help-28

[regal forge]

Apparantly they want us to identify whether it is a region or not

[sharp dawn]

Jps

[regal forge]

according to notes definition region: if it is an open set together with none, some or all of its bounary points

[wary walrus]

oh ok

[sharp dawn]

C come le ven diagram des ensembles

[wary walrus]

well we got a closed set here

[regal forge]

cool, so just closed

[wary walrus]

it has no interior points

so i don't think it can be called a region then

[regal forge]

so just closed and not polygonally connect and etc.

[wary walrus]

yep

[regal forge]

okay thanks

[wary walrus]

and obviously not bounded so not compact

[regal forge]

I think I understand

can I ask you more when I reach to other parts and mobius transformatinos

[wary walrus]

uh

i just started studying those

so i don't think i'll have the required knowledge

[regal forge]

So you've done the polygonally connected right?

[wary walrus]

well i know what it means

[regal forge]

okay thanks

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[sharp dawn]

Bro tu connais quelques choses sur l'extension algebrique

[sharp dawn]

J'en ais mare de déterminer les polynôme minimales pour les valeurs alpha engendre dans un K algebre

[wary walrus]

2-3 définitions et résultats mais pas plus

[sharp dawn]

Frer j'ai un K[x] UN ss K ev de C engendre par alpha^n

Et n a valeur dans N

La question me demande de trouver les polynômes minimale pour valeur de alpha donné

Tu c l'astuce?

Le poly k-min est note Pk(alpha)

Et il est irréductible

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[sharp dawn]

@wary walrus je rédigé ma solution

[sharp dawn]

,reopen

[sharp dawn]

Au Fuck la photo est encours toute tordu 😭😭😭

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[warped sapphire]

how do i even simplify this

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[brisk obsidian]

Multiply by the conjugate of the numerator.

[warped sapphire]

ok

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[full forum]

Please don't occupy multiple help channels.

[vocal crescent]

I did this work since I thought it might be helpful but not sure where to go from here.

maybe something like this?

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[blissful nova]

looks good

[full forum]

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[tropic portal]

Notation question

[tropic portal]

Suppose I have a set ${A_1, A_2, ..., A_n}$ and I want to sum over all the possible combinations of 3 that can be made with this set. How would I write that with maths notation

[glossy valve]

Realist

[narrow ermine]

Can you give an example of one combination just so I can see what you mean by a combination?

[tropic portal]

(A1, A2, A3)

[narrow ermine]

Ok

[tropic portal]

order doesnt matter

[narrow ermine]

Well you have n things to choose from

And order doesn't matter

So the operation you are looking for is a binomial coefficient $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

[glossy valve]

Azyrashacorki

[narrow ermine]

This gives the number of choices when picking k elements from a pool of n of them

That's not what you were asking though was it

[tropic portal]

It wasnt but thanks for trying

I'm trying to write a random variable as the sum of indicator functions

[narrow ermine]

I doubt there is a specific notation for summing over all such combination

s

[tropic portal]

ok

[narrow ermine]

Someone more knowledgeable in that field can probably give a more definitive answer

[tropic portal]

I have a probability question though based on this

[glossy valve]

Realist

Realist

Realist

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[white jungle]

i need help on starting this

[rain sundial]

Use chain rule

[white jungle]

how would i use chain rule?

new to the chain rule

[torn jolt]

https://www.youtube.com/watch?v=wl1myxrtQHQ

[rain sundial]

d/dx(sin(x+sin2x))= d(sin(x+sin2x)/d(x+sin2x) ×d(x+sin2x)/dx

[white jungle]

im going to give it a go agaib

[empty sapphire]

A nested chain rule is just you keep going inward and inward until you can't take a derivative anymore. So in your case, it'd be

[wide sundial]

I would write (g+h)’f’

For my next line

Unpack the rules one by one

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[twin wolf]

Did I correctly anti differentiate (2x+3)^1/2?

[light sonnet]

Take the derivative to check

[twin wolf]

So I guess I didn’t do it right?

I don’t understand

[light sonnet]

How did you do that?

[twin wolf]

Using differentiation

[light sonnet]

You were suppose to take the derivative of that

[twin wolf]

What?

[light sonnet]

To check your answer

[twin wolf]

Oh

I don’t really understand the process of anti differentiation could you explain it to me

[light sonnet]

It's just the reverse process to taking the derivative

[twin wolf]

Yeah I get that but it’s easier to say it like that than to actually do it

Like what should I be looking for during anti differentiation

[light sonnet]

For the problem you are doing, you're doing a technique called u sub

[twin wolf]

So how does that work?

[light sonnet]

So given f(x) = (2x + 3)^(1/2), you basically did u = 2x + 3, correct?

[twin wolf]

Yeah

[light sonnet]

The proper form is $\int (2x + 3)^{1/2} dx$

[glossy valve]

dldh06

[light sonnet]

So u = 2x + 3
Meaning that $\int (u)^{1/2} dx$

[glossy valve]

dldh06

[light sonnet]

Do you understand up to there?

