 89 to me looks like it's Prime

it's not going to factor anymore we

can't really take the square root of

either of those two numbers so I would

simply say that the distance is the

square OT of 178

The Distance Formula in 3-Dimensions

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Distance Formula in 3-Dimensions. In this video, I show the formula to find the distance between two points in 3-dimension space.  It is just a slight extension of the 2d case, so it will be hopefully easy to remember!


all right in this video I'm gonna do

another example of solving a quadratic

equation by completing the square and

again the first thing I like to do is

put all my X's on one side and all my

numbers on the other so I'm going to

start off by subtracting 12 X from both

sides so I'm going to stick the negative

12 X in the middle there's my -1 so if

we subtract 12x on the right side we

would be left with 0 and then I'm just

going to add 1 to both sides so we would

get 3x squared minus 12x equals positive

1 and again what I like to do on these

the first thing I like to do is always

make the coefficient on the x squared

into a 1 again you kind of need the

coefficient to be a 1 before we can take

1/2 of the coefficient in front of the X

so it's kind of a common mistake to try

to do that step you know to do 1/2 a

little too early so I'm simply going to

divide both sides by 3 if I divide the

left side by 3 I would have to divide

each term by 3 so there I am dividing

everything by 3 on the left side we

would be left with x squared we have

negative 12 over 3 which is simply going

to leave us with positive excuse me a

negative 4 we still have our X leftover

and on the right we have our positive

1/3 so again I'm going to do the step

now of taking one half of the

coefficient in front of the X which is

going to be negative 4 if I take 1/2 of

that I get negative 2 and again I take

this number and I square it so negative

2 squared is going to be positive 4

and this is the number that I add to

both sides so I have x squared minus 4x

I'm going to throw in a positive 4 on

the right side we had our 1/3 and again

I'm going to add a positive 4 to that

side as well so I'm going to add 4 to

both sides again the left side is now a

perfect square x squared minus 4x plus 4

is going to factor as X again whenever

you take one half of the coefficient

whatever that number is that's how the

left side will be written as a perfect

square so you can check if you multiply

X minus 2 times X minus 2

you'll get x squared minus 4x plus 4 on

the right side I'm going to get common

denominators so you can think about

fours being 4 over 1 we can multiply top

and bottom by 3 and get 12 over 3 so on

the left side we're now left with X

minus 2 quantity squared and on the

right side left 1 plus 12 which will be

13 thirds and again now the idea is we

take the square root of both sides we

have to stick our positive and negative

on the right side and we've got X minus

2 equals will have a positive square

root of 13 over 3 and I kind of think is

this fraction simplify or could I kind

of break up the square root make either

part simplify and it doesn't look like

to me if that it simplifies at all so

I'm just gonna leave it alone that'll be

one of the equations we'll have to solve

and the other one will be X minus 2

equals negative square root of 13 over 3

and now all we have to do is add two to

both sides so we'll get positive square

root of 13 over 3 plus 2 as one of our

solutions and then our other solution

will be negative square root of 13 over

3 plus 2 and again now we've got our two

solutions to the original quadratic

equation

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Completing the Square to Solve Quadratic Equations: More Examples - 5


okay here we're going to do some

examples of uh writing decimals as mixed

numbers and we'll reduce those uh the

fractions that are left over and again

we do this the exact same way as simply

writing you know any decimal uh as a

fraction so I'm going to write 2.8 over

one again we we like to not see decimals

in our fractions uh that's how we tend

to write them so to get rid of the

decimal place in the numerator we'll

have to multiply by 10 over 10 well 2.8

* 10 again that just shifts the decimal

place that'll give us 28 in the

numerator and then in the denominator 1

* 10 is 10 we could start saying well 10

goes into 28 how many times etc etc um

or we can reduce first okay I think I'm

going to go ahead and reduce my

fraction is what I'm going to do again

these are both even and so so when I see

a a fraction where the numerator and the

denominator are both even I just

immediately divide by the top by two

which will give us 14 and the Bottom by

two which will leave us with five there

could have even been a factor uh you

know some factor in common larger than

two in this case it wasn't but again to

me if I see they're even I just start

chopping it down by two because it's

kind of mindless and I can do it quickly

and it just gives me smaller numbers to

deal with okay well we're left 14 over5

I don't think that's going to reduce uh

as far as our fraction goes anymore and

now we can just do you know we can do

our our long division and find the

remainder I think we can do this one

hopefully uh I think hopefully we can do

it in our head let's see so five goes

into 14 I guess two whole times without

going over right because 2 * 5 would

give us 10 we would need an extra four

to get up to 14 and then we just have

the five left over so 14 over five is

going to leave us with 2 and 4 fths as

our mixed number and again notice by

reducing at the beginning our fraction

will already be reduced here at the

end

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Writing a Decimal as a Fraction, Example 3


okay in this video we're going to look

at an example related to the limit laws

so just a couple problems here and

suppose that we're given that the limit

as X approaches two of our function f of

X that equals four and the limit as X

approaches two G of X that equals

negative five we want to compute each of

these following limits so Part A we've

got the limit as X approaches two of

three times f of X minus two times G of

X over f of X Part B again we'll do the

limit as X approaches 2 of f of X times

G of X plus G of X so let's do Part A

here first for the limit laws maybe

we'll go through this one and bits and

pieces if you have a limit of a fraction

it basically says you can take the limit

of the numerator over the limit as X

approaches two of the denominator okay

well there's also some more rules that

say okay if you have addition and

subtraction it says you can look at the

limit individually so we'll have the

limit as X approaches two of three times

f of X minus the limit as X approaches 2

of two times G of X again in the

denominator I'm going to write it a

little closer we've got the limit as X

approaches two of our function f of X

another one of the limit law simply says

if you have constants being multiplied

by your function you can factor those

out so we have three times the limit as

X approaches two of the function f of X

minus I'm going to pull the two out

front two times the limit as X

approaches two of G of X well again all

of that is over

the limit as X approaches 2 of f of X

and now we're in a position to start

plugging in the given values so the

moral of the story is you can basically

just plug things in a lot of people

nobody I mean very few people write all

this stuff out when doing limits so

let's see we said the limit as X

approaches 2 of f of X that was given to

us as

for so I'm just gonna plug that in so

we'll have three times four minus two

times well when we do the limit as X

approaches to G of X we said the limit

as X approaches two of G of X is just

negative five so let me plug that in

there as well and again we just said the

limit as X approaches 2 of f of X that's

giving us the value 4 and now we're in

business we can just do our arithmetic

so 3 times 4 is 12 negative 2 and

negative 5 is positive 10 all over 4

let's see that's 22 over 4 which we

could reduce we could make that 11 over

2 or I guess that's what 5 and 1/2 so

that would be our limit the the answer

to the limit question in Part A

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Limit Laws to Evaluate a Limit , Example 1.  In this video, I go through all of the steps using the limit laws to evaluate a limit.


