help-0

[median oar]

Actually

Do you understand what I mean by this

Yeah

If the question is the purple one

That’ll be a<x<b

If the question was the grey one

It’ll be x<a or x>b

Where a,b are the roots of the quadratic

Look at the quadratic right

If I asked what x values has the curve under the x axis

That’ll be between -3/4 and 1/2

No don’t do that

I’m showing you how you can do it without graphing

If I asked what x values has the curve under the x axis

That’ll be between -3/4 and 1/2

That’s the question you just did

[median oar]

That corresponds to the purple part

That’s in the form a<x<b

In this case it’s -3/4<x<1/2

So this answers the question: for what values of x does it satisfy 6x² + 5x - 4 < 0

(That’s your original question)

Ok now we can ask a different but similar question

If i instead asked: for what values of x does it satisfy 6x² + 5x - 4 > 0

(The equality sign has been flipped here)

We are looking for what x values, has the curved above the x axis?

No don’t try it

If we look at this graph, this new question is the grey one

It’s very hard to glean the answer in this form

[median oar]

Look at the equation for the grey one

That’s the new question

Now look at the graph

It’s on the left and on the right

That is in the form x<a or x> b

In this case it’s x<-3/4 or x>1/2

[median oar]

This is the answer to the question

[median oar]

.

So that’s what I mean by Then determine whether you have a<x<b or x<a or x>b

Those are the 2 types of answer forms you can get (aside from degenerate and non 2 solution quadratics)

Also if you were asked what x satisfies x² > 0

It’ll just be all real x except 0 since 0 !> 0

Now, the methodology to solving these inequalities without graphing is first finding the roots, then finding which form you need based on the inequality in the question, and whether the parabola is face up or face down

Step 1: find the roots

You could think of this like that

Or you can think of the x² = y graph

All x except 0 is above the x axis

That’s the root

Now we want it above the x axis

And we want it for a face up

That means it’s in the form a<x or x>b

Where a, b are the roots

(Roots in this case is 0 and 0, a double root)

Hmm

Which part

[median oar]

This jump?

Don’t solve it

Ok

A little

Consider this parabola

Let’s call it f(x)

Find x values that satisfy f(x) > 0

This means x<a or x>b

Yeah?

That’s when the curve is above the x axis

[median oar]

That’s this

And if we ask f(x) < 0 it’ll be inside, a<x<b

That is exactly what f(x) > 0 means

That can also be true

In which case we can also ask y>0

If you choose it to be true then it can be true

f(x) can equal h

Or s

Or 𝚽

Or λ

I do study it at uni so I suppose so

We use x and y because it’s convention to do so in 2D

We could use a and b for our axis

We could have 3a² + 2a - 4 = b

We’d then write f(a) = b

But also it could be g(a) = b

It’s because y is not necessarily true

But f(x) indicates it’s a function of x

The quadratic

(Note g(x) also means it’s a function of x, the f is just convention)

It’s usually not very well explained

A lot of people (including me) struggle with the concept of algebra itself

Of how different letters can mean all sorts of things

Anyhow, yes

Suppose this

If we are asked f(x) > 0

Notice that it’s no longer x<a or x>b

Yep

So asking about f(x) > 0 or f(x) < 0

And whether the parabola is face up or face down

Will flip the form of the answer

That’s why it’s the 2 things you have to consider when doing these questions

You don’t have to graph it

But you do have to correctly deduce the form of the answer

Yes

That’s why

[median oar]

.

It doesn’t matter

Use any preferred method to solve for the roots

Factorisation works as well if it can be done

I wouldn’t try factorise that

4 and 6 has too many factors and 5 is too close to 4 and 6

It looks like there’ll be a lot of possible options when factoring by inspection so I wouldn’t waste my time doing that

I’d just slam it into quadratic formula and call it a day

Yeah speed wise I’d recommend quadratic formula

Yeah it’s not too bad

If you’re slow doing it you practice your arithmetic

Quadratic formula is not scary

Cubic and quartic formulas are scary

You do practice problems

I’m sure you can find a ton of question banks for arithmetic online

Then do more practice

Practice problems will increase speed

A thorough understanding of all the concepts will also allow you to find shortcuts

👍

[lone heart]

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[rich basin]

How do I should that there exist a vertical asymtope at x = 0?

[chrome salmon]

Is there?

[rich basin]

isn't there when you graph it

[chrome salmon]

I don't think so

[chrome salmon]

Is that asymptote?

[rich basin]

no

[chrome salmon]

Then what is it?

[chrome salmon]

Ok

[rich basin]

so how do we shouw that it is not differentiable at 0?

[chrome salmon]

You can show if right and left side differentiation are not equal

[rich basin]

what are the left and right differentiation?

[lone heart]

@rich basin Has your question been resolved?

[median oar]

If you differentiated f(x) you’d find that the derivative DNE at x = 0

It’s not defined since you have /0

[lone heart]

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[loud skiff]

Hi so for part A, can I just subsitute (x-1) into the maclaurin series for e^x?

[loud skiff]

I remember my teacher saying something about not being able to subsitute monomials into macularin series, only monomials

@mellow saffron

ooops

<@&286206848099549185>

[chrome salmon]

(x-1)²

[loud skiff]

yes thats what i meant

[chrome salmon]

Alright

[loud skiff]

Would that work?

