help-27

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[cyan axle]

How does the equation for a derivative change when I’m trying to find the second or third deravative?

[cyan axle]

Like using power rule as an example

X^2= 2X

So what changes when I’m looking for the second or third derivative

[quick peak]

what is your question exactly?

[cyan axle]

Like how do I find the second derivative

How does my formula change

[quick peak]

it's the same

[supple knot]

You differentiate the derivative

[cyan axle]

How do I do that?

2X but that’s 0

[supple knot]

No

[cyan axle]

How does it work then?

[supple knot]

Just use power rule again

And constant multiple rule

,tex .diff rules

[cyan axle]

There is no power though

[woven radish]

dr. matlab plot

[supple knot]

Yes there is

[cyan axle]

2X?

[supple knot]

x = x^1

[cyan axle]

=x

So just x

It’ll just end up as 2(x)

[supple knot]

x = x yes

[cyan axle]

yes

[supple knot]

2x = 2(x) yes

[cyan axle]

Yeah

But what next

[supple knot]

What are you confused about

[cyan axle]

What I’m suppose to do next

[supple knot]

Do next for what

[cyan axle]

There’s no slope of a straight line so idk what to do next to find the second derivative

[supple knot]

.

[supple knot]

.

[cyan axle]

nx^n-1

[supple knot]

All straight lines have a constant slope

[cyan axle]

So 2X

No that dosent work

Can I even apply the power rule

2x^1.

2x^1-1

Still 2X

[supple knot]

No

1-1 = 0

[cyan axle]

Yeah

So it’s gone

[supple knot]

x^0 = ?

[cyan axle]

X

[supple knot]

No

x^1 = x

[cyan axle]

So 0

[supple knot]

No

,calc 3^0

[woven radish]

Result:

1

[cyan axle]

Okay

But ^1 is still 1*

[supple knot]

No

[cyan axle]

Mb it’s 1

[supple knot]

x^1 = x

[cyan axle]

But I can just ignore it then

Yeah

Still x

[supple knot]

?

[cyan axle]

1

[supple knot]

Yes

[cyan axle]

2x^1-1

2x^1

But I can ignore the ^1

Still 2X

[supple knot]

.

[cyan axle]

Dammit

I’m coming off working on physics for like 8 hours cut me some slack

2x

Bro it’s still 2X

[supple knot]

.

[supple knot]

1-1=0

[cyan axle]

Yes I know addition

Hopefully

So you want me to say 2x^0

[supple knot]

.

[cyan axle]

Chill

Shush

[supple knot]

?????

[cyan axle]

Lapse of judgement

[supple knot]

Am I being mean to you

[cyan axle]

No

Okay to find the second derivative of x^2 I need the first derivative

2x

And the second derivative is getting the derivative of that

So it’s like

F(2)=x^2 ?

Shiiii…..

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[graceful ore]

HELPPPPPPP

[devout snow]

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[graceful ore]

NO

BRO I TYPO

LOOK at channel 30

🙏

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[pallid ravine]

did i do this right?? im doing a circuit and i dont see my answer choice anywhere so can anyone tell me what i did wrong?

[wicked edge]

the slope of the tangent line at c is $$\frac{f(b)-f(a)}{b-a}$$, not c

[woven radish]

Cycadellic

[wicked edge]

so, now that we know $f'(c)$, we need to find c

[woven radish]

Cycadellic

[wicked edge]

just find the derivative of f and set it equal to -1.3...

then, the solution which is in [-3,-1] is c

[wicked edge]

didnt see the ' in the problem at first glance, so since we already know f', we set that = -1.33..., then solve

[pallid ravine]

why set to -1.333?

Arent the end points the two um points i use to find the slope of the secant line

I thought its -2? No?

And set equal to f’(x) given?

@wicked edge

[wicked edge]

ill be more concrete

we know $$f'(c)=\frac{f(b)-f(a)}{b-a}, c\in[a,b]$$ and $$f'(x)=\frac{-3-x^2}{x^2}$$

[woven radish]

Cycadellic

[pallid ravine]

yes

[wicked edge]

you found $$f'(c)=\frac{f(-3)-f(1)}{-3--1}=-1.33...$$

[woven radish]

Cycadellic

[wicked edge]

therefore, $$f'(c)=-1.33...$$

[woven radish]

Cycadellic

[wicked edge]

but, we know f'

so, $$\frac{-3-c^2}{c^2}=-1.3...$$

[woven radish]

Cycadellic

[pallid ravine]

But f(-3) is 2?

and f(-1) is -2?

[wicked edge]

remember that we are given $$f'(x)=\frac{-3-x^2}{x^2}$$

[woven radish]

Cycadellic

[wicked edge]

therefore,
$$f'(c)=\frac{-3-c^2}{c^2}$$

[woven radish]

Cycadellic

[wicked edge]

but, by the mean value theorem, $$f'(c)=\frac{f(b)-f(a)}{b-a}=-1.33...$$

[woven radish]

Cycadellic

[wicked edge]

so, i have shown to you that
$$\frac{-3-c^2}{c^2}=f'(c)=-1.33...$$
then, it must follow that
$$\frac{-3-c^2}{c^2}=-1.33...$$

[woven radish]

Cycadellic

[wicked edge]

make sense?

i can type it more clearly if you need

[pallid ravine]

No ❌ wait its not saying the answer is -4/3 tho

thats like the answer to the previous quesrion

answer to #7 that brought me there

[wicked edge]

oh man

my bad

[pallid ravine]

ur good

[wicked edge]

not sure what that was about

[pallid ravine]

i mustve fucked up then cuz i think the closest answer is - root 3, but it leads me back to the beginning so i messed up somewhere else

😣

[wicked edge]

you messed up on the last step

[pallid ravine]

what is wrong there?