[twin wolf]

Yeah I think so

I’ve only used the F(x) notation for anti derivatives haven’t used integral yet but I think I get it so far

[light sonnet]

So notice how that integral has both u and x in it but it needs to be uniform, meaning it should have one letter, in this case u, because we replaced the 2x + 3 with u

So to convert that dx in terms of u, you want to take the derivative of u = 2x + 3

So du/dx = 2

Then you rearrange that so it's in terms of dx
du/dx = 2 -> du/2 = dx

So far so good?

[twin wolf]

So we take u’ to be 2?

[light sonnet]

Yes because of what was said here

[twin wolf]

Okay yeah

[light sonnet]

Does this part make sense?

[twin wolf]

Hmm don’t think so

Could you elaborate please?

[light sonnet]

you want to take the derivative of u = 2x + 3
In terms of x, hence the derivative of u = 2x + 3 is du/dx = 2

Then you rearrange that so it's in terms of dx

Basically setting it equal to dx

So that's why du/dx = 2 -> du/2 = dx

[twin wolf]

So if du/dx = 2 then du = 2dx

du/2 = dx okay I get that algebra. But is there a reason we want to rearrange so it’s equal to dx?

[light sonnet]

So notice how that integral has both u and x in it but it needs to be uniform, meaning it should have one letter, in this case u, because we replaced the 2x + 3 with u

$\int (u)^{1/2} dx$

[glossy valve]

dldh06

[light sonnet]

You have both u and x in there

And you just only want one kind of letter

[twin wolf]

oh okay

[light sonnet]

Because the original was $\int (2x + 3)^{1/2} dx$, it's pointless to go back to x (and because we're doing u sub)

[glossy valve]

dldh06

[twin wolf]

Right

[light sonnet]

So have du/2 = dx and the integral $\int (u)^{1/2} dx$, you substitute now, you know what dx is in terms of u

[glossy valve]

dldh06

[light sonnet]

$\int (u)^{1/2} \frac{du}{2}$

[glossy valve]

dldh06

[light sonnet]

Basically what I did there was replace the dx with du/2 because of the statement of du/2 = dx

[twin wolf]

Yeah

[light sonnet]

So the 1/2 is a constant, so that gets pulled out

$\frac{1}{2}\int (u)^{1/2} du$

[glossy valve]

dldh06

[twin wolf]

oh you can do that?

[light sonnet]

Now you just integrate like normal, reverse power rule

[light sonnet]

Which part? The 1/2?

[twin wolf]

Yeah

[light sonnet]

Yes because it's a constant

Same logic with derivatives, you can pull out constants, and you can do the same with integrals

$$\frac{1}{2}\int (u)^{1/2} du$$
So you integrate and get $\frac{1}{2}\left[\frac{(u)^{3/2}}{3/2} + C\right]$

[glossy valve]

dldh06

[twin wolf]

Wait so what do we do with the du again?

[light sonnet]

Basically it's integrating $\frac{1}{2}\int (x)^{1/2} dx$

[glossy valve]

dldh06

[light sonnet]

Wdym

[twin wolf]

Like we have a du next to the u^1/2

[light sonnet]

.

[light sonnet]

You just integrate in terms of u

Meaning you leave things with u

[light sonnet]

Like this

[twin wolf]

When 1/2 is outside of the integral does that mean we don’t need to integrate that

Like this?

[light sonnet]

$\frac{1}{2}\left[\frac{(u)^{3/2}}{3/2} + C\right]\$
That's what you have, but remember that at the beginning we did u = 2x + 3
So you want to plug that back in to be in terms of x again so
$$\frac{1}{2}\left[\frac{(2x+3)^{3/2}}{3/2} + C\right]\$

[glossy valve]

dldh06
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[light sonnet]

Yes exactly

[twin wolf]

Ah okay

[light sonnet]

This is the last step and you're done

You can simplify if needed

[light sonnet]

Because if you notice here the (1/2)(2/3) = 1/3

[twin wolf]

Okay I see

[light sonnet]

Then you end up with $$\left[\frac{(2x+3)^{3/2}}{3} + C\right]$$

[glossy valve]

dldh06

[twin wolf]

What about the 1/2?

[light sonnet]

.

[light sonnet]

$\frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$

[glossy valve]

dldh06

[twin wolf]

Oh

Yeah

But what about the constant C, we don’t distribute it to that to make 1/2*C?

[light sonnet]

No

Because 1/2 * C is a constant too

Meaning you can just write C in its place

[twin wolf]

Ah okay

[light sonnet]

Now take the derivative of this and you'll see that you'll get your original problem

[twin wolf]

Okay

[light sonnet]

But I gotta go, look up u sub

[twin wolf]

Will do, thanks for the help

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[torn jolt]

What does assume the initial displacement have to do with this?

[onyx glen]

the object starts at position 0

[torn jolt]

I got it right but if I used initial displacement = 0 then I would be wrong

[onyx glen]

but you DID use initial displacement=0...

[torn jolt]

Where

[onyx glen]

by not adding anything to the integral

[tawdry iron]

0 to 5

[onyx glen]

otherwise you'd have $s(5) = s(0) + \int_0^5 v \dd{t}$

[torn jolt]

?