[chrome salmon]

Yes

[lone heart]

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[terse cargo]

find a formula for these sequences

[chrome salmon]

First is not hard

[terse cargo]

what about the second one

[keen plinth]

second is not hard

[terse cargo]

How do I do that one

[chrome salmon]

Think

[tacit arch]

some combination of n, 1/n, -n, (-1)^n

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[alpine sable]

I realize this is kinda basic question but how does this argument even make sense?

a multiple is any integer (not restricted to positive integers) that's being multiplied to the number

[fallen verge]

Can there be a multiple of a number between 0×x and 1×x?

[alpine sable]

no

[fallen verge]

Since we know that b>1, then b-1>0

we can also say b>b-1>0

So b-1 is in between two consecutive multiples

Therefore, b-1 cannot be a multiple of b

[alpine sable]

i don't get what you mean by b-1 is in between two consecutive multiples

[fallen verge]

Consecutive multiples are like 6 and 12 (in terms of 6) or 27 and 36 (in terms of 9)

[alpine sable]

mhm i get that

[fallen verge]

Basically you know that a multiple of a number cannot be between two consecutive multiples

[molten pivot]

Try factoring out the b:
b-1 = b(1-1/b)
When is 1/b an integer?

[alpine sable]

ah that make sense

[alpine sable]

hmm wait i get the multiple part but i don't see how you're tying things together post that

"Since we know that b>1, then b-1>0
we can also say b>b-1>0"

This is fine

[fallen verge]

Perhaps try dividing by b

Maybe that will make more sense

[alpine sable]

ah wait you mean b is a multiple

and zero is another multiple

[fallen verge]

Yes

b=1×b and 0=0×b

[alpine sable]

and b-1 can't be in between two consecutive multiples, wait why so?

yes

oh then it won't be an int

[fallen verge]

Exactly

[alpine sable]

okay cool ill try to work with that logic to eliminate other choices and see in the question

thanks

[fallen verge]

To be honest, its easier to find the answer rather than use process of elimination

[alpine sable]

Yeah i mean i just picked a bunch of numbers and kinda solved it

but i was curious how you'd do it mathematically

[fallen verge]

Yeah you can do that

For proofs, a lot of disproofs are just "i found an example"

[alpine sable]

Oh lol so you think i should just resort to numbers instead of actually trying the mathematical way?

it felt like cheating by just using numbers lmao

[fallen verge]

For example one of the shortest proofs is a counterexample to a theorem, lemme see if i can find it

[alpine sable]

i tried a = 3 and b = 2 and another combination and got b-1 quite trivially but it didn't seem right to do that

[alpine sable]

Interesting

[alpine sable]

It feels like i'm leaving the hard part for the examiner lmao to verify it holds for all reals because i'm just bashing

[fallen verge]

Basically it disproved the fact that a sum of fifth powers must have at least 5 terms to sum to another fifth power

That was the whole proof

[alpine sable]

Lol, i guess that should be inspiration to resort to bashing

[fallen verge]

Sorry i went off on a tangent, can you explain this pleade

[alpine sable]

oh i took two integers a and b and eliminated all choices which weren't a multiple of either a or b

a = 3 and b = 2 eliminates like 3 of the options and another combination i took was 5 and 15 i think

to eliminate one more leaving b - 1

[fallen verge]

Yeah you can do that

[alpine sable]

Right

Also isn't their argument kinda flawed

[fallen verge]

Which point?

[alpine sable]

This one

[fallen verge]

Why do you think so?

[alpine sable]

Hmm actually maybe not i'm high

Because of the restriction of a and b being greater than 1

[fallen verge]

Dont forget about a>b

[alpine sable]

Right

How would you do this question though?

Using the argument you mentioned above?

[fallen verge]

Yeah, b-1 is sanwiched between consecutive multiples

Of both a and b

[alpine sable]

But how would you identify it like instantly

I guess the conditions imposed should be the give away

[fallen verge]

I guess the instinct comes with practice

[alpine sable]

where do i find questions of this sort? I don't mind more difficult questions i guess

Number theory?

[fallen verge]

Knowing b<a, so if you have spmething less than b, it is a multiple of neither

I guess it qualifies as n5

Nt*

[alpine sable]

Hmm

Okay i'll try to look into number theory then

Thank you~

[fallen verge]

No problem

[alpine sable]

.close

[lone heart]

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[queen raven]

question

[queen raven]

in a composite function

the domain of the outside function would be the range of the inside function, yes?

but say if
f(x) = 1/x^2
g(x) = 3x+2

the range of g(x) is R but the domain of f(g(x)) would have to be R excluding -2/3

since there would be an asymptote at that point

[marsh rapids]

Yes

,w domain of (1/(3x+2)^2)

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[clever pagoda]

wow you guys are super good at helping, i didn't know that "it's not hard" was good advice

[keen plinth]

if you couldnt tell, that was sarcasm after the asker somehow figured out the first question after faf said it wasnt hard

[clever pagoda]

point still stands

[marsh rapids]

Words of encouragement work as well

[clever pagoda]

lmfao

[queen raven]

what would the range of f(g(x)) be ?

f(g(x) would be x-4

so the range would be R, but since the domain of g is R+ wouldnt the range of f(g(x)) be R+ as well?

