help-49

[near brook]

oh i see where i messed up

i thought 4 was the VA

ok lemme js fix that answer rq

[shell wigeon]

Oh so you meant -3 and 1 instead of -4 and 2?

[near brook]

well for 2 i meant 1

and yea -3 for -4

[shell wigeon]

Yeah that's what I said

[near brook]

mb

im stressing out rn

[shell wigeon]

Alright

[near brook]

Increasing Int : ( -3, infinity) u - Infinity, 1) (-infinity, -1)
Decreasing Int : (- infinity, -3) u (-1, infinity) u (1, - infinity)

[shell wigeon]

Ok, why (-3, infinity)?

[near brook]

cause its going up right here

[shell wigeon]

Ok that's not how intervals work

You are only concerned with the x-coordinates

[near brook]

i thought you guys said that increasing interval is where the graph is increasing

[near brook]

but we dont have any coordinates tho

[shell wigeon]

We have an entire set of coordinates, the x-axis

[near brook]

oh thats what u meant

okay okay go on

[shell wigeon]

Do you understand what a function's domain is exactly?

[near brook]

yes a bit

[gusty zephyr]

yo gng

fuck of maths

[shell wigeon]

Tell me

[near brook]

whatever x cannot equal

[shell wigeon]

No, the opposite

[near brook]

um

[shell wigeon]

A function's domain is the set of inputs (values of x) that the function accepts

[near brook]

oh okay

[shell wigeon]

Yes, this domain is all of R, except 1 and -3

That's what this notation means

[near brook]

righttt okay okay

[shell wigeon]

An interval is also a set, just written differently

[near brook]

oh okayy oh okay thats why theres infinity basically for XER

[shell wigeon]

For example, the interval $(1, 5)$ is equivalent to ${x \in \mathbb{R} \mid x > 1 \land x < 5}$

[grand pond]

Nel

[shell wigeon]

Does that make sense to you?

[near brook]

okay okay yea

[shell wigeon]

Another example: $(-\infty, +\infty) = \mathbb{R}$

[grand pond]

Nel

[near brook]

rightttttt okay okay

[shell wigeon]

R is just the interval from -inf to +inf

And R is, of course, also a set

Now, the interval(s) of increase of a function is the interval on which the function is increasing

[near brook]

alr

[shell wigeon]

So if you give me (-3, +inf), I'm going to understand it as f increases from -3 all the way to +infinity

And I'm not talking about the y values here, only the x values

(although you need to know that the y values increase, of course)

[near brook]

wait so infinty from where

from -3 to -4 and so on for -3 to -2 and 0 and so forth

[shell wigeon]

Wdym

[near brook]

so for -3, inf where is it going

up or down

[shell wigeon]

For all x > -3

[near brook]

wait i realize it depends on the interval werre putting it into

mb

[shell wigeon]

(-3, +inf) = {x in R | x > -3}

[near brook]

right okay okay

[shell wigeon]

If you understand that better, can you try again?

[near brook]

okay soooo

Interval of Increasing : (-inf, -3)u (inf, -1)u (1,infinity)
Interval of Decreasing ( - infinity, -3) u(-1, infinity) u(2, infinity)

[shell wigeon]

(inf, -1) is not valid since +inf > -1

Also you're still using -4 and 2...

[near brook]

broo its hard on PC okay! 😭

[shell wigeon]

What is?

[near brook]

looking at the graph

[near brook]

wait nel did i get the -inf,-3 right tho

[shell wigeon]

Yeah, f is indeed increasing on (-inf, -3)

[near brook]

WOOOO

[shell wigeon]

You're gonna have to choose though, you also said it's decreasing on that same interval

[near brook]

is this part decreasing?

[shell wigeon]

I think you need to review quite a lot of definitions

[near brook]

u right'

[shell wigeon]

A function is increasing if the y values increase as the x values increase

In other words, the graph goes up, when read from left to right

A function is decreasing if the y values decrease as the x values increase (graph going down, still reading from left to right)

[near brook]

WAITTT so my (-inf, -3)

it represents that right

[shell wigeon]

All the way to the left end of the number line, yes

[near brook]

OHHHH

okay okay W start

[near brook]

anyway can i ask u abt this

i thought the right side is the postive infinity side

[shell wigeon]

Did you mean to write (-inf, -1)?

[near brook]

no +inf

[shell wigeon]

When reading just "inf", I assume +inf

An interval is a set of values between two bounding values, where you write the lower bound first and the upper bound second

+inf cannot be the lower bound if -1 is the upper bound

[near brook]

sooo basically it wouldve been -inf, -1 still

[shell wigeon]

I can't read your mind

[near brook]

im so confused

i thought these follow the exact rules

unless it was -3 to -1

which it isnt cause it doesnt connect to each other

[shell wigeon]

I'm really not sure what you mean

In green is the interval (-inf, -3); in purple is the interval (-1, +inf)

Those are just examples

You can clearly see that f is increasing in the green part, but hopefully you realize that's not the only part where f is increasing

[near brook]

oh

[shell wigeon]

So... can you try one more time?

After that I'm gonna have to go

[near brook]

oh okay soooo

Increasing Int (-inf , -3) u (-3,-1)
Decreasing Int (-1, 1) (1, infinity)

?

[shell wigeon]

Yes

[near brook]

like all of it?

[shell wigeon]

Yes, that's all

[near brook]

oh okay the -3,-1 doesnt make sense

before u go can u explain wh its -3, -1

[shell wigeon]

Why?

It's clearly going up in that part of the graph

[shell wigeon]

It's the part left uncolored here

What is not making sense to you?

[near brook]

ik but its not connectedddd

like its not apart of each other like -inf, -3

[shell wigeon]

I don't understand

[near brook]

its fine dw abt it ill prob just figure it out in the next 6 hours alone

[shell wigeon]

Are you talking about the discontinuity at -3 or the point where it changes at -1?