[glossy valve]

Ann

[onyx glen]

btw you missed the dt

[torn jolt]

Oh yeah

But they ask for the distance travelled in the first 5 seconds the initial displacement shouldn’t affect it right

[onyx glen]

yes

thats why they tell you to take it as zero

[torn jolt]

But even if it wasn’t zero it shouldn’t affect it

Because they don’t want you to add on the initial displacement they want the distance travelled in the first 5 seconds

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[crystal cobalt]

how to disprove the last one?

[crystal cobalt]

i know the answer is D but yeah still

[cursive badger]

2x-y<y

=

[onyx glen]

dispove 2x - y < 2xy ?

[cursive badger]

2x<2y

[onyx glen]

<=> not =

[crystal cobalt]

yeah more like showing it doesn’t always have to be true

i could try numbers but what’s the “slick way”

if there is any

[onyx glen]

i do not see any slick way

[crystal cobalt]

so i just simplify a bit and test with numbers?

[onyx glen]

yeah ig

[crystal cobalt]

okok thanks

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[twin wolf]

Im finding a formula for the sum $\sum_{i=1}^{n}\left(i^{3}-4i\right)$

[glossy valve]

water beam

[twin wolf]

I wanna know if im correct

so i rewrote $\sum_{i=1}^{n}\left(i^{3}-4i\right)=\sum_{i=1}^{n}\left(i^{3}\right)-4\sum_{i=1}^{n}i$

[glossy valve]

water beam

[twin wolf]

and then

$\frac{n^{2}\left(n+1\right)^{2}}{4}-4\left(\frac{n\left(n+1\right)}{2}\right)=\frac{n^{4}}{4}+\frac{n^{3}}{3}-\frac{7n^{2}}{4}-2n$

[glossy valve]

water beam

[twin wolf]

did I do anything wrong here?

[left bone]

it's the right idea, we can double check arithmetic with WA

[twin wolf]

alright

[left bone]

,w sum from n=1 to k of (n^3-4n)

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by tuzis

[twin wolf]

hm

doesnt look like the same thing

[left bone]

the only difference is the denominator in the n^3 twrm

they have n^3/2 and you have n^3/3

[twin wolf]

ohhh

i meant to have

2 as a denominator

didnt see i put 3 there

okay cool

[left bone]

👍

[twin wolf]

thanks

[left bone]

looks good then nice work

[twin wolf]

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[eager crag]

when can i use y=x to find f(x)=f-1(x)

[neon basin]

You mean f^-1(x)?

$f^{-1}(x)$

[glossy valve]

USS-Enterprise

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[eager crag]

yes

[neon basin]

Graphically, $f(x)$ and $f^{-1}(x)$ are reflections about the y=x line.

[glossy valve]

USS-Enterprise

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[royal halo]

I need help w/ no.7

[royal halo]

I do not know how to start off

[hot herald]

start by writing the equation that represents that first statement

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@royal halo Has your question been resolved?

[royal halo]

/close

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[royal halo]

.reopen

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✅

[royal halo]

is this right?

of what i did

[hot herald]

not quite, you need to be careful with your subtraction

(also write equations on a single line)

[royal halo]

this is confusing

[hot herald]

you need to subtract the whole angle, not just whatever term happens to appear first

[royal halo]

let e try againt

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[lusty leaf]

2 secs late bud

😭 sowy

[formal scarab]

that is like 10 pixels

what's the exponent

[lusty leaf]

x/3

Limit tends to infinity

[formal scarab]

$\lim_{x\rightarrow\infty}(\frac{7x+10}{1+7x})^{x/3}$

[glossy valve]

chlamydia

[lusty leaf]

Yep

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[runic oxide]

2x-1 = |x+10|

[runic oxide]

answer is 11 and -9 or 11 and -3

?

[proper plank]

second

unless? hm

nah i think so

[hollow wharf]

You can plug into the original to check

[proper plank]

|x+10| = ±(x+10) so u just need to plug those two in

[hollow wharf]

Yes it’s the second pair

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[misty raven]

this a trapezium

[misty raven]

?

[odd fox]

Ye

[misty raven]

k ty

[runic oxide]

tysm!

[proper plank]

yw

[proper plank]

do you mean between 0 and T

if so yes

[misty raven]

i meant overall

@proper plank

[proper plank]

between 0-11.8 you have 5 sides no?

u can split it up

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[tacit blade]

hello!

[tacit blade]

can someone help me check and see which answer is correct

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[eager hinge]

why f1 is equal to Tension

[eager hinge]

/rotate

[teal whale]

,rotate

[eager hinge]

.rotate

[eager hinge]

can someone explain

help

why f_1=T

.close

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[spring vigil]

what is happening

[formal scarab]

it wants you to move the arrow to show x>27/6?