[marsh rapids]

Yes

[queen raven]

but answers say [-4, ∞)

is it because domain is 0 so when x=0, y=-4?

so range is [-4, ∞)

[marsh rapids]

Isn't that the range/image ?

[marsh rapids]

Oh nvm I read domain again

The range is [-4, +inf[ yes

[queen raven]

ok thanks 👍

why does g(f(x)) not exist?

ik that the range of f does not subside in domain of g entirely but isnt there points where it does

from 0 to ∞ ?

[marsh rapids]

There are but you'd need to restrict f's domain for it to work

Rigorously

[lone heart]

@queen raven Has your question been resolved?

[lone heart]

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[zenith olive]

hi

[zenith olive]

10 pls

[pliant cedar]

draw a picture and use some trig i guess

[zenith olive]

💀💀💀

[gray isle]

picture is not accurate

the path of 950km should begin at the end of the 800km part

[zenith olive]

righ tyeah

[zenith olive]

okk uhh

updated

pic

[gray isle]

why aren't you using a ruler for this

(or straight edge / side of textbook)

[zenith olive]

cause im not bothered

[gray isle]

anyway, now that you have a more accurate diagram,
can you try continuing from here

[zenith olive]

what am i mean tto find

wait is the answer 275.1

it is

.close

[lone heart]

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[zenith olive]

ty

[lone heart]

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[vagrant sleet]

can some help me with a dual simplex task? what am i doing wrong?😭

[vagrant sleet]

the first part is the function i need to minimize, i wrote it all in matrix form and then swapped the places of the vector c andb and then i transformed that into a max problem, and then i had to turn it back into a min problem aka multiply all with -1 , but on the end i again get minus values on the right side; am i missing something? am i doing something wrong? i was just following the lecture slides...

[lone heart]

@vagrant sleet Has your question been resolved?

[mortal trellis]

Not sure why you want to turn the dual problem again into a min problem?

Either way you don't have to change the conditions, only the target function

That aside, you are treating the conditions the wrong way

Both the primary and dual problem should have the same number of conditions

(if you include free conditions)

Your primary problem has an equality but you are ignoring that when you write it down in matrix form and just write >=

[vagrant sleet]

cus in the lecture when we had a max problem we swap everything and make it into a min problem and then back again into a max problem (?)

when i have a >= i added a new variable

cus all of them should be =?

this is what the problem looks like normally wirtten down

[lone heart]

@vagrant sleet Has your question been resolved?

[lone heart]

@vagrant sleet Has your question been resolved?

[mortal trellis]

https://i.imgur.com/1NyydzJ.png

this is the notation that we used if that helps you

note that we have 0 S x and y R^D 0. the order of the 0s and the variables switch

that essentially gives an extra minus in the inequalities

[lone heart]

@vagrant sleet Has your question been resolved?

[vagrant sleet]

hmmm okayyy

i don't know now how i should do this then

look, if i do the dualisierungsschritt one time in it i again get a negative value on the right side

or am i still doing it wrong?

or is it whatever if i have a negative value on the right side?

cus in the lecture slides it says if there is a negative value on the right side then we can't do the regular simplex we have to switch it into a dualisierungsproblem

[mortal trellis]

I think you are confusing something

the right side can be negative

are you thinking of the simplex table?

(also dont ignore the bot)

[lone heart]

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[violet glade]

Hi! I need help with this geometry problem.

[slender quiver]

how would you solve it

[violet glade]

The only thing I got is the fact that the length of the square is smaller than sqrt(10)

[slender quiver]

i might be wrong but

lets put the square on all sides

uh

as an example

then

make a octagon

[violet glade]

like this?

[slender quiver]

oh yeah

wait

it isn't a a perfect octagon

[violet glade]

yeah

[slender quiver]

but its really easy to find the sides

[violet glade]

Is it a formula that I should use or something?

[slender quiver]

ask someone else

[violet glade]

ok

[slender quiver]

the area of the octagon is 7 blue sqaures

[violet glade]

Yes, but I don't know how to find the area because the only thing I got is the radius of the circle and I'm not sure if there is any link between the radius and this.

.close

[lone heart]

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[slender quiver]

it tells you how to do the problem

[cinder spindle]

ah you're right

cheers

🤡

...

.close

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[trail sorrel]

do not know where to start

[plain flame]

substitute f(x) and f(-1) in the limit

then you can just calculate the limit

[chrome salmon]

Or if you want to do it fast

Then find f'(-1)

[lone heart]

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[flat remnant]

we have the first picture. we do integral and get the 2nd. then because we know y(pi) = 1 we get the 3rd picture. but why in the 3rd picture is cos(2x) not negative?

[lone heart]

@flat remnant Has your question been resolved?

[flat remnant]

<@&286206848099549185>

[chrome salmon]

Think

What do you think about it being wrong

[flat remnant]

?

[chrome salmon]

The photo being wrong?