[wicked edge]

arithmetic mistake

[pallid ravine]

oh

[wicked edge]

f'(-1.33...)=-2.6875, not -2

[pallid ravine]

so whats c

[wicked edge]

so, i agree that

with x=c

[pallid ravine]

Yes

[wicked edge]

so, multiply both sides by x^2, and simplify, make quadratic and find the root

and dont forget $\pm$

[woven radish]

Cycadellic

[pallid ravine]

ohh

+/- root 3?

[wicked edge]

good so far

[pallid ravine]

but poditive isnt in interval

[wicked edge]

right

[pallid ravine]

so just - root 3

[wicked edge]

and since we know a quadratic only has two roots, youre done

[pallid ravine]

got it

ok thank u sm

[wicked edge]

np

[pallid ravine]

it brings me back to #1 so i have to check my work previously idk what else i got wrong but now i get

God bless

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[placid verge]

I am not sure if I did this right

[crystal drift]

check your c

and e/f

and i

[crystal drift]

if lim from right side = lim from left, then that should equal the lim

[crystal drift]

i think you just made a small mistake here

[crystal drift]

theres a hole at (3,-2)

[placid verge]

do i switch e and f

[crystal drift]

yup

[placid verge]

i would be undefined?

[crystal drift]

yup

[placid verge]

is c 1

[crystal drift]

that's also right

[placid verge]

thanks

[crystal drift]

np

[placid verge]

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[analog hamlet]

How do you find axis of symetry from vertex form?

[charred sleet]

for a parabola?

it’s symmetrical about the x coordinate of the parabola

[analog hamlet]

Like how do I find it from this equation??

Does the question I am asking make sense?

[unique spire]

$a * (x-h)^2 +k $ where (h ; k) it the coordinate of the point of interest

goddammit

[analog hamlet]

So would (1,3) be my vertex?

Or (-1,3)

[unique spire]

h=-1

[analog hamlet]

Ohhh I see

So would I plug in -1 into the function to get my y int?

[unique spire]

this is how it looks

[analog hamlet]

Sweet, thanks

[unique spire]

since a is negative, there's a max
if a was positive there would be a minimum

[analog hamlet]

Yeah that makes sense

[unique spire]

to find the zeros of a function, solve $0 = a * (x-h)^2 + k$

[woven radish]

nico

[analog hamlet]

3?

[unique spire]

the zeros would be $x=(-1)\pm \sqrt{3}$

[woven radish]

nico

[analog hamlet]

Ohhhhhh, I see

So what would be my y int?

[unique spire]

what do you mean by y int?

[analog hamlet]

Y intercept

[unique spire]

it you're talking about the y intercepting the x axis then it the zeros of a fonctions

where simplified :$h\pm \sqrt{\frac{-k}{a}}=x$

[woven radish]

nico

[analog hamlet]

I'm a little stuck on finding the y intercept now

[unique spire]

i think by axis of symetry: they want the value of h
by vertex : they want the value of k
by y-intercept : they want the value of x where y=0, so the zeros of the function

[analog hamlet]

My zeros are -1 and 3

Do I just add them?

Y intercept = 2

[unique spire]

you sure about that?

y=-(3+1)^2+3=-4^2+3=-16+3=-13 not y is not equal 0 at x=3

$y=a(x-h)^2+k\
0=a(x-h)^2+k\
-k=a(x-h)^2\
\frac{-k}{a}=(x-h)^2\
\pm \sqrt{\frac{-k}{a}}=\sqrt{(x-h)^2}\
\pm \sqrt{\frac{-k}{a}}=x-h\
h\pm \sqrt{\frac{-k}{a}}=x\$

[woven radish]

nico

[analog hamlet]

Are you sure it isn't 2? The line intercepts the y axis at positive 2

In the graph you showed me as well

[unique spire]

oh if you want where f(x) intercept the y axis then solve $y=a(0-h)^2 +k$

[woven radish]

nico

[analog hamlet]

Oh perfect

Thank you very much!

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[sweet veldt]

how to get from the letting w= line to the last line

i understand the rest

[oblique gale]

Isn't $\sqrt{(x-h)^2}=|x-h|$

[woven radish]

qbibubi

[sweet veldt]

room is occupied

[oblique gale]

Oh I'm sorry didn't notice!

[sweet veldt]

also for this one

actually nvm i get how to solve this but im still a bit confused on the last part after doing the substituion with s= in this case or w= in the initial question

like how to go from that step to the final particular solution

<@&286206848099549185>

[sweet veldt]

@slow burrow

from the characteristic line, bx - ay

in what way am i plugging that in after doing the subtition with w to get the final particular solution

is it litearlly just plugging it into w?

wait

[woven radish]

chef

chef

[sweet veldt]

did i do this right

.close

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[fallen mason]

I need help regarding problems b-d on this homework problem. I think a is correct ( I got 6 Ohms for my equivalent resistor) but my professor didnt explain divider rules so well.

[devout snow]

@fallen mason Has your question been resolved?

[devout snow]

@fallen mason Has your question been resolved?

[fallen mason]

Or are they all in series with each other?

[devout snow]

@fallen mason Has your question been resolved?