[spring vigil]

yeah

oh damn i just realized i put the circle on the 5 not 4.5

andy yeah ik the arrows backwards

[spring vigil]

this should be right

[formal scarab]

yeah

[spring vigil]

careless mistake

[spring vigil]

i was panicking cause i was on my last submission thank you though

./close

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[dense edge]

could I somehow manipulate that top example into an answer that uses a different method than below? or not really

[fast peak]

well you could u-sub

[dense edge]

oh

which method would you recommend for this problem?

[dense edge]

u-sub from here?

[fast peak]

yes

[dense edge]

could you also u-sub from the original?

[fast peak]

yes

[dense edge]

OK

so that would save time

it sounds like u-sub would be the fastest?

i would like to avoid any trig identity bullshit if possible

not sure if it's possible tho..

[formal scarab]

they both end up with -cscx

[dense edge]

like the simple ones are fine but I don't know if my brain can effectively store all of these for an exam

[fast peak]

remembering stuff sucks. and in the end the derivatives all boil down to expressing stuff as sin and cos and using product/chain rule

so they can all be solved using ibp or u-sub

[dense edge]

so you can literally take any trig question, no matter how complex, and break it down to simple sin/cos using u-sub and/or integration by parts?

[fast peak]

in principle, yes

[dense edge]

that would make my life a heck of a lot easier. it's what you do too? or do you memorize all this shit

[fast peak]

I look it up if I need it. because I don't have to write tests on that shit anymore

but it's what I would do if I had to

[dense edge]

i wish exams would give trig tables, but since they don't I think I should use the u-sub / ibp methods to simplify trig expressions into extremely simple sin and cos only

[onyx glen]

i think it's a matter of practice to cement them in your mind

than straight memorization

[dense edge]

you know all of them by heart? or still sometimes forget a few

[onyx glen]

all of what

[dense edge]

these

[onyx glen]

line #6 is wrong btw int tan(x) dx is not sec^2(x)

[fast peak]

only students know stuff like that by heart

[onyx glen]

lines 7 and 8 are redundant

as are most of these integral-derivative pairs

derivatives of sec and csc i don't bother memorizing cause i have never found a use for sec where 1/cos would not do

[dense edge]

oh woah, you are right. i copied these from somewhere else, i gotta remember where..

might have been organic chemistry tutor.. i tend to find the most mistakes on his channel

[onyx glen]

why do you have the derivatives of cot and -cot on separate lines too

it's strange to me

why not also include 2 cot, 3 cot, 4 cot and 42069 cot as well

[dense edge]

lol, i think it was just for practice, and for different +-

i may be missing some identities too

[onyx glen]

weak justification

[dense edge]

why? i have all my notes here it's just to practice and write them down https://web.goodnotes.com/s/4MqLSJxlN9qvdabDYidlN8#page-8

[onyx glen]

it takes up space

[dense edge]

and they sync on the web in real time with GoodNotes on iPad

[onyx glen]

it shows you have not grokked the linearity of the derivative

[dense edge]

i will probably make flash cards out of these identities for furhter review

hoping to make it through a good chunk of these today

i watch at 2x speed but with notes it about averages out to same speed

[spice orchid]

you need to spend way more time on the first couple tbh

rather than trying to get through as many as you can

[dense edge]

yeah, I'm re-watching the first one

3B1B helped me out a lot as well yesterday

in our course we learned Riemann Sums first day

good for visualizing the approximations, but we don't really use it in calculus, as we want the exact values

but I suppose the formula still holds true, if n is approaching infinity

is there any reason in particular that FTC Part 2 is taught before FTC Part 1?

like why not switch indefinite to Part 1, and definite to Part 2?
since definite is taught after indefinite

[onyx glen]

augh my internet

which part is which again

[dense edge]

mine too, I think Discord bottleneck

the message takes a while to send, I see a grey text for about a minute and then it appears

[onyx glen]

yeah can you remind me which part of the ftc is called #1 and which one is #2?

[dense edge]

wait, i thought FTC 2 is when there is a definite integral? but it's when limits are constant

FTC 1 includes a variable for a limit

[onyx glen]

first line is stinky

should be $\dv{x} \int_a^x f(t) \dd{t}$

[glossy valve]

Ann

[dense edge]

maybe my prof has been teaching this wrong, I swear I heard definite integrals are FTC 1
indefinite integrals are FTC 2

[onyx glen]

"definite integrals are FTC1" is oversimplification

[dense edge]

we learn FTC2 before we learn FTC1

that appears to be quite common

[onyx glen]

idk like

it's probably bad naming that got stuck

fifty-fifty chance and all

[dense edge]

lol, another minor/annoying math thing that is set in stone now

[dense edge]

why not f(x)?

why f(t)

oh you started with d/dx

hmmm

I gotta think about that..

so this is wrong?

[wicked aurora]

Yes

x cannot both define the interval and move across it

[dense edge]

amazing. yet this is first result on Google Image search...

Google should probably fix this and give less PageRank to CueMath

"the math expert"

https://tenor.com/view/disappointed-disappointed-fan-seriously-what-are-you-doing-judging-you-gif-17485289

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[bright jewel]

I need help on this

[bright jewel]

I don't understand why if I add 0.5333 and 0.6500 it doesnt give me what it lists is my school thing messed up or am I?