[near brook]

yes

YES

[shell wigeon]

... which one

[near brook]

mb the -3

the first one

[shell wigeon]

Right well that's why we write (-inf, -3) u (-3,-1) and not just (-inf, -1)

-3 is not in the domain, so we skip it

It doesn't matter that it's not connected, we only care that f increases there (in each interval)

[near brook]

ohhh okay

[shell wigeon]

Similarly we write (-1, 1) u (1, +inf) instead of (-1, +inf) since 1 is not in the domain

[near brook]

ohhh okay

[shell wigeon]

Alright I'm off, maybe @frozen talon can take over if you have any more questions (the range? your teacher made a mistake there, so...)

[near brook]

no its fine ill prob js figure it out

gn guys

.close

[midnight plank]

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[final oracle]

p(x) and q(x) are polynomials with a degree of 3 that satisfy:
p(0) = q(0)
p(1) = q(2)
p(2) = q(4)
p(3) = q(6)+30
Find the value of p(4)-q(8).

[visual tiger]

What have you tried?

[final oracle]

I'm still typing bro wait

[dense bronze]

just joined, hi

[final oracle]

I found that p(x)≥q(x)
So I can say that
p(x) = ax³ + bx² + cx + d
q(x) = ex³ + fx² + gx + h

p(0) = q(0), thus h=d
p(1) = q(2) -> p(1) - q(2) = 0
p(2) = q(4) -> p(2) - q(4) = 0
p(3) = q(6) + 30 -> p(3) - q(6) = 30

First equation:
p(2)-p(1)=q(4)-q(2)
8a+4b+2c-a-b-c = 64e+16f+4g-8e-4f-2g
7a+3b+c=56e+12f+2g

Second equation:
p(3)-q(6)=30
27a+9b+3c-216e-36f-6g=30

[scenic wyvern]

hi, welcome to the server! to chat, please head to #discussion!

[final oracle]

Since a and e are both on x³ then a=e.k for some coefficient k

[lyric charm]

i think you're going about this in a difficult way

[visual tiger]

Ok, so that's one way to do it ig... there's much simpler jic

[lyric charm]

yeah

[final oracle]

Simpler how?

[visual tiger]

As a hint, you're not supposed to consider polynomials p and q

[lyric charm]

let f(x) := p(x) - q(2x)

[visual tiger]

Well the cat's out of the bag ig

[lyric charm]

then

• f is of degree ≤ 3
• f(0) = f(1) = f(2) = 0
• f(3) = 30
• goal is f(4)

yeah i guess it is

meow

[final oracle]

f(x) = x³-x²+4x?

[visual tiger]

f(1) is not 0 here

[final oracle]

Oh wait

Yeah I forgotten

[visual tiger]

Coefficient method probably isn't the easiest again. You can try to find factored form of f

[lyric charm]

is it not?

p(1)=q(2) is stated

oh do you mean his f doesnt satisfy

[visual tiger]

Yes

[final oracle]

f(x) = x³-3x²+2x?

Wait that does not satisfy f(3)=30

[visual tiger]

Yep

So how do we change f accordingly

[final oracle]

Idk, using system of equations?

[visual tiger]

Not really

You just need to multiply f by a good coefficient

it won't change f(0),f(1) or f(2) since they stay at 0

But it'll change f(3) to what you need

[lyric charm]

the thing is f being a cubic with roots 0, 1, 2 actually leaves us with one more degree of freedom you havent considered. you dont know that f(x) = x(x-1)(x-2) but rather that f(x) = kx(x-1)(x-2)

and this k is what you're to find out using f(3)=30

[final oracle]

Didnt know that property of a cubic equation

So I can just multiply the equation till I get 30?

So just multiply by 5?

Since f(x) = kx(x-1)(x-2)
f(3)=k(3)(3-1)(3-2)=30
k(3)(2)(1)=30
k=5?

[visual tiger]

Yes

[quasi goblet]

i have an idea maybe would work
you see that that the input in the function q is 2*the input in the function p so you can write the functions as:
p(x)=ax^3+bx^2+cx+d
q(2x)=ex^3+fx^2+gx+h
and from the first given we know that d=h
now define a function of difference f(x)=p(x)-q(2x) and want p(4)-q(8) so we want f(4)
f(x)=(a-e)x^3+(b-f)x^2+(c-g)x=x(ix^2+jx+k)->(just replace the constant to make it more simple)
and from the givens we have x=0,1,2 are roots so solve for i,j,k then find f(4)

[visual tiger]

Well that's what we did

[final oracle]

f(x) = 5x³-15x²+10x?

So now just input f(4)

[visual tiger]

But not even needing to solve for i,j,k, we can just write f(x) = kx(x-1)(x-2) as we know 0,1,2 are roots of f which is cubic

And solve only for one coefficiebt

[quasi goblet]

i did not see cuz i was typing🫠

[visual tiger]

Sure, but isn't it easier to compute f(4) with its factored form

[final oracle]

Oh yeah, forgot about that

f(x) = 5x(x-1)(x-2)
f(4) = 5(4)(3)(2) = 120

Thanks

[quasi goblet]

yes that better

[final oracle]

.solved

[midnight plank]

Channel closed

Closed by @final oracle

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[midnight plank]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
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[pine salmon]

Why is the answer to life, the universe, and everything 42 if 37 appears everywhere.

[pine salmon]

?

[slender walrus]

help channels aren't the place to ask such philosophical questions.
please use #chill or #discussion

[woeful turret]

🤔

how is the answer to everything 42

[slender walrus]

HGTTG

[graceful drum]

hitchhikers guide to the galaxy

[woeful turret]

"The Hitchhiker's Guide to the Galaxy is a comedy science fiction franchise created by Douglas Adams"

if its comedy fiction why are people taking it seriously

[wary epoch]

This seems to be not a serious question

.close

[midnight plank]

Channel closed

Closed by @wary epoch

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[wary epoch]

All hail the Imperium

[midnight plank]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
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• Type the command .close to free the channel when you're done.
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[boreal ingot]

Pls help. I got -1/sqrt(x^2 -a^2)

[boreal ingot]

I've done answers A and E both are wrong

[wispy mortar]

are u sure about the -1?

