okay in this video I want to talk about

integration using inverse trig functions

and this will be the first example so

kind of the two most useful formulas are

here at the bottom so recall that the

derivative of arcsin is one over the

square root of 1 minus x squared well

that means if we integrate that function

1 over the square root of 1 minus x

squared will just get arc sine plus C

back and the same thing for arctangent

okay so in our first problem here let's

integrate 1 over the square root of 1

minus 4t squared so typically when I see

something like this you know I'm just

thinking well lots of things really but

if I you know if I know it's an inverse

trig function well I'm just kind of

thinking about which one it looks like

the most well certainly of the two that

we just had listed it looks the most

like the integral of 1 over the square

root of 1 minus x squared which we said

was arc sine of X ok so basically what I

have to do is I have to make my integral

you know look like this formula and the

way to do it is usually just going to be

usually just a little bit of algebra and

I mean basically it boils down to a u

substitution is really you'll see the

the gist of a lot of the problems so

really it says you want one - some

something squared well I do have that in

a sense I have 4t squared but I'm going

to rewrite 4t squared as some quantity

being squared well I guess I could

simply rewrite that as 2t squared right

square the 2 you'll get the four square

the T certainly you get the T squared so

what we're going to do at this point is

this is where the U substitution comes

into it so we'll simply let u equal to T

what's inside the parentheses so D u

would be 2 DT divide both sides by 2

we'll get 1/2 D u equals DT so now if we

just simply rewrite everything the DT

again oh or place that with the

one half D you and now underneath the

square root

we simply have one minus u squared and

now we've made our integral through u

substitution look just like the formula

so it says we'll simply get one-half arc

sine of U plus C but again since we

started with T's in this problem let's

finish out with T's so we'll get one

half arc sine of 2t plus C and that's

your answer

so that's all there is to it in this

case it's just basically knowing a

formula and just doing a little bit of

algebra you know I guess you could

certainly skip this algebra step but to

me it makes it clear you know what

you're substituting in so that's why I

like to put that step in there but all

right I'm going to do some other ones

you know a couple other ones definitely

at least one with a definite integral so

the follow up example too should be

posted as a video response at the bottom

if you scroll down just a little bit and

you can check that one out if you're

interested

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Integration Using Inverse Trigonometric Functions - Ex 1.  In this video, I give two formulas and show how to solve a problem with a bit of algebra and a u-substitution.  Example 2 (the other video) is a definite integral.
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all right so one more example here

factoring some trig Expressions so here

we've got two 2 cosine ^ 2 X plus 2

coinx minus

24 okay the first thing I always do in

factoring is look for you know the

greatest common factor well two will go

into all of those so we can certainly

Factor the two out and then we would

have cosine 2 X plus cosine x minus well

I guess 2 * -12 will give us our 24 back

but I think we can factor a little

further um at this point and to me what

I see at this point is um to me I see

something that looks very much quadratic

okay so when we think about the cosine 2

X plus cosine x - 12 that to me really

says hey you've got something squared

plus that same thing minus 12 and I

would think you know how does x^2 + x -

12 Factor well that's going to

factor I think rather nicely so we need

two numbers that multiply to -2 but add

up to pos1 I think we could use

pos4 and -3 well that's how x^2 + x -2

would Factor well we don't have x^2 + x

- 12 we have cosine 2 X plus cosine X so

all I'm going to do is just replace my

x's with cosine

X's so the same way to factor

um to factor this expression

instead of just X we'll have cosine x +

4 and then we'll have cosine

x uh minus 3 and notice again if you

distribute all this stuff out um if you

do distribute out we'll get cosine x *

cosine X which will be cosine s we'll

get -3 cosine x + 4 cosine X which would

give us our positive cosine X pos4 and

-3 will leave us with our -12

so to me this looks like a nice

factorization for this original

expression so I don't see much else we

can do with it so um that to me looks

like a good place to stop

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


all right in this video we're going to

do an example of expressing a complex

number in trigonometric or Polar form

and the idea is okay we've got some

complex number Z we can write that as uh

X+ Yi if the real part is X and the

imaginary part is y so this uh this this

complex number uh we can make an angle

with the Positive x-axis and that'll

simply be our angle Theta if you think

about the distance out to that point the

distance to the point is the modulus but

we're going to call that R you can kind

of think about that as being a radius in

a sense well uh we could come up with

some relationships we could do cosine so

cosine would be cosine of theta would be

adjacent X over H the hypotenuse which

is R if you solve that for X we'll get x

= r cosine Theta the same way we could

solve for y and get yal R sin

Theta if you replace the X with r cosine

Theta and the Y with r sin Theta you can

Factor the r out and it basically says

we can rewrite our complex number Z as R

* the quantity cosine theta plus I sin

Theta so um let's try to make some sense

out of one of

these suppose we want to uh we're going

to convert this complex number 2 + 2i

into Polar form okay so the first thing

I'm going to do is just make a little

sketch here so over two up

two okay so there's

our there's our complex number Z = 2 + 2

I so now to go back to our formulas um

again the first thing I'm going to

figure out is R which is our modulus so

R is just going to be the

modulus and to get the modulus we just

take uh each part the real part and the

imaginary part and we Square them add

them together in square root so we'll

get 4 + 4 which is8 so we'll get the

square root of 8 uh we can rewrite that

though as the < TK of 4 * thek two which

will give us 2 < tk2 so now we've got

our r

value well we don't know our angle Theta

here just yet uh but we can figure that

out just using again a little bit of

trigonometry so the ad Adent side has

length two um the opposite side also has

a length of two if we do tangent OFA we

would get opposite over adjacent so 2

over two which is going to be one well

tangent of Pi

over4 or

45° that'll give us one so now we know

our angle

Theta and now we're in business we can

rewrite our complex number it says we

write r R so the R value we said was 2 <

tk2 um and then we simply take cosine of

theta which is cosine of

45° plus I * s

of 45 degrees and now uh this is what

we've got what we want this would be our

final solution

here okay we've now uh taken our complex

number and put it in polar form

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Expressing a Complex Number in Trigonometric or Polar Form, Ex 1.  In this video, I show how to write a complex number in polar form. Example 1.


okay in this example we have two

rectangles that are similar and we want

to find the value of x so our small

rectangle has a height of six and a

width of 10 and our larger rectangle has

a height of 21 and a width of X and the

way I'm going to do this is I'm going to

set up a proportion so what I'm going to

do is I'm going to set up my proportion

by looking at the small

length over the longer

length

and I'm going to make that equal to the

same thing the small

length over the long length and I'm

going to do this for each rectangle so

okay so for my smaller rectangle well

the small side small length is six the

long one is 10 so there's my

fraction my proportion and I'm going to

do the same thing with the other side so

the smaller length L is 21 and the

longer one is X and that's what we don't

know so we want to determine that value

X okay so now we've got an algebra

equation to work with and what I'm going

to do is I'm going to do cross

multiplication so what cross

multiplication

says it says if you have one fraction

we'll call it A over B equals another

fraction C over D it says you can turn

this into equivalent uh equation by

cross multiplying we'll take a *

D and then on the other side we'll do B

* C so that's why they call it cross

right it looks like a little cross and

you're multiplying so I'm going to do

the same thing I'm going to take six and

multiply that by X well that's just

going to give me 6X and then I'm going

to have 10 ultip

21 now you could go ahead and multiply

10 * 21 and get 210 I'm not going to do

that and the reason is is um it's to me

it's going to be easier just to go ahead

and simplify without multiplying so I'm

trying to solve for x since I'm

multiplying by six I'm going to start

off by dividing both sides by six well 6

/ 6 is just 1 x and now let's see we can

we can uh cancel out some things so I'm

going to write uh so first off I'm going

to write 6 as 2 * 3

notice we could write 10 as 2 * 5 that's

still 10 and 21 is 3 * 7 so what I'm

trying to do is cancel out common

factors and so notice we could cancel

out a two and a two we could cancel out

a three and a three now be careful it

looks like we've crossed everything out

in the denominator notice if we multiply

by one that's not going to change the

value right 2 * 3 * 1 is still the same

thing thing is 2 * 3 so there is still

an imaginary one there if you will so

what we're left with is 5 * 7 which is

35 over 1 well 35 over 1 is just 35 and

that's going to be our value for x and

again that seems uh you know intuitively

that seems kind of reasonable to me um

you know if it has a height of six and a

width of 10 well the width isn't quite

double right it isn't quite double if

you double six you would get 12 so if

you double 21 you would get I guess 42

so to me intuitively this length should

be something a little bit smaller than

42 and hey that's what we're finding so

just a quick little intuitive check as

well

GED 'Most Missed' Math Practice Problems , Code Q.3.b,  #3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The GED Testing Service keeps a list of the most missed math topics and can be seen here: http://www.gedtestingservice.com/uploads/files/126de0283c94ff6323b3b5b3f310da5d.pdf
This video shows how to work a sample problem using their coding system that is shown on the left using the link above.