[flat remnant]

idk what youre getting at

[chrome salmon]

The last photo being wrong

,w y' = sin(2x) , y(pi)=1

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[flat remnant]

yeah thats what i think it is
but the picture has cos(2x) positive

[chrome salmon]

And I'm telling you the possibility that it may be wrong which it is

[flat remnant]

oh
you said that in such a confusing manner

[chrome salmon]

My bad

[flat remnant]

ty for help

.close

[lone heart]

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[chrome salmon]

Let me say that it is wrong

[lone heart]

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[alpine sable]

Where did I go wrong?

[alpine sable]

In solving this radical equation

[vale wigeon]

is the last line 3 + 4sqrt(x+2) - x meant to be complete or were you not done writing it

[alpine sable]

It’s meant to be compelte

[vale wigeon]

so you intentionally didn't mean for it to be one side of an equation with 0 on the other side.

[alpine sable]

No I did

[vale wigeon]

[vale wigeon]

first you say the line is complete, now it turns out that there is a =0 missing...

but ok,

[alpine sable]

Sorry

[vale wigeon]

you don't seem to have made any mistakes thus far.

so keep going.

[alpine sable]

But the answer is 23

Where do I even go from here

I am so confused

How will this ever lead to 23

[vale wigeon]

answer-chasing will get you nowhere.

you squared both sides once.

with some more prep you can do it again.

[alpine sable]

Prep?

[vale wigeon]

preparation.

[alpine sable]

So don’t jsut square it right away?

I’m confused

[vale wigeon]

do not just square it right away, because that'll only make your life too difficult.

get the radical alone first.

[alpine sable]

How do I do that’s

[vale wigeon]

well you could add x-3 to both sides, perhaps.

[modest osprey]

Bro...

Can you please type the question or am i just seeing that √x+2 can be cancelled?

[vale wigeon]

yes keep going

[chrome salmon]

First kniw the values of x where the left hand side is defined

[vale wigeon]

and try not to get tied up with self doubt

[vale wigeon]

"where did i go wrong" implies you think you DID go wrong somewhere

[alpine sable]

@vale wigeon

[quasi vector]

your subtraction is wrong

[alpine sable]

23

[vale wigeon]

9 - 32 is not -22.

[alpine sable]

Got it thanks

[unreal idol]

.open

[quasi vector]

!help

[lone heart]

Please read #❓how-to-get-help

[unreal idol]

!help

[lone heart]

Please read #❓how-to-get-help

[vale wigeon]

read #❓how-to-get-help for instructions on how to open your own channel.

[unreal idol]

ok

[quasi vector]

.close

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@normal elk Has your question been resolved?

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[queen quiver]

Does this question require sine rule maybe>

[lone heart]

@queen quiver Has your question been resolved?

[lament helm]

hey

question is asking "what is B",right?

@queen quiver

[queen quiver]

Yes

[fierce yarrow]

I got this, if it means something

[lament helm]

can you wait I will send the solution

[queen quiver]

No problem

[lament helm]

hey demi

a,b and c belong to A,B and C, right? did question give this info?

@queen quiver

[queen quiver]

It doesn't say so but I assume so yeah

It says so for the other questions

[lament helm]

hmm

I understand

[lone heart]

@queen quiver Has your question been resolved?

[queen quiver]

no

[lament helm]

I am still looking

[lament helm]

I found 1=-1 with this question LOL

I am still trying

xd

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@queen quiver Has your question been resolved?

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[azure forum]

can someone explain where the hell this 2 is coming from

why wouldnt you just combine the two terms

o wait im dumb nevermind '

[long axle]

U gotta make a common denominator

[azure forum]

🤣 yea idk why i just blanked like that

was so confused looking at that

[long axle]

Lol

[azure forum]

my b

[long axle]

U good

No worries

[azure forum]

.close

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[fierce lark]

hello Im confused as to why g) in this question is necessarily true, from my understanding of the delta-epsilon definition of a limit inorder for a limit to exist there has to exists a 0<delta st. |x-c|<delta and all epsilon must be |f(x)-L|<epsilon

[fierce lark]

so if my definition is correct, wouldnt g be false because |f(x)-5|<.2 would mean there could be an epsilon .1< and still <.2 that satisfies the limit

[rose sigil]

g is just a weaker statement than the given one

if something is less than 0.1, it's also less than 0.2

[fierce lark]

ah I was thinking about it the other way around

that makes sense then ig

and just to make sure I understand the delta part of the limit,

its essentially just asking if its possible for there to exist a single delta that would fit the original limit correct?

ex:

in f) 0<|x-3|<1/4 would be true because 1/4<1 and so there has to exist a delta that satsifes (at least the first part) of the limit

[rose sigil]

in f) 0<|x-3|<1/4 would be true
umm idk what to say about that

and like everything else you said was too vague, i don't wanna say yes :c

[fierce lark]

what exactly was I vague with that I can clarify?

[rose sigil]

everything here basically

[fierce lark]

so, what im trying to say is that the D-E limit proof says that there exists a single delta for all epsilon that satisfy a certain range, so using that wouldnt it be true in the case of these questions as long as |x-3|<c where c<1 is true that the delta part of the proof would be true? since |x-3|<1?

I think Im wrong but Im really not sure where

[rose sigil]

wdym single delta?