[boreal ingot]

Ya

[wispy mortar]

this is wrong

there is no -1

[boreal ingot]

WHAT

BROOOOOOO

OMMGGGGGGGGGGG

[wispy mortar]

yes

[boreal ingot]

This is through my uni aswell omfg will be complaining to the higher ups

[wispy mortar]

no way 😭 this is a resource from uni?

[wispy mortar]

bte

btw u shouldn't use direct proofs

try to derive them yourself to double check

[boreal ingot]

I looked online and it seems that you're right

[boreal ingot]

YES

Absolutely unbelieveable

I've spent like 20 mins trying to figure out how I've been getting this wrong

[wispy mortar]

they probably had a typo from the line but thats unacceptable ngl

[boreal ingot]

Writing an email complaining as we speak

Thank you

[wispy mortar]

hope u win the case

[boreal ingot]

Are the rest of these right out of curiosity ?

[wispy mortar]

yess

just the middle one

[midnight plank]

@boreal ingot Has your question been resolved?

[midnight plank]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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[mystic ivy]

God this is to hard I’m going to cry

[fallow scarab]

,tex .alg lesson

[grand pond]

riemann

[signal holly]

you serious 💀

[mystic ivy]

Yes

It’s hard

[signal holly]

🙏

[mystic ivy]

Is this a helping sever or just a piss take

[signal holly]

anyway lets work it step by step

[mystic ivy]

I struggle with maths and it’s hard for me

[signal holly]

first divide both side by 2

[mystic ivy]

I tried that

Yep

[fallow scarab]

you can just read this if you don't wanna get bullied by them

[signal holly]

write what you got

[mystic ivy]

What number do I divide by 2

Isn’t it 8

[signal holly]

The LHS and RHS

[signal holly]

Divide the left hand side and the right hand side of the equal sign by 2

[mystic ivy]

Yeah I did it

[signal holly]

think of it like a weight balance

if two plates have the same mass

they are balanced

[mystic ivy]

I got 1 and 4

[signal holly]

I am asking for the equation

[mystic ivy]

2x + 2 = 8?

[signal holly]

Also the thing is the equation has 1 variable and degree 1

so there can be only 1 answer

[signal holly]

why 8?

[mystic ivy]

Wait

[signal holly]

in question it is given 10

[mystic ivy]

This one?

I got the wrong one

[signal holly]

☠️

[mystic ivy]

2x + 1 = 10

Isn’t that the equation

[signal holly]

no multiply 2 to both x and 1

[mystic ivy]

1

[signal holly]

2 is outside the bracket

so distributive property

[signal holly]

you remember distributive property?

[mystic ivy]

Haven’t learnt that

[signal holly]

ohh

which grade are you in btw

[mystic ivy]

8

In uk

[signal holly]

ohh

[mystic ivy]

Secondary schools here don’t teach that

[signal holly]

anyway distributive property is

[mystic ivy]

I’m literally good at maths with other stuff this one’s a little bit hard because I have loads of notes but no examples on this type of question

[signal holly]

a(b+c)=ab+ac

[mystic ivy]

Writing thwt down

That

Is that exam level questions

[signal holly]

you got this?

[mystic ivy]

Let’s continue with the question

Alrigjt

Alright*

[signal holly]

lets start simple

like 2(1+2)

[mystic ivy]

Here’s a example how we do it

[signal holly]

what does it simplify to

2(1+2)=?

see you would genrally add first

dont do that

[signal holly]

yes good

those are correct

do like that

[mystic ivy]

Gimme a sec

[signal holly]

are you done?

[mystic ivy]

4

I got it wrong many times bc I took one away from ten

[signal holly]

no

no

[mystic ivy]

Instead of diving by 2 in 5 times tables

I got it right tho

[signal holly]

when you divide

[signal holly]

no

what did you do bro

[mystic ivy]

Divided by 2

Got 5

5 minus 1

4

[signal holly]

firstly when you divided why did you leave out the +1?

[mystic ivy]

You meant to do opposite

[signal holly]

2(x+1) divided by 2 is (x+1) not x

[signal holly]

bro I am just correcting the process

[signal holly]

think of like 2(3+1) divided by 2 would be (3+1)

not 3

[mystic ivy]

Kk

[signal holly]

because 2(3+1)=2*4=8 which divided by 2 is 4=(3+1)

[mystic ivy]

Ty for helping

[signal holly]

ok

[midnight plank]

@mystic ivy Has your question been resolved?

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[orchid saddle]

Im genuinely confused on the phrasing of this proposition

[orchid saddle]

Let me try and explian what i understand

So there is a set A and what [[2]]^A does is it maps functions from A to either 0 or 1. Now the POWER set of A maps the subsets of A to either a 0 or 1. So say I had the set A = {a,b}. The power set would be, empty set, {a}, {b}, {a,b} right

then when the mapping takes place

u can get smth like {0}, {1}, {0,1} respectively?

such that 0 corresponds to a

and 1 corresponds to b?

im so confused

but how do u know that 0 correspond to a and 1 to b?

[obtuse basin]

[[2]]^A is a set of functions, it doesn’t really map anything, so it would be more correct to say [[2]]^A is the set of functions that map elements of A to either 0 or 1

Same thing with the power set, it’s the set of subsets of A, it doesn’t map them

This is just regrading the wording though

[orchid saddle]

do we know whch elements go to 0 and which to 1

[orchid saddle]

so what exactly is this proposition talking abt

[obtuse basin]

No, what you get are indicator functions of those subsets

Can you explain what the indicator function of a subset is?

[orchid saddle]

well i know that an indicator function like say ID_B is one which if B is a subset of A and say we took out a subset of A, it would output a 1 if that element in the subset we took out was also an element in B but 0 if it wasnt

[obtuse basin]

You mean took out an element* of A?