all right in this video we're going to

do a few examples related to finding 2x

two determinants of a matrix so the idea

is if your Matrix a if you have the

entries a b c d to calculate the

determinant of that Matrix we simply

sort of do like a little cross

multiplication we take a * D and then

subtract away the product of B * C if

this determinant equals zero we say that

Matrix a is a singular Matrix so just

some

terminology um so a couple questions

here true or suppose we want to know if

this Matrix -2 -10 4 -20 um is that a

singular Matrix or not well again all we

have to do is just simply calculate the

determinant so the determinant of a here

will take -2 * -20 we subtract away the

product of 4 *

-10 well -2 * -20 is pos4 it looks like

we have a -4 and a -10 which will be

another positive 40 so since we're

getting 80 out for the determinant and

80 well that's certainly not equal to

zero this Matrix is not singular so by

saying it's singular we would say that's

a false statement okay definitely not

singular suppose we want to calculate a

determinant here generically we've got a

= 4X -3 and Y well again to calculate

the determinant of this we just sort of

cross multiply here here so I'm going to

get 4 *

Yus I'm going to take -3 *

X well we can simplify that and say it's

4 y uh +

3x one last Quick one here suppose we

know that the value of the determinant

of this Matrix is 24 um we've got -2 6 W

-2 6 3 and W if we know this determinant

of this Matrix is 24 we want to figure

out the value for w

well again to calculate the determinant

we'll take

-2 * W and then we'll subtract away the

product of 3 * 6 and again we know that

this determinant has to equal 24 so now

we just got a nice little equation we

can solve this is -2 W - 18 = 24 well we

can add 18 to both sides if we add 18

we'll get postive 42 on the right if we

divide both sides by -2 will be left

with W = -21 so we would basically have

to have a -21 hanging out in the bottom

right corner of our Matrix so that this

determinant equals 24

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Determinant of a 2 x 2 Matrix - A Few Basic Questions. In this video, I show how to find the determinant of a 2 x 2 matrix, and do a few related problems.
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okay in this video we'll do one more

example of multiplying and dividing

complex numbers that are given to us in

polar form and actually not quite

because the bottom one isn't going to be

given to us in polar form uh well the

numerator of the bottom problem so we've

got 3 + 3 I over cosine of 90 Dees plus

I sin of 90° so recall uh we can write a

number in polar form we can write zal R

cosine of theta plus I sin Theta um to

get R remember we just take the real

part and square it and add to that the

IM imaginary part squared and then we

square root that so for 3 + 3 I went

ahead and uh graphed that real quick so

there's a our 3 + 3

I well to get our r value we would

simply take 3^

2ar + 3 2ar well that's going to be 9 +

9 or 18 we can always write 18 as 9 * 2

we can pull the square root of 9 out is

a three so we would be left with 3 < tk2

for our value for R to figure out our

value for Theta

here I just think about making a little

right triangle well the adjacent side is

length three the height is also uh

length three the opposite side so it

says tangent of theta would is be the

opposite over the adjacent or 3 over one

Excuse Me 3 over three which gives us

one and tangent of 45 degrees is going

to equal one so let's see um I think we

can rewrite the numerator now it says

this would be the same thing as our r

value again which we said is 3 < tk2 so

we' have 3 < tk2

time cosine of the angle which we said

is

45° so cosine of

45° plus

I uh times s of 45°

okay that's all over uh we can even

write this as 1 time cosine of

90° plus I of

90° but now we've seen uh from the other

examples all we do again uh when we're

dividing we simply divide the R1 by the

R2 and then we subtract the

angles so in this case if we take 3 <

tk2 over 1 we'll just be left with 3 *

the root of two and then we would just

have cosine of

45° -

90° + I * s of

45° -

90° and uh at that point we've basically

got our solution so we can always clean

it up a little bit more 3 < tk2 this

will be cosine of

-45° plus I s of

-45° and I think I would stop leave it

there need one more set of parentheses

now I'll leave it there um and now we've

got uh the polar form for our our new

number after we perform the division

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Complex Numbers: Multiplying and Dividing in Polar Form, Ex 2. In this video, I give the formula for multiplication and division of two complex numbers that are in polar form.


all right so just another example of

estimating the sum of an alternating

series so here suppose we uh estimate

the the series Nega 1 the n + 1 over 3

to the N by using uh this this expansion

so notice this is n = 1 2 3 4 five so in

this case the air is at most uh the size

of the absolute value of the first

neglected term well again if we filled

in the next term um you know so any = 1

2 3 4 5 if we plug in Nal 6 into our

into our formula we would get well let's

see negative - 1 to the 7th that's just

Nega 1 then we would get 3 to the 6th so

3 raised to the 6 power uh

729 and let's see if we take one and

divide that by

729 I'm getting

37 1 74 when I throw that into a

calculator so it says our error is at

most the value 0.001

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Alternating Series - Error Estimation #2.  In this video, I sum up the first few terms of an alternating series and then I find the maximum error involved.


all right in this video I'm going to do

another example of finding a polom that

satisfies certain conditions so in this

case we want a polinomial f ofx of

degree

3 okay it's got zeros at 2 and -2 so

that means I know that xus 2 is going to

be a factor and x + 2 is also going to

be a factor the leading coefficient is

pos1 we have this extra bit of

information that if you plug -4 in we

get 30 out right now if we multiply this

out this would only be degree 2 so that

means there's another factor in there um

let's call it xus C so this the C value

that's the number if I knew that I would

basically have my polom so I have to

work to figure out this value C and

again I'm just going to

use um this condition it says if you

plug -4 in we get 30 out so I'm going to

put 30 on the left side

and then everywhere there's an X I'm

just going to plug in

-4 and then I'm going to do a bit of

arithmetic again to solve for my value C

and then I'll I'll know my

polom so let's see I'm going to get rid

of the one I mean one times anything

doesn't change it so -4 and -2 is -6 -4

+ 2 is going to give us a -2 we still

have our -4 - C left over so let's see

this is 30 -6 and -2 is pos2 we still

have4 minus c um a couple different ways

we could go about doing it here I'm

going to just divide both sides by 12

you could distribute out first it'll all

get you to the same place so if we

divide by 12 um well the TW will cancel

on the right side 30 over 12 what's the

biggest number that goes into both of

those um I think think six will go into

both of those so 30 / 6 would be 5 30

div excuse me 12 / 6 would be 2 on the

right side we would simply be left with

this value -4 minus

C well okay I'm going to add four to

both sides so that'll give me negative

excuse me 5 Hales + 4 over 1 = C we can

multiply top and bottom by two so

that'll give us 8 + 5 which is 13 / 2

and that's going to give us C well if we

multiply both sides by 1 it says -3/2 is

going to be our C value and now we know

our polinomial I can basically just go

back up to my f ofx and fill it in so it

says the polom would be x - 2 * x + 2 x

- -3 would give us x + 13 and now we've

got our polom uh that satisfies the

given conditions

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding the Formula for a Polynomial Given:  Zeros/Roots, Degree, and One Point - Example 2.  If you know the roots of a polynomial, its degree and one point that the polynomial goes through, you can sometimes find the equation of the polynomial.  In this video, I show how!


okay in this video we'll do one more

example of solving a linear equation so

again the first thing I want to do is I

tend to work sort of inside out so I'm

going to try to clean up the stuff

inside the brackets um basically try to

get rid of all the parentheses and all

the brackets so our negative is out

front our six will leave alone you could

think about this as being negative 1 so

if we distribute 1 * positive 7 is -7 -1

* POS X ISX

on the right side if we distribute our

positive 3 3 * 2x is going to give us 6X

3 * -5 will give us

-5 um I'm going to keep simplifying

inside the brackets so 6 - 7 that's ne1

and then we still have our negx and

again brackets are the same thing as

parentheses it's just sort of a you know

a way to help us visually break it up um

if you have 80 sets of parentheses it

can be confusing to look at 10 - 15 is

going to give us -5 so I'm combining my

like terms on the right side we still

have our positive 6X left over if I

distribute again you can think about

this as being Nega - 1 so -1 * 1 is pos1

-1 * X is postive X we've got --5 + 6x

still on the right side and again I'm

just going to try to put um all my

numbers on one side all my variables on

the other

I'm going to start by subtracting

X from both sides that'll leave us with

positive one on the left we've got -5

plus okay so 6x - x is going to be

5x and last but not least to move the

neg five over I'll have to add five to

both sides and 1 + 5 is going to give us

six we still just have our 5x left over

on the right side and since we're

multiplying by five to get rid of that

we'll divide both sides by five and then

we'll get X = 65s as our solution to the

original equation

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


Okay, here we have the question

it says, "If a divided by b equals two,

what is the value of four times b divided by a?"

So we have the answer choices  zero, one, two, and four.

Now, we can certainly do some algebraic manipulation,

and, you know, solve for a or b, and then substitute it in here, and do stuff like that.

Don't do that. I definitely wouldn't do that on this question.

I would just pick two that-I would just pick a value of a and b, so that when I divide I get two,

and then I would plug them in here.

So, notice,

You know, if, well a equals two, and b equals one.

Right, we would have a divided by b, that would be two over one...

that equals two.

Okay, so these values of a and b satisfy this original, you know, requirement.

Just substitute those values now into four b over a.

So, four times b over a, that would equal four times,

Again, we said b is equal to one,

a is equal to two,

that would be four over two,

which would give you two, and boom, there's your answer.

So again super quick,

nothing tricky in that one,

Again, you know, you can certainly do some algebraic manipulation to come up with this solution.

But again, to me that would be overkill,

It would be so much faster just to pick some values that work, and you've done this, you know,

in probably less than ten seconds, and you're off to the next one.