[wicked aurora]

How did you get 39?

[bright jewel]

by adding all known frequencys

[wicked aurora]

I don't think that is how you are supposed to do it

It doesn't say no 3 is any kind of total

I think you should compute the relative frequecy first and use that to find the frequency

[bright jewel]

ok It was timed and I would like to understand it better

It said its 21 instead of 39

how did it go to that because someone in the past on this discord told me to add em all up

to find the unknown frequency

[wicked aurora]

Okay, there is only one possible value of the third relative frquency. Find it

[bright jewel]

?

thats 0.3500

but I'm asking how did they find 21 for the frequency

I know how to do the rest entirely

I just don't understand fully the frequency stuff

ik RF and CRF

[wicked aurora]

Can't you determine frequency from RF?

[bright jewel]

I'm not sure thats my question rn

ik for sure its 21 and everything else is correct

to determine RF you need frequency

[wicked aurora]

So, relative frequency seems to be proportional to the frequency, which you also can tell by their names

[bright jewel]

so are we both lost on how to solve it or you know and your trying to explain? I've melted half my brain cells already trying to figuire out other stuff within the lesson today sorry if I'm getting confused here

[wicked aurora]

the second alternative

sorry for the delay

[bright jewel]

no worries

also I got it now I think

I forgot a step 39 is correct in a sense

[wicked aurora]

You need to solve a proportion

[bright jewel]

I just forgot to do 60-39

to get 21

[wicked aurora]

Oh, haha

You could also multiply 0.35 by 17/0.2833

[bright jewel]

oh

ty tho

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[chrome idol]

I am not sure how much proof would be needed for this.

By definition of bounded, $(a_n)$ is bounded if there exist $M >0$ such that $|a_n| < M$ for all $n \in \mathbb{N}$. Negating this would give: "$(a_n)$ is unbounded if for all $M > 0$, there exist $n \in \mathbb{N}$ such that $|a_n|> M$" which means that either $a_n > M$ or $a_n < -M$"

[glossy valve]

KN

[chrome idol]

actually, it should. It is proof by definition.

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[stray raven]

hello

can anyone help me with the proof

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@stray raven Has your question been resolved?

[gritty rose]

Use the hint

[stray raven]

i didnt quite get how to start to move with it

can yoy help me

[gritty rose]

Well what's the theorem and proof

Understand that one first

[stray raven]

what he means by that f(theta) = Summation(- infinity to infinity) f hat(n) e^-intheta converges to f(theta) for the condition mentioned at the end of question

[gritty rose]

Uh

Your text is confusing

Just screenshot the thm and proof

[stray raven]

or the limit tend to infinity Nth partial sum f theta converges to f theta

[gritty rose]

Which part of that exactly don't you get?

[stray raven]

nothing its clear i just dont know how to syaty

start

[gritty rose]

Jeeze uhhh

[stray raven]

this is the theorem that gave the corollary

[gritty rose]

Go back to the beginning where you do understand something

[stray raven]

is it actually easy?

lol

[gritty rose]

Do you know why uniformly continuous implies the limit is continuous

No it's not easy necessarily

Helping you is hard because you're not being precise what you want help with

[stray raven]

yeah because lim x tends to c f(c) exists

for any c on interval

and using uniform contionus

for all |x-c|< theta

f(x)-f(c)< epsilon

which implies limit exist

[gritty rose]

That's not the right definition for your theorem

You have a sequence of functions

Find the definition that applies to the sequence converging uniformly

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[torn jolt]

Please I need help with this

[torn jolt]

I never use limits in ivt so I’m confuse

<@&286206848099549185>

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@torn jolt Has your question been resolved?

[torn jolt]

<@&286206848099549185>

[next steeple]

rebonjour (rebonsoir ?)

f est continue et admet pour limite + infini et - infini dans l'intervalle ]a;b[

ok

pour la question 1), que veut dire que f(alpha)f(beta) < 0 ?

||l'un est strictement positif, l'autre strictement négatif||

[torn jolt]

Ça veut dire qu’ils ont deux signes opposé

[next steeple]

exactement

[torn jolt]

Donc je dois montrer cela ?

[next steeple]

oui

[torn jolt]

Je dois poser un f(x) qui vérifie ce truc ?

Non attend mdr

[next steeple]

tu sais que f(x) -> + infini quand x tend vers alpha

et f(x) -> - infini quand x tend vers beta

et f continue

[torn jolt]

Alors f(alpha) positivé et f(beta) négative ?

[next steeple]

hm non

x est dans l'intervalle ]alpha;beta[

alpha et beta sont donc exclus de l'intervalle

[torn jolt]

Ah oui c’est vrai

Le prof il avait dit qu’il faut utiliser la définition de la limite dans cet exo mais on a jamais vu ça car je suis en terminal

du coup j’ai chercher mais la définition en l’utilise l’orque la limite = un nombre

Pas l’infinie

[next steeple]

ici

ça veut dire qu'il existe x dans ]a;b[ tel que f(x) soit aussi petit que l'on veut, mais aussi f(x) aussi grand que l'on veut

puisque f(x) tend vers les deux infinis

[torn jolt]

oui je comprend

Je dois trouver cet alpha et ce bêta qui montre cela ?