[devout snow]

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[orchid leaf]

If $Y = \emptyset$ but $X \neq \emptyset$, then there are no functions from $X$ to $Y$.
If $A \times B$ and every element of $A$ is related to no elements of $B$, then $R$ is the empty set.

[woven radish]

Forsaken

[orchid leaf]

Are there no functions of this sort because of the definition of a function? While a relation can have the second set empty and exist as the empty relation?

[stone stump]

well for a function you need that every element x in X appears in some pair (x,y)

for a relation you dont need that

[orchid leaf]

That is what I mean

Is it necessary though that every element x is related to some y? I have seen some definitions of function as necessitating only that there is at most one element from the second set. So there can be none?

[stone stump]

well in that case I wouldnt call it a function. a partial function perhaps

[orchid leaf]

I see so for it to be a total function it can't be at most one but exactly one

[stone stump]

yes

[orchid leaf]

thanks

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[cosmic gust]

Find all triples (a, b, c) such that:\
$a,b,c \in \mathbb{Z^+}$\
$a^3 + b^3 + c^3 = (abc)^2

[woven radish]

Oğuzhan
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cosmic gust]

How can I solve this?

[long sundial]

Is this for a particular course or competition maths?

[kind prairie]

.

[cosmic gust]

It's from a shortlist, I can't solve it

[long sundial]

If it's for a particular course there might be a taught method to solve in which case I'm no help.

Otherwise I'd test some low numbers and try get an idea on what works and breaks. Usually gives an idea for the proof

[cosmic gust]

Hmm okay

[long sundial]

For instance if a=0 you know b=0 and c,=0

[cosmic gust]

No it's positive integers

[long sundial]

Then for a=1 what solutions exist (if any)?

[cosmic gust]

Hmm let me try that

Ask it somewhere else, this is not the appropriate place to ask

And I don't think you are allowed to do that

[long sundial]

<@&268886789983436800>

[devout snow]

@cosmic gust Has your question been resolved?

[devout snow]

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[rotund tundra]

In the definition of existence,
f is a function defined on a deleted neighborhood of x0.
I don’t get why does it have to be a deleted neighborhood,
what does that mean
And why f can be defined on a neighborhood including X0

[rancid grove]

existence of what?

[devout snow]

@rotund tundra Has your question been resolved?

[rotund tundra]

Existence of limits

[rancid grove]

okay. the definition says that it CAN just be defined on a deleted neighborhood

being defined at x0 is not a requirement for a limit to exist

sure it can, but it does not have to

[rotund tundra]

Oh so I understood it wrong, Thanks.
I thought that it has to meet the requirement of being in a deleted neighborhood.

But can I use the definition if I am not referring to necessarily a deleted neighborhood?
Even tho that’s the definition’s assumption?

[rancid grove]

yes!

[rotund tundra]

How come?

Just asking to specify that in my answer so there will not be holes in my proof

[rancid grove]

you can just remove x0 on the neighborhood you consider

consider the deleted neighborhood U around x0 for which f is defined

so now you have the requirement for the existence of an accumulation point

also because with limit, you do not care about what happens AT the exact point but what happens as you approach it. so the numbers around are more important than the number itself

[hollow elk]

Need help

[rotund tundra]

And I don’t care about the neighborhood here
So I can just consider the question in the deleted neighborhood?

[rancid grove]

do not take over other people’s help channel

[rancid grove]

yes

i do not get this though. what are you trying to prove??

[rotund tundra]

I disproved:
if for every g:R->R the limit of [f(x)g(x)] at x->x0 exist
Then the limit of f(x) in x->x0 is 0

With a counter example

[rancid grove]

does the limit of g have to exist?

is that an assumption?

[rotund tundra]

Technically according to the assumption yes

[rotund tundra]

If assumption
Then something

[rancid grove]

hmm weird statement but okay i think what you showed might be correct.

did you mean that you proved that the individual limits exist?

that’s what you have on the first part here?

[rotund tundra]

You don’t think I need to show or state that I assume something that isn’t exactly based on the definition?

[rotund tundra]

That’s my whole answer techinicly

[rancid grove]

ohhh the first half is incorrect

just because the limit of the product exists, the individual limit does is incorrect

can you show me the full statement of what you’re proving?

[rotund tundra]

Why is the first half incorrect?
I based that on
Exist * something = exist

Then the something has to exist

[rancid grove]

you dont know if the limit of f even exists

it doesnt say that limit of f exists

there are a lot of counterexamples for your problem. you can even go as far as neither limit of f nor g exists.

[rotund tundra]

But we have that the multiplication of both exists

[rancid grove]

yes that can happen

[rotund tundra]

Is that possible that
(Doesn’t exist) * (doesn’t exist) = exist?

[rancid grove]

yes

[rotund tundra]

Crazy
Just saw that
Exist / DNE = DNE

[rancid grove]

limits are weird

for your problem, no need to prove anything. just write a counterexample

[rotund tundra]

So if I assume f and g doesn’t have a limit
Then that’s a straight counter example since
Since When saying lim f(x) = 0 the lim is existence which is 0

[rancid grove]

yes

[rotund tundra]

Ok thank you so much
Wouldn’t have realized idk how multiplication between works even without u

[rancid grove]

lmao you are welcome! 👌

[rotund tundra]

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[winged parcel]

How do I solve this? 1,(2) being 1,2222222222... and the answer is 5/18, I just need to understand how to get this result

[worn galleon]

The fraction 1/9 has a satisfiying form as a decimal
You can use that to help you out

[winged parcel]

Just checked, noticed that the fraction 1/4.5 is 0.222222.... , still not sure how to use that to solve the problem

[worn galleon]

Well you want to find 1.222222222..., preferably as a fraction so you can evaluate it with the other fraction

And you know how to write 0.222222222... in fractional form now
And you also know how to write 1 as a fraction
And you can add fractions together
The rest I think you can do from there

[winged parcel]

Ah alright I got it now

thanks 👍

.close

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[meager river]

i Have an algebra question

[meager river]

I am just a high school student, but this question stumped me. I even tried looking it up and i still have no solution.