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! SAT Math - Question 8, 2016 Edition and Beyond


all right in this video we're going to

start looking at some examples related

to the integral test so in part A here U

we've got the series from Nal 1 to

Infinity of 1 over 3 to the n and again

we're just going to figure out whether

these series converge or diverge so um

certainly different ways to do this one

other than the integral test but uh

again just so we can talk about it using

the integral test so if you think about

the function 1 over 3 X that's the same

thing as well 13 all raised to the X and

recall an exponential function

um y = 1 3 to X would look uh you know

like our uh the graph I have here so

certainly we can use the integral test

on this because it's continuous it's

positive and it is decreasing so this is

certainly a a problem that we can use

the integral test

on so we've got the series from 1 to

Infinity of 1 over 3 to the N uh

correspondingly I'm going to think about

the integral from 1 to Infinity of 1

over 3 to X DX and then I'm going to uh

decide whether this improper integral

converges or diverges so I'm going to

replace my Infinity with a t we'll do

the limit as T goes to Infinity um the 3

to the X I'm going to bring that

upstairs and make it 3 to

thex so when when we integrate this

again you could just do a u substitution

let U equal

x after we do our U substitution we'll

get 3 to

thex um we'll have to divide by the

coefficient we'll end up getting a

negative one and recall for exponentials

you also have to divide by the natural

logarithm of the

base so from one to

T so here we've got the limit as T goes

to Infinity um so let's see I'm just

going to pull the negative out front so

there's my minus sign um let's see we

would have one over so we would have 3

to the X but that would turn into a t

times the natural logarithm of three

minus we would have one over so we would

have 3 to the 1 when we put it uh back

in the denominator 3 thex again times Ln

of

3 the only thing I'm really interested

in is whether this is a convergent or

Divergent improper

integral well you know 1 over 3 * Ln of

3 this is definitely some finite number

so no problem there uh the only thing

I'm thinking about is what happens with

the limit well as T goes to Infinity 3

to the T would get very very very large

so then we would have one over a very

very big number well one over a big

number goes to zero so we've got 0 minus

some finite number this uh improper

integral is definitely going to equal

some finite number and that tells us

it's

convergent and then so also since the

improper integral is convergent we can

also immediately conclude that uh the

series from 1 to Infinity of one 1 over

3 to the N is also

convergent again you know probably the

easiest way to adjust this one just to

point it out real quick in case uh maybe

you didn't think about it again this is

just a geometric series as well 1 over3

raised to the N so certainly you can

recognize it as being a geometric series

um which is what I would do and say it

converges because the r values between

negative 1 and one um and then I would

be off to the next problem but again

just to talk about it just to illustrate

the use of the uh the integral test here

Integral Test to Evaluate Series, Ex 1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Integral Test to Evaluate Series, Ex 1.  In this video, I show how to evaluate a series by using the integral test.


all right in this video I want to talk

about satisfying the domain of a

function or equivalently we're going to

find the domain typically anytime I see

a domain question there's two big things

I think I have to avoid these two big

restrictions if there's any fractions

floating around we have to omit any

values of X that give us zero in the

denominator so what to solve that

equation and if there's any square roots

or really any even powered root we don't

want any negative numbers to appear

underneath that radical after we kind of

simplifying do all the arithmetic so two

questions let's find the domain of these

two one over x squared minus 7x minus 30

and then B we've got the square root of

2x plus 3 so to me kind of the idea the

values that are bad or the values that I

must omit for this first equation I

don't want any value of x that makes the

denominator 0 so I'm basically thinking

if I said you know what makes the

denominator x squared minus 7x minus 30

I'm thinking what makes that equal zero

well we just hopefully factor this so

let's see I need to know I need an X and

an X to get my x squared I need two

numbers that multiply to negative 30 so

I know I'm going to need a negative and

a positive those two numbers when I

multiply have to get negative 30 but I

have to add up to negative 7 so I think

if we use negative 10

and positive three those two numbers

will fit the bill and again I want this

to equal zero so if we set X minus ten

equals zero we'll get x equals ten and

we'll get x equals negative 3 is

solutions so our domain is going to be

all values of x all values of x except

for x equals 10 and x equals negative 3

we could write that in interval notation

by simply saying it's from negative

infinity up to negative 3 then we have

to skip over negative 3 and go up to 10

and then we have to skip over 10 and go

all the way to positive infinity so sort

of two different equivalent ways the

other one here we have the square root

of 2x plus 3 so again to me the values

that are good in this case is the way I

think about it it's good if whatever

underneath the square root is greater

than or equal to zero that's what we

need to happen we only want numbers that

are greater than or equal to zero

whatever is under the square root well

2x plus 3 that's what's underneath the

square root so this now produces an

inequality I have to solve unfortunately

this is just a linear inequality so it's

not too bad I can subtract 3 from both

sides and get 2x greater than or equal

to negative 3/2 and it says X needs to

be greater than or equal to negative 3

halves so again this is my domain in

this case and an interval notation we

can write that as brackets negative 3

halves up to infinity and that would be

the domain for our our radical or square

root function

❖ Finding the Domain of a Function Algebraically (No graph!) ❖

Finding the Domain of a Function Algebraically - Step-by-Step Examples!

Description: In this video, I walk through finding the domain of a function algebraically—without using any graphs! We'll cover two key examples.

In the first example, we analyze a rational function where we need to exclude values of 𝑥 that result in a zero denominator. This helps us determine the valid domain of the function.

In the second example, we solve a linear inequality to ensure that we don’t end up with negative numbers under a square root.

Each of these examples is fundamental and widely applicable, appearing throughout algebra, precalculus, calculus, and more. Whether you're learning about domains for the first time or need a refresher, these examples will help you master finding domains algebraically.

What You Will Learn:

How to find the domain of rational functions by excluding zero denominators
How to solve linear inequalities to ensure non-negative square root values
The foundational skills for determining the domain of a function without graphing

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

#maths #algebra #domain #squareroot #rationalfunctions #roots #graphing #domainofafunction #mathhelp #algebrabasics #functiondomain #rationalexpressions #findingdomain #inequalities #functionanalysis #algebratutorial


all right in this video i'm going to do

another problem

uh we're going to solve x to the two

thirds minus nine equals zero

and you can kind of think about this as

in a sense is

sort of like being a quadratic um

and the idea is we can really rewrite x

to the two-thirds as

x to the one-third quantity

squared because again if it's in

parenthesis you would multiply these and

get to the two-thirds power

minus nine equals zero and again you

know i don't know if it makes it any

easier or not a lot of times people will

do a little substitution

maybe we can let y equal x to the one

third

and then it does look very much like a

quadratic we could actually say this

original equation

is quadratic in x to the one third

but if we rewrite our x to the one third

squared minus nine equals zero using the

substitution we would simply get well

y squared minus nine equals 0

and again we could factor both sides or

we could add 9

to both sides in this case y squared

equals 9

and then to get our solution we would

take the square root of both sides

remember you have to put a plus or a

minus a plus and a minus on one side

whoops i'm jumping ahead of myself

we still have the square root of 9 which

is simply

going to give us 3 so it says y has to

equal positive or negative three but

again we have to remember our

substitution that we did here

because we let y equal x to the

one-third

so what that would tell us then it says

x to the one-third either has to equal

positive three

or x to the one third has to equal

negative three

and since we're raising the left side to

the one third power we're taking a cube

root really

the way that we get rid of that is we

cube both sides

so we'll raise both sides to the third

power

x to the one third cubed is simply x to

the first

three cubed is going to be three times

three which is nine times another three

which is 27

and likewise we have x to the one third

so if we

cube that and we also cube the right

side as well

negative 3 times negative 3 times

negative 3 is going to give us

negative 27. so now we have our

solutions to our

original equation

Solving Fancy Quadratics - 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 !! Solving Fancy Quadratics - Example 2.  In this video, I solve problems using the same techniques (after a relabeling) that we use for 'regular' quadratic equations.


all right so just one more problem here

using the arc length formula so in this

problem we've got a tree that's 1,500

yards from an observer and that subtends

an angle of 2 degrees we're gonna use

the arc length formula to estimate the

height of the tree okay so here's a

here's you or somebody and so suppose

you know suppose we could almost

represent you as a little dot and a tree

we've got this tree that's way far away

or pretty far away you know fifteen

hundred yards is a good little ways and

to get to the top of the tree so let's

make it a little Christmas tree here

we'll holiday festivities so there's our

little tree or a little pine tree and

the idea is to the top of the tree

you're gonna make a little a little

angle here of two degrees okay so the

idea is you know the tree assumedly is

standing straight up and down you know

the the circle that you're really

thinking about is definitely curved so

that's why we're approximating the

height of the tree here but the idea is

if we can just find that arc length that

arc length is going to be approximately

equal to the height of the tree so

that's the basic idea we just have to do

the same thing as before just use our

arc length formula so let's see we said

that s was equal to R times theta R is

the radius well in this case this is

1,500 yards away you can imagine that's

just sitting on a circle so the radius

will be 1500 again we've got to convert

theta into radians so let's see pi

radians is equivalent is equal to 180

degrees if we divide both sides by 180

we'll get PI over 180 radians is

equivalent to one degree and now if we

just simply multiply both sides by two

we'll get two pi over 180 radians equals

two degrees so that's simply going to be

PI over 90 radians is going to be the

same thing as equivalent to two degrees

so for our angle theta we'll plug in PI

over 90 and now I'm just going to do the

arithmetic so let's see we'll do 1500

over 90 and I'm going to approximate pi

by using 3.14 let's see here let me get

my calculator so we've got 1500 divided

by 90 I'm getting that to be sixteen

point six repeating times 3.14 so let's

see when I multiply that by 3.14 I'm

getting this to equal 52 point three

repeating but again we wanted to round

this to the nearest yard so this is

pretty close to 52 yards so we would say

the height of that tree give or take a

little bit is 52 yards using our arc

length formula

Arc Length Formula - 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 !! Arc Length Formula - Example 2. In this video, I use the arc length formula to estimate the height of a tree.