[hot bluff]

think he means this

[rose sigil]

yes but what does that have to do with what you are saying?

like

delta part of the proof would be true

what does that mean?

[fierce lark]

(delta part of proof) |x-c|<delta, (epsilong part of proof) |f(x)-L|<epsilon

[rose sigil]

also is there a reason you are adding the word single?

[fierce lark]

is that not correct?

wouldnt sayign there "exists" a delta mean that there has to be AT least 1

[rose sigil]

yea, and if there's one then there's infinitely many

[hot bluff]

yes, there's one delta that can define epsilon, but there are infinitely many deltas WITHIN the tolerance (interval) of your epsilon

[rose sigil]

it seems like you're like looking at it in a weird way

[fierce lark]

would it be fair to say that the delta represents the p in the limit interval (c-p, c+p)

[fierce lark]

off of here

[rose sigil]

umm no

[fierce lark]

oh

[rose sigil]

that's an extra requirement in the definition, in addition to the epsilon delta stuff

[fierce lark]

I mis read

I think I understand what yall are saying now

[hot bluff]

does a picture help?

[fierce lark]

FOR ALL the epsilons there has to be an x in the delta interval that essentially corresponds to the Y value in the epsilon?

[rose sigil]

no

[hot bluff]

there's only one epsilon. because epsilon is the tolerance (or how exact) your result is.

[rose sigil]

umm no, for all epsilons was a good thing to start with

the rest didn't make much sense

also what's the Y value in the epsilon?

[fierce lark]

I drew a picture to show what I meant idk if thisll help

white says : All must line up with an x

green says: in this case epsilon dosent line up in the delta interval its false

I essentially tried to show that there is an epsilon that DOSENT have a delta existing that satisfies the limit proof

[rose sigil]

idt this picture is the right idea

[fierce lark]

what would you say is incorrect or missing

brb 2min sorry

[hot bluff]

if $|x-c|<δ$ then you are within the interval that would get your output in the range of epsilon. that picture shows an x where $|x-c|>δ$

[ocean seal]

b0ngl0rd

[rose sigil]

i also gotta go

[fierce lark]

yes exactly

so Im trying to say in this this case, this limit is not true

but with a smaller epsilon interval it would be true

[hot bluff]

smaller?

[fierce lark]

horriblie drawing, but this is what I mean

[hot bluff]

if that was the case than your delta doesn't fulfill the definition of epsilon-delta.

g is true

it has given a larger epsilon than the initial definition in your problem

same delta, larger epsilon still fulfills the definition

[fierce lark]

ahh true

[hot bluff]

it just means your delta is more precise than needed

twice as precise if we are exact

[fierce lark]

so if I were to say that epsilon is 1000 with that same limit it woudl still be true because all the x values c+delta and c-delta go the Ys that are L+epsilon or L-epsilon

so then there is never an epsilon that is too large, however there could be a delta that is too big!

so then this would be a correct graph of a false delta epsilon limit

[hot bluff]

i think you might be overthinking. think of it like error tolerance. in statistics it would be a p-value or confidence level. how precise your limit approximation is is determined by epsilon, which in turn you find a delta which fulfills the epsilon's interval around L. too big of a delta just means its not precise enough for your epsilon. its not really important to think about deltas that are too big because they are useless if they don't fall within epsilon.

[fierce lark]

ah I think I was unclear, I was trying to say were false in relation to the question above where it asks if the limits are "necessarily" true

thats my faut

[hot bluff]

well we know g is true. which other one were you referring to

[fierce lark]

nothign in particular, I was just trying to say, if we followed the same limit above and we came up with sum like this where delta was much larger then epsilon then it wouldnt be necessarily true?

tbh I think it would be better if I tried an example?

[hot bluff]

we can only infer based on what is given. so using only what is given, we can determine what is true to an extent, or necessarily true within the confines of the epsilon-delta definition of a limit.

if the delta could possibly produce a y value outside the interval around L (epsilon), then it would not fulfill the epsilon-delta definition and could be considered false.

[fierce lark]

yup that makes perfect sense

so thats why in the case d)

although the delta epsilon limit holds its not technically correct because there can be x value sin the delta range that are outside epsilon

[hot bluff]

for d we know its false, because now our delta can produce values outside epsilon, yes. so the epsilon-delta definition does NOT hold true.

$|f(x)-L|<ϵ$ would not be satisfied.

[ocean seal]

b0ngl0rd

[fierce lark]

then for b, the delta range is .8 and so we thereore know since for values in epsilon <.1 satisfies delta values <1 it has to ALSO satisfy delta values <.8!

[hot bluff]

do you know the answer key for b?

[hot bluff]

the way i would prove b is solving the inequality for x. both the original and b's

so

the original we get 2 < |x| < 4 and b we get 2.2 < |x| < 2.8

(2.2, 2.8) lies within (2,4) so yeah it works out

[fierce lark]

then finally, for g) it has the same delta range but a larger epsilon range so why is it necessarily true?

[hot bluff]

as long as delta produces values within epsilon the definition holds true

[fierce lark]

is it just me does it feel like the delta-epsilon defintion kinda works backwards makes more sense

[hot bluff]

we start with an epsilon first

epsilon begets our delta

epsilon is our "target," so to speak. delta is our guide to hit it.

can't hit a target if you don't know what you're aiming for.