The question is "Find the equation passing (1,-3) and perpendicular to 4y + 2x = 1

[mighty knoll]

Have you tried anything?

[meager river]

The thing that stumped me was i have to type it in y=mx+b form but m and b have to be integers

I was getting 0.5x - 3.5

[mighty knoll]

m and b don‘t have to be integers unless your question states so

[meager river]

it does

[mighty knoll]

Well then it would be useful to post the whole question

[meager river]

yeah one sec

[winter patrol]

how did you get 0.5 for the slope
it doesn't seem you identified the slope and/or applied properties of perpendicular lines correctly

[arctic temple]

also there can only be one line passing through 2 distinct points

[meager river]

finally figured out how to use macs picture editor lol

[meager river]

i popped it into desmos too

it worked

[winter patrol]

not sure what you're looking at but
y = 0.5x - 3.5
won't be
and doesn't look perpendicular to 4y + 2x = 1

[winter patrol]

you should be able to see that the angle between those lines isn't 90°

[meager river]

https://tenor.com/view/oh-my-gosh-i-think-you-might-be-right-kristen-bouchard-katja-herbers-evil-gif-23151947

[winter patrol]

first identify the slope of:

4y + 2x = 1

[meager river]

just 2 right?

[winter patrol]

no

how are you getting 2

[meager river]

-2?

[winter patrol]

how?

please show me how you're reaching that value

[meager river]

rise/run

[winter patrol]

otherwise i'm going to assume its a guess esp if its wrong

[meager river]

you go down 1 and over 2 so 2/1

[winter patrol]

that's not how rise/run works

you're treating down as +ve,
and doing stuff like run/|rise|

[meager river]

so what should i do instead

[winter patrol]

use proper signs

do rise/run

up is positive, down is negative

you go down 1
the "rise" of the given line will be -1

[glossy dew]

Perpendicular slopes multiply to -1

That's a hint

[winter patrol]

over 2
run will be 2

[meager river]

yes. -1/2 right?

[winter patrol]

yes. the slope of the given line will be -1/2

and the apply properties of slopes of perpendiclar lines to get the slope of the desired line

[glossy dew]

Youd get -x/2 = -1

if x is the unknown slope

[winter patrol]

shouldn't use x for slope here

bad to use the same variable to represent multiple things

[meager river]

but my slope has to be an integer.

[winter patrol]

yes and it will be

do you know the relation between slopes of perpendicular lines?

[meager river]

so the slope of the first one was 2x so, the negative reciprocal would be -1/2. Now i do not know how to make that a whole number.

[meager river]

no

[winter patrol]

firstly slope shouldn't contain x

[meager river]

ok

[winter patrol]

secondly we just established that the slope of the given line is actually -1/2

[meager river]

yes.

[glossy dew]

yeah mb bro

should've used m

[winter patrol]

and you literally just mentioned the relation between the slopes of perpendicular lines

the negative reciprocal

[glossy dew]

Yeah

[winter patrol]

the slope of

4y + 2x = 1
is -1/2

what's the negative reciprocal of -1/2

[meager river]

2?

[glossy dew]

Yep

now substitute that in the slope intercept form

y = mx + b

[meager river]

y = 2x + ?

[glossy dew]

Yeah we can still keep it as b

We'll find out what it is

[meager river]

ok

[glossy dew]

Now the question says it passes through a point right?

(1, -3)

So that point must be a solution of the line equation that we are trying to find

[meager river]

-5?

[glossy dew]

You're saying the value of b?

Wait I'll check

Yes, you are right

y = 2x - 5

That's the answer

[meager river]

YES!

thank you!!!!!!!

[glossy dew]

Now you can close the channel using .close

[meager river]

.close

[glossy dew]

So that someone else can use it

[devout snow]

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[devout snow]

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[still dawn]

for this question, how do you know what to u-sub? i was thinking of doing sinx for my u-sub but the solutions said u-sub was cosx

[misty crest]

well notice that cosx is in the denominator so letting u=sinx wont cancel it

if we let u=cosx then du=-sinxdx

so one of the sinx cancels

and were left with -sin^2x/u du

but

sin^2x can be rewritten in terms of u

which is cosx

@still dawn

does this make sense

[still dawn]

so i guess u kinda have to think ahead when u choose ur u-subs?

[misty crest]

yea

so sin^2x=1-cos^2x

[still dawn]

right right

[misty crest]

=1-u^2

but there’s a negative so

it’s really u^2-1

so (u^2-1)/u

=u-1/u

integrate this

[stuck field]

I think u = cos(x) was enough help already, don't solve it knief.

[misty crest]

he knew u=cosx already

[stuck field]

Then sin^2(x) can be written in terms of u is probably enough.