okay in this video I just want to do a

quick little discussion about Newton's

method again and how it you know may not

work so again that was the Newton's

method formula um hopefully you've seen

the other examples on how to compute

things so uh again this is kind of the

basic geometric idea so given the

following equation an initial guess we

want to know why Newton's method would

be bad so suppose we want to find the

Zer or the roots to uh this equation x -

2^ 2 - 1 and we start with the first

guess of two and again you want to plug

in a number that gives you something

close to zero well if we plug two into

this um if we plug two into this we'll

get well 2 - 2 we'll get 0^ 2 - 1 we'll

get - 1 and in one of the other videos

negative-1 was actually uh you know kind

of uh would have been our so we started

with an initial guess and we got

negative one out and again that's close

to zero and we started using it and

everything worked out great so um what's

the issue why would uh x equal to be bad

here so I'm just going to graph this

function real quick uh remember x^2 is a

parabola um xus two on the inside

actually shifts it two to the right okay

so there's xal 2 and then the minus one

would move it down uh one unit so that

would be the vertex of our Parabola and

it's going to cross the uh you know it's

going to cross the x-axis somewhere I

think we can almost uh maybe you can

almost figure out the answers here uh

um but suppose again you know um just

talking about geometrically why I can go

wrong what Newton's method does is

somehow when you start at your first

guess what it does is it looks at a

tangent line at that point and the

tangent line you know hopefully you pick

a

point where the tangent line will

eventually hit the X AIS because if that

t line hits the X AIS actually that

place where it hits so suppose instead

of using two we use some other value

maybe like

2.6 um so 2.6 I'm going to get a tangent

line that looks something like that so

if 2.6 if that was my first guess this x

intercept uh Newton's method that's what

it actually does it picks out the x

intercept so whatever this x intercept

is when we go through Newton's method

and we do this iteration and compute our

formula for little x sub 2 xub 2 is

magically going to be that point and

then the idea is well maybe I could go

up here and find that point on the graph

and I could find a new tangent line and

again we're trying to get close to X

intercepts on the real graph so notice

by looking at this new value I'm getting

even closer to the true x intercept and

that would be our third guess etc etc

etc well the point is if we take uh if

we try to use two as our guess if if you

think about the tangent line there the

tangent line there is just going to be

horizontal it's going to be a flat line

so the issue is the tangent line in this

case well it never ever actually hits

the x- axis so you never would actually

get yet a new um a new a new guess okay

um and you can try you know going

through the formula and plugging it into

the derivative and kind of see what

happens but geometrically this is the

issue if we picked it at x equals 2

unfortunately our tangent line again

never hits the x- axis and that means we

never get our second iteration guess and

our or I guess technically we would do

one iteration to get x sub 2 but we

never get our new guesses because the

tangent line never hits the

xaxis

Newton's Method - How it Can FAIL - More Examples Part 3 of 3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Newton's Method - More Examples Part 3 of 3.  Here I give the geometric idea behind Newton's Method and show how it can go wrong and fail to yield an approximation.  Newton's method does not always work!


okay this is a sample SAT problem taken

from the Princeton Review and it says if

X over 2 is less than 12 and y equals X

minus 4 then X X Y could equal which of

the following and this is a pretty

typical just plug and chug problem so

again the first coordinates are x

coordinate the second ones are y

coordinate we just have to have to test

which one of these pairs works so if we

plug X into the first one certainly 5

over 2 is less than 12 so that that is

satisfied well if you plug these same

values into the second part it says y is

9 and if we plug in x equals 5 well 9 -

excuse me 5 - 4 does not equal 9 so it

says that this solution when in fact did

not work if we plug in the second one it

says well it has to satisfy 19/2 being

less than 12

well 19 is smaller than 20 20 divided by

2 is certainly less than 12 so that

criteria is satisfied and let's see so

it says if we plug in y equals 23 does

that equal X which is 19 minus 4 and

again we're subtracting so this

condition is not satisfied so the second

one couldn't be an answer let's plug in

20 over 17 so we have to ask ourselves

this 20 over 2 less than 12 well

certainly it's 10 so that works let's

see if we plug 17 in for y does that

equal 20 minus 4

well again 20 minus 4 is 16 so that

doesn't work let's see if we plug in 22

we have to ask ourselves is 22 divided

by 2 less than 12 well 22 divided by 2

is 11 so that condition is satisfied and

then we have to plug in 18 for Y and we

have to ask ourselves does 18 equal

let's plug in 22 minus 4 well 22-4

does equal 18 so that will be

solution for this problem just a very

typical plug and chug type of problem

SAT Math Question #4

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! A sample SAT math question taken from a practice exam of the Princeton Review.
For more free math videos visit http://JustMathTutoring.com


alright as promised here's gonna be one

more example about solving some absolute

value inequalities so in this case we've

got twelve minus two times the absolute

value of two minus and then in

parentheses five X minus six all of that

greater than or equal to four yea so the

first thing again I typically try to get

the absolute value by itself I'm gonna

do I'm gonna subtract 12 from both sides

so that's gonna leave me with negative 2

and then absolute value inside the

absolute value I'm going to distribute

the negative so I would have negative 5x

and then a negative and a negative would

make positive 6 if I do 4 minus 12

that's going to give me negative 8 on

the right side the next thing I'm gonna

do is divide both sides by negative 2 so

on the left the negative twos will

cancel out inside the absolute value

again I'll have a positive 2 and a

positive 6 which is a positive 8 minus

5x and again since we divided by a

negative number our inequality flips

negative 8 over negative 2 is positive 4

and again for this type of inequality I

just remove the absolute value and also

make the quantity inside the absolute

value greater than the negative of the

number that we had okay so now I just

subtract 8 from each side so on the Left

negative 4 minus 8 is negative 12 we've

got negative 5x left in the middle 4

minus 8 is going to be negative 4 and

again now simply to get the X by itself

I'm just going to divide everything by

negative 5 again remembering that we

have to flip our inequality in this case

so I'm going to divide by negative 5

divided by negative 5 divided by

negative 5 okay so we've got a positive

12 fifths on the left again my

inequality flips direction the negative

fives will cancel leaving me with just X

again the inequality flips and then we

would have positive four-fifths

equivalently we could write this as

four-fifths less than or equal to X less

than or equal to 12 fifths and again if

you want to write this using interval

notation we'll use brackets from 4/5 up

to and including twelve fifths

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving Absolute Value Inequalities, MORE Examples - Example 2


Hey and in this limit example again

we'll try to plug in H equals zero

notice if we do plug in H equals zero

we'll get 4 squared which is 16 minus 16

so again we'll get 0 on top and clearly

we're getting 0 on the bottom so we have

to do something algebraic to try to

figure out what's going on here notice

the only thing that I see right off the

bat at least is there's this 4 plus h

squared so we could always cancel or

excuse me expand that out and see if

something cancels out so again remember

this is 4 plus h times 4 plus h so we'll

get 16 plus 4 h plus 4 h that'll give us

plus 8 h plus h squared minus 16 all

over H well the 16s will cancel out and

I'm left with

8h + H in the numerator but since they

both have an H I can factor an H out of

there then I have 8 plus h leftover

divided by H and again now I can cancel

out my H's and this H in the bottom is

what's you know certainly giving me some

problems anyway because I'm dividing by

0 so I'm left with the limit as H

approaches 0 of 8 + H and again we go

back to our old trick of plugging and

chugging if you plug in H equals 0

we're left with 8 and that is our

solution

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Calculating a Limit by Expanding and Simplifying


all right in this video I'm going to do

an example of solving an equation that

involves two absolute value Expressions

so at first it's not going to look like

there's two but wait just a second here

so suppose we have the absolute value of

the fraction p + 1 / 3 - P = 12 well

there's only one absolute value in there

but to get started if you have the

absolute value of a fraction you can

break this up as the absolute value of

the numerator over the absolute value of

of the denominator this is an equ

equivalent algebraic way to break up

absolute value of

fractions equals 12 and what I'm going

to do is I'm going to multiply both

sides by this quantity the absolute

value of 3 minus P so that's just going

to move over so I've got the absolute

value of p + 1 equal 12 * the absolute

value of 3 minus

P at this point what we do is we create

two separate equations to solve we take

p + 1 = 12 * the quantity we just

basically uh remove the bars on both of

them you can imagine they're both in

parentheses the other equation that

we're going to solve is we're going to

remove the bars p + 1 =

12 but on one of these um we're going to

stick a negative in front of

it so instead of just writing 3 minus P

like we did the first time I'm going to

write 3 - p but there's a negative

associated with it and now this is going

to give me my two equations that I have

to solve so on the left we have p + 1 12

* 3 is 36 12 * P will be -2 p and then

what I'm going to do is I'm going to add

12 P to both

sides so that'll give me 13 P at the

same time I think I'll subtract one from

both sides so that'll give me 13 P = 35

and if I divide both sides by 13 that'll

give me that P = 35 13 as one of my

Solutions and the other solution here

this is p + 1 we could write that as -2

* 3 - P so now if I distribute things

out we have p + 1 -12 and 3 is - 36 -12

and P is POS 12 p and the same thing so

let's add 36 to both sides that'll give

me positive 37 I'm going to subtract 1 P

from both sides that will give me 11 P

and now if I divide both sides by 11

I'll get 37 / 11 = P as my second

solution to this absolute value equation

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving Absolute Value Equations Containing TWO Absolute Value Expressions - Example 1