all right in this video I'm going to do

another example involving direct

proportion so again if we say y varies

directly as X we can write that as yal

KX so in this problem uh the way simple

interest works on a bank account is it

varies directly as the amount of

principal in the account so I'm going to

use I for the interest and maybe P for

the principle so it says the interest

varies

directly um and is the amount of

principal so we can write that statement

as i = k * P they tell us a little bit

of information here they tell us if

$1,400 earns 84 in interest we want to

know how much money a 5,000 uh if you

had $5,000 in the bank how much would

that earn an interest so okay the first

thing we're going to do is solve for K

so they tell me that um so the interest

is $84 I'm going to plug that on the

left K is what we're trying to solve for

and that was on an initial principle of

1,400 so to solve for K I'll divide both

sides by

1,400 and you know this is one I think

I'm just uh going to go ahead and use a

calculator on it so 84

/400 I'm getting that to be

0.06 so it really says you're earning 6%

um in in regards to simple interest but

okay um so really it says the equation

we should be using it says i = k which

is

06 times the principal so okay well to

answer the question is to figure out the

interest we'll just simply take 06 and

multiply that by the new principle which

is

5000 and 006 * 5000 uh should simply

just be 300 which it is

so it says if you're earning obviously

the same amount of Interest simple

interest it says whereas $1,400 would

make you $84 in interest this tells us

that uh if you put $5,000 in the bank

you're going to earn uh under this

scenario $300 in interest

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Direct Variation / Direct Proportion - Example 2


all right in this video we're going to

look at finding the derivative of a

vector function and to do this all we

have to do is simply take the derivative

of all of this stuff basically involving

T so we'll take the derivative of the

stuff in front of I the stuff in front

of J and the stuff in front of K so when

we take the derivative of cosine PI T

we'll get negative sine times pi T but

then we'll have to take the derivative

of the inside so we'll pick up a pie so

we'll have negative pi times sine of pi

times T all of that times I plus for the

second part we can use the quotient rule

so we'll get the denominator the

derivative of the numerator will just be

e to the t minus the numerator times the

derivative of the denominator which will

be 2t all over the denominator squared

which is T to the fourth times J and

then if we take the derivative of 4t

cubed

we'll get twelve T squared times K and

that's now our derivative so I think I'm

going to do maybe just a little bit of

simplification so we have negative pi

sine of PI T times I I guess we could

factor we could factor a T out of the

numerator which would cancel out with a

T in the denominator we could even

factor in e to the T out as well so we

could factor out e to the T and a T and

then in parentheses we would have a T

and then I guess we would need a minus

two that's all over t to the fourth

times J and then our 12 T squared I'm

just going to leave that alone so the

last thing I'm going to do is just

simplify this part so negative pi times

sine of PI T times I plus okay so we can

cancel out the T that will give us T to

the third in the bottom we still have e

to the T times t minus two all of that

is being multiplied by j

plus our 12 t squared times K and now we

have our derivative so again nothing

crazy all we're doing is just simply

taking the derivative of each little

part so kind of three derivative

problems combined here into one

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Derivative of a Vector Function - Another Ex 1. In this video, I show an example of finding the derivative of a vector function.  It is pretty simple: just take the derivative of each part!


in this example we're gonna take five

numbers that are written in percent form

and we're gonna turn those into decimals

and this is easiest pie if you remember

how to do it and all we have to do when

we rewrite a number as we rewrite a

percentage as a decimal all you do is

just replace the percent sign with a

think about putting your decimal there

but oh actually it's not in the correct

spot we always move it two places to the

left so 26 percent I'm gonna put as the

decimal places point two six and I'm

even just gonna put a zero in front so

26 percent is zero point two six and

again the reason why we're doing that we

saw that we could write a percentage as

a fraction by just dividing it by 100

and if you divide something by 100 you

in effect move the decimal place two

places but as a shortcut that's all

we're doing so okay

125 percent let's get rid of the percent

move the decimal place two spots so it

says we would get 1.25 so 125 percent is

equivalent to one point two five two

percent well it feels like there's only

one place to move it let's put some

zeros in front so if we get rid of the

percentage move it two places to the

left put our decimal sign right there

we would have 0.02 fourteen thirty five

percent 1435 same thing get rid of the

percentage and move it two places

that'll be 14 point three five and last

but not least if we have 0.34% same

thing we still have to move with two

places to the left so let's see if we

took the decimal place and moved it two

places to the left we would now have

zero point

let's see still two zero

three four that would be the equivalent

decimal form

Writing a Percentage as a Decimal ( 5 examples)

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Writing a Percentage as a Decimal (5 examples)


all right in this video we're going to

start talking about roots of unity which

just basically means we're taking

complex roots of the number one so we've

seen how to take uh complex Roots the

first thing we can do is start off by

writing our complex number in polar form

and we can write one as just one *

cosine of 0° plus I * s of

0° uh we can then use our our formula

that says we would get cosine of really

uh the original which would be 0 deg +

2i K N the same thing we'll get S of 0 +

2 piun * k n so we can use this formula

to calculate the roots of unity and

another thing geometrically uh roots of

unity are spread evenly around the unit

circle so that'll kind of make it easy

geometrically to find them so uh suppose

we've got this example here so uh so

again we know that the the roots of

unity are spread evenly around the

Circle we want to calculate the value of

the root to which the arrow is pointing

at okay so if you think about going back

to this

formula um if you plug in notice if you

plugged in k equals 0 you would get 0

degrees um so you'd have cosine of zero

the uh the point um at at on the unit

circle uh on the positive x-axis that's

always going to correspond to K = 0 zero

and there'll always be a point right

there and then uh everything's spread

evenly uh hopefully it looks spread

evenly this would correspond to K = 1 k

= 2 ah okay well this is K = 3 k =

five six and seven notice if you count

the number of roots there's one two 3

four five six seven eight so what that

really means is uh we need to use this

formula we just need to calculate the

value of this formula where Nal

8 and again that's the number of roots

if you count and then we said where k

equals 3 so K = 3 so that's all we're

going to do now we'll just plug it into

this formula it says we would get 1 *

cine of 2 pi * K which is 3 over n which

is 8 plus I sin of the same thing we'll

get 2 piun * 3 over

8 and if we clean this up a little bit

that's just going to be cosine let's see

2 over 8 would be uh pi over 4 so we

would really have 3 pi over

plus I of 3 pi over

4 well cosine of 3 pi over 4 is is going

to be < tk2 over

2 s of uh 3 piun over 4 is going to be

positive < tk2 over 2 so now we've got

our answer it's going to be < tk2 over

2+ < tk2 over 2 *

I

Roots of Unity, Example 1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Roots of Unity, Example 1.  Here is give a little discussion about finding roots of unity and the unit circle.


all right so here we're going to do an

example of finding a second derivative

so we'll use the derivative formula at

the

bottom um so here we're going to find

the second derivative uh for x = t minus

t cubed and then y = 2T +

5 so the first thing we'll have to do is

well find the first derivative so the

first derivative dy over DX that's going

to be dy over DT all over DX over DT so

the derivative of y with respect to T is

just going to be two the derivative of x

with respect to t uh so the derivative

of 1 T would be 1 minus uh the

derivative of T 3 will be 3

t^2 all right so now to get the second

derivative it says we have to take the

derivative with respect to T of our

first

derivative and then we divide that by

the original DX over DT we said DX over

DT uh was just 1 - 3T squared so notice

nothing really happens in the

denominator when you take the second

derivative you really just kind of uh

you know uh taking the derivative of the

first derivative in the numerator and

then again just leaving the dxdt part

alone all right so um if we take the

derivative of uh so I'm just going to

kind of do the the numerator off to the

side

so we could write this as 2 * 1 - 3t^

2st again I'm just focusing on the

numerator for the moment and just

bringing the denominator

upstairs well we would use the chain

rule in this case so uh we would get -2

leave the inside Alone 1 - 3t^ 2 uh

subtract one from the exponent and then

we would have to take the derivative of

the inside which is going to give us-6 6

T all right so when we simplify -2 and -

6t will give us POS 12T and then we can

put the 1 -

3t^2 squared back in the

denominator so this is just our

numerator after taking the derivative so

12T over 1 - 3t^ 2^ squared well all of

that is being divided by 1 - 3t^ 2ar so

again if you think about uh the

denominators being over one if you flip

and

multiply when we flip and multiply we'll

just be left with

12T over 1 - 3t^ 2^ 2 * 1 over 1 - 3t^ 2

so I would probably write The Final

Answer here is 12T over 1 - 3 t^ 2 but

uh

so we've got a squared one and a one to

the first power so we can write all of

that to the third power and now we have

our second

derivative

Derivatives of Parametric Equations, Another Example #2 - Second Derivative

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Derivatives of Parametric Equations, Another Example #2.  In this video, I give the formulas to find the first and second derivative of a parametric curve, and then I find the second derivative.