[next steeple]

tu dois motnrer qu'ils existent

[torn jolt]

Ta pas une idée comment le montrer je suis vraiment confuse ?

[next steeple]

tu peux remplacer x par alhpa et beta

[torn jolt]

Ouii bah en faite j’essaye d’utiliser tvi

[next steeple]

ah oui, ça marche aussi

[torn jolt]

car avec la 2 question c’est forcément ça

[next steeple]

oui

en fait pour la première question, puisque tu peux prends x tel que f(xà soit aussi petit que tu veux

tu peux dire que tu prends f(x) très petit négatif

enfin, je sais pas combien tu dois faire avec la rigueur

[torn jolt]

ahh je vois

Mais le problème que j’ai ici c’est que l’indice que le prof m’a donné c’est d’utiliser la définition de la limite

Mais je vois pas comment je peux l’exploiter icic

[next steeple]

dire que f(x) tend + infini quand x tend vers a

avec f continue

avec x dans l'intervalle ]a;b[

ça signifie que pour tout réel M, il existe x dans cet intervalle que f(x) >= M

[tribal oxide]

Oh un random

[torn jolt]

En mode je dois utilise cet définition

[next steeple]

cocuou

[next steeple]

ok, ben tu peux prende des e arbitraires

[torn jolt]

Vous êtes tous français ici haha

[torn jolt]

Epsilon ?

[next steeple]

oui

[torn jolt]

Et le l ?

[next steeple]

le l c'est la limite non ?

[torn jolt]

Non icic dans l’exercice c’est l’infinie

Mais attend t’es en quel classe ?

T’es en terminal toi aussi ?

[next steeple]

oui oui, c'est l'infini

bon, désolé, je dois partir

[torn jolt]

Okay c’est pas grave

[next steeple]

je pourrais pas t'aider plus que ça

[torn jolt]

merci pour l’aide quand même

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[torn jolt]

No

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[lucid sierra]

if log(3) = a and log2(10) = b then how do i simplify log 675 (you have to put both a and b)

[lucid sierra]

ive got into
log 675
log (3^3)
3A + log (5^2) and then idk what to do

[fathom saddle]

675 = 3^3?

Does 675 factor into 3s and 10s lol

[formal scarab]

675=3^3*5^2

you've used a=log3, so what about b

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[digital fractal]

Would it be differentiable if x = 0, f(0) = 5?

[digital fractal]

I'm hesitating between B and C here

[teal nimbus]

Have you computed the left and right hand limits of the difference quotient?

Or tried to at least

[digital fractal]

yes

I plugged in -1 and 1

but since it's in the absolute value

it would be 4

limit from left and right both exist

but are not equal to 5

[atomic blade]

Wdym plugged in -1 and 1

[digital fractal]

limit of 0 approaching from the left and right

yea that's what I tried

[atomic blade]

I mean you'd need to do f(-1 + h) ... versus f(1 + h) ...

[digital fractal]

I actually did lim from the left and lim from the right

-1+h and 1+h would give 4+h for both

[coarse elbow]

What do the points -1 and 1 have to do with this?
Did you try calculating and simplifying the limit in the question?

[atomic blade]

Show me your work

[digital fractal]

well it's asking for x=0 , so I tried getting the limit from the left and right

[atomic blade]

Because you'd need to define the limit differently for each side

[coarse elbow]

The expression you need to look at is $$\frac{f(h) -f(0)}{h}$$

[glossy valve]

Morrow

[atomic blade]

Since |x| is equivalent to x for x > 0, likewise equivalent to -x for x < 0

[digital fractal]

hold on I'm a bit confused here

all I did was f(x)=5-|x|

f(0)=5-|0| = 5

f(0+h)= 5 - h

[coarse elbow]

Not quite

h can be negative

can't just remove the absolute value

Unless you're looking at h > 0 first

in which case that's fine

And then you check for h < 0 afterwards

[teal nimbus]

Yeah but to do that you may assume some properties about the "derivative" of f, which is what you're looking for already...

Like the derivative could vary a lot around 0, so that may be the best approach for any function if you don't know how it behaves

[digital fractal]

but all I'm looking for is whether the function is differentiable or not

[coarse elbow]

Well continue this line of thought then

[digital fractal]

ok so would I just do 5 - h -5 /h?

which would be -h/h

but that doesn't really help right?

[coarse elbow]

And that's -1

If we're assuming h > 0

[digital fractal]

yes

[coarse elbow]

(specifically h =/= 0)

[digital fractal]

because it cannot be 0 right

[coarse elbow]

yeah we're first looking at h -> 0 from the positive direction

And we found that the limit is -1

We need to check the case where h -> 0 from the negative direction, and see if we get the same limit

[digital fractal]

right

• • 1 / -1

[coarse elbow]

lol discord formatting

[digital fractal]

so positive 1 / - 1

which would still be -1

so the limit does exist at -1

as long as it is not 0

[coarse elbow]

I think you messed up there

[digital fractal]

did I?