[misty crest]

he wasn’t responding

[still dawn]

i already did the question

i was just wondering how u go about choosing the proper u

[misty crest]

ok

[still dawn]

ty though

[stuck field]

Preferably you want the terms in your integral can all be broken down into something simpler, an integral that you can do much more easily. And much of the terms should really be just something like "du" when you're choosing u. In this case, when you did u = cos(x) you were able use sin(x) = du and rest all too was easy to write in terms of u.

[devout snow]

@still dawn Has your question been resolved?

[devout snow]

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[restive river]

hi can someone help me solve this using log

[restive river]

like ik it should be log(base 4)16√128 but how do i further solve

[winter torrent]

often with these it's helpful to prime factorize EVERYTHING

and reduce that square root

[devout snow]

@restive river Has your question been resolved?

[restive river]

Simply the second question such that it's 4^x = 4^....
Now you can compare the exponent, it will get you the answer

It doesn't necessarily need log but you can use it I guess

[devout snow]

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[hot steppe]

can someone help

[stuck field]

write it as exp( ... )

[restive river]

what tools can you use

are you allowed to use taylor series and/or L'hopital

[hot steppe]

l'hopital

my friend solved it but i have no clue why he solved it the way he did

he basically just put e^ln(sin^2(2x))/ln(3x)

and then he replaced x with 0 but i know that part i just don't know where he got the "e" from

@restive river

[devout snow]

@hot steppe Has your question been resolved?

[woven radish]

themadchessplayer

[hot steppe]

I don't get it

why is there a L there

[unkempt narwhal]

L is what we want to calculate.

[woven radish]

themadchessplayer

[hot steppe]

would it be possible to explain this with the limit i sent here since that would be easier

@unkempt narwhal

[hot steppe]

omg thank you very much i understand it now

[crisp niche]

yvw

[hot steppe]

but wait

what if a<0

[crisp niche]

it is not possible if you compute limits of your form

the base of the power, must be positive

the function is called: power-exponential real function, that exists only if the base of the power is positive > 0

[hot steppe]

ok again thank you so much the a^b=e^blna helped a ton

[crisp niche]

smiles

[hot steppe]

is that a rule or something?

[crisp niche]

it is the basic formula, which is direct form from the logarithm definition, it is being taught in every course, where one is taught the logarithms

[hot steppe]

so do i use it only when ln is a power of something?

[crisp niche]

you should formulate your question in different way: you should ask, when you may use it, and the answer is:

if you have an expression of the form [ f ( x ) ] ^ g(x)

when you calculate:

limits

or

derivatives

[hot steppe]

so even if it was sin(2x)^cos(x) and then i would have been e^cos(x)*ln(sin(2x)

[crisp niche]

correct

please write this formula in your exercise book, to not lose it in the future

[hot steppe]

ok thank you so very much

[crisp niche]

🙂

[hot steppe]

@crisp niche you still here?

[crisp niche]

for now yes )

[hot steppe]

lim x->infinity

that part

[crisp niche]

let me look at it,

[hot steppe]

ok

[crisp niche]

the result is , surely, 0, but you should write it better, i show you the correct style:

[hot steppe]

it wasnt me

i wanted to ask

why did they remove 16

[crisp niche]

look :

that is elementary, methodological, solution

[hot steppe]

why did you add an extra x^2 and then put 16/x^2-1

[real pewter]

he factored x^2 out

basically divided evry term by the highest power of x

[crisp niche]

I took the x^2 out, to make it possble be simplfied with numerator

[real pewter]

allows you to notice how the functoin behaves asymptotically as x tends to infinity

[real pewter]

you basically divide every term by the highest power of x (x^2 in this case). that either reduces every x term to a constant or a term like a/x^n which would tend to 0 as x tends to infinity, and the constants would not vary. this way you can calculate the limit easily

[hot steppe]

ohhhhhh

so we divided x with x^2 and 16 with x^2

but why -1

[real pewter]

there are other ways to approach this, for example if the denominator and numerator are both polynomials, theres 3 cases:

1. numerator has the highest power of x, meaning the function diverges
2. denominator has the highest power of x, meaning the function goes to 0
3. the numerator and denominator have the same order of x, meaning the function tends to the numerators coefficient of that power of x over the denominators coefficient of that powrer of x. (example: (3x^2+x)/(2x^2+1) considering only the highest power of x coefficients, we get 3/2 which is the limit as x goes to infinity. the sign would vary if x goes to negative infinity only if the power of x is odd

[real pewter]

thats dividing -x^2 with x^2, giving -1

an easy trick (dont think this is how it works, this is only a short cut) is to only consider the highest powers of x on the numerator and demoniator

[hot steppe]

wouldn't 16-x^2/x^2 be the same?

[real pewter]

in this case, you get x/x^2 which is 1/x. "substitue" infinity, and it becomes 0, which si the limit. if you got something like x^2/x it would = x. sub infinity and you get infinity, which is the limit, meaning it diverges. if you had 3x^2/2x^2 you get 3/2, which is the limit

[real pewter]

no because the highest powers of x on the numeraor and denominator are both x^2

[hot steppe]

so you can't divide the highest powers of x with each other?