[orchid saddle]

wait

so we take out an element in a

Oh wait

yh

[obtuse basin]

Yes, that’s correct

[orchid saddle]

so we take out an element of A and it will let me knwo if its in the subset B or not

[obtuse basin]

So 1|({a}) would be the indicator function of {a} as a subset of {a, b}. If you want to describe 1|({a}), it’s better to do it with its mappings

Which are 1|({a})(a) = 1 and 1|({a})(b) = 0

[orchid saddle]

sorry im not understanding notation lemme digest this

ahh yes

i get it

so say we had 1|({b}) and the set {a,b,c} then it would only put a one if b was pulled out

alright so how does the proposition work with that idea

[obtuse basin]

What do you mean by putting and pulling out here exactly?

[orchid saddle]

like it would output a 1

show a 1

idk how to say it

[obtuse basin]

a is not an element of {b} though

So it would output 0

[orchid saddle]

for my example

i understand identity now

[obtuse basin]

Okay, can you explain to me what a bijection is?

[orchid saddle]

well if something is surjective and injective then its bijective

and injectove is

if f(a) = f(b) then a= b for a certian function f to be injective

and for sujective

there exists a b in B such that for all a in A f(a) = b?

[obtuse basin]

For every b in B* there exists such an a in A

[orchid saddle]

oop sorry

yes

lemme try understand that

[obtuse basin]

Right, so injectivity of 1| would mean that if indicator functions coincide, the subsets must be equal

And surjectivity means that if I were to give you function A -> [[2]], it can be interpreted as an indicator function of some subset of A

[orchid saddle]

oh so we're saying if the subsets output the same "array" of 1 and 0 then they are identical?

[obtuse basin]

Yes

Any ideas on how to show that?

[orchid saddle]

wait im not understanding this

[obtuse basin]

So here the codomain of 1| is [[2]]^A and the domain is P(A)

[orchid saddle]

ahh wait

so if u give me an array of 1 and 0 then i can somehow get back the order of subset?

[orchid saddle]

yep

[obtuse basin]

The sentence “for every element of codomain, there is some element of domain mapping to it” translates to “for every function A -> [[2]], there must be some subset of A with exactly that function as its indicator finction”

[orchid saddle]

ahhhh yes

i see

[obtuse basin]

That’s the idea, but do keep in mind that what I’m giving you is a function rather than an array

[orchid saddle]

yep yep, just for my own understanding

well to prove the bijection, i can prove its injective first

[obtuse basin]

Yup, any ideas on how to do that?

[orchid saddle]

if say i have 2 subsets of A

say B, B'

wait no

lemme go back to fundamentals, f(a) = f(b) then a = b right

so

yes so if i have 2 subsets B and B'

if i assume that identity_B = identoty_B'

then B = B' ?

[obtuse basin]

Yes, that’s what you want to show

[orchid saddle]

hm

[obtuse basin]

Should say assume here rather than show

[orchid saddle]

ah yes sorry

can i have a hint?

😭

wait

lemme think abt this acc

well if i say i have an element in B, sy b we have that Identity_B (b) = 1

and i assumed that identity_B = identoty_B'

so tht means

identity_B' (b) = 1

also

wait

im confusing myself

[obtuse basin]

You’re on the right track

[obtuse basin]

Now what does this tell you?

[orchid saddle]

well if they both equal 1

im just going in a loop surely

[obtuse basin]

Well, what does this tell you according to the definition of identity_B'?

[orchid saddle]

b is also in B'

ohhh

[obtuse basin]

Yup

[orchid saddle]

ohh so B = B'

[obtuse basin]

So every element of B is an element of B'

[orchid saddle]

yesss

i seee

[obtuse basin]

Not yet

[orchid saddle]

oh wair

surjectivity

[obtuse basin]

What you’ve shown is just this

[orchid saddle]

oh do i need to do a converse?

[obtuse basin]

You also want to show that every element of B’ is an element of B

[obtuse basin]

Yes, exactly

[orchid saddle]

like if I assume b is an element in B' then id_B' (b) = 1

[obtuse basin]

However, you can notice that reversing the statements in this reasoning gives you a valid proof of that

[orchid saddle]

yes its basically same

but reversed

[obtuse basin]

Yup

[orchid saddle]

what im having triuble understanding is

[obtuse basin]

So now we know B = B' and 1| is indeed injective

[orchid saddle]

why do i make this assumption in the first place, the only reasoning i got from this is from the f(a) = f(b) implies a=b for injectivity

what acc is the purpose

[obtuse basin]

That’s how implications work

P implies Q means that assuming P, you can show Q

[orchid saddle]

so we assume that the identites are true then show theyre true?

[obtuse basin]

We assume that the identities are equal and then show the inputs are also equal

[orchid saddle]

oh wait we assume that f(a) = f(b) then show that a=b

ah yes

i get that

for sujectivity tho

so going back to the beginning, For every b in B there exists an a in A st f(a) = b

so

in this case

for every [[2]]^A there exists a subset of A such that..

f(A) = 1?

nah

[obtuse basin]

For every function f: A -> [[2]], there exists a subset B of A such that 1|(B) = f

[orchid saddle]

identity of B = f? what does this mean

so from the identity output we can work out what inputs we had?

[obtuse basin]

Functions being equal means that, besides their domains and codomains being the same, their outputs are always the same for each input

So identity_B(a) = f(a) for all a in A

[orchid saddle]

wait but Identity_B(a) only outputs 0 or 1 right

[obtuse basin]

Yes

[orchid saddle]

but surely f(a) outputs other things?

ohhhhhh

wait

no

f(a) also outputs {0,1}

[obtuse basin]

Yup

[orchid saddle]

so we're saying that what erer element we put in f, we get the same output from the identity?

hm but to prove surjectivity

OH WAIT

[obtuse basin]

Yes

[orchid saddle]

what if i use idea of preimage

is that valid?

to show sujectivity

[obtuse basin]

I believe that only changes how you’re wording the proof

The ultimate goal is to describe a subset of A whose identity equals f

[orchid saddle]

ill try so

wait idt what im thinking makes sense

if i say B is the subset right then if a is an element of B

so B being like the preimage of 1

then Identity_B (a) = 1

and so like f(a) = 1

but if a is NOT in B then identity_B (a) = 0

and so f(a) =! 1

[obtuse basin]

You might have meant a being the preinage of 1 under identity_B

[orchid saddle]

so f(a) must be 0

[orchid saddle]

yess this

i need to consolidate pre images 😭

is that valid

[obtuse basin]

Yeah, so how can you describe B using f given that logic?