all right in this video we're going to

look at the fundamental theorem of

calculus part two and I think this is

the part probably a you know most people

use it maybe they don't recall that it

was called this so what the fundamental

theorem of calculus part two tells us it

tells us how to calculate a definite

integral so if we want to integrate

little f of X DX from A to B it says

what we do is we we find capital f of X

which is an antiderivative and then a

lot of times we'll just make a little

bar here that's the notation kind of

saying well now we've found an

antiderivative we still have to evaluate

that from A to B and to evaluate this

what we do is we calculate F of B minus

F of a so we take the upper limit of

integration and subtract away what's

called the lower limit of integration so

again all this condition says is that

capital F is an anti derivative so

pretty straightforward you know now you

just have to just know antiderivative so

I've picked one here completely at

random sort certainly a pretty basic one

so we're going to calculate the definite

integral from 1 to 3 of X plus 2 well

the first thing we have to do is find an

antiderivative of X to the first and

what we do when we have X to a power is

we add 1 to that power so I'm going to

get X to the second and then divide that

by 2 and then I need to find an

antiderivative of the constant 2 well

2x would be an antiderivative of the

constant 2 and you can always check you

have a correct antiderivative by simply

taking the derivative and you should get

back your your original function back

and what I'm going to do now since we

found an antiderivative I'm just going

to write well from 1 to 3 and what I'm

going to do now is just simply

everywhere there's an X I'm going to

plug in a 3 so I'm going to have 3

squared over 2 plus 2 times 3 so this is

like my F of B part or in this case it's

going to be F of 3 and then what we do

is we subtract away okay so now

everywhere there's an X I simply plug in

1 so we would have 1 squared over 2

two plus two times one so here we're

subtracting away our F of a which in

this case again is just F of one and now

we just have to do the arithmetic so I

don't think this one is a too terribly

bad to compute let's see so 3 squared is

9 so we have 9 over 2 plus 2 times 3

which is 6 minus well let's see we would

have 1/2 plus 2 so let's see that's 9

halves plus 6 minus 1/2 minus 2 when we

distribute the negative let's see 9

halves minus 1/2 that's going to give us

8 halves positive 6 minus 2

let's see positive 6 minus 2 that's

going to give us positive 4 well 8 over

2 is just 4 so we have 4 plus 4 4 we get

the value positive 8 so in this case 8

would be the value of our definite

integral but again all it says all the

fundamental theorem of calculus part 2

says I shouldn't say all it says it's

super super super important it says to

calculate a definite integral find an

antiderivative plug in your limits of

integration subtract and you'll have

your answer

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Fundamental Theorem of Calculus. Part 2.  The FTC part 2 simply tells that to evaluate a definite integral, we find an antiderivative, plug in the limits of integration and subtract.  Simple! (Which is what makes it soooooo amazing!)


okay here we're going to graph the

system of equations y equals negative 2x

plus 7 and y equals negative 2x minus 5.

so notice both of these are linear

equations they both have the same slope

of negative 2. the only thing that's

different is the y-intercept one has a

y-intercept of positive seven and the

other one has a y-intercept of negative

five so i'll graph negative two x plus

seven first so let's see one two three

four five six

seven there's our y-intercept

and now the slope again is negative two

so down two over one down two over one

so again this was going to make

obviously a nice rough little sketch

so if we play connect the dots here

you can play connect the dots and get

one of our lines

you could always even find the

x-intercept if you wanted to to make an

even better graph

so there's y equals negative two x plus

seven

and now the only thing that'll be

different for our other line is that

it'll have a y-intercept

of

negative five but then the same thing

the slope is negative two so down two to

the right one down two to the right one

these should obviously be parallel lines

let's see here

hopefully they'll look relatively

parallel

i think pretty close so looks good to me

so there's our other line y equals

negative

minus 5. so in terms you know the

question obviously doesn't ask this but

if we tried to find a solution to the

system of equations we could immediately

say that there are no solutions and we

can deduce that from the graph

since the lines never intersect we can

simply say that there's no solution to

our original system of equations

hey it didn't ask that

but a lot of times that's why we do

graph our system of equations to try to

see if there is a point of intersection

or not

Graphing a Simple System of Equations

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Graphing a Simple System of Equations


all right so just another applied trig

problem here suppose you've got a hiker

and uh he's hiking up a 12 degree slope

he hikes at a constant speed of 3 miles

hour we want to know how much altitude

he gains over the course of 5 hours of

hiking all right so I'm just going to

make a little right triangle here's our

uh angle of

12° um if the guy is walking 3 miles per

hour for 5 hours he's walked a total of

15 miles so that means the hypotenuse is

just going to have length 15 all we're

really trying to figure out is this

change in

elevation U maybe we'll represent that

with

Y so now all we have to do is just use a

trig function so let's see um in

relation to our angle 12° we have the

opposite side we know about the

hypotenuse so the trig function that

relates the opposite to the hypotenuse

would be sign so simply s of

12° is going to equal the opposite which

is y over the hypotenuse which is 15 and

now all we have to do is simply solve

for y so I'm going to multiply both

sides by

15 so it says 15 s of

12° is going to be uh his elevation

change uh let's throw this in a

calculator just to make some more sense

out of these numbers uh again make sure

your calculator's in degree mode I'm I'm

getting s of

12° to be roughly equal to uh

208 after I

round so if we simply multiply that by

15 um I'm getting

3.12 so we can simply say that his

elevation gain is roughly

3.12 miles so a lot of elevation gain

there so um but again you know no big

deal here again all you're doing is just

making a little right triangle and just

thinking about you know which trig

function you need and then it's just a

matter of punching in a couple things

into the

calculator

Trigonometry Word Problem, Example 2

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Trigonometry Word Problem, Example 2. In this video, I use a bit of trigonometry to find the elevation gain of a hiker going up a slope.


okay in this example we're going to find

the amplitude and period of the given

function and then we're also going to

describe any phase shifts or vertical

shifts when compared to the

corresponding basic trig function so

we've got y = /4 cosine of 12x + < / 2 -

1 so to get the amplitude probably the

easiest part we'll just take the

absolute

value of the number out front - 1/4 and

that'll give us 1/4 as our amplitude

now to get the period we're actually

going to have to rewrite this a little

bit and inside the parentheses we want

the coefficient on the X to be a

positive

one so what we can do well we can simply

Factor the 1/2

out so then 1 12 * X would give us well

12 x and then 12 multiplied Pi would

give us our Pi / 2 and the minus one is

still out there

this is going to be our B value when we

do the period so 2 pi ided the absolute

value of B so in this case we'll have 2

Pi / the absolute value of2 well that is

just 1/2 2 pi ID a half that's going to

give us 4 Pi same thing as 2 pi multipli

2 so there's the period and the

amplitude what else did we want U we

wanted the vertical and the horizontal

or the phase shift as well if we look

inside the I used brackets here if we

look inside the brackets or the

parentheses that's where we're going to

get the phase shift so in this

case we have X+ Pi that's me that means

uh if we compare it to the graph of just

cosine uh well really we should be

actually comparing it to cosine of 12x

it's going to be shifted to the left

um to the left Pi units and then the

vertical

shift in this case since we have a minus

one out there the the vertical shift

will equal negative 1 or it'll be one

unit downwards

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding Amplitude, Period, Horizontal and Vertical Shifts of a Trig Function EX 1


