precalculus

[viscid thistle]

you can like ask questions aboout different subjects

ee tok and cas as well

[calm whale]

ye I've kinda done all of that already

😄

[viscid thistle]

ooh. 2nd year?

[calm whale]

yes

[viscid thistle]

do u think the ib is harder than a level

[calm whale]

yes definetely

hm lets put it this way

if you're smart academically + orginized it doesn't matter

if you're really organized (as in doing your best) but are not so bright then IB will be easier (more rewarding)

[viscid thistle]

blackout bruh

[calm whale]

if you're a dumb and lazy shit then you'll die either way

[viscid thistle]

talks about organized while playing chess

[calm whale]

🤔

[viscid thistle]

hmm did u take/are taking further maths hl?

[calm whale]

I'm thinking alisher alright I'm just taking my time

I am taking HL maths/physics/econ

wanted to do HL chem

[viscid thistle]

how is maths hl compared to further maths alevel

[calm whale]

but I couldn't fit it in 😢

both require a lot of thorough preparation

[viscid thistle]

which one is harder though

[calm whale]

🤔

they are about the same

mmm in a-levels you have more time to prepare

on 3 subjects

[viscid thistle]

some do 4 right?

like how we can do 4 hls in ib

[calm whale]

IB is more about preparing adequately in a short amount of time

uuuh 4hls in ib

I wouldn't advise

I had this mate did 4 hls and got a 45

but generally a nono

@viscid thistle can you spot the mate in 1

no in a-levels you start with 4 and then you can drop 1

[viscid thistle]

damn my school requires me to do English lit and History of the Americas both at HL, so ill have to do 4 HLs to get maths and physics

[calm whale]

honestly if I could restart I would 100% do a-levels

[viscid thistle]

...

[calm whale]

Qxf2 is mate in 1

as I can't say I enjoy sl history/spanish or english

[viscid thistle]

ye

[calm whale]

english is alright but the other two 😶

cancer REE

[viscid thistle]

I made a mistake putting that horse there

[calm whale]

you made a mistake

taking on f6

[viscid thistle]

that too

[calm whale]

that was what lost the game basically

pushing g6 or maintaining the tension was the way to go

@viscid thistle every school requires english lit and a langl; you dont need HL physics to get into an engineering degree

[viscid thistle]

yes but mine makes it HL

[calm whale]

maths is what matters tbh 🤔

o lol tell them to fuckoff

[viscid thistle]

wtf is HL physics

like "BLOODY HELL MATE, BLIMEY, FUUCKOFFF" right? is that how u do it in england?

[calm whale]

mmmm

[viscid thistle]

you broke the maths brady

[calm whale]

you'd just give em the 2 fingers

mmmm if you're going for cambridge does that means you have 150+ iq right?

[viscid thistle]

lol didnt take a test

same

but cambridge requires physics HL

[calm whale]

that's a joke d:

for engineering probs ye 🤔

unfortunate

you;ll get fucked by your IA's lol

[viscid thistle]

yeh ill become suicidals

how about maths

[calm whale]

both maths and physics IA are cancer

[viscid thistle]

do you need 150+

in maths

yeh they are the 2 hardest ib

not reallty lol

[calm whale]

they aren't hard, they are just cancer

[viscid thistle]

just some hard work

@viscid thistle are you asian

[calm whale]

clearly not

his doge is

[viscid thistle]

of course not mr caucasian

i was sarcastic

im an asian yeh

why did you delete ur chat alisher?

[calm whale]

what do you do in pre ib

[viscid thistle]

I understood you meant irl so why ask

[calm whale]

I'm 100% sure that this is a waste of time

I legit don't see how you're going to cambridge and taking that

[viscid thistle]

no no no no

im just like year 10 dude

im not a year 12 kid mate

does cambridge have scholarships

[calm whale]

you're 15?

something like that 🤔

[viscid thistle]

yeh

2 months to 15

@viscid thistle does cambridge have scholarships

[calm whale]

they do

[viscid thistle]

@calm whale So we use the same textbook for IBs, just go in less depths, so when we reach 16-17 doing the IB, we can test for SLs only in 1 year. @viscid thistle Very few for foreigners

did ramanujan have one 🤔

[calm whale]

what can you do

maths-wise?

complex numbers?