[orchid saddle]

like if its not 1

then it will be 0 as its the only other element in [[2]]

which makes it surjective

[orchid saddle]

well if B doesnt contain a then it

[obtuse basin]

<@&268886789983436800>

[orchid saddle]

💀

hm

say B contains a then identity_B (a) = 1

if it doesnt contain a then identity_B(a) = 0

so like f(a) is either 0 or 1

which is the domain

idk if that makes sense

[obtuse basin]

Technically you’re getting close, the codomain of f being [[2]] and the implications you mentioned tell you that the converses are also true

Now replace identity_B with f and what you get is that

f(a) = 0 implies a is not an element of B and f(a) = 1 implies a is an element of B

[orchid saddle]

yep

oh which make up the domain

[obtuse basin]

So, B is the set of all a in A such that f(a) = 1

And this finishes the proof of surjectivity

Which also finishes the proof of bijectivity

[orchid saddle]

ahh this makes sense yea

damn that was nice proof

tysm ❤️

genuinely made me understand

.close

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[valid spire]

heyy,
guys why if x²=y <=> x=sqroot(y) or x=-sqroot(y)
but sqroot of x is always positive ?
like per exemple : sqroot of 4 is always 2, but 2 squared and -2 squared are both 4.
why ?
( that's what i found in reddit and google )

[hearty rune]

the sqrt function only returns positive values
so if you have y=sqrt(x), y will always be positive for valid x

when solving y=x^2 however, because (-x)^2=x^2 -x and x are solutions

sqrt(4)=2 only

if y^2=4 then y could be 2 or -2

[slender walrus]

sqrt(x^2) = |x|,
(not x, which is where the misconceptions come from)

[valid spire]

sorry but u just reapeted what i just said
i'm asking why

[hearty rune]

that's how the sqrt function is defined

theres no further why to that part of it

[slender walrus]

assuming y is non-negative,
thing the sqrt of both sides gives
sqrt(x^2) = sqrt(y)
|x| = sqrt(y)
and from that the result you posted

[hearty rune]

when it was defined they would have to choose the positive or negative root
otherwise it wouldnt have been a function (2 y values for x values)
so we chose to only return the positive one sqrt(x^2)=|x|

[valid spire]

i didnt understand why sqrt of 4 is only 2 even if 2 and -2 squared are both 4
as i know the sqrt is the value that if u square it, it gives u the value under root

[valid spire]

i didn't understand u

[hearty rune]

If sqrt(4)=2 and -2

then sqrt(x) wouldn't be a function, because we would have multiple outputs to one input

so the sqrt function was chosen to return the positive root only - so it would be one-one

and so defined sqrt(x^2)=|x|

[valid spire]

ow, i didnt learn this :((

wait

if sqrt of x can't have 2 solutions
why if we have x²=y we have 2 solutions for x

[hearty rune]

because x^2 is still one-one
its not a problem if two x values return the same y value

but one x value cant have two y values

(for it to be a function)

[valid spire]

aaa ?

i didn't understand

[valid spire]

x can have 2 values cuz it's squared okey i got it
but why sqroot of x can't have 2 values

[hearty rune]

,w graph y=x^2

[grand pond]

Results provided by WolframAlpha

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

Requested by aldrnari_

[hearty rune]

,w graph y=sqrt(x), y=-sqrt(x)

[grand pond]

Results provided by WolframAlpha

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

Requested by aldrnari_

[hearty rune]

alright, just that top graph

thats what it would look like if sqrt(x) returned both the positive root and negative root

but you can see that its not a function, because for one x value, we have two y values (except at 0)

[hearty rune]

here we can see every x value has a single y value, so its a function

[valid spire]

tysm for ur efforts to explain to me, but i didn't understand

sm1 explain to me guys !

?

[valid spire]

x²=y isn't a function ?

[hearty rune]

its not, no

wait sorry

i read that as x=y^2

y=x^2 is a function

[valid spire]

if it is x=y², isnt a function ?

[hearty rune]

no, because take x=4, we have y=2 and y=-2
so its not injective

[valid spire]

tysm

[midnight plank]

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[wary ferry]

Idk what im soing

[subtle blaze]

What happened to the dx

[wary ferry]

Oh

Its hiding on me

Gotta find it

.clsoe

.close

[midnight plank]

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[midnight plank]

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[dapper lance]

Help please

[dapper lance]

With number 12

[remote flame]

I believe you need to put 26 under a root

[lament creek]

what's i26 though

did you mean to put a root there?

[dapper lance]

Idk

[remote flame]

imaginary

[dapper lance]

Yea

[cerulean oyster]

:clueless:
26i is 26 times the imaginary unit.

[dapper lance]

That

[remote flame]

lmfao

[dapper lance]

But its wrong 😑

[lament creek]

I know what i is, just curious why it's written that way

[remote flame]

...no?

[dapper lance]

On formative

[remote flame]

well his answer is wrong

yes

but the answer has an imaginary number

[lament creek]

but the discriminant is positive for number 12, no?

oh wait nvm

[remote flame]

lol

[dapper lance]

Wait can you help me with these instead because I think that problem is broken

X squared plus 2x +11=0

Using quadratic formula

[lament creek]

maybe it wants you to leave the answer in root form instead of evaluating it to the imaginary unit

[dapper lance]

Ok

Oh I got it right

[lament creek]

nice

did you just leave it in root form? is that what they want?

if it is, we'll leave all future answers in root form

[dapper lance]

The answer was x= -1+or minus i square root 26 over 3

[lament creek]

ah then you missed a root sign

[lament creek]

as for this, I presume it's $x^2 + 2x + 11 = 0$?