[fierce lark]

ok, just 1 final thing because Ive taken up alot of your time, in this case epsilon is 2 and in the original case epsilon is 1 so then wouldnt the values from 1-2 have no x values aroudn the delta that work for them

and the defintion says all for all epsilon

[hot bluff]

any delta that satisfies ϵ_1 will satisfy ϵ_2 given ϵ_2 > ϵ_1

[fierce lark]

oh

I think I completely get it

for some reason

[hot bluff]

all this case did is make our "target" bigger. if we trained to hit a 10 inch wide target, making the target 20 inches wide will do nothing to our aim. we still hit our target.

[fierce lark]

when I was thionking for all epsilon I thought it mean ALL y values in the epsilon range, however it means for all epsilon there exists a y value in the epsilon range

that clarifies everything I think

thank you very very much!

[hot bluff]

np

[fierce lark]

🫡

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[ember wren]

how can i find the area of the grey area?

i know i need to do integrations but idk which method of integration

[vapid shuttle]

it looks like the area under the red curve, above x=0, and bounded on the left by the green curve and on the right by the blue curve right?

I'd split it into two different areas

area on the left and on the right of the y-axis

well

On the left hand side for example

you will need the area red curve yet above the green curve from x=0 to their first intersection

you will also need the area under the blue curve but above x=0

from its first intersection

that will give you half of those two triangular-ish areas

[ember wren]

wait can you screenshot the graph and shade the area?

[vapid shuttle]

a couple notes about this

finding the area of the orange will require you to either split the bounds

or minus an area of the purple region

because if you don't split the bounds at that intersection of red and green (about y=2.5) ?

see how that will give you the purple area twice

other than that, finding the yellow area should be good, same with the purple

then just double everything once you are done since the graph is symmetric

[ember wren]

hm

so you split it like this?

but how do you the orange area tho

like which method do you use

[vapid shuttle]

double integral, the region is bounded in y by the red curve and the intersection line (y=2.5?)

then you will have to split again twice

the intersection of the red curve with the green at those two points

it will only give you the area accurately in between those intersection

then you'd have to make another integral to find the area to the left of the leftmost intersection and to the right of the rightmost intersection

[ember wren]

i am new to integration

how do you perform double integration?

i only learnt the basic in class but the teacher wants us to go home and learn new methods by ourselves

[vapid shuttle]

I recommend youtube for that

[ember wren]

can you just like

shade out different individual areas i need to find?

then ill find my way to calculate them

[fierce yarrow]

I think there is an easier way

but correct me if Im wrong

[vapid shuttle]

there probably is, I must go though. Goodluck

[ember wren]

here

you mean like this right?

[fierce yarrow]

yes

[ember wren]

yeah i thought of that too

but since this is like an essay i am doing

i am trying to research for more maths

so i can do more writing about what i have researched

[fierce yarrow]

yeah I get you

[ember wren]

is there a different way you can think of thats not too complicated?

[fierce yarrow]

hmm

let me think

I dont think there is way thats not too complicated because of this two little parts

[ember wren]

which parts?

[fierce yarrow]

accidentally sent zoomed picture

but you can see what I mean

[ember wren]

hm

is there a method to find..area between 3 curves and x axis?

it sounds crazy but like

like that

[fierce yarrow]

if you want to make it a little bit more interesting you should find these two coordinates

[ember wren]

i can try, then what do i do after?

[fierce yarrow]

yes, but its inverse process

you find the whole thing

than that part

[fierce yarrow]

you get integral of green function from (-10.2, 0.5) to (1, 0.5), then subtract green integral from those new coordinates

then you get this

aaand

then you can subtract blue function from (-1, 0.5) to (0, 0.5)

and you can get

what you wanted

[fierce yarrow]

.

just not blue

but green

but its the same process

for blue

[ember wren]

oh

[fierce yarrow]

is it okay?

[ember wren]

so you find this

and in order to eliminate the red part

you use enclosed area but 2 curves?

thats region is between green and red curve right?

so you find the big green(-10.2 to 0), subtract the tiny blue under, then subtract the red on top

[fierce yarrow]

yes, but I just realized

you can’t subtract the red part just like that

cause as you said its 2 curves

i think you should make it in two parts

[ember wren]

wym two parts

cant you find area between 2 curves?

[fierce yarrow]

you make straight line (dark blue)

then find integral of blue

and integral of light blue

[ember wren]

OH

[fierce yarrow]

thats the point, I’m not sure

[ember wren]

basically setting up a new x-axis?

[fierce yarrow]

yes

cause

as you can see

there are two different Y coordinates

and when you were finding integral of some function, you would be on the same Y

right?

[ember wren]

yeah

but i saw these

so i thought you can find area between them in one go

[fierce yarrow]

yes but its subtracting

you find integral of pink

then integral of purple

and pink-purple

is the gray part you are looking for

[ember wren]

yeah ig

[fierce yarrow]

ohhhh so yes

yes

[fierce yarrow]

to find red part

[ember wren]

yeah yaeh

thats what i meant

[fierce yarrow]

you need to find big green integral - red integral in this new coordinates

[ember wren]

yeah

but online i only see people do like

area between curve and curve

or area between 2 sin curves

i havent seen anyone giving an example of between a curve and sin curve

[fierce yarrow]

but keep in mind that green part must be in the same coordinates as red part

[fierce yarrow]

like this one you just sent ?