all right so one more example here of

simplifying products of binomials

involving trig functions and the one

that we I think left undone from the

previous video was tangent of theta plus

cosine squared theta plus sine squared

theta multiplied by tangent theta minus

cosine squared theta minus sine squared

theta so of course you can start

multiplying this all out but hey don't

do that

anytime you see cosine squared and sine

Squared's again I always think is there

any way to sort of bring this identity

into it so cosine squared theta plus

sine squared theta equals 1 well hey

that's exactly what we have sitting here

the first factor so certainly we can do

that notice in the second term we can

actually I think factor and negative out

of there so let's let's maybe do that so

I'm going to rewrite this whole

expression so we have tangent of theta

maybe we can already replace our cosine

squared plus sine squared we'll go ahead

and replace that with our positive 1

okay so just regrouping all that stuff

and replacing it with a plus one and

then on the right side we have tangent

of theta notice we could really factor

the negative out of there just out of

the cosine squared in the sine squared

if i factor the negative out I would

actually have to make that a positive

sine squared theta but again the same

thing as before now we've got cosine

squared theta plus sine squared theta

so I'm gonna make my brackets into

parenthesis it doesn't really matter if

they're all the same here so then we've

got tangent of theta minus well cosine

squared plus sine squared equals 1 again

we're gonna use negative 1 because well

there's a negative you know from before

and now what I'm going to do is just

simply multiply this stuff out so we're

gonna have tangent of theta times

tangent of theta which is going to give

us tangent squared theta will have

tangent times negative 1 that's gonna

give me a negative tangent theta will

have a positive 1 times tangent theta so

that's just plus tangent theta and then

we have plus 1 times negative 1 so

that's just gonna be a negative 1 well

the negative tangent theta and the

positive tangent theta are just gonna

cancel out we're left with tangent

squared theta minus 1 and again now I

think about this identity that relates

tangent that involves tangent squared so

there's one at least that says tangent

squared of theta plus 1 equals secant

squared theta and I think well you know

is there really much of anything to do

you know can I somehow can I somehow

manipulate this tangent squared theta is

there any way to what's a bit of a says

is there any way can I manipulate this

identity tangent squared theta plus 1

equals secant squared theta is there any

way I can sort of manipulate that to

make it say you know tangent squared

theta minus 1 well if you multiply both

sides by negative one that's going to

make the negative tangent squared minus

one so that's not what we need

you could always you know replace

tangent squared maybe with you know

secant squared minus one but again I

don't see how that's really gonna help

things simplify down very much at all so

I think I would probably just leave it

right here and I would consider that to

be simplified

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Simplifying Products of Binomials Involving Trigonometric Functions, Ex 2


okay in this video we're going to start

looking at converting between polar and

rectangular equations and again these

are kind of the uh the equations that

relate our polar coordinates to our

rectangular coordinates so we're going

to find the rectangular form of each of

these given polar

equations

so kind of the idea is so okay so here

we've got R = 3 * cosine of

theta well to me uh um we could always

on the left side we could even solve for

r on the

left we could have that R equals the

square root of x^2 +

y^2 but the problem is if I plug that on

the left I still don't know what to plug

in on the right so if only there was an

extra R hanging out on the right we

would have R cosine Theta and then I

could uh replace that with an X well

that's what I'm going to do I'll just

multiply both sides by R so by

R then we'll m multiply the right side

by R so we have uh r^2 equals I'm going

to write it as 3

R cosine Theta but now we can just start

using our formula so R 2 is the same

thing as x^2 +

y^2 r cosine Theta that's the same thing

as X so if you wanted to we could write

this as x^2 - 3x +

y^2 equal 0 and hey uh now we''ve turned

it into uh an equation involving

rectangular coordinates or cartisian

coordinates so that's all there is to it

um I would just

say so the solution here would just be

x^2 -

3x + y^2 =

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Converting Between Polar and Rectangular Equations, Ex 1.  In this video, I take one polar equation and write it in rectangular (cartesian) form.


Okay, in this video, I want to do another example of working  with fractions and,

what we're gonna do is we're gonna add or subtract some fractions that have different denominators,

and also have some variables present. So I think there's two main things that you have to remember

when you're adding or subtracting fractions. Uh, those two things would be

that you have to common denominators; that's the first thing.

So notice neither of my examples we have common denominators, so we'll have to fix that a little bit.

The other thing is once you do have common denominators, the arithmetic occurs in the numerator.

It happens in the top part of the fraction, so let's do our first example here.

We have 3x/5 minus 2x/15, so definitely different numbers in the denominator.

And, what I first think of is I think what's the greatest, excuse me, what's the least common multiple?

What's the least common multiple of 5 and 15? What's the first number that both of those divide into evenly?

And the least common multiple of those two numbers would be 15.

So, I think well, I want a 15 here in my bottom fraction on the first side, and I think

well, how do I do that?

The only real way to, to change the denominator is to multiply it by something.

So, I think in this case, we would have to multiply the denominator by 3,

but remember if you do it to the denominator, you also have to do it to the numerator.

And this is all valid because really we're just multiplying by the number 3 over 3 which is 1.

If you multiply something by 1, it doesn't change its value.

So we may change how it looks, but numerically we would still, we would still get the same value.

Okay, so 3 times 3x would be 9x, and hey now I've got my 15 on the bottom.; that's why we did it, minus 2x/15.

And now I can just do the arithmetic in the numerator. I do 9x minus 2x. 9x minus 2x would be 7x/15.

And if you wanted to, you could, uh, pull the x out to the side and write it as (7/15)x.

Alrighty, in my next example here we've got 3y/4 and 5y/24,

so the first thing I'm gonna have to do again is get common denominators.  So I think

what's the least common multiple of 4 and 24?

and in this case I believe the least common multiple would be 24.

So we would have to multiply the numerator and the denominator by 6.

Again, the important thing is you gotta multiply the denominator by 6 to get the 24, but as a result

we also have to do it to the numerator.

So on top, we get, uh, 18y/24 plus 5y/24

If we add 18y and 5y, we simply get 23y/24.

And again, we could write this as, we could pull the y out to the side 23/24 times y.

And now we have combined our fractions.

Fractions:  Adding and Subtracting Fractions with Unlike Denominators

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Fractions:  Adding and Subtracting Fractions with Different Denominators and containing variables!
Two examples!


Okay, so here we've got a little graph above that displays

the total cost, C, in dollars of renting a boat for h hours.

So, my graph's a little rough, and I've labeled

so this would be the point (0,5), there on the vertical axis.

It goes through (1,8), (2,11), (3,14), (4,17).

Just to give you some more points.

So, the first question here is, what does the C-intercept represent in the graph?

Is it 'the initial cost of renting the boat',

'the total number of boats rented',

or 'the increase in cost to rent the boat for each additional hours'?

So, again, it tells us that the vertical axis represents the total cost in dollars.

So, the idea is if you go, you know, if you rent a boat  for zero hours,

maybe had plans to go and you don't get out there.

It says, it's still gonna cost you five dollars, after being on the boat for one hour, it's jumped up to eight dollars.

So, the idea is the initial cost is gonna be this five dollars.

So, answer choice A is clearly correct here. It's 'the initial cost of renting the boat'.

It's not 'the total number of boats rented', right?

It has nothing to do with the vertical axis, it has nothing to do with the number of boats.

It tells you the cost.

Again, 'the total number of hours the boat is rented', that's answer choice C,

again, the vertical axis has nothing to do with the number of hours it has to do with the cost.

And D says 'the increase in cost to rent the boat for each additional hour'.

Well, no 'cause, again, the C-intercept has a value of five

and if we go an additional hour,  it's only jumping up to eight,

two hours, it's only jumping up to eleven.

So, it's only jumping up by three dollars per hour.

Definitely wouldn't be answer choice D either.

Again, if I saw this on the test, answer choice A. Boom and I would be on to the next question.

SAT Math - Question 15, Section 2, 2016 Edition and Beyond

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Section 2 - A Calculator IS allowed!


okay here we're gonna look at solving a

few different equations more

specifically what are known as linear

equations and we'll have to do a couple

a couple steps to solve them so in Part

A here we have negative 4x minus 14

equals 10 typically the way I always

thought about it is first I want to get

for an equation where there's an X to

the first power or variable to the first

power I want to get the term involving

just X all by itself so if there are

other X's I'll try to combine them so on

the left side I see two terms I see one

involving X and then I see just a number

and then on the other side I see just a

number so to get the the X's basically

by itself since I've subtracting 14 I'm

going to add 14 to both sides so on the

left side that's gonna leave me with

negative 4x equals well let's see 10

plus 14 is going to be 24 and now what I

need to do at this point since I'm

multiplying by negative 4 to get the X

by itself I'm just simply going to

divide both sides by negative 4 so

negative 4 divided by negative 4 is 1x

let's see a positive over a negative is

a negative

24 divided by 4 will give us 6 so the

solution to our first equation will be x

equals negative 6

Solving Two - Step Linear Equations

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving Two - Step Linear Equations.  Here we look at solving linear equations where we have to perform addition / subtraction as well as division / multiplication to solve.


all right in this video we're gonna do a

couple questions related to finding the

period of a tangent and a secant

function things have been changed here a

little bit so we've got three tangent of

2 times the quantity X minus 1 plus a

and then in Part B we have negative 4

secant of 4 times the quantity X plus PI

minus the square root of 2 so recall the

formula before we have been using for

the period when we were doing sine and

cosine we were using 2pi divided by the

absolute value of b but that's because

sine and cosine had a period of 2pi

recall the tangent just plain old

tangent X has a period

of just pie okay so okay the next has a

period of just pie so instead of taking

two pie and dividing it by the absolute

value of B all we have to do is just

take PI and divide it by our absolute

value of B well again our B value is

just whatever is the coefficient on the

X inside the brackets so you can imagine

if you distribute it you would have 2x

or again equivalently this is gonna be

our B value so we would have the

absolute value of 2 which is just 2 so

our period would actually be just PI

over 2 in this case and again just kind

of the effect that would have

remember the the asymptotes normally for

tangent or at PI over 2 and negative PI

over 2 and then of course there's others

now our asymptotes will get pulled into

PI over 4 and also negative PI over 4 so

this is definitely not the graph of this

one but this would be the graph of

tangent of 2x again just a little a

little segment of it there so again

that's the effect it's just compressing

the graph by a factor of 2 to get the

period for our second one here

well secant is really just 1 over cosine

cosine does have period 2pi secant also

has period 2pi so now we go do you go

back to our original formula that we saw

before 2pi divided by the absolute value

of B so we'll get 2pi over in this case

4 well 2 over 4 it looks like again hey

we're getting a period of PI over 2 so

only thing to remember again tangent and

cotangent have a period of Pi sine

cosine secant cosecant all have a period

of 2pi and then it's the same thing just

dividing by the absolute value of your B

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Basic Questions Related to Tangent, Cotangent, Secant, Cosecant, Ex 3. In this video, I just address a few basic questions about Tangent, Cotangent, Secant, Cosecant.