[grand pond]

Mikya

[lament creek]

if it is, it's the same trick

[dapper lance]

Yes

That’s the answer I need to solve

[lament creek]

throw this into the quadratic formula the same way you did q12

[dapper lance]

Ok

[midnight plank]

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[midnight plank]

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[spiral forum]

hey

[azure oracle]

hi

[lament creek]

do you have a question?

[azure oracle]

👀

[spiral forum]

yes

i have a question

[lament creek]

that's the question from #help-19

[spiral forum]

ye we have the same question

[lament creek]

have you read the helper's explanation in #help-19?

[spiral forum]

ye but that didnt help me @lament creek

[lament creek]

then what's your question?

or, if you have the exact same problem as the OP there, what did you not understand?

(about the helper's explanation)

[spiral forum]

i was just geniunely confused on how horizontal asymtotes, work here cuz here's what im thinking. 1/sinx , the denominator has a higher exponent than numerator, then why wouldn't the the horizontal asymtote be at 0.

[lament creek]

sin(x) is not a polynomial function.

[spiral forum]

ohhhhhhhh

so what rules do i apply then here?

to find the ha

[lament creek]

take the limit of 1/sin(x) as x grows without bound (or simply consider what happens to the graph as x keeps getting bigger)

[spiral forum]

so i would do 1/sin(100) or something liekt hat

[lament creek]

you'll find that the graph oscillates between the section above sin(x) and the section below sin(x) forever

depending on the sign of sin(x)

key point is that the graph never approaches a fixed value as x keeps getting bigger

[spiral forum]

hmm i see whatchu mean

[lament creek]

therefore, there is no HA

[spiral forum]

so always i do as x apporaches infinity, what's the value gonna be bigger or smaller right?

[lament creek]

no. the concept of HAs is that they represent some fixed value that the function tends to as x grows without bound

keyword: fixed

[spiral forum]

oh, so if it has more than 1 value it doens't have a fixed ha

[lament creek]

if it oscillates, it definitely doesn't have a HA

which is the case here

[spiral forum]

ok make sense

also do u know any other type of questions where a equation is not a polynomial function

like maybe cosx could work

thank you so much

this helped me a lot

[lament creek]

trignonometric functions, logarithmic functions, exponential functions

at least three

do you still have questions? if not, you can .close the channel, and welcome to the server!

[midnight plank]

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[rare maple]

I have found a way for just 2 circles and any given distance

[surreal moon]

This is the Problem of Apollonius. It is not trivial

Isaac Newton did it in his Principia Mathematica Philosophiæ Naturalis Principia Mathematica

[rare maple]

Hmmm

[lyric charm]

isn't principia russell's work

or is one of us confusing themselves

[surreal moon]

Oops

[rare maple]

Newton wrote one too

[surreal moon]

you right

me wrong

[rare maple]

I think

Why was this question on the mo server

[surreal moon]

Math Olympiad?

[rare maple]

Yes

[surreal moon]

It's on that level

you can do it with basic geometry

It's just really hard

You have to be pretty bright to see that path without help

[rare maple]

Does it involve inversion or just normal constructions

[surreal moon]

iirc, you can do it with just properties of hyperbolas

(yes, using just compass-straightedge constructions)

[rare maple]

I will give it another try then

[surreal moon]

good luck. I don't even remember the whole proof 🙂

[midnight plank]

@rare maple Has your question been resolved?

[wary valve]

@rare maple maybe try ET function

[midnight plank]

@rare maple Has your question been resolved?

[rare maple]

I have made no progress till now

I will probably attempt this later

.close

[midnight plank]

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[timid topaz]

Hey

[pearl hull]

Hii

@timid topaz ya want to ask anything?

[timid topaz]

yes po, I need help

[raven zenith]

sure, just ask your question there will always be people who can help you, and it's faster if you post it right away

[placid ridge]

Ok. Whats your question?

[midnight plank]

@timid topaz Has your question been resolved?

[small jasper]

1f: Missing curly brackets. Should be {1,3,5,7,9}.
1g: good.
1h: good.
1i: good.
1j: good.

2a: good.
2b: good.
2c: good.
2d: good.

3: I'm pretty sure they want curly brackets for the enumerations. Ex. For part a, {6,7,8,...}.

a) I don't see why x would have to be even. x being an integer should be fine.
b) Omit the x = part of your inequality in the set builder - x is not equal to one singular value, but rather can take on an interval of values (as shown with your inequality). Also, does the grade have to be an integer? Because your enumeration and set builder notation conflict. Finally, if the maximum grade is 100, then your enumeration should be written up to 100 (writing 99 ... suggests you can have all integers more than 99, including values like 101).
c) Same advice for omitting the x = as in part b holds. Enumeration should be fine.

[midnight plank]

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[leaden seal]

would i figure this out by checking the second and third derivatives

[subtle zinc]

what's a point of undulation

[leaden seal]

2nd der = 0

and 3rd der = 0

[bold peak]

Interesting

If so yes

[subtle zinc]

what's point of inflection

[bold peak]

That's what you would do

[subtle zinc]

i just wanna check if she knows

[bold peak]

Ah

[subtle zinc]

cuz like obvi ur not gonna check that with the 4th derivative

[bold peak]

Mbmb

[leaden seal]

when 3rd = not zero

[subtle zinc]

yeah so to answer ur original question

no

u check that with 5th derivative

[leaden seal]

ok

what does it mean by concavity doesnt change

[subtle zinc]

7th deriv < 8th deriv

[bold peak]

Huh

[autumn canopy]

Zah is back

[autumn canopy]

Are you sure?