[viscid thistle]

yes

polar

[calm whale]

modular algebra?

[viscid thistle]

parametric

no not modular yet

[calm whale]

number theory?

[viscid thistle]

we dont do number theory at all, even at maths hl, my teacher doesnt choose that as an option

[calm whale]

discrete mathematics? -strong/inductio?

oh okay

[viscid thistle]

oh man I am 14 and all I learned so far is a bit of calculus and set theory

how do I die

[calm whale]

integration/derivatives?

implicit explicit?

[viscid thistle]

yeh but not until the end of the year, implicit explicit yes

derivatives and limits

yeh me too @viscid thistle

oh

[calm whale]

eq-n of vector lines?

[viscid thistle]

you guys know wat? im playing dota and enjoying my life for the last day of thanksgiving

see yah

[calm whale]

matrices?

[viscid thistle]

yes matrices

yes vectors

[calm whale]

don't play dota D:

plane eq-ns?

Lhopital? Optimization?

[viscid thistle]

not yet, end of year

see yah im playing dota

[calm whale]

🛑 dota

[clever inlet]

🤔

[calm whale]

🤔

@clever inlet I hate you

¯_(ツ)_/¯

[clever inlet]

Rip

Dota is bad

[calm whale]

yes

[clever inlet]

My friend picked up dota

[calm whale]

League is superior 🤔

[clever inlet]

He comes to school like twice a week now

If even that

We had a low attendance meeting at school

And he missed that too

[calm whale]

wait what grade is that

[clever inlet]

11/12

[calm whale]

hmm that's unnaceptable really

[clever inlet]

I don't think he can pass the year

Cause you need 85% attendance

And he's at like 20

[viscid thistle]

WHAT THE HECK GUYS

HOW IS LEAGUE SUPERIOR

it's fun

dota is boring and slow

"outplays" are through micro decision making

not mechanics

.....

league is flashy af

no

@viscid thistle r u an american maths student

no

not american, not a math student

lel what r u

[calm whale]

what

you just said MICRO DECISION MAKING, NOT MECHANICS in the same sentence

🤔

micro = mechanics

macro = map movement

[viscid thistle]

Canadian business student

micro is not mechanics

microdecision making*

[calm whale]

micro is mechanics

this isn't debatable

it's a given, a fact if you'd like

[viscid thistle]

microdecision making involves positioning in lane, positioning in teamfights,

macro is rotations, vision

micro also involves lane management early game

but this is a calc chat, mb.

Did you take like maths 31 in high school, if you are Alberta-rian?

Ontario

and I didn't focus on math and shit in hs

i went a super standard path

😮

[inland forum]

i need help bros

with trig

anyone willing to help

shooot me a private message

[viscid thistle]

@inland forum I can help u in 2 weeks lol

[inland forum]

now preferably

[fast tinsel]

This is the method we use to solve a sin(x) + b cos (x) = c

I just can't wrap my mind around the fact that someone came up with this, how do you even get the idea 😮

[willow bear]

tbh this is a slightly roundabout way of going about this

[fast tinsel]

Is there a better one?

[willow bear]

hmm

well

okay so let's take the expression a sin(x) + b cos(x)

and let r = sqrt(a^2 + b^2)

then we can rewrite it as

=tex r \left(\frac{a}{r}\sin(x) + \frac{b}{r}\cos(x)\right)

[willow bear]

and now the cool thing about a/r and b/r is that

(a/r)^2 + (b/r)^2 = 1

so they are the cosine and sine of some angle

we can call that angle phi

and get

=tex r(\cos(\varphi) \sin(x) + \sin(\varphi) \cos(x)) = r \sin(x + \varphi)

[fast tinsel]

Wow that's really interesting

Is this a more common/used method?