all right in this video I'm gonna do

another example of solving an equation

that involves rational exponents so I

should have probably a little lowercase

Y so let's solve for y here so we've got

4 times the quantity y squared minus y

plus 1 to the 1/4 the exponent it's only

applying to the stuff in parentheses not

to the 4 and on the right side we have

16 y all being raised to the 1/2 so what

I do is again I basically look at the

denominators so the denominator is a 2

and a 4 and what I try to do is I try to

find the least common multiple of those

numbers that is I'm trying to find the

first number that both 2 and 4 divide

into evenly and the least common

multiple of the numbers 2 & 4 that is

the number 4 that's the first number

that both of them divide into evenly

when I determine this number what I do

is I raise both sides to exactly that

power so I'm just exactly rewriting the

left side but now I'm gonna write it all

to the fourth power and on the right

side I've got 16 Y to the 1/2 again I'm

gonna raise all of that to the fourth

power okay so on the left side again we

have to be a little careful because this

exponent has to get applied to both the

4 and the factor y squared minus y plus

1 so we'll actually get 4 to the fourth

so again if I just multiply the 1 in the

4 but on the y squared minus y plus 1 if

I take the 1/4 times the 4 that's just

going to give me to the first power

on the right side notice we're gonna get

okay so now notice the 16 why's in

parentheses okay so I'm just gonna take

16 Y and I'm gonna take 1/2 times 4 and

that's gonna give me to the second power

okay so now we can keep simplifying here

a little bit more so 4 to the fourth is

actually 16 squared which is 256 then we

have Y squared minus y plus 1 left over

well conveniently on the right we have

16 squared or 256 we also have to square

the Y so the first thing I'm going to do

here this is starting to type well this

is now certainly a quadratic equation

I'm going to divide both sides by 256

just to get rid of those coefficients so

on the left side we've got W squared

minus y plus 1 on the right we're simply

going to be left with a W squared and

now this is nice because if we subtract

I keep saying W if we subtract Y squared

from both sides so minus y squared minus

y squared we'll be left with negative y

plus 1 equals 0 and now I can simply add

Y to both sides and we're going to give

our solution again you should

technically whenever you're dealing with

fractional exponents it's a good rule of

thumb to go back and check these

solutions but I think if you go back and

check you will in fact find that y

equals 1 is a solution

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving an Equation Involving Rational Exponents - Example 2


all right in this video we're going to

do an example of solving a trigger

equation using the quadratic formula so

we're going to uh do got three different

examples here let's try to find uh all

Solutions in the interval from 0 to 2 pi

for these and the first one will be

cosine ^ 2 x - cosine x + 1 equals z and

you know you can certainly try to uh

sort of factor this out I think you'll

convince yourself that you can't really

Factor it out um I'm just going to

rewrite this as cosine x uh 2 - 1 *

cosine

x + 1 = 0 and the way the reason I'm I'm

writing this is this is what we could

call a quadratic equation uh not an X

but an cosine X so kind of think you

know sort of similarly uh suppose this

just said x^2 - 1 * x + 1 equals

0 this is called a quadratic equation in

X this would be a quadratic equation in

cosine X now normally uh you know since

we can't find solutions by factoring at

least I can't figure out how to factor

it we can always use the quadratic

formula so remember X = B plus or minus

the square < TK of b^2 - 4 a c over 2 a

now that's what we would do um if we had

X here and again remember a is the

coefficient in front of the x²

B is the number in front of the X and C

is just the

constant the exact same thing uh holds

true for this equation uh this quadratic

equation in cosine x uh the a the B and

the C are still going to be the same the

only difference is instead of Simply

writing x equals blah blah blah we would

have well now the thing being squared

isn't just X but it's cosine X so

equivalently what we would do is we

would plug in cosine

X but now we get the coefficients in the

same way I look at uh sort of the cosine

to the first Power the number in front

of that would be my B value so I'll get

negative

um1 plus or minus the square root of

again b^ 2 which would be -1

2ar - 4 * a which again would just be

positive

1 times C which is going to be the

constant which is just POS

1 all over 2 * a which we said is

1 and now what we do is we just have a

new equation involving cosine X that we

would have to solve but I think in this

case at least uh so we get a negative

and negative which will make positive 1

I think underneath the radical Well -

1^2 is 1 we'll get 1 - 4 which would be

-3 well we can't uh I mean we can use

complex numbers uh you know to find Sol

Solutions but what we want to do is find

real valued Solutions so since we have a

negative number underneath the square

root uh we would simply say that this

equation has no real uh

Solutions we'll look at an example uh I

believe the next one certainly for sure

um so we we'll definitely see at least

an example somewhere where you know you

do get a solution and hey what do you

have to do from there because in that

case you're going to have to usually

typically use inverse cosine or invers

sign the inverse of whatever your trig

function is uh you'll have to think

about you know all solutions so a little

trick here but again this is kind of the

basic idea um that I want to point out

in this example it's still a quadratic

the coefficients you know still think

about the coefficients the same way

again just instead of putting x equals

blah blah blah you'll have to put you

know whatever trig function is being

squared you'll have to put that on the

left side

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all right in this video I want to do an

example of finding the inverse of a

function if one exists and if it doesn't

exist we'll justify that as well so in

our first example here we'll use f ofx =

8 - x^2 over 5 so the first thing uh

that we'll do when we find an inverse I

always just write it you know instead of

F ofx I write y so we have 8 - x^2 over

5 = Y and then um the Y I replace that

simply with an X and everywhere that

there's an x uh in the original function

I'll replace that with those with y's so

here there's obviously only uh One X to

replace but you could end up with

certainly more than one X that turns

into multiple

y's okay so all we're going to do here

now is simply try to solve for y so I

can multiply both sides by five to get

rid of my denominator that'll give me 5x

on the left = 8 - y^2

um I'm going to do two steps at once

here I'm going to add the y^2 to the

left side so now we would have positive

y^2 the 8 he's still hanging out on the

right and the 5x I'm going to subtract

that from both

sides so I'll have y^2 = 8 - 5x and now

what I have to do to get rid of the uh

the square is I'll take the square root

of both

sides but remember when you take the

square root you have to put a positive

and negative on um one of the sides and

that's true for any even powered root

we'll get y on the left equal

posi 8 -

5x okay and this immediately tells me

that in fact there is no

inverse and again kind of the idea is it

says for two distinct inputs you know if

you plug in something for x you're going

to get two different uh basically two

different values you know if you plug in

zero you would get the square OT of 8

you would get positive negative square

root of 8 and that says for one input

you're getting multiple outputs and that

means you don't have a function and uh

that's what we want we want when we find

the inverse we want that to be a

function so we'll say no inverse

function and another way to see this

really quickly um if you think about it

you could always break this fraction up

as 8 fths - x^2 over 5 or I can write

that as 1/5

x^2 and if you remember your graphs here

um the graph of x^2 is just a

parabola um netive x^2 would actually uh

flip it upside down the positive 8 fths

would move it up so actually the graph

would look something like that unless

I'm

mistaken and again now the idea is this

function is going to fail that

horizontal line test and remember if you

draw a horizontal line and it hits your

graph in more than one place that again

is another uh way to determine that your

function has no inverse so kind of a

quick graphical way uh but here also you

know the important algebraic way

Finding the Inverse of a Function or Showing One Does not Exist (Without a Graph!), 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 !! Finding the Inverse of a Function or Showing One Does not Exist, Ex 2.
In this video, I justify algebraically that a function does not have an inverse.


okay in this video I'm going to do

another example of finding the inverse

of a function so in this case we have G

ofx = -5 + 5x over -3 - 3x so I'm going

to write y = -5 + 5x over -3 -

3x um and now to find the inverse I'm

going to switch out I'm going to turn my

Y into an X and everywhere I have an X

those are going to become y's

and again now my goal is to try to solve

for y and hopefully get a single uh

unique function here which will tell me

it has an inverse the first thing I'm

going to do is I'm going to get rid of

the denominator I'm going to multiply

both sides by -3 - 3 y so on the left

side we'll have well -3 - 3 y * X on the

right side we've got our -5 + 5 y over

-3 - 3 Y and I'm going to

multiply well again I'm multiplying by

that denominator so that it'll simply

cancel

out um at this point you can think about

the X's being out front if you'd rather

move it out front it's multiplication so

you can kind of put it in any order you

want I'm going to distribute that X so

I've got

-3x - 3x y = -5 + 5 y so again our goal

here is to solve for y um what I'm going

to do is I'm going to put everything

that has a y on the same side so we have

our -3 XY notice we have a y on a term

involving y on the right side I'm going

to add 3xy to both

sides at the same time this neg five

that's on the right since he doesn't

have a y I'm going to add five to both

sides so on the left side we would have

5 - 3x the 3x y would be canel out with

the uh positive 3x y on the right side

the 5 and the POS 5 cancel and now we

have 5 y +

3xy so last but not least uh what we're

going to do here um to try to get y by

itself is I can Factor the Y out both

terms have a y I would need 5 + 3x in

the

parentheses um and okay at this point we

simply just divide out the 5 +

3x 5 +

3x and then that's then going to give us

our inverse function so it says on the

right side now we would be left with one

y um I think I said on the right side we

would have just one y on the left side

we have 5 -

3x um over 5 + 3x this is going to be

our inverse function notice we're not

getting like any positive negative

square roots like in our previous

example so it says the inverse function

of that

original would have the formula uh 5 -

3x over uh 5 + 3x

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding the Inverse of a Function or Showing One Does not Exist, Ex 3. In this video, I find the inverse of a rational function.