all right in this video we're going to

look at an example related to the

derivative of the square root function

so suppose we are given that the

derivative of square root of x is going

to be one over two times the square root

of x we want to know what's the

derivative of square root of x over

eight so just to kind of justify this a

little bit again if we use the

definition of the derivative on our new

function we would have the limit as H

goes to zero and then we would have the

square root of x plus h over eight minus

the square root of x over eight all over

H so again I'm just using our definition

f of X plus h minus f of X all over H

and again we're just taking the limit of

that as H goes to zero so I'm just using

the definition of the derivative on our

new function square root of square root

of x over eight well notice what we

could do is in the numerator we could

factor that one-eighth out right we

could just pull the 1/8 out and then we

would have well one eighth times the

square root of x plus h minus the square

root of x all over H and the idea here

is is that the limit depends on H okay

so this constant 1/8 we can actually

just factor that out front so we would

have 1/8 times the limit as H approaches

0 of the square root of x plus h minus

the square root of x all over H but if

you think about it now this last part

this last part would just be the

definition of the derivative for the

square root of x function okay so you

know if you think about if we started

with the function say f of X equals

square root of x if we wanted to find

the derivative we would get the limit as

H goes to 0 square root of x plus h

minus square root of x all over H and

that's what we have and you know in

new expression so what I'm trying to say

is we can actually rewrite this this

thing this would just simply be equal to

1/8 times the derivative of the square

root of x function but we're already

given that the derivative of the square

root of x is going to be 1 over 2 square

root of x so now all we have to do is

just multiply the 8 and the 2 and we'll

get 1 over 16 times the square root of x

so the moral of the story is when you're

kind of multiplying or dividing by a

constant if you know the derivative of

one of them so in this case we knew the

derivative of square root of x is 1 over

2 root X it turns out that the

derivative of square root of x over 8 we

just take the original one and we just

multiplied it by 1/8 that's all we did

in this case so so that's all there is

to it okay so again we could have sort

of just jumped and I could have said hey

just take this one and multiply it by

1/8 but again just trying to give you a

little bit of an argument as to why it

is so we can just factor the 1/8 out and

again what we're left with would just be

the definition of the derivative of

square root of x but again we know it so

we can plug that in and we can get our

new derivative

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Limit Definition of Derivative, Square Root Example.  In this video I use the limit definition of a derivative to find the derivative of a function involving a square root.


okay in this video I'm going to do some

examples about factoring the difference

of two squares sometimes people will say

it's factoring uh the difference of two

perfect squares um and the formula that

we're going to use it basically says if

you have something squared minus

something squared you can put that in

parentheses by just taking the things

that are being

squared and one set of parentheses make

it a minus and the other set make it a

plus that's all you got to

do um be very careful F if you have a^2

plus B2 uh if the exponents are squares

this doesn't Factor you can't do

anything at all to this so people will

often try to make it do something

somehow Factor it but uh it it's it's

not correct so be very careful about

that if the exponents are both threes it

does factor in that case so just some

you know little nuances that to uh be

aware of but anyway okay so just a

couple basic examples here uh 9 x^2 - 49

and then uh 36 y 4 -

100 all you have to do and we can even

rewrite it first um we want some we want

to rewrite the first term as something

squared well to write nine as something

squared we would need a three and then

to get x2 we just need an X so if you

multiply 3x by 3x we get 9^ 2 and then 4

49 we can write that as 7^ s i I

typically skip this step honestly but uh

just to illustrate so this is our

a this is our B and hey it says we have

a 2 minus b^ 2 well that factors it says

we take a which is 3x and it doesn't

matter the order in which you put the

negative and the positive uh in the

parentheses so we'll take 3x - 7 and 3x

+ 7 and now we have have it factored so

typically what I always do um kind of

the shortcut for me at least the way I

think about it anytime I see something

with two terms so here's one term here's

the other term if there's a a minus in

between them it always factors basically

I just take the square root of the first

term which um I I guess you know maybe

maybe you've seen it square roots a

little bit maybe you haven't worked with

them a lot but the sare < TK of 9 uh x^2

would just be 3x and 3X and the square <

TK of 49 will just be seven and seven

and then I just stick a minus and a plus

in between them so maybe we can do kind

of the same idea uh on this one 36 y 4us

100 so I recognize there's two terms

there's a minus in between so I say hey

that's going to

factor uh I think what number times

itself is 36 well I would need a six and

a six what would I have to multiply by

itself to get y 4th we would need y^2

and

y^2 and then I think well what number

multiply what's the square root of 100

uh it's just going to be 10 and then I

just stick a negative and a positive

inside of there and again now I have it

factored okay so nothing nothing worse

than that and again you can distribute

these out and make sure in fact that

everything cancels out and you do get

this original expression

back

Factoring the Difference of Two Squares - Ex 1

Factoring the Difference of Two Squares - Step-by-Step Guide
In this video, we'll explore how to factor expressions using the difference of two squares formula. We introduce the formula: A^2 − B^2 = (A−B) (A+B), show how to identify the terms A and B, and work through example problems to demonstrate how to apply the formula effectively.

What You Will Learn
The formula for factoring the difference of two squares.
How to rewrite expressions in the form A^2 − B^2

 Step-by-step examples to illustrate the factoring process.

Techniques for recognizing and factoring more complex differences of squares.
Factoring the difference of squares is an essential skill in algebra that can simplify polynomial expressions and solve equations more easily. By the end of this video, you'll understand how to break down and factor these types of expressions using the formula effectively.

If you find this video helpful, please like, comment, and subscribe! Share it with classmates, teachers, or anyone studying algebra and factoring techniques.

Support my work on Patreon: https://www.patreon.com/patrickjmt?ty=c

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all right so here we're going to look at

another example using the root test so

in Part B we've got from 1 to infinity

negative 1 to the n times n to the N

over 3 n to the third plus 1 the first

thing I'm going to do is rewrite this a

little bit we can break up the

denominator instead of 3 raised to the

end of the third plus 1 power we can

write that as 3 raised to n to the third

times 3 to the first power ok again if

we have like bases we would add the

exponents so we would certainly get all

of this stuff right back and the 1/3

you know the over 3 you can think about

that as being a 1/3 which is just a

constant so I'm going to factor that out

and again just the reason for doing that

is it's nice if the powers or just if

the exponents are sort of just to the

nth power or into an exponent it'll be

much easier I think to simplify things

down yeah I don't think it'd be the end

of the world if you didn't do this but

to me it'll be a little cleaner so I'm

just going to look at this part now you

know if this series converges when you

multiply it by 1/3 it converges if it

diverges when you multiply it by

one-third it's going to diverge so we'll

take the limit as n goes to infinity

we'll take the absolute value of

negative 1 to the N in race to the N and

then 3 raised to the N to the 3rd power

again well the nth root we'll take that

to the 1 over N power well when we take

the absolute value the only thing that

would happen is the negative 1 to the N

would be gone so we would have N to the

N over 3 to the N cubed raised to the 1

over N but let's see if we take N to the

N power and raise that to the 1 over N

power again we just multiply exponents

so n times 1 over N is going to give us

an end to the first in the numerator and

the denominator we're going to have

let's see so we would have n cubed

divided by n so that would leave this

with 3 N squared in the denominator

well the derivative of this is going to

be is going to be well I guess we should

say first the limit of this is going to

be infinity over infinity so we can

certainly use l'hopital's rule here so

if we take the derivative of this if we

think about this as being you know just

in terms of derivatives well the

derivative of end of the first would be

one this is like an exponent or an

exponential function 3 raised to some

function well when we take the

derivative it'll repeat itself so we'll

get 3 to the n squared if we get the

natural logarithm of the base but then

we also have to by the chain rule take

the derivative of the exponent so we'll

get 2 to the N well now as n goes to

infinity the denominator is definitely

going to infinity the numerator is just

1 so this limit is going to equal 0 so

now we've justified that our limits 0

well that's certainly less than 1 and

that would tell us that our series

converges so again down in the bottom

all I'm using is the the formula a to

the f of X if we have you know a raise

to something the derivative of that is a

raise to the f of X times the natural

logarithm of a times F prime of X that's

how I'm getting again the derivative of

the bottom just in case you've forgotten

this little formula

The Root Test - Another Example, #2

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Root Test - Another Example, #2.  Just another example showing that a series converges or diverges using the root test.