[coarse elbow]

f(h) = 5 -|h| = 5 - (-h)

[digital fractal]

it's -h/h right?

[coarse elbow]

So f(h) - f(0) = (5 + h) - 5 = h

[digital fractal]

oh yes I did then

[coarse elbow]

So it's h/h

[digital fractal]

right

so it's 1 from the right and -1/-1 = 1 from the left

[coarse elbow]

-1 from the right

[digital fractal]

1/1?

[coarse elbow]

from here

-1 for h > 0, and 1 for h < 0

[digital fractal]

ok so the limit does not exist then

because it has to be equal from left and right

[coarse elbow]

Yeah, the limit does not exist

[digital fractal]

ok but how do I test differentiability then?

I know it's non-differentiable if there is a sharp corner

[coarse elbow]

Well what's the definition of differentiable?

[digital fractal]

if the left and right limit exist?

wait no

if left and right limit = f(x) right?

but if left and right are not equal, then it can't differentiable

[coarse elbow]

It's's just that the limit you calculated exists (if we're talking about differentiability at x=0 specifically)

And we found that it doesn't

[digital fractal]

and it doesn't

so the answer should be B then

[coarse elbow]

No, B says that the left and right hand limits don't exist

they do

[digital fractal]

oh they both exist but the limit does not

[coarse elbow]

Yeah

[digital fractal]

ok it's very specific

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[torpid oxide]

Hello this might be an easy question because I just started calculus but how can I say that this is equal to positive infinity without a calculator ?

[silk rock]

x → 1⁻ ⇒ x < 1
Then
x² < 1² = 1
x² - 1 < 0
1 - x² > 0

Also, remember that in Limits, 1/0 = ±∞

[torpid oxide]

Ok I understand now thank you so much

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[pine rampart]

are these traces of a cone or a hyperboloid in one sheet?

[pine rampart]

the traces in x=k make me think its a hyperboloid in one sheet because the smallest radius the ellipses get still has some area (isn't a point like it would be in a cone)

but the traces in z=k make me think its a cone because of k=1, the intersecting lines

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[muted kite]

hii!, can someone double check my solution if I did it correclty? I was supposed to find the arithmetic series. Thank you!

[tulip marlin]

use a calculator. 30 terms isnt many.

[muted kite]

hmm? which item are you referring to if may I ask?^^

[tulip marlin]

the ones which are computing the 25th term or something

you just go 2, 5, 8, 11, 14, ... for example

add 3 each time until u hit the 25th one

[muted kite]

I did itt, it was -523

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[muted kite]

hello?? can someone confirm

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[late gorge]

how do i make a function that gets 4 times larger and it starts at 20k

[late gorge]

so

it would be 80k at 2 time

and then 320k

is it like $4^x$ something?

[glossy valve]

孙悟空

[late gorge]

because

[rapid rain]

Up to a multiplicative constant

[late gorge]

what does that mean?

[rapid rain]

Like $C\cdot 4^x$

[late gorge]

$20000*4^x$?

[glossy valve]

孙悟空

rafilou2003

[late gorge]

so if

[rapid rain]

If 20k is your money at x=0 then Yes

[late gorge]

1= 20000, 2: 80k, 3:320?

ok thank yoiu

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[livid nova]

how did he do this help

[reef dust]

same base denominator

they wrote 2/3 as 4/6

because u can only add or subtract fractions with the same denoninator

then they simplified the denoninator

by dividing both the numerator and the denominator by 3

[livid nova]

oooohhhhh

ty very much

[half mist]

He muliplied 2/3 num and denominator 2 times

so he got same denominators

[livid nova]

oh ok

this pls

@half mist he went ofline cna u help

[reef dust]

what do you notice about the numerators

[livid nova]

2 add

eveytime

[reef dust]

expand on that

wdym

[livid nova]

oh numeraotr

sorry i read id denomitaor ...............

so

its weird patern in numeraotr

[reef dust]

1 3 7 13 21

from 1 to 3

what is added

[livid nova]

2

[reef dust]

3 to 7

[livid nova]

4

[reef dust]

7 to 13

[livid nova]

6

[reef dust]

13 to 21

[livid nova]

8

[reef dust]

now do you see it

+2, +4, +6, +8

whats next

+?

[livid nova]

10

[reef dust]

so 21 + 10

that is ur numerator

as for the denominator the pattern is pretty obvious it just adds by 2

[livid nova]

oohh ty

one more question

[reef dust]

its a quadratic equation

are you famikiar with those?

[livid nova]

uh i dont think so

maybe i am but dont knwo name

[reef dust]

its in their working out

they did middle term split

if you arent familiar with it, learn about quadratic equations and middle term split first

i have to go for now though

[livid nova]

oh ok thank you

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[reef iron]

Why are these not equal?

[gritty rose]

cos(0) = ?

[reef iron]

1

[primal thicket]

the integral of cos isn't cos?