[willow bear]

it's more transparent imo

[fast tinsel]

I see

Will definitely be showing this to my teacher, thanks a lot

[rugged marsh]

hey

[willow bear]

👀

[rugged marsh]

i was asked to find the asymptotes of cot(x)

[willow bear]

uh huh

[rugged marsh]

i wanted to ask, if tan(x) was undefined, will cot(x) also be undefined?

[willow bear]

nope

[rugged marsh]

why?

1/(undefined) is defined?

[willow bear]

no because cot is defined as cos/sin, not 1/tan

[rugged marsh]

oh, i thought it was 1/tan(x)

anyway, in general, is 1/(undefined) defined?

[willow bear]

no

[rugged marsh]

why?

can you give an example of "undefined" execept division by 0?

also, does this net neutrality thing happen in only the U.S.?

@willow bear

[willow bear]

it seems to be US only, yes

[viscid thistle]

Dunno if Logarithms are supposed to be "pre-calculus", but this is a fun challenge...

[viscid thistle]

@SpriteFan274#1529 yes, it is pre-cal

[viscid thistle]

@SpriteFan274#1529 lel can u give us the solution, i combiened it, put the first term into log base 5, stick it together but i dont know how from there

[hoary lantern]

thas jsut unnecessary

[brisk root]

I have pondered this question for sometime, our teacher is a stickler for details yet provides few for the question. Aside from assumption for the altitude of the antenna, is there another possible way to solve this? If so, please solve.

"A radio antenna is 80 feet tall. The angle of elevation from the top of the antenna to a landing airplane at an altitude of 640 feet is 3° 40'. Find the line-of-sight distance from the top of the antenna to the airplane."

[patent beacon]

What's the assumption?

Well, the question defines a right angled triangle, and you can use trig to solve it

[brisk root]

My assumption is the radio antenna base is at an altitude of 0. I may have over looked something. Did you account for this altitude or have I missed something simple (again)?

Sorry for the delay, I was expecting a notification.

[brisk root]

That is based on the assumption that the altitude is 0 for the base of the antenna. However, since you have arrived at the same diagram as me(though a little less to scale haha) I will use that. I appreciate it, truly.

[brisk root]

"A boat is sailing due east parallel to the shoreline at a speed of 10 miles per hour. At a given time, the bearing to the lighthouse is S70 E, and 15 minutes later the bearing is S63 E. The lighthouse is located at the shoreline. What is the distance from the boat to the shoreline?"

I have a solution, however many of my classmates and I share different answers. Does this look correct?

[nimble oxide]

Can someone explain to be how-to find the domain of composites.

[willow bear]

of what? compositions of functions?

[nimble oxide]

Yes, composite functions.

[willow bear]

mh

do you have an example we could work through or do you just want a set of general guidelines?

[willow bear]

aight so

in general, when finding the domain of a composition like $$f \circ g$$, you'll want to do the following

[willow bear]

start with the set of real numbers (R)

then find the domain of g

what might be easier to find, and what i recommend you find, is what isn't in the domain of g

in other words, what values of the input to g will "break" g

(i'm not yet talking about your specific example)

once you've done that and gotten your new, restricted set (which isn't the domain yet!) you'll need to find what values of the input to f will break f

and then solve the equation(s) or inequality(ies) g(x) = <whatever you found will break f>

that might sound confusing

[nimble oxide]

No, I believe I understand.

[willow bear]

so here's a diagram if you need

the "g" and "f" boxes represent their respective functions, and D(f) and D(g) stand for the domains

[nimble oxide]

Ah, so the domain of g and f are irrelevant to the domain of f comp g?

[willow bear]

no? they are anything but irrelevant lol

you need to know 'em so that you know when your composition fails

[nimble oxide]

Oh, gocha!