all right here we're going to look at

some derivatives involving the number pi

so sorry about my little my little

handwriting there did not like my a so

in Part A we're going to find the

derivative of 5 sine of 8 pi X and the

thing to remember is you know pi is just

a number don't do anything different

than you would any other times people

sometimes see PI and they want to kind

of do I don't know different things so

so here we'll leave the five that just

just gets carried along so the

derivative of sine is going to be cosine

but here we're going to have to use the

chain rule so we'll leave the inside

alone and then we have to take the

derivative of the inside well again if

you know if it just said 8 X the

derivative of 8x would just be 8 but

that's not our number the derivative the

number we do have is 8 times pi so the

derivative of that will just be well 8

times pi and now if we multiply 5 times

8 pi will be 40 times pi and then we

have cosine of 8 times pi times X

leftover and that's our derivative so

maybe let's do Part A Part B here as

well so here we've got 4 pi times cosine

of 3 PI x times sine of 3 PI X again

we've got a constant out front

we'll just sort of leave that alone and

multiply it back at the end so let's

leave the 4 pi alone well now I've got

to use both the product rule and the

chain rule so the derivative of cosine

will be negative sine

let's leave the 3 PI X alone then we

have to take the derivative of the

inside which would be 3 times pi and now

we'll leave the sine of 3 PI X we'll

leave that part alone then we'll put the

plus sign in between there and now we'll

kind of do the same thing so now we can

leave the cosine of 3 PI X alone well if

we do the derivative of sine 3 PI X

again the

we're what use the chain rule so we'll

get cosine of 3 PI X and then the

derivative of the inside will just be 3

times pi all right so I think we've got

it all in there so let's see in terms of

simplification now we've got 4 pi okay

we've got 3 pi in here I'm going to pull

that out front

we've even got a negative so let me make

that negative 3 PI then we have sine of

3 PI X times sine of 3 PI X I'm going to

write that as sine squared 3 PI X then

on the second our second term again we

can pull the 3 pi we have cosine of 3 PI

x times cosine of 3 PI X we can write

that as cosine squared of 3 PI X and I

actually think we could factor out the 3

pi and maybe use an identity on this as

well but for now let's just leave it

like that we'll keep that as our

derivative so again just remember pi is

a number and that's all I'm doing is

just treating it like an another number

so nothing different still product rule

chain rule type problems

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! More Derivatives Involving Trigonometric Functions, Ex 1.  Just some more derivative problems involving trigonometric functions.


all right in this video we're just going

to do a little introduction to the

trigonometric functions and uh basically

the trigonometric functions are defined

in terms of a right triangle and the

idea is if you have some little angle

here Theta uh this is the little symbol

Theta kind of across from that that's

what we're going to consider that to be

the opposite side and that's going to

have some length associated with it this

will be the adjacent side at the bottom

and the uh the the longest side is the

high hypotenuse of the right triangle

the way that we Define s of theta is

it's just going to be whatever the

opposite length is over the length of

the hypotenuse cosine of theta is the

adjacent length over the hypotenuse

length tangent of theta is the opposite

length over the adjacent length and then

cosecant is basically one over s which

is going to flip it so cosecant of theta

will be the hypotenuse over the opposite

secant will be the hypotenuse over the

adjacent C will be the adjacent over the

opposite so what do we mean here um so

we've got a little triangle we want to

find s cosine and tangent for this angle

D okay so s of

D we've got cosine of angle d uh s s of

angle D cosine of angle D and tangent of

angle D well it says simply to get the

value of sign so again here's our our

angle D kind of in the bottom left Cor

corner it says sign we just simply take

the opposite length over the hypotenuse

length well the opposite length is six

the length of the hypotenuse goes in the

denominator so that's 2 < TK 13 and

that's that's basically it that's your

answer um you can always reduce this uh

6 / 2 is 3 over the < TK of 13 um you

could rationalize this denominator by

multiplying top and bottom by the s < TK

of

13 and if you do that in the numerator

we be we'll be left with 3 < TK 13 and

then in the denominator Ro < tk3 over <

tk3 will just leave us with

13 okay so sometimes people like you to

rationalize the

denominator okay cosine of this angle D

it says to figure out cosine all we do

is just take the adjacent length over

the hypotenuse length well the adjacent

length length is four the hypotenuse

length is 2 < TK 13 again we could

divide uh 4 over 2 is going to be 2 over

the < TK of 13 and then we could do the

same thing as we did before we could

multiply top and bottom by the square

root of 13 and that would give 2 < TK 13

over

13 okay last but not least tangent it

says we use the opposite side which

would be six over the adjacent side

which is four and 6 over 4 just reduces

to three Hales so we've now got our

answers um s of the angle D is 3 < tk3

over 13 cosine of the angle D is 2K3

over 13 and tangent of the angle D is

simply going to be equal to three Hales

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Trigonometric Functions: The Basics! Example 1. In this video, I define all of the trigonometric functions in terms of a right triangle and then do a few specific examples where I find some different trig values.


Alright, so just a couple more examples here on multiplication of integers.

And here, I'm going to do negative three times four times negative five.

And then, five times negative one times eight.

So,  again, I'm just going to multiply these pair-wise.

So, two at a time.

Let's see, I've got a- So if I do negative three times four...

Let's see, I've got a negative times a positive, that's definitely going to give me a negative...

Three times four would be twelve.

So negative three times negative four will be negative twelve, and then we'll just drop down our negative five.

Well, now we have a negative times a negative which is going to be a positive.

And then, twelve times five will be sixty.

So, it says our final solution here will be positive sixty.

Part B: We've got five times negative one.

Well, that's going to give us negative five.

We have a positive times a negative so that will be a negative.

Five and one is five. Again, we'll just drop down our positive eight.

Now we have a negative times a positive, which is a negative.

Five times eight is going to be forty.

So our solution, five times negative one times eight...

When we multiply those together, we'll be left with negative forty.

And, again, you know- It's certainly one of the most common mistakes to make...

Is our signs, you know, positive and negatives.

And, it's easy to do, you know.

But, obviously, it's a big difference, right?

Negative forty is a much diffrent number than positive forty.

So, one thing I always try to be careful about is - you know - it's easy to mix up signs mistakes.

So, a lot of times if I have a - you know -  problem on homework...

If I was messing it up, a lot of times that would be the first place I would check.