[gritty rose]

oh that's the mistake he made

[primal thicket]

lol okay i thought i was going crazy good

[reef iron]

oh I see idk what happened there

Thanks

[primal thicket]

once u use sin it should flip the signs

[reef iron]

yh ik

the mistakes I make are always simple things I for some reason miss

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[inner finch]

I could use a hint for what I should research?

I know the equation for tangent planes is

And so I thought the intersection was when f(x, y)=g(x, y)

So, I thought to put the two tangent plane equations equal to each other and then you'd somehow end up with the equation buuut no

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[inner finch]

Nope..

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@inner finch Has your question been resolved?

[inner finch]

Nope..

[fast peak]

#help-5 message

see the discussion I linked there

[inner finch]

Yes, he's my friend 😂

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[spark field]

We haven't really talked about how the amplitudes, period, or midline truly affect the graph yet, but it's on my homework assignment. Could I get some help with this?

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[spark field]

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[spare ingot]

Do holes in a rational function get included in the domain?

[spare ingot]

For example, would the -2 get included or not?

because when simplified all the way, x+2 isn't in the denominator

[rugged harbor]

No, it's not in the domain, even though it simplifies. You are studying the first function, not its simplified version
For example f(x) = x/x has as domain R - {0}

[spare ingot]

gotcha

[rugged harbor]

[spare ingot]

so the domain would be (-inf, -2)U(-2,2)U(2,+inf)

[rugged harbor]

In this case yes, well done

[spare ingot]

thanks 🙂

[rugged harbor]

You did everything

[spare ingot]

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[frail berry]

just a simple question real quick. How do I calculate the fourth number, when I know the average of four numbers, but I am only told three numbers?

[light sonnet]

Do you know the formula for average?

[frail berry]

Im not sure what you mean, but I know how to calculate the average of numbers given

[light sonnet]

There's a formula

I'm trying to get the OP to recognize this

[last gazelle]

sorry boss

[frail berry]

I do not speak english as my first langauge so please dumb it down

[light sonnet]

What was your first language

[frail berry]

Finnish

[light sonnet]

Did you learn a formula for average?

[frail berry]

this paper does not have any formula for it

I can give you the numbers if that would help to make it easier

[light sonnet]

But did you learn one?

[frail berry]

No

[last gazelle]

oh you have three numbers?

[frail berry]

Yes

[last gazelle]

do you know the average?

[frail berry]

but I have an average for four numbers

yes the average for the four numbers is 8

[last gazelle]

^ then do you know the formula for average?

[frail berry]

I only know how to count the average of numbers given

if thats what you mean

[last gazelle]

first what three numbers are you given?

[frail berry]

5, 10 and 12

[light sonnet]

Do you know how to calculate average? Were you shown a method to find average?

[frail berry]

yes Ill try to write it with the keyboard

ill use the numbers I said as an example

__1 * 5 + 1 * 10 + 1 * 12 __
3

[last gazelle]

that's the average of the three numbers but we are given the average of four

[frail berry]

yes I do not have the formula for that

[last gazelle]

lets says we took the average of 5,10,12, and some other number x

[frail berry]

It was not taught nor is it on the paper

[frail berry]

yes

[last gazelle]

how would you do this ^?

[frail berry]

I think like so

1 * 5 + 1 * 10 + 1 * 12 + 1 * x
4

[light sonnet]

The process for finding average for 3 numbers, 4 numbers, even 100 numbers is the same, you just increase it

[light sonnet]

Yes and that equals the average given

[last gazelle]

yep! now that's the average of those four numbers, and we also know it must be 8

finding that number now is easy! 🙂

[frail berry]

sorry I do not understand? I dont understand how I can calculate it when I do not have the fourth number

[light sonnet]

You have an equation

Do you know how to solve equations?

Like for example 3x + 4 = 19

Can you find x?

[frail berry]

yes it is 5

[last gazelle]

yep! but now we have $\frac{5+10+12+x}{4}=8$

[light sonnet]

Denominator is 4

[last gazelle]

oh yeah good catch

[glossy valve]

Stumpman

[frail berry]

okay I understand now

thank you 👍

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[stray oasis]

How do i do b

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[green saffron]

Set them equal

Like x^2-x-9=2x-5

[stray oasis]

Ye

what else

[elder pumice]

Move one equation to the other side such that it’ll be something like
ax^2 + bx + c = 0

[stray oasis]

x^2+x-14=0?

No

x^2+3x-4=0?

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[cobalt bay]

anyone know how to find the equivalent resistor of this problem?

[light sonnet]

Do you know your resistors in series and parallel rules?

[cobalt bay]

yes

i cant figure which one is in parallel or in series

so r6 and r5 are in series

[light sonnet]

Move the wires

[unkempt sparrow]

Start by the obvious ones and replace them with an equivalent.

r3 and r4 are in parallel. That will give you a new resistor. then r2 and this new one will be in series. You replace r2 and the combined r3, r4. Now this one is in parallel with r1...etc.

[light sonnet]

You can move things around like this

[cobalt bay]

ohhh