So for the example I gave, 5 and 0 wouldn't be in the domain, correct?

[willow bear]

uh

no

i'm pretty sure g(5) is in the domain of f 😛

[nimble oxide]

Oh....but f(5) would make it invalid,....

[willow bear]

f(5) would break yes

you don't want g(x) to equal 5

[nimble oxide]

So why would 5 still be included?

[willow bear]

f(g(5)) = f(1/5) = (1/5 + 4)/(5 - 1/5)

and that does not end up with dividing anything by zero

[nimble oxide]

O.....

[willow bear]

it might be helpful to rewrite the definition of f as f(u) = (u + 4)/(5 - u) to distinguish the inputs to f and g

[nimble oxide]

and if you solve that it'll be (1+4x)/(5x-1)

[willow bear]

define "solve"?

also, you're definitely missing parentheses there

[nimble oxide]

simplifying the composition would be a better wa for explaining it

[willow bear]

also, the simplification of $$\frac{\frac{1}{x} + 4}{5 - \frac{1}{x}}$$ to $$\frac{1 + 4x}{5x - 1}$$ ignores the fact that $$x$$ could not be $$0$$ before the simplification

[nimble oxide]

This is giving me a headache. 😅 ugh

OHH nevermind I get it, the domain of f comp g can't equal (1/5) becasue you have to set the denom. equal to 0. with the answer being

[willow bear]

, the domain of f comp g can't equal (1/5)
contain, not equal

the domain is a set

[nimble oxide]

(-infin,0) U(0, (1/5)) U ((1/5), infin)

[willow bear]

yup looks correct

[nimble oxide]

ugh, finally.. domain gets me everytime.

[tired dagger]

could anyone help with this problem please?

kinda confused as to how I'm supposed to set this up

[willow bear]

looks like law of sines could be useful here

try obtaining all the angles in the triangle

[tired dagger]

literally just tried that and got the wrong answer

[willow bear]

care to show what you tried?

[tired dagger]

it was on an older problem and the computer seems to change the numbers every time you get it wrong

[willow bear]

i mean you might have screwed up on the arithmetic

[tired dagger]

I subtracted 43 from 180

to get the second angle

is it from 180 or subtract from 90?

[willow bear]

well what do 43° and the angle you're finding have to add up to?

[tired dagger]

honestly don't remember. I assumed 180 because I recall my professor mentioning something regarding subtracting from 180 on the coordinate plane

[willow bear]

do you know what a 180° angle looks like

[tired dagger]

yeah and I just realized the mistake I made

sorry. Should have paid attention and noticed it earlier

[languid wind]

how would i approach the following question:

solve for x in the domain 2pi:

3sin2xcosx + 3sinxcos2x = 1

[calm whale]

you would use somehitng called double angle identities

google it

[languid wind]

is 2cosx and cos2x the same thing?

[calm whale]

it is not

that's why you have DOUBLE ANGLE IDENTITIES

[languid wind]

3sinx2cosx +3sinx(cos^2x - sin^2x) = 1

[calm whale]

something like that

but not quite

alomst there

[willow bear]

@languid wind this is literally sin(2x+x) but expanded

@calm whale nope

this literally screams for an ANGLE SUM IDENTITY

[timid plank]

hmm

what is the domain of f(x) = tan(2x+3)-1 (the restricted one to where it can be one to one)
i got (-pi-6)/4< x < (pi-6)/4

i am confused because when I found the inverse things didn't add up

because the domain of f(x) should be the range of f-1(x) but it is not

[willow bear]

what did you get as f^-1(x)?

[timid plank]

(arctan(x+1)-3)/2

[willow bear]

uh huh

yeah, the range of that is ( (-pi-6)/4, (pi-6)/4 ) as required

[timid plank]

hmm

ok

but then the range of the inverse would be -pi/2 < (y-3)/2 < pi/2

and solving that doesn't result in the