Multiplication / Multiplying Integers

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Multiplication / Multiplying Integers.  Here I look at a couple of quick examples dealing with multiplication of integers.  The only things we have be really careful of is the sign we get after multiplying.  Otherwise, it is the same as multiplying all positive numbers!


all right so another example here of

writing

a decimal negative 4.145

we're going to write that as a mixed

number in simplest or

the most reduced form so

this one will be a little bit more

obnoxious just simply because

there are more digits after the decimal

place

nothing worse than that okay same idea

let's get rid of the old

decimal uh sign so it looks like we'd

have to move it three times so i'm going

to multiply by

1000 over 1000.

all righty let's see so that's going to

leave us with negative four

one four five negative four thousand one

hundred forty five in the numerator

over one thousand

well let's see so again lots of

different ways that we could go about

doing this in my other example i reduced

first here i think i'm going to go ahead

and make it into a mixed

number

and the reason why i'm going to do that

well i'll show you

i mean we could immediately i know that

five goes into the top and the bottom i

could start dividing by five again there

could be some factor

larger than five that works but i know

for sure five works and that can help me

make the numbers smaller another way i'm

recognizing to kind of make you know the

fraction sort of smaller

well again i know we have a negative

over a positive so it's a negative

a thousand would go into 4145 exactly

four times without going over

well we would have 145

over a thousand left so as a mixed

number we can write this fraction

as negative four and 145 over a thousand

well again i know that this fraction

reduces 145 over a thousand

but again it's easier to me to reduce

it's easier to me to reduce

than this original thing so again kind

of you know play with it it doesn't

matter how you do it as long as you

simplify eventually you'll

get the same answer so kind of off to

the side i'm just going to reduce 145

over a thousand

again i know that 5 will go into both of

these numbers

so let's see 5 times 2 is 10 again times

20 would be a hundred times 200 will

give us a thousand

in the denominator let's see five times

30 would be 150 but that's kind of one

multiple

too big so instead of 30 we'll need one

smaller which i think would be

29 all right so again we can just cancel

out our fives

so 145 over a thousand we can write that

as 29

over 200 again now you could maybe think

about factors uh let's see

anything in common that will reduce well

29 is a prime number

so again this will be a reduced fraction

so we can write our final mixed number

as negative four

and 29 over

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Writing a Decimal as a Fraction, Example 4


okay so here we're just going to look at

a little review of uh some series

notation and in this case what we're

going to do is we're going to sum up

this finite series from i = 1: 4 of -1

to the I * I sared so again all the

summation notation says you know

wherever the the bottom number starts

everywhere you know there's an i we plug

in that number so we'll have -1 to the 1

* 1^ 2 then we just put our plus sign

and then we just increase to the next

whole number so positive2 so then -1

sared and then we're going to have 2^

squared and then again we increase to

another uh the next whole number so1 to

the 3 then we'll have 3^ SAR and then

another plus sign for again our

summation notation and then the last

thing you do uh you plug in the last

number that's like your stopping point

so we'll have1 to 4th

time 4^ 2 and now this is what we have

to evaluate so 1 s is 1 it looks like we

get negative 1 um we'll have 2^2 which

is four and it'll stay positive then it

looks like we'll have I guess

A9 and then it looks like a positive 16

so let's see um so negative 1 + 4 that's

three uh 9 and positive 16 that's going

to leave us POS 7 so to me it looks like

our finite sum in this case is just

going to be

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Intro to Summation Notation and Infinite Series, Ex 1. Forgot about summation notation? Here is a little refresher example! Notice the ' i ' has NOTHING to do with complex numbers. : )


all right so here a little word problem

to go with our uh integer examples so

suppose it's uh -4° at 600 a.m. at your

home so nice and nice and chilly and

then let's suppose uh it goes from -4

degrees and the temperature increases

20° but then by 6: p.m. it falls by 14°

so kind of makes sense right it's cold

in the morning gets warmer and then it

cools off towards the end of the day we

want to know what's the temperature at 6

p.m. so let's see it was -4° at the

start of the day or at least at 6:00

a.m. um and then well we would have to

add 20 because we're going to increase

the temperature by

20° but then we would have to subtract

away 14° because well the temperature

Falls so what I'm going to do here um

we've got -4 + 20 - 14 I'm just going to

combine my first two numbers here -4 and

positive 20 and again um these are uh

opposite signs equivalently you can

think about this as being positive 20 -

-4 so again remember our uh these are

commutative we can just switch the order

on these and we'll still get the same

thing so I'm going to use that little

property so 20 minus 4 that's the same

thing as

16 then we'll have to still subtract

away the 14 well 16 minus 14 is going to

leave us with two so it says at 6: p.m.

it looks like the temperature is going

to be uh 2° C so a little bit warmer

than it was at 6: a.m. but uh still not

weather you'd probably want to be at the

uh you know if it was two degrees

Celsius you probably wouldn't want to be

at the beach right so okay so again it

increases you know so from -4 degrees if

we add 20° that means you know at 3 p.m.

it's 16° C but then it drops by 14° so

that's going to leave us with 2° C total

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Subtraction / Subtracting Integers - Word Problem.
Here I look at a little word problem dealing with adding and subtracting integers.


all right here we're just going to do a

few more examples related to the power

rule so um just a couple different

derivatives so i think most of the

derivatives we saw in the other examples

were kind of like part a

and in part a we've got x to the 11th

well when we take the derivative

again that number just comes out front

so

we'll get 11 times x and then we

subtract

1 from the exponent well 11 minus 1

would just give us 10 and hey we get 11

times x to the 10th that's our

derivative and that's all we have to do

for part a

this is true you know it doesn't have to

be a nice whole number

so in part b we have g of x equals x

raised to the

2.137 well

same rule still applies so if we take

the derivative

whatever the exponent is that's what

comes out front so we have

2.137 times x and then we subtract 1

from the exponent well 2.137 minus 1

would just give us 1.137

and there's our derivative so same idea

hey you got decimals

to me i say who cares you know you still

do the exact same thing

so for part c we have h of x equals x

raised to the square root of five so

again when i look at this i say well

i've got a variable to a number the

exact same rule

gets used as before so the square root

of 5 will come out front

and then we have x raised to the new

power well you know it's kind of

you know in the others you can take one

away and kind of get a nice new clean

number um you know and in this case well

the new exponent would just be square

root of five minus one and i would leave

it like that you know i can't really you

could always approximate this with a

calculator but then you're doing exactly

that you're approximating so if you want

to keep the exponent exactly like it

should be we can just write it again

just a square root of five minus one and

that's all there is to it so sometimes

people get thrown off by the uh you know

the

not whole number exponents i guess but

again

it's still the same type of function you

still just have a variable to a number

so this power rule still applies just

like before

Calculus : Basic Power Rule Examples

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Calculus : Basic Power Rule Examples.
Disregard the intro about the proof; these are videos that are licensed to a company so that title goes with their chapter!


all right here we're going to start

looking at some examples related to the

chain rule and the chain rule says

basically when you have a composite a

composition of functions F of G of X it

says to take the derivative of a

function like this you take the

derivative of the outside function and

you leave the inside when alone and then

the chain rule part you multiply by the

derivative of the inside part so that's

the chain rule so this is probably

certainly I would say the most used

differentiation rule around it just

seems like every problem you use the

chain rule on it at some point

so not literally every problem but a lot

of problems so let's look at a few

different examples here so we've got f

of X equals 6 times X cubed plus 2x

raised to the 100 power let's take the

derivative of that one first alright so

so it's kind of like before you can kind

of a the 6 when we take the derivative

is just going to come along and then we

can multiply by the derivative at the

end so to me the way I think about it is

I kind of start from the most outside so

to me the thing on the most out side is

this exponent of 100 and I think if I

just had like X to the 100 power I would

get well 100 X to the 99 power well we

don't just have X but we just leave all

of that part alone okay so the 100 will

come out front we'll leave all the stuff

on the inside alone we'll take 1 away

from the exponent and now we have to

take the derivative of the inside part

so again that's kind of with a chain

rule saying here it says whatever sort

of on the most inside at the end we take

the derivative of that stuff so the most

inside part to me is the X cubed plus 2x

so when we take the derivative there the

X to the 3rd will give us 3x squared and

then the derivative of positive 2x will

just be positive 2

and that's it that's now our derivative

so again you can clean this up a little

bit

six times 100 is 600 I'm going to write

the 3x squared plus 2 part next and then

the X to the third plus 2x race to the

99th power will be left over

❖ More Chain Rule Examples ❖

📚 Understanding the Chain Rule: A Basic Example 🧮

In this video, we explore the Chain Rule, one of the fundamental techniques in calculus for finding derivatives of composite functions. This tutorial is designed for beginners and provides a clear, step-by-step approach to using the Chain Rule in a simple example.