[ember wren]

yeah

my green is a sin curve and red is just a quadratic

wait actually

nvm

i understand

[ember wren]

yeah this is the first time i see this

i got it now

[fierce yarrow]

good

[ember wren]

so between this method

and drawing a new x axis

which one is more "interesting"

xd

[fierce yarrow]

😆

i think that more interesting would be drawing x axis

but only if that can work

[ember wren]

yeah i will try that

[fierce yarrow]

and Im not sure if you can do it that way

it was just my idea

[ember wren]

we already subtract the blue part

so lets do smth else for red

[ember wren]

i think you can, you just need to drag the functions down

like instead of making a new axis

just move/translate them downwards

until they reach the x axis

[fierce yarrow]

hmmm, but I think that then you would lose a little bit of integral

[ember wren]

wym "lose integral"

[fierce yarrow]

like, you would lose surface of the grey part

[ember wren]

oh like

just move down temporarily

to find the red

then for the grey keep the original function

[fierce yarrow]

hmmm

you can try that

and see

if its gonna work

[ember wren]

yeah ill try

tysm

[fierce yarrow]

np

[lone heart]

@ember wren Has your question been resolved?

[lone heart]

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[north hemlock]

what is the thing before "Problem" supposed to mean?

[north hemlock]

is this a math thing?

[alpine sable]

[last ether]

Chapter or module

As far as I'm concerned

[alpine sable]

Usually problem means question

[north hemlock]

ah so not math

[alpine sable]

so question 2?

[north hemlock]

what is the thing before "Problem" supposed to mean?

[alpine sable]

oh ok

[last ether]

The comma?

[north hemlock]

true!

the comma

yeah wtf is a comma

[last ether]

[alpine sable]

The comma ofc

[north hemlock]

anyways rip, my prof is trippin

there are no problems on that apge

page* so gg

[alpine sable]

ok so chapter 8.4 maybe

GG

[north hemlock]

.close

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[wary stream]

No it's a space

[last ether]

Legal code

Oh silcrow

[wary stream]

Yep

[wary stream]

All I looked up was overlapping s

And got that

,w silcrow

[ocean seal]

Results provided by WolframAlpha

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

Requested by dldh06#2659

[last ether]

Yeah I forgot it's called a silcrow

[wary stream]

Good trivia fact

[lone heart]

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[primal galleon]

I'm struggling with a painting problem and would really appreciate some help. The problem goes like this:

"If it takes 6 painters 2 hours to paint 4 walls, how long does it take 4 painters to paint 8 walls?"

I know that the answer is supposed to be 6 hours, but I'm not sure how to get there. I tried using the formula for work time:

Work Time = Work / Work Rate

[lone heart]

@primal galleon Has your question been resolved?

[sour dove]

@primal galleon

A way to think about this is like this. We can create a ratio of man hours worked divided by the number of walls to get a rate:

[ocean seal]

MellowDramaLlama

[sour dove]

Then, we can use that rate with our current known information and cancel out units:

$\frac{ \frac{3 \text{ painter hours}}{1 \text{ wall}} \times 8 \text{ walls}}{4 \text{ painters}}\\$.

[ocean seal]

MellowDramaLlama

[sour dove]

We then see we can cancel the wall unit in the numerator with the fraction in the denominator, as well as the painter unit. Which then leaves us with :

$\frac{ \frac{3 \text{ \sout{painter} hours}}{1 \text{ \sout{wall}}} \times 8 \text{ \sout{walls}}}{4 \text{ \sout{painters}}} = \frac{24 \text{ hours}}{4} = 6 \text{hours}\\$.

ugh that command won't work

$\frac{ \frac{3 \text{ \sout{painter} hours}}{1 \text{ \sout{wall}}} \times 8 \text{ \sout{walls}}}{4 \text{ \sout{painters}}} = \frac{24 \text{ hours}}{4} = 6 \text{hours}\$.

[ocean seal]

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

[sour dove]

hope that helps!

[primal galleon]

So we get $$\frac{\frac{3\ \text{painter hours} }{1\ \text{wall} } \times 8\ \text{walls} }{4\ \text{painters} }$$ $$=\frac{24\ \text{painter hours} \ \times \ \text{walls} }{4\ \text{painters} } =6\frac{\text{painter hours} \ \times \ \text{walls} }{\text{painters} }$$

[ocean seal]

dgh

[primal galleon]

how do we end up with hours alone

[sour dove]

dimensional analysis

notice how painters cancels out in teh num and the denom?

[primal galleon]

here?

[sour dove]

yep! The painters and the walls cancel all out and we're just left with hours

[primal galleon]

so painter cancel with painter, walls cancel walls, and we're left with hours

[sour dove]

bingo bango!

[primal galleon]

thanks

[sour dove]

here's a video on it

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

[lone heart]

@primal galleon Has your question been resolved?