in these examples I want to simplify

some expressions with fractional

exponents and notice in these examples

at least they all have common

denominators I'm also going to do the

extra step of converting those into

radical notation as well just so

hopefully you'll see how to do both

things alright so here's my examples X

to the 1/4 times X to the 3/4 well

remember our algebraic property that if

you have like bases being multiplied all

we simply do is add the exponents so we

have to take 1/4 plus 3/4 but that just

gives us X to the 4 over 4 which we can

simply write as X to the first or just X

and since its X to the first you know I

wouldn't write that as a radical at all

I will just leave that alone but the

next one okay the same idea we have X to

the 2/10 multiplied by X to the 7/10 so

again we'll just add two tenths and

seven tenths and that will give us X to

the 9 over 10 and if you wanted this as

a simplified rational exponent well then

we have at 9 over 10 doesn't reduce as a

fraction but if we also wanted to write

it say in radical notation we could do

that so again X goes underneath the the

denominator which is the 10 that becomes

the root

and then we simply raise all of that to

the ninth power ok so again my my nines

have a tendency to look like A's but

that's supposed to be a nine there so

alright last one here we have X to the

3/8 times X to the two eighths well if

we just again add three eighths plus two

eighths will give us X to the 5/8 and

I'm going to rewrite that one more time

again using radical notation so it

becomes the eighth root of x all raised

to the fifth power

ok so again a useful trick I think for

sure being able to convert from rational

exponents into radical notation a lot of

times when you have to simplify again

you know numbers to fractional exponents

this at least is how I think about them

this is how I compute them I think about

roots and then exponents so alright I

hope these examples make some sense and

help you out

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Multiplying Variables with Rational Exponents - Basic Example 1


so just another quick uh sequence

question here so does the sequence uh Ln

of 1/ n Cub uh does that converge is

that true or false well again I think

this is you know not not too crazy to

evaluate you know basically as you know

um as n

increases and again

uh we're technically only using uh you

know like a

whole numbers when we plug into our

sequence so but the idea is n increases

certainly 1 over n the 3 this is going

to approach zero so as n increases 1

over n the 3 that will approach zero um

and if you think about the graph uh you

know natural

logarithm here's our natural logarithm

graph um you know as as as the input

gets closer closer to zero if you just

want to think about it in terms of

limits uh almost you know as as one over

n the3 approaches uh zero we should say

from the right side um the natural

logarithm simply goes down to negative

Infinity so I would say false uh the

sequence actually

diverges and the sequence would uh

diverge it would just approach negative

Infinity

Sequence Example : Converge or Diverge

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A simple example of deciding whether a sequence converges or diverges.


alright one more example here of writing

a fraction as a decimal so here we're

gonna do negative 13 over 5 if you

wanted to you could make this into a

mixed number and then take the the you

know the fraction that's left over and

write that as a decimal again I'm gonna

do it just exactly like we did our last

example and I'm simply going to just do

long division I know that there's a

negative in there so I'm gonna tack that

you know I'm gonna include that back in

here in a moment but for now I'm just

gonna say well 13 divided by 5

I'm just gonna worry about you know the

actual digits I'm gonna do the same

thing I'm gonna add in some zeros and in

a few zeros so now I'll simply say well

let's see you know five won't go into

one but five will go into 13 it'll go in

there twice without going over two times

five is ten and again we just subtract

13 minus 10 is gonna leave this with

three we just dropped down our zero so

now I'm looking at the number 30 and

they say well five will go into 30 how

many times or again 30 divided by 5

would give us exactly in this case 6 6

times 5 is 30 hey that's perfect

we've got no remainder that tells us

that our decimal place now terminates

and we've got our solution so it says we

can write negative 13 over 5 as again

let's not forget our negative that's

negative 2.6 so that one I think was

even a little bit easier than the last

one since we only had to do a couple a

couple steps

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 Fraction as a Decimal, Example 2.  In this video, we use long division to write a fraction as a decimal.


all right in this video I'm gonna do

another equation involving two absolute

value expressions so again it first

initially certainly look like there's

only one but algebraically again we can

break up this fraction on the left as

the absolute value of 2x plus five over

the absolute value of 3x plus 9 equals 6

what I'm going to do is I'm going to

multiply both sides by this absolute

value of 3x plus 9

so it cancels on the left so we're left

with the absolute value of 2x plus 5

equals 6 times the absolute value of 3x

plus 9 and again what we do at this

point to find our solutions we just

create two equations the first equation

I just basically imagined the bars turn

into parenthesis so I'm just replacing

the absolute values with parentheses

I'll have to solve that equation the

other equation what I do is I remove the

bars and on one of the absolute values

okay so this is not correct just yet I

have to stick a negative on one of the

absolute values or the other I think it

was sticking on the one on the left just

so it looks a little bit better when I

write it

so notice the only distinction we remove

the absolute values in both of them one

of the equations has a negative present

okay so to solve the first one we can

just get removed removing the

parentheses 2x plus 5 if we distribute

the 6 we'll get 18 X 6 and 9 is positive

54 let's see I can subtract 2x from both

sides that would give me 16x on the

right plus 54 now I can subtract 54 from

both sides let's see I think 55 minus 54

would be negative 49 equals 16x

and if we want to simply get the X by

itself we can divide both sides by 16 so

that'll tell us that negative 49 over 16

is one of our solutions

to do the other solution we're going to

do the same thing the first thing I

would do is distribute my negative so

I'll get negative 2x minus 5 equals 18x

again plus 54 let's see if I add 2x I'm

going to get 20 X on the right plus 54

and again if I subtract 54 from both

sides now negative 5 minus 54 is

negative 59 equals 20 times X again if I

want to solve for X I'll just divide

both sides by 20 and that'll say

negative 59 divided by 20 equals x is my

other solution

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving Absolute Value Equations Containing TWO Absolute Value Expressions - Example 2


all right in this video we're going to

do an example of finding the height of

an object um and we're just simply going

to use some basic trigonometry to help

us do this so suppose um suppose you're

you're just chilling out you know

watching a balloon go up in the air

because you got nothing else to do

suppose uh that you're able to measure

this distance from you this horizontal

distance from you to the balloon and

that's 2,000 ft suppose also you've got

you know you happen to have some

surveying equipment laying around and

you get a little angle of 30° to the

balloon we want to know what's the

height of the balloon so this is

certainly very useful practical um

normal types of questions out there in

engineering um you know you know anytime

people are making buildings or roads or

definitely doing all this stuff just

making measurements and angles so

certainly very practical math here so um

okay again to me we're trying to figure

out um this height so here's my little

angle that I'm given in this problem so

2,000 ft that would represent um stuff

about the adjacent side um and this Ang

the the height would represent sort of

the uh the opposite

side so I'm thinking which of the trig

uh trig functions relates opposite sides

and adjacent sides well there's actually

two of them uh but the one I'm going to

use here is just tangent so remember

tangent um of an angle is defined to be

the the ratio the opposite

side to the adjacent

side so in this case we're going to have

that well tangent of

30° that's going to equal again the

opposite side so we'll just call it

maybe o for opposite um over the

adjacent side but we know that's 2,000

actually let's don't use o for opposite

because it looks uh it looks like a zero

to me so let's just call this height Y

and we'll make this a y and everything

will be a little less confusing so try

to use some better

notation okay so we have tangent of 30°

that equal y over

2000 again simply to get the Y by itself

we can just multiply both sides uh by

2000 so if we multiply both sides by

2,000 uh we'll get the 2,000 on the left

just like we did but if we multiply y um

on the right they're just going to

cancel out so we get the Y by itself so

that's all we're doing there and now

again this is something that I would

just want to uh plug into a calculator

so again at this point make sure your

calculator is in degree mode so let's

see tangent of

30° um let's see so I'm going to

approximate this

now so tangent of 30° uh is

approximately

577 so that's roughly going to be the

height of our uh balloon here so now if

we simply take 577 and multiply that by

2,000 I'm getting

1154 so it says the height of this

balloon is going to be roughly equal to

Ft

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding the Height of an Object Using Trigonometry, Example 1. In this video, I find the height of a balloon by knowing a horizontal distance and an angle.