Google gives me "Second derivative is zero and curvature does not change sign"

But third derivative = 0 is not equivalent to 'curvature does not change sign'

[leaden seal]

hmmm

[autumn canopy]

Which would be the opposite of a point of inflection

Where curvature does change sign

[autumn canopy]

Can you check your definitions?

[leaden seal]

yh okay

[autumn canopy]

Is it this?

[leaden seal]

for odulation

[autumn canopy]

Right

Aka "curvature does not change sign"

[leaden seal]

so is d correcT?

[autumn canopy]

Haven't looked at it yet

What did you get for the POI?

[leaden seal]

(0,0)

and POI=0

[autumn canopy]

Right

f'(x) = 5x^4 + 1 and f''(x) = 20x^3, left to 0 it's negative and right to it it's positive

i.e. curvature does change at 0

i.e. inflection point

meanwhile the third derivative is zero, yes. This shows that it can still happen that it's an inflection point, third derivative zero does not imply that it's an undulation point

@leaden seal

What does hold though, is that third derivative nonzero and second derivative zero => point of inflection

[midnight plank]

@leaden seal Has your question been resolved?

[leaden seal]

5x^4+1 is always positive

and 20x^3 is 0

[leaden seal]

idk what you mean by left to zero is positive

[autumn canopy]

I said left to 0 f'' is negative

Like plug in x = -0.1

(-0.1)^3 is negative

and 20 * that is also negative

[leaden seal]

so 2nd derivative is negative

so changes from positive to negative

[autumn canopy]

At 0, f''(0) = 0

It changes from negative to positive though

Because a little left from 0, f''(x) < 0 and a little right to it, f''(x) > 0

[leaden seal]

sadly i dont understand this i will read up on it now

[autumn canopy]

f'' describes the curvature, right?

If it's positive, it's convex (curved upwards), if it's negative, it's concave (curved downwards)

If we say the curvature changes at one points, we mean it goes from convex to concave or from concave to convex

And this means a little left from the point, it has to have the old curvature, a little right to it, the new one

(because in the middle, at the actual point, it changes)

[leaden seal]

for this question would it be cos(very large number)e^very large number

[midnight plank]

@leaden seal Has your question been resolved?

[flat veldt]

,w plot cosx × e^x from -10000 to 10000

Yup

[flat veldt]

Hmm not sure about that cos(very large number)
Remember cos theta is oscillating between -1 and 1

Yeah so
What happens to the e^x part

As x->∞

@leaden seal

[midnight plank]

@leaden seal Has your question been resolved?

[leaden seal]

becomes infinity?

[flat veldt]

Yeah it becomes really large so you can say that cos(x) dictates whether the function is above the x axis or below

Right?

[leaden seal]

yh

[flat veldt]

Okay so we know that as x->∞ y will oscillate between...

[leaden seal]

-1 and 1

[flat veldt]

No

y = e^x • cosx
a=e^x
b=cosx
a->∞
-1≤b≤1

[leaden seal]

og

oh

so infinity?

[flat veldt]

It oscillates between "infinity and -infinity"as x increases

This is the end behaviour at the right

At the left what do you think will happen

[leaden seal]

wdym at the right?

are you talking about the graph

[flat veldt]

Mhm

As x->∞

Just to be sure have you done the domain question

[leaden seal]

(infinity,-infinity)?

[flat veldt]

Sorta the lesser value stays at the left (:

[leaden seal]

is it like substituing infinity into cosxe^x

[flat veldt]

For the domain?

[leaden seal]

yes

[flat veldt]

No you look for points where the function is defined. If that is too tedious then look for the set of points that the function is undefined and exclude them from the remaining points that's the domain

[leaden seal]

and how would we do that for cosxe^x

[flat veldt]

e^x and cosx are defined everywhere so their products will be too

[leaden seal]

and how do we know if its not defined?

[flat veldt]

By examining it's properties
There are properties of some common functions
Like polynomials are defined everywhere
c^x is defined everywhere c≠0
then cosx and sinx are defined everywhere

[leaden seal]

and defined everywhere means negative infinity aswell?

[flat veldt]

It means you won't get something like 0 in the denominator or √(negative number) when you evaluate the function everywhere

[leaden seal]

and o find the limit or end behaviou

we see what happens to yas x approaches +/- infinty?

[flat veldt]

Yes

Basically
Just
$lim_{x -> \infty}$ f(x)
And
$lim_{x -> - \infty}$ f(x)

[grand pond]

Clover 🍀🦋

[flat veldt]

If the limit doesn't exist then you can explain why

[leaden seal]

so for this do we sub infinity for x?

or a very large number

[flat veldt]

Yes
But you might still need to do other stuff afterwards

Okay so what are the end behaviours

[leaden seal]

-infinity

+infinity

[flat veldt]

Umm no

As x approaches infinity f(x) oscillates rapidly

Remember

[leaden seal]

ok

[flat veldt]

Then what about x -> -∞

@leaden seal

[leaden seal]

hmmmm

[flat veldt]

e^-∞ cosx

[leaden seal]

0?

i think im gonna come back to it tmro

its just not clicking are there any videos on it?

[flat veldt]

Yes 🥳

although cos causes the sign to change ±0=0

Why sad

[leaden seal]

i just understand

DONT

i need to learn it somehow

[flat veldt]

Relax and try again

Try doing some precalc questions from calculus early transcendentals

They help to identify limits and evaluate stuff easier 🙂

[leaden seal]

i need a video to watch

[flat veldt]

Oh okay

Organic chemistry tutor and khan academy have videos on it

[leaden seal]

okay thanks

.close

[midnight plank]

Channel closed

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[flat veldt]

I think I'm hallucinating

[midnight plank]

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[native wharf]

Wondering if this would be a ocrrect answer for this

x²/x(x+2) the answer said x/x+2(also said answers may vary so)

[native wharf]

(please ping when u can help)

[simple field]

the two expressions are equivalent for x != 0, you can cancel one of the x's on top and bottom of your expression to get the other one

[native wharf]

but

will i still be correct?

if i didnt

[simple field]

would need to see the full questions to know

but probably fine tbh

[native wharf]

@simple field this is the full question

this too is apart of it

[simple field]

ok i would say then maybe your answer wouldnt be accepted

because you were given an explicit form for the function to be in and yours isnt in that form

[native wharf]

oh I see

how would the proper answer be?