[lone heart]

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[flint inlet]

how you integrate this without getting such a large value x^2/sqrt(7-x)

[gray isle]

wdym by

without getting such a large value

[flint inlet]

−
2
(
−
x
+
7
)
1
2
(
3
x
2
+
28
x
+
392
)
15
+
C

oops

like this

I have and it wont accept this is a valid answer

[gray isle]

what's "it"

what exactly are you entering into the system

[flint inlet]

Im so sorry I just messed up with the input some where

stupid mistake

[lone heart]

@flint inlet Has your question been resolved?

[lone heart]

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[gusty gale]

Can i have help with my algebra 2 question?
<@&286206848099549185>

[carmine cedar]

Area of a trapezoid = (1/2) * ( height) * (sum of the two parallel sides)

In this case it'll be 0.5 * 8 * (10+15)

[gusty gale]

it doesnt work

i think its asking for the equation with no numbers, just the varibales

that the equation needs

@carmine cedar

[carmine cedar]

..this is the equation

A = 0.5 * h * (b1 + b2)

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@gusty gale Has your question been resolved?

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@gusty gale Has your question been resolved?

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[chrome salmon]

What is CN? Cartoon Network?

[azure forum]

critical numbers

and cartoon network

yes

[slate monolith]

What

Critical points are where the function is staying still

The function is decreasing or increasing everywhere else

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[cunning locust]

When they say "smallest," are they referring to the y value?

[marsh rapids]

they're 3 real numbers. You can compare them

[cunning locust]

but i they didnt provide a formula so this is a bit tricky for me

i know that f' would have its critical value touching the x axis

[marsh rapids]

the information on f's behavior gives inequalities. Use them

[cunning locust]

f'' im not too sure tho

use them how i dont have f'(X) < 0

i only have the graph f(X)

[marsh rapids]

no but you have f' >= 0

[cunning locust]

we want smallest x value

or biggest x value

[marsh rapids]

smallest

[cunning locust]

so f' > = 0?

[marsh rapids]

that's what "f is increasing" tells you

[cunning locust]

could you graph these

im having trouble understand what theyre asking for

[marsh rapids]

you have 3 real numbers
f(-1), f'(-1), f''(-1)
Using what you know about f, which one of these 3 numbers is the smallest ?

There's no graphing involved

unless you want to go and plot f' and f''

which is imo harder than this question

[fast lichen]

f(-1) you know the exact value of

[cunning locust]

idk because i dont have the other values

heres how i see it, lmk if this is wrong

[molten pivot]

How about this:
What does the sign of the 1st derivative tell you about a function?

[fast lichen]

f'(-1) and f''(-1) atleast you have the sign

[cunning locust]

that it becomes smaller, for example:

f(x)=5x

f'(x)=5

f''(x)=0

[fast lichen]

the sign tells you if its increasing/decreasing

for the 1st derivative

[cunning locust]

you mean the "-" in front?

[fast lichen]

no the sign of the value of f' evaluated at x=-1

its an increasing function they tell you

what can you say about the sign of the first derivative

when they tell you f is increasing

is it positive or negative?

[cunning locust]

that it is also increasing like in my Picaso painting

[fast lichen]

no can you just answer my question

[cunning locust]

are they referring to the y value?

[fast lichen]

im asking you about the sign

not whetehr its increasing

is it positive

or negative

[cunning locust]

since f(-1) is a negative number, then f'(-1) will be be a positive number

[fast lichen]

ok good

now they also tell you that f is conave up

what does that tell you about f''

is it positive or negative

[fast lichen]

ok your explanation is

completley wrong

first off f(-1) isnt even a negative number

f(-1)=0

2. this has nothing to do with the value of the 1st derative

[cunning locust]

for some reason i thought the sign flips

[fast lichen]

why would you think that

[cunning locust]

after the first deriviative

ive done some experiments

[fast lichen]

forget your experiements

do you remember yesterday

we solved a problem

[cunning locust]

but its my lifes work

jk

[fast lichen]

of finding the interval in which a function decreases?

what inequality did we solve

[cunning locust]

yes i remember, it was when x<30

[fast lichen]

well that was the answer we ended up with

what fact did we use

about f'

[cunning locust]

i dont even remember what i had for lunch yesterday

jk

i dont remember koter

can you remind me

[fast lichen]

if a function f is increasing on an interval then f'(x)>0 on that interval

if a function is decreasing on an interval then f'(x)<0 on that interval

this should make sense intuitively

f' is velocity

if its positive you are increasing

if its negative you are decreasing

[cunning locust]

ok

that does make sense

[fast lichen]

so f(-1)=0 just by looking at the graph

they tell you its increasing so f'(-1)>0

and there is a similar fact about a function being concave up or concave down regarding the 2nd derivative

try figuring that out. its also intuitively maybe a bit less so

depending on how familiar you are with acceleration

[cunning locust]

f(-1)>0
f'(-1)<0
f''(-1)>0 ?

smallest in this case would be f'(-1)

[fast lichen]

how did you just put say that f'(-1) is both positive and negative

[cunning locust]

fixed it

[fast lichen]

ok. why do you think f'(-1)<0

[cunning locust]

because you said so