all right in this example we're going to

use some half angle identities to

evaluate some trig expressions so these

are the identities the half angle

identities notice for cosine and sine

there's kind of a choice positive or

negative basically that's just you're

going to have to think about the

quadrant that your angle is in and then

determine if it's positive or negative

so for example suppose we want to figure

out sine of twenty two point five

degrees well all I'm going to do is I'm

simply going to rewrite a sine of twenty

two point five notice we could rewrite

that if you double twenty two point five

you get 45 degrees well so I could write

twenty two point five degrees as 45

degrees divided by two okay

well twenty two point five degrees this

is definitely going to be in quadrant

one now I'm just going to go back to our

identity it says well once you have sine

of your angle divided by two it says

again you have to determine whether it's

positive or negative but again since

twenty two point five degrees is in

quadrant one I know that we'll use the

positive square root it says all you

have to do is you take one minus cosine

of whatever value is on top so that will

be 45 degrees

then it says we simply divide all of

that by two and that will give us our

value so I think we can clean this up a

little bit so we've got the square root

of 1 minus well cosine of 45 degrees

that's square root of 2 over 2 all over

2 and I think what I'm going to do just

to get rid of the fractions I'm going to

multiply you know my the top of my

fraction by 2 and the bottom of my

fraction also by 2 and if we do that

again I don't know not the end of the

world probably if you don't do that

we'll get 2 times 1 which is 2 minus the

square root of 2 all over 4 couldn't

even rewrite this a little bit more we

could take the square root of the

numerator so we would have 2 minus the

square root of 2 again all of that's

being square rooted

and then we could take the square root

of four individually and the square root

of four is simply going to give us two

and to me I would say you know I would

probably leave it at this at this place

there's not a lot else you can do in

terms of the simplification here

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Half Angle Identities to Evaluate Trigonometric Expressions, Example 1.  In this video, I give some half angle identities and show how they can be used to solve some trigonometric equations.


all right so a couple more examples here

of adding whole numbers and here I've

got everything written uh vertically

already so in part A we're going to add

546 and 85 546 and 85 and in Part B

We'll add 289 and 76

together so again I just move right to

left let's see 6 + 5 that's going to

equal

11 so again the one's digit I just leave

in the one's column and then the 10's

digit which is well also one that's what

I carry to the next column so there's my

one getting carried to the next column

um so let's see 1+ 4 that's five 5 + 8

well that's going to give us

13 so in this case I'm going to drop my

three down and now I'm going to carry

again the digit the one over to the next

column and 1 + 5 is going to be six so

it says 546 and 85 gives us 631

631 let's see 9 plus six that's going to

be 15 so again I'm going to put the five

down carry the one let's see 1 + 8 is 9

9 + 7 that's going to give us 16 so

again I'm going to drop my six down and

again carry yet another one and then 1 +

2 is going to be three you can think

about the you know the space as just

being a zero 1 plus 2 is 3 so it says if

we take 289 and add 76 we're going to

get 365

Adding Whole Numbers , Example 2.  Three Digit Plus Two Digit

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Adding Whole Numbers , Example 2.  Here we look at two examples of adding a two digit and three digit number together.


all right so one more little example

here so suppose we approximate uh theare

otk of 1 + x^2 near x = 1 with the

polinomial < TK 2 +22 * X - 1 again we

want to know uh what the remainder

is okay so it says the remainder

associated with a degree one

polom it says well that's going to be uh

we'll have to take the second derivative

evaluated at Z over 2 factorial and then

we'll have x minus our C value which is

positive1 again raised to the second

power well all we have to do is take a

couple derivatives

here and then uh replace our x with our

Z and uh that's it

so F ofx okay so that's 1 +

x^2 uh to the 12 power so the first

derivative will get 12 * 1 + x^2 to the

-2 power and then we have to do the

chain rule so *

2x so the 1/2 and the two will cancel

and we'll be left with x * 1 + x^2 to

the - one2 power well we can do yet

another derivative so our second

derivative we'll have to use the product

rule so uh when we take the derivative

of 1x we'll get one we'll have 1 plus

plus x^2 -

12 plus we'll have to take another

derivative so let's leave the X Alone um

we'll have to do the chain rule so the -

one2 will come out front we'll have 1 +

x^2 to the

-3 and then the derivative of the inside

will give us

2x so let's see if we can't clean this

up a little bit so 1 + x^2 to the - one2

um it looks like the 1/2 and the two

will cancel we'll still have a we have

an X and an X so

X2 then we have 1 + x^2 -3

power I'm going to factor this I'm going

to pull out the 1 + x^2 to the -3

power pulling out the smaller of the two

exponents on the 1 plus

X2 well if we multiply that by 1+ x^2 to

the first when we distribute We'll add

the exponents and that'll give us our

first term back and then we'll just Min

- x^2 when we distribute to get our

second term

back so let's see we can put the 1 + x^2

to the three

Hales in the denominator then we would

have 1 + x^2 - x^2 which is just going

to leave us

one all right so now it says um

therefore our second derivative

evaluated at

Z would be 1 over 1 +

z^ 2 to the 3es

power and now I'm just going to replace

that back in my uh

expression so if we plug that in it says

we'll have one over 1 + z^2 to the 3

power we would have a 2 factorial in the

denominator um so two factorial is just

the same thing as multiplying by two and

then we would be left with our x -1 2

and now we have our expression so again

it's really just a matter of just taking

derivatives um plugging in that

expression simplifying the factorial and

that's really all you have to do

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Taylor's Remainder Theorem - Finding the Remainder, Ex 2. In this example, I use Taylor's Remainder Theorem to find an expression for the remainder.


so in this question we're given the

system of equations x + y = 9 and x + 2

y =

-25 and we want to know according to the

system of equations what is the value of

x so we don't have to figure out you

know the complete solution X and Y just

X so what I'm going to do in this one is

again you could use uh substitution to

solve this if you want

to I think I've done at least a couple

examples um you know of these sat

questions where we've used substitution

let's use elimination by addition and

since I want to solve for x to me that

tells me well let's

eliminate let's eliminate the variable Y

and that way we'll be left with an

equation involving only X and then we'll

be able to solve

it okay so I want to do elimination by

addition so I notice I have a positive Y

and a positive 2 Y and I think well if I

had a NE 2 y here if I add --2 Y and

posi 2 y I would be left with 0 Y and

then I would have an equation involving

only X so to do that I'm going to

multiply both sides of my first equation

by -2 so I've got to do the left side

and also the right side that's going to

give me um a new system of equations

that I can then solve using elimination

by addition okay

so okay so if I distribute on the left

I'll have

-2X um -2 * POS Y is -2

y 9 * by -2 will give us POS 18 I'm not

doing anything to the second

equation and now I'm going to do my

elimination by addition

so -2x + x that's going to leave me with

x -2 y + 2 y well that's 0 Y and that's

what we wanted um 18 + -25 that's going

to be -7 and now if I just multiply both

sides by

-1 or equivalently I could divide by-1

I'll be left with positive X on the left

side and POS 7 on the right side so it

says the value of x that would you know

um part of the solution to this equation

the x value would be equal to pos7

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! SAT Math - Question 18, 2016 Edition and Beyond.
Section 1 - No Calculator allowed!


all right in this video I want to do

another uh factoring of a perfect square

trinomial and in this video I'm going to

end up using the second video a^ 2 - 2 a

+ b^ 2 it says we can factor that as the

quantity A- B * Aus B again multiply out

the bottom one you'll see you'll get

this right back and in this video I'm

going to factor 81 x^2 - 36x y + 4

y^2 again what I do in these is uh I'm

not so much looking at that and just

going wow that is a that is a perfect

square trinomial um to me in my head I'm

thinking well I got to factor it so uh

there's three terms so I know I'm going

to make two sets of parentheses here and

again at least you know one thing I'm

hoping that would work so to get the 81x

squ um I could use a 9x and a 9x you

know again there's certainly other

numbers that multiply to give us 881 but

this is a lot of times my first guess to

get the 4 y^2 I'm either going to need a

positive 2 Y and a positive 2 y or

they'll both have to be negative but

since the term in the middle is

negative they'll both have to be

negative and in my head I'm thinking you

know is that right well if you multiply

it out 9x and 9x is 81 x^2 which works

9x and -2 y will give us a -8 18 XY on

the inside we'll get another -8x y if we

combine those hey we'll get our - 36x Y

and then on the outside we'll get a pos4

y^2 so again you can really check that

this

original if we think about our a as

being uh 9x and if we think about our b

as being

um the

2y it says really you can think about

this original problem of being in the

form a^ 2 - 2 * a *

B plus b^2 and if you rewrite uh you

know if you were to fill this in using

these values we would get this original

back and again it says it factors as a

minus B * a minus B and that's what we

just did so again you know I I'm not

really noticing this thing I I don't

notice that if I double you know the

first and the last and hey everything

works out to be the middle to me I'm

just thinking you know again one thing

that kind of Clues me in to try this is

you know 81 is a nice is is a is a

perfect square four is also a perfect

square and again it just makes me think

probably you know not always but at

least that's a good first place to

check

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Factoring Perfect Square Trinomials - Ex 2


all right here we're going to look at an

example of dividing a whole number by a

mixed number and again what I'm going to

do is I'm just going to make both of

these into single fractions and then I'm

going to take my division and write it

as

multiplication so I'm going to write 10

is 10 over 1 we'll leave the division

alone for just a moment well we'll make

this into a fraction 5 * 2 is 10 + 3 is

13 we still have the five in the

denominator

well I leave my first fraction alone

division becomes

multiplication and then we

flip our fraction over and at this point

well I think are there any common

factors well 10 and five in the

numerator there's only a one and a 13 in

the

denominator uh nothing has uh uh we

can't reduce any of these fractions

immediately so I'm just going to go

ahead and multiply so 10 * 5 will be 50

1 * 13 is just 13 and of course we could

leave it as a fraction we could write it

as a mixed number again so let's see uh

so if we do 50 divided by 13 let's see I

guess 13 * 3 would be

39 we could subtract that would leave us

with 11 so we could write it as a mixed

number by saying it's going to equal

three 11 is our

remainder divided by 13

so either one of those would be correct

again just the improper fraction or as a

mixed number

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Dividing Fractions and Mixed Numbers, Ex 2