[real pewter]

so if you divide by x^2 you get (16/x^2 -1)/(1)
limit of 16/x^2 as x -> inf is 0, the limit of -1 and 1 is just -1 and 1 so the function goes to -1/1 which is -1

[real pewter]

wait i think i misunderstood you

[real pewter]

yes, you are supposed to divide every term in the function by the. highest power of x

in this case, he just factored out x^2 which yields the same result

induvidually dividing x, x^2 and 16 by x^2 yields the same result

[crisp niche]

to make it look more elegant )

[real pewter]

when you have only constants and terms like a/x^n finding the limits is simple, because constants dont vary, and a/x^n always goes to 0 as x goes to infinity

and dividing by the highest power of x leaves you with only constants and terms in the form a/x^n

[hot steppe]

im so sorry because i might sound dumb right now but is x/16-x^2=x/x^2(16/x^2)=x/16

[real pewter]

it would be better if you could write/type this out, as you seem to have made a mistake

did you factor x^2 out or divide all the terms by x^2?

[hot steppe]

wait

[real pewter]

also dont be sorry, its allg

math is a bitch

[hot steppe]

it really is

[hot steppe]

ok this is better

[crisp niche]

but the other photo of mine looks better, hence i used previous wway, to make it look nicer

becasue it is how we write books

[hot steppe]

ok i understand it now

[crisp niche]

eng is notmy native one, though )

[hot steppe]

and why do we do this

like the whole thing

[real pewter]

it reduced every term to either a constant, or a term in the form a/x^n
then you consider the effect of the limit on each term
if its a constant, say a
as x -> inf, a is just a
if its a/x^n, as -> inf, a/x^n goes to 0
that way you can eaisly calculate the limit

notice here how shes calculated the limit for each term

[hot steppe]

nono i meant why we found 0 for

[real pewter]

ohh

well do you mean the application of limits in the real world? or

[hot steppe]

no i mean what do we need this in the problem i showed earlier

[hot steppe]

.

[crisp niche]

asymptotes

[real pewter]

does your question require you to draw a graph?
either way, the limit is simply a property of the function

[hot steppe]

yes

[real pewter]

graphically its an asymptote yes

you will see the graph tend to y = 0 (the x axis)

you see, the graph tends towards the axis

asymptotes become very important when sketching

[hot steppe]

wait

what if it wasn't 0

[real pewter]

heres one that tends to x = 1

[hot steppe]

ohhhh

[real pewter]

and heres one that diverges

it actually has an oblique asymptote, see how it approaches being a straight line

[hot steppe]

ok

now

[real pewter]

it would approach a line y=mx+b which you can calculate

some just shoot off to infinty

like the x asymptote

[hot steppe]

can someone explain how y^i works

[real pewter]

thats y', or y prime

the derivative of y

the function you have is in the form f(x)/g(x) meaning youd use the quotient rule to differentiate it and find its derivative

[hot steppe]

wouldnt derivative of y be -2x

[hot steppe]

oh nvm

[real pewter]

uh wait

there

[hot steppe]

nono nvm

[real pewter]

easier to see

[hot steppe]

yeah

[real pewter]

does that make sense?

[hot steppe]

yes

[real pewter]

cool

[hot steppe]

so why would the derivate of y equal to that

the one the person in the photo has equalled it to

[real pewter]

so check this

[real pewter]

.

your function is in the form f(x)/g(x)

[hot steppe]

did they continue using that

?

[real pewter]

f(x) = x and g(x) = 16-x^2

[hot steppe]

yes i know

[real pewter]

so when you use the chain rule formula

[hot steppe]

but they continued using that rule

[real pewter]

you get this

[real pewter]

wdym? they used it and then simplified

[hot steppe]

yeah so this is simplification?

[real pewter]

this is the chain rule substitution

[hot steppe]

ok i got this part

[hot steppe]

im talking about this

[real pewter]

numerator becomes 16 - x^2 + 2x^2 which = 16 + x^2

so the numerator is 16+x^2

[real pewter]

thats just

[real pewter]

this

simplified

they simplified the numerator

by expanding the brackets

[hot steppe]

yeah so they simplified the chain rule part

[real pewter]

yeah

[hot steppe]

ok

would you please tell me how did he simplify it

[devout snow]

@hot steppe Has your question been resolved?

[devout snow]

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[stark coral]

I’ve been at it for probably over 30 mins and I still can’t prove this

[worn galleon]

What have you done so far?

[stark shuttle]

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[stark shuttle]

!show

[devout snow]

Show your work, and if possible, explain where you are stuck.

[stark coral]

i ended up getting it

but its still confusing

1 sec

I don’t understand why we have to use the binomial square formula on the numerator but the difference of 2 squares on the denominator, but if I used diff of 2 squares on the numerator and on the denominator, it wouldn’t work, even though it’s the same thing?

The numerator and denominator could cancel out to 1 if I used difference of 2 squares on both the numerator and the denominator

Yet it also (1-cosx)/(1+cosx)?

[devout snow]

@stark coral Has your question been resolved?

[devout snow]

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[restive river]

does anyone know of other shorthand notations for the trig functions?

[restive river]

feynman came up with these but i want a notation for doing antiderivatives fast, and these are easy to mistake for eachother

[supple knot]

Whatever it is, it's nonstandard and you'd be the only one in your class to use them

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[jolly sierra]

hi

[jolly sierra]

.close

[devout snow]

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[strange fractal]

Hi

[strange fractal]

On the right side of the equation you can see the a^(n+1) and - b^(n+1). Did they come about because v=n was used for a in the first sum and 0 was used for b in the second sum?

[timber pebble]

sorry, which step are you confused about?