🚀 What you’ll learn:

The definition and purpose of the Chain Rule in calculus.
How to identify composite functions and when to apply the Chain Rule.
A detailed walkthrough of a basic example, highlighting each step in the differentiation process.
Tips for simplifying your derivative and understanding the relationship between the functions involved.
Whether you’re new to calculus or brushing up on your skills, this video will help you gain confidence in using the Chain Rule effectively.

Be sure to check out my other videos for additional insights into calculus concepts, including implicit differentiation, product rule, and more!

👍 Don’t forget to like, subscribe, and hit the notification bell for more math tutorials! 🔔✨

Watch now to master the basics of the Chain Rule and enhance your calculus skills!

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all right so just a little word problem

here involving fractions so suppose to

make a one pound of cookies to make a

pound of cookies one of the ingredients

that you need is 3 and 1 12 ounces of

sugar so suppose we want to make uh not

so many cookies we only want to make a

four of a pound of cookies and we want

to know how much sugar should we use so

we have to kind of scale things down a

little

bit well the idea is what we're going to

need in terms of

sugar you know so for one pound of

cookies we needed 3 and2 oz well now

we're only going to need

1/4 we're only going to need 1/4 of the

original amount of sugar so we're going

to need a fourth of the 3 and 1 half

ounces okay so we need uh a quarter of

the three and A2 so anytime you see the

word of in a problem that almost always

uh stands for multiplication and that's

what we'll have to do we're going to

have to take 1/4 of 3 and a half ounces

okay so one quarter so again we've seen

how to make uh mixed numbers into

fractions so 3 * 2 is 6 plus one is

7 over two we leave the same denominator

and now we just multiply there's no uh

no common factors to get rid of well 1 *

7 is 7 uh 4 * 2 is 8 so let's see uh

before we needed uh 3 and2 o so now

we'll need

78 um of an

ounce okay so we'll need a little bit

less than 1 ounce of sugar uh to make

the this this quarter pound of cookies

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Multiplying Fractions and Mixed Numbers, Example 4


alright so here another example we're

going to find an angle given the value

of a trig function and in this case

suppose we know the tangent of theta

equals 1 and if fate is again an acute

angle in a right triangle we want to

figure out what the angle theta is equal

to so again I'm going to put theta right

down there so remember tangent of theta

that's the opposite the the the ratio of

the length of the opposite side to the

ratio of the length of the adjacent side

so the opposite side here you know you

can think about since we know tangent

theta equals 1 we can just write that as

1 over 1 so I'm simply going to make the

opposite side have length 1 and the

adjacent side also have length 1 but at

this point we really actually have

enough information already to figure out

theta okay so the idea is if you know if

this is let's maybe think about degrees

if this is theta degrees so opposite the

side theta degrees it has length 1 well

that means since this side has length 1

that means this angle up here would also

have to be equal to theta well if we

take theta plus theta well we know that

this other angle by default that has to

equal 90 degrees and we know if we add

up all the angles inside of a triangle

that equals 180 degrees well one theta

plus 1 theta is going to give us 2 theta

we can also just simply subtract 90

degrees from both sides so if we

subtract 90 degrees we'll get 2 theta

equals 90 degrees and now if we simply

divide both sides by 2

we'll get that theta equals 45 degree

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding an Angle Given the Value of a Trigonometric Function - Example 2.  In this video, I deduce the angle associated with a given trigonometric value.


okay in this video I'm going to look at

finding the average value of a function

on an interval and as far as stuff that

you see dealing with integration and

calculus course I would say this is

probably one of the more straightforward

things to do here so the formula to find

the average value of a function f ofx on

the interval A to B what we do is we

just take 1 over B minus a so one over

the length of the interval and then we

multiply that by the integral from A to

B of f ofx DX so typically when you see

these problems at least when I saw them

there really wasn't a lot more to doing

these problems and just using that

formula so just a quick example here so

we're going to find the average value of

the function f ofx = x^2 + x + 1 on the

interval from 1 to

3 so a lot of times they just write it

abbreviated as FV so you know obviously

the average value of F and what we do is

we just take one over well and this

case a our a value is going to be one

our B value is simply going to be three

so we'll take one over the difference 3

minus 1 and then we integrate from 1 to

3 of our function so we've got x^2 + x +

1 DX and that's now all we have to

calculate here

so well 1 over 3 - 1 that's just 12

let's integrate the function x^2 + x + 1

so the anti-derivative of x^2 well we'll

get x to the 3 over 3 the

anti-derivative of X is simply x^2 over

2 and then the anti-derivative of one

will simply be X and again we have to

evaluate this from 1 to

3 okay so now it's just sort of plugging

and chugging so if we substitute 3n

we'll have 3 3r / 3 which is 3^ 2 or 9

plus again if we substitute 3 in we'll

get 3^ 2 or 9 /

2 then we'll add to that three so

there's our upper limit of integration

minus let's just substitute in the lower

limit so if we plug in ones we'll just

have 1/3 + 12 +

1 and again now it's just a matter of

simplification so let's see um if we

take 9 + 3 that's 12 and if we subtract

1 12 - 1 is going to leave us with 11 so

that takes care of those terms 9 Hales

minus 12 that's going to give us eight

halves and well eight halves is just the

same thing as four so that'll take care

of those two terms and then we simply

have minus

1/3 okay so let's see 11 + 4 is 15 so we

would have 15 over 1 - 1/3 so if we get

common denominators that's going to be

45 over

3 -

1/3 so that's going to be 12 * well 45/3

- 1/3 that's just uh

44 over 3 and 44 ided 2 is going to

leave us with 22 over 3 and that's our

average value so again nothing too

terrible at all

um you know to me the only thing that

could really make a problem like this at

all too terribly difficult which just be

you know if if the function that you're

calculating the anti-derivative of is

just more messy but other than that it's

just one over uh the length of the

interval calculate your definite

integral and multiply those values

together and that'll give you the

average value

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Average Value of a Function on an Interval Using Calculus.
In this video, I give the formula to find the average value of a function on a closed interval and work out on simple example.


all right here we're going to do one

more limit problem here we're going to

look at the limit as n goes to infinity

of n raised to the negative n to the

third power the first thing I'm going to

do is just rewrite this so n raised to

negative n to the third okay we can

always put that over one well if we can

make negative exponents positive by

sticking that exponent in the

denominator so now we have 1 over N

raised to positive n to the third and

you know if you think about it again

we're letting n go to infinity so

certainly for n values greater than or

equal to one I claim that well in raise

to n to the third power that's always

going to be whoops almost got my

inequality backwards and raised to n to

the third power that's always going to

be greater than or equal to n to the

first I mean imagine if you plug in one

you'll have one to the first on the left

and one to the first on the right if you

plug in two for example though you're

going to have 2 raised to the power of

two cubes so 2 to the 8th is certainly

greater than or equal to 2

again if you plug in 3 you know 3 raised

to the 3 cubed or 3 raised to the 27

power certainly greater than or equal to

3 so I think this is a pretty clear I

don't think many people would argue with

that well that tells us that if we take

1 over this number that's going to be

less than or equal to 1 over N again you

know our fraction now the fraction on

the left has a bigger denominator well

if a fraction has a bigger denominator

in the same numerator that fraction is

going to be smaller

well again since we're using positive

numbers and a some positive number

lifeWise went over n raised to n to the

third that's always going to be greater

than or equal to zero so now we can

bring in our squeeze theorem and we can

simply look at the limit as n goes to

infinity of the left side which is 0

well that limit certainly is 0 likewise

if we look at the limit as n goes to

infinity

the right side well the limit as n goes

to infinity of 1 over N that's also

definitely 0 so what this now tells us

is that by the squeeze theorem so by the

squeeze theorem the limit as n goes to

infinity of 1 over N raised to n to the

third must also equal 0

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Squeeze Theorem and Absolute Value Theorem, #3.  Here we find the limit of another sequence by using the squeeze theorem.