[midnight plank]

@native wharf Has your question been resolved?

[lament creek]

the proper answer seems to be the one given by the answer key

well at least, it's one of them. yours doesn't work because you have a squared term in the numerator (and what could be expanded to be a squared term in the denominator)

the question explicitly wants both the numerator and denominator to be linear, so

[midnight plank]

@native wharf Has your question been resolved?

[native wharf]

some other answers to other questions have squared

[lament creek]

can you show those questions (together with this instruction so I can confirm it's under the same question) and show the answers to those questions?

[native wharf]

yep

im gonna reopen my textbook rq

@lament creek

[lament creek]

well, I kind of asked for two things 😅

(btw you don't have to ping, I'm lurking anyway)

ah I see the problem. c) is impossible without a quadratic on both top and bottom (but which then simplifies to be linear on both)

d) is flat out impossible with a linear denominator

[native wharf]

so wait whats going on 😭

[lament creek]

faulty question I suppose

or you can gungho and ignore the linear requirements

[native wharf]

😭

[lament creek]

😅

I mean, you yourself have seen that they ignore their own requirements

(and they have to - c) and d) are outright impossible without originally tossing a higher-order term in either the numerator and denominator of the function)

at least c) does simplify down to linear / linear, but there's no hope for d)

[native wharf]

Ok this makes sense what ure saying

thank u for the help

❤️

.close

[midnight plank]

Channel closed

Closed by @native wharf

Use .reopen if this was a mistake.

[midnight plank]

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[cerulean oyster]

We already have done this one

lmao

[worldly pine]

yes but

idk

i want to solve this with a,everyone can solve by this, method

[lyric charm]

such method does not exist

[worldly pine]

i didn't grasp your method conceptually...
what if they made some changes in the question...i will be left blank then

i am close to finishing this topic..
i want to understand this chapter..but afraid of giving too much time to it

[cerulean oyster]

The simple idea behind it is that:
By dividing the whole average into two parts, where they share exactly one element
And then getting them back together, you can find exactly what element is shared

Since its counted twice

@worldly pine

[worldly pine]

can we call an average a middle number

solving for mean gives middle number?

there is one more question like this..in it they asked for 'What is the middle number that is counted twice'

[cerulean oyster]

To be honest, this is more of a theorical thing that goes outside of what the problem is about
But the arithmethic mean (aka average) is actually a poor way to approximate the center value (in some cases)

[cerulean oyster]

And also, the middle number we are talking here is one element from the average, not the average

[cerulean oyster]

I could try walking you through the idea of this problem again if you want

[worldly pine]

please

[cerulean oyster]

First of all, do you understand what an average/mean is?

[worldly pine]

sum of observations/ no. of observations

central value of the series

[cerulean oyster]

Okay, so

[worldly pine]

you know what i am going to cram these few questions into my head

[cerulean oyster]

Mb, i was in the bathroom and ran out of battery 😭

Lets just say I give you a list of numbers

{6,9,9,1,4,3,1}

You could find the mean of that list, right?

[worldly pine]

yes

[cerulean oyster]

But, also, if you wanted, you could find the mean of a subset of those numbers

Lets say, the first four elements

or the last four

or every other number

Whatever you want

[worldly pine]

ok

[cerulean oyster]

{6,9,9,1,4,3,1} -> Mean = ~4.71

Lets just say i take the first four numbers
6,9,9,1 -> Mean = 6.25
And last four numbers
1,4,3,1 -> Mean = 2.25

I hope you can see that these two subsets share exactly one element.

the last 1 in the first
the first 1 in the second.

[worldly pine]

yes

[cerulean oyster]

Well, you can obviously see what the element is yourself because i just invented the numbers myself

But, ill state to you that
If we know the mean of all the 7 numbers (~4.71)
The mean of the first 4 (6.25)
and the mean of the last 4 (2.25)

Then we can find exactly the number in the middle without any extra information

[worldly pine]

4.71 × 7 = 6.25 × 4 + 2.25 ×4 - x

right?

what next

here calculation is tedious

[cerulean oyster]

This is the idea, yes, theres a mechanical way to do it, which is what me and Ann were trying to explain

But youll have to do arithmetic either way

[cerulean oyster]

btw, that isnt tedious

[worldly pine]

in the question that i provided, it is

[cerulean oyster]

$74 \cdot 33 = (72.8 + 77.2) \cdot 17 - x$

[grand pond]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[cerulean oyster]

thats a really basic operation

and if youre not comfortable doing that either with paper or mentally, you should be practicing

Once you do that, youll find the number shared (in the middle)

[worldly pine]

we have to find mean again after that

[cerulean oyster]

This has x = 108, which is the element in the middle

then to exclude it from your previous mean:

$\frac{74 \cdot 33 - 108}{32}$

[grand pond]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[cerulean oyster]

And this will be the mean/average of the 33 numbers without the middle element (so 32 total)

[worldly pine]

okay

not so long ig

[cerulean oyster]

statistics is riddled with long arithmetic operations if you are tasked to do it manually

[worldly pine]

ok...i will do this way..
and focus on other topics with equal intensity

wait

[worldly pine]

? don't get it

[cerulean oyster]

What?

[worldly pine]

this

[cerulean oyster]

The mean/average is part of statistics

If you have to do calculations by hand, you will usually have to do long operations

[worldly pine]

My exam want to do it manually

[cerulean oyster]

Then youll have to do long operations

No way around it

To be honest, this method is extremely short

Calculating means by hand can easily take a few minutes if you have a lot of numbers to go through

[worldly pine]

the photo i gave previously is, so called by the teacher - deviation or weighted average..., is an easy way to approach these type of problems

but i am not getting it

do i share it here again?