[strange fractal]

About the last

[timber pebble]

it would appear they are peeling off the last term of each sum

[strange fractal]

Its the same like the step before but in the last one you can see first a^(n+1) that is added and the last term -b^(n+1)

[timber pebble]

$\sum_{\nu = 0}^n a^{\nu + 1}b^{n-\nu} = \sum_{\nu = 0}^{n-1} a^{\nu + 1}b^{n-\nu} + a^{n+1}b^{n-n}$

[strange fractal]

To get a^(n+1), I have to do n=v in the first sum?

First sum

[woven radish]

jan Niku

[timber pebble]

i guess, i dont like the use of 'do'

doing nothing but pulling off the last term

[strange fractal]

And at b

Its not the same

[timber pebble]

nah, its the same deal

just from the other end

[strange fractal]

hmhhmhmhm

[timber pebble]

were plucking off the first term

[strange fractal]

so we set v=0?

[timber pebble]

$\sum _{\nu=0}^n a^\nu b^{n-\nu+1} = a^0b^{n-0+1} + \sum _{\nu=1}^n a^\nu b^{n-\nu+1}$

[woven radish]

jan Niku

[timber pebble]

its not really set, again, its just pulling terms out

just like say we had uhh

$\sum _{i=0}^5 i$

[woven radish]

jan Niku

[timber pebble]

we could write this as $0 + \sum_{i=1}^4 i + 5$

[woven radish]

jan Niku

[timber pebble]

and pull out the first and last term

[timber pebble]

doing exactly the same thing here

[strange fractal]

Would this be true? $0 + \sum_{i=2}^5 i + 1$

[woven radish]

maizz01

[timber pebble]

yea, youre pulling off the first two terms?

so this is fine

[strange fractal]

yea ogey

[timber pebble]

youre not really doing anything other than applying commutative and associative property

[strange fractal]

yes

I have last one question

other task

[timber pebble]

yea im not sure what the hell this means

ah, no

just write out some terms

youll see

its actually just a reindexing, i think

but, write out some terms

if you cant sort it ping me after you try, but i think youll see

[strange fractal]

Maybe its important to know, the task Substituted l=v-k.
I had that topic never before. Just started in.
As you can see l begins at 0 so I thought if v=k then l=v-k/k-k= 0 but I didnt get the v-1

Above

[timber pebble]

there are some typical techniques

and important results

when youre working with sums

[strange fractal]

I know that I can split sums

And put them together

[timber pebble]

the one referenced here is $\sum _{i=0}^n i = \sum _{i=1}^n i = \frac{n(n+1)}{2}$

[strange fractal]

thats it xD

Yea I see

[timber pebble]

you can derive this yourself, if you want more practice

[strange fractal]

Yesterday, I tried to understand without knowing this rule/technique whatever. It seems like the gauss sum

n(n+1)/2

or not xD

[woven radish]

jan Niku

[timber pebble]

sorry i had i not n

[strange fractal]

np

[timber pebble]

typically youd use geometry

well IDK typically

thats how i thought of it

[strange fractal]

I see

[timber pebble]

lemme see

https://www.youtube.com/watch?v=PzvISL7ajSs

this video does a great job of it

you can maybe convince yourself with enough scribbling you could have landed at the same result

[strange fractal]

Interesting

yea yea i see

Now I understood the formula

[timber pebble]

there arent too many of these "standard" sums so you can pick them up quickly

[strange fractal]

True true

[timber pebble]

theres maybe like sum of n, n^2, n^3

basel problem

not a ton come to mind

[strange fractal]

xD

Thats a topic that makes math interesting tho

[timber pebble]

OH geometric

of course

$\sum x^n =\frac{1}{1-x}$

[woven radish]

jan Niku

[timber pebble]

classic

[strange fractal]

There is difference between finite and infinite series in geometry

I noticed that yesterday

[devout snow]

@strange fractal Has your question been resolved?

[devout snow]

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[sleek hamlet]

Hey guys I just need help with 7b

[supple knot]

,tex .point slope

[woven radish]

dr. matlab plot

[sleek hamlet]

What

[supple knot]

Slope is the same as gradient

[sleek hamlet]

So?

[supple knot]

Did you read this

[sleek hamlet]

Idk what it means

[supple knot]

Which part

f(x)?

[sleek hamlet]

The part where m(x-a)+b

[supple knot]

Why is it confusing

m is slope

a is the x coordinate of some point

b is the y coordinate of that point

[sleek hamlet]

If a is the x coordinate what’s x

[supple knot]

x is a variable

[supple knot]

Same x in number 7

You just have numbers and variables for a, b, m

[sleek hamlet]

I still don’t get it

Why does the equation work

[winter patrol]

did you try rearranging the equation to slope intercept form?

[devout snow]

@sleek hamlet Has your question been resolved?

[sleek hamlet]

I tried

[winter patrol]

can you show your attempt

[sleek hamlet]

K

[supple knot]

you didn't multiply both sides by 12 correctly

$r + s = \frac{t}{12} \Ra 12(r+s) = t$

[woven radish]

dr. matlab plot

[supple knot]

what's 12(r+s) = ?

[sleek hamlet]

Ohh ok

[sleek hamlet]

Omg actually marry me bro thanks so much for ur help ❤️

[devout snow]

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[proper terrace]

im confused on what its trying to ask

[restive river]

I

Have a question

[thin fern]

It's probably because you unnecessarily wrote "a_n = "

[rustic jetty]

It seems like you put $a_n$ before your answer, is that visual or did you actually do that?