okay in this video I just want to make a

quick comment on some uh notation and uh

talk about a definition of What's called

the gradient vector and I did a couple

examples of directional derivatives and

recall basically the directional

derivative um in the direction so if you

take um to compute the directional

derivative of a function that's

differentiable in two variables um in

the direction of the vector the unit

Vector AB it says we take the partial of

X multiply it by a the partial of Y

multiply partial with respect to Y

multiply that by B um what I want to

point out is well we can actually simply

write this as the product of two vectors

I can take the partial of

X um the partial of Y as my

components and then I'm just going to

dot this by the vector

AB okay so no big deal but we actually

give this first Vector a name and it

ends up uh being important and it's

What's called the

gradient okay so we could call this the

gradient of the function f and this is

often times abbreviated as a little

upside down triangle with a with the F

attached to it so that's it this is

what's called the gradient so um it says

the gradient of

f again it's just simply the partial of

um with respect to X the partial with

respect to Y and again we can simply um

even rewrite this as the partial of f

with respect to

X um I plus the partial of f with

respect to y j um and that's the

definition of our gradient so I just

want to point this out because I'm going

to be using this notation um talking

about the gradient in some other videos

okay so all right nothing nothing deep

but I just want to point that out before

I moved along so again if you have any

questions about the notation or anything

feel free to post comments

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Gradient Vector - Notation and Definition. In this video, I talk about the directional derivative, and the corresponding definition of the gradient vector.  Nothing deep here!! Just a small observation and some notation!
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all right in this video I'm going to do

another example solving an absolute

value inequality it's going to be at the

type where we have a quantity greater

than or equal to some number where that

number is greater than or equal to zero

and what we're going to do again is

we're basically just going to set up two

inequalities that we have to solve

so the first inequality basically just

says remove the absolute value so 8

minus 2 D over 9 greater than or equal

to 5 or it's going to turn out the other

solution will be when the quantity

inside 8 minus 2 D over 9 is less than

or equal to the negative of that value

and that's going to give us our two

inequalities now that we have to solve

so the first thing I'm going to do is

multiply by 9 on each side of the first

inequality that will cancel out the

nines on the left and just leave me with

8 minus 2 D greater than or equal to 45

I can subtract 8 from both sides so I'll

have negative 2 D greater than or equal

to 45 minus 8 I believe that's going to

give us a positive 37 and now we can

divide both sides by negative 2 again

remembering that we have to flip the end

of quality now because we've divided by

a negative number so we'll have 37

basically negative 37 over 2 we'll do

the same thing on the other inequality

just multiply both sides by 9 that again

will cancel out the nines on the Left we

have 8 minus 2 D less than or equal to

negative 45 and again if we just

subtract 8 and subtract 8 I'll have

negative 2 D less than or equal to in

this case I'm going to get negative 50 3

so negative 45 minus 8 and then again

I'm simply going to divide by negative 2

by negative two and it says D has to be

greater than or equal to 53 over two so

it says as long as D falls in one of

these two intervals of numbers that will

give us our solution

so again graphically just just maybe see

it on a number line so here's negative

37 over two it can be less than or equal

to that or 53 over 2 that's positive it

can be greater than or equal to that so

in terms of writing this using interval

notation we can say our solution is the

interval negative infinity up to

negative 37 over two parentheses again

excuse me brackets brackets we've used

brackets here because we can include it

or again brackets 53 over two to

positive infinity again we always put

parentheses around the infinities but

again certainly brackets here and now

we've got our solution to our very

original absolute value inequality

Solving Absolute Value Inequalities - Example 3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! math algebra absolute value inequality solve example


all right so in this example I just want

to do uh another another problem of

writing some compound inequalities and

again all compound inequality typically

means as you have multiple inequalities

we just want to write it in interval

notation so again when I think about or

um I'm going to use everything that I

shade so okay so it says X has to be

less than five so I'm going to put an

open circle there because we don't use

it but everything less than five would

be well everything to the left all the

way out to

Infinity the other inequality says X has

to be greater than 5 but again not

including five so that would give me

everything to

the to the right of five with uh again

all the way out to Infinity so if I want

to describe every you know everything I

have shaded using interval

notation it says we'll go from negative

Infinity up to but not including

five and we can use use this interval

sometimes people will write the word or

sometimes you'll see people write this

symbol this Union so it says we can go

from negative Infinity to five that

takes care of everything on the left or

it says we can kind of skip over five so

again I'm going to put parentheses to

indicate that we don't use it all the

way up to

Infinity so when we write this interval

notation it's really just a fancy way of

saying use every every single number

works except for the number five live

Writing Compound Inequalities Using Interval Notation - Example 2

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okay here we're going to do a limit

problem involving the absolute value

theorem so here we're going to find the

limit of the sequence x sub n = -1 the n

+ n^2 over n^2 and again intuitively to

me what's going to happen you know if

you plug in larger and larger values for

n let's forget about the first part for

a moment as we plug in larger and larger

values for n you know we're going to get

some big number squared over some big

number squared it's the same number so

this ratio you know if it wasn't for the

negative 1 to the end we would just get

one well what's going to happen um you

know the negative one to the N all

that's either going to do to the

numerator is either subtract one or add

one so again to me for large values of n

the the numerator and the denominator

are almost going to be exactly the same

value which makes me think this limit

should equal zero to justify this again

we're going to use the absolute value

theorem

so remember that says if the

limit of the absolute value of a subn

equals 0 or in this case X ofn if that

happens that tells us that the sequence

uh without the absolute value also

equals zero so we're going to use this

little result and the way that we're

going to do this is well we're going to

take the limit as in goes to Infinity of

-1 n + n^2 all over n^ 2 and all we're

going to do is just break this up using

some algebra so we can write this as the

limit as goes the limit as n goes to

Infinity of 1 n over n^

2 plus n^ 2 over n

2 so we'll have the limit as n goes to

Infinity again of -1 to the N over n^2

well n^2 over N2 that's just going to

give us

one

so again we can kind of look at these

individually and then the limit as n

goes to Infinity of just one well the

limit as n goes to an Infinity of one is

just one the limit of a constant is the

constant so the only thing we really

have to think about now is What's

Happening Here well again I think it's

pretty easy to argue the top is always

going to bounce back and forth between

negative one and positive one the

denominator is going to get arbitrarily

large so to me this this first limit

should certainly go to zero but again to

bring in this absolute value

theorem notice

um so notice that the limit as n goes to

Infinity if we take the absolute value

of1 to the N over n^2 so I'm just

looking at the first part well that's

just going to equal the limit as in goes

to Infinity if we take the absolute

value of the numerator uh again you know

without the absolute value we're either

getting positive 1 or negative 1 uh when

we raise it when we take the absolute

value of it it'll just become postive

one over n^ 2 but again uh this limit as

n goes to Infinity the denominator will

get arbitrarily large one over a big

number is going to be zero so hey in

fact by the absolute value theorem this

first limit is actually going to be zero

so really we're left with 0 + one so

again in this case our limit will simp

converge to one

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Squeeze / Absolute Value Theorem for Sequence : Ex 2
Here I look at an example where I use the squeeze and or absolute value theorem to find the limit of a sequence.