[woven radish]

Dork9399

[thin fern]

this channel is occupied

[rustic jetty]

!occupied

[devout snow]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[proper terrace]

its correct

thanks

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[south dust]

I had a bonus question on my alg 2 H quiz and it was simply "x/3 = -9/x" I got "x= 3i√3" do you think this was right? what is did was got rid of the fractions by multiplying their denominators and ended up with x^2=-27 and i simplified using the square root method and got my answer. Did i mess up anywhere?

[south dust]

to go more in-depth with my work
x/3=-9/x
multiply by 3/1 on both sides and get:
x= -27/x
multiply by x/1 on both sides and get:
x^2 = -27
square root both sides:
√-27 = i√27
√27/√9 = 3
√9 = 3 also
giving me:
x = 3i√3

does this make sense, did i go about it wrong?

no one is going to answer?

[tender cobalt]

when you square root both sides you have a ±
so x^2 = -27 turns to x = ±√-27
from this, you should have x = ±3i√3

[devout snow]

@south dust Has your question been resolved?

[devout snow]

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[tender cobalt]

the a0 z here looks like a typo - shouldnt it be a0 instead?

[devout snow]

@tender cobalt Has your question been resolved?

[dim turtle]

na it’s fine

[devout snow]

@tender cobalt Has your question been resolved?

[supple knot]

Yea it's fine. Why is this channel still open

[tender cobalt]

they said "na"

.close

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[pulsar tundra]

i'm stuck on this question, i found this work online, but i'm not sure why x² + x², and (2)(x) are square rooted

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[lean robin]

how is this wrong?

did i distribute wrong?

[winter patrol]

show work

[lean robin]

was getting my phone*

[low nexus]

i'm pretty sure your work is correct?

[winter patrol]

-7i + 8i isn't -i

[low nexus]

(1 + i)(8-7i) = 8 - 7i + 8i + 7

oh wait sorry didn't realize you wrote that

[lean robin]

oh its + i

thanks cant believe i couldnt see that for a while thank you

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[magic sigil]

someone help me pls.

[woven radish]

charlie d amagiglio

[magic sigil]

im trying ti find the beta values for this regression function

i found the values but when i do the system of equation and substitution to find B1, I'm not sure if im correct

[woven radish]

charlie d amagiglio

charlie d amagiglio
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[magic sigil]

this is my beta one^^

and my beta zero is this

anybody gonna help me

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[cerulean hollow]

So if g(7) = 9 and f(9) = 16, then (f ∘ g)(7) =

I'm so confused, how do I do this?

[winter patrol]

do you understand (f o g) function compostion notation

[cerulean hollow]

f o g is f(g(x)) no?

[winter patrol]

$(f \circ g)(x) = f(g(x))$

[woven radish]

ℝαμΩℕωⅤ

[winter patrol]

apply that here

[cerulean hollow]

so

16(9(7))?

[winter patrol]

no

how are you getting that

start by replacing x with 7

[cerulean hollow]

oh

do I replace g with 7 as well?

[winter patrol]

no

g is not a variable

nor are you told g has the value of 7

[cerulean hollow]

wait, so what do I do after f(g(7))?

[winter patrol]

use the info given in the question

[cerulean hollow]

OHHHHH

I see

crazy how I missed that

since g(7) = 9, when its f(g(7)) that is f(9) which is 16 right? @winter patrol

[winter patrol]

yes

[cerulean hollow]

aight thx

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[hidden hinge]

Similarities is that they are all autonomous

[weak cove]

should be easy to graph their direction fields then

[hidden hinge]

ok

[weak cove]

which are going to be invariant under translation

[hidden hinge]

invariant under translation in the 't' direction right?

[weak cove]

if that's what you want to call the independent variable sure

[weak cove]

mhm

[hidden hinge]

how do i use these direction fields to answer the question/?

[weak cove]

well you don't want the whole direction field for the second part

you've desrcibed their similarities and differences from the direction fields in general

or atleast you know how to

right?

So moving on to part 2, each direction field, we plug in this initial value problem and we get just 1 solution curve, no longer an entire field of solution curves

which you can directly examine to answer that part of the question

[hidden hinge]

yes thats true

[weak cove]

So do you understand how to proceed then?

[hidden hinge]

sorta, so im going to compare two lines

[weak cove]

two curves

both functions of t

[hidden hinge]

and the lines go through the point (0,1) where (t,y)

How would i solve this problem without using the direction field? or is that what im supposed to do

[weak cove]

You no longer need the direction fields for the second part

y'=y^2

y(0)=1

is a diffeq you should be able to solve

there is one unique solution

likewise with the other two equations

[hidden hinge]

its my first week in linear algebra

i actually don't know how to solve it

😅

[weak cove]

$$y'=y^2$$
$$\frac{dy}{dx}=y^2$$

[woven radish]

Austin

[weak cove]

If this doesn't help

this is called a seperable differential equation

so search a video on how to solve those if you don't recognize the name

they're very easy to solve

[hidden hinge]

ok

[weak cove]

and then plug in the initial condition

y(0)=1

[hidden hinge]

to solve for c?

[weak cove]

and repeat for the other two equations

[weak cove]

yes

[hidden hinge]

ok will do ty

[weak cove]

No problem

[hidden hinge]

🙏

.close

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[sinful mango]

Zinc has a density of 7.10 g/cm^3. If a cylinder of zinc weighing 36.2g is completely immersed in a graduated cylinder that originally contains 33.79 mL of water, what will be the new water level? Please and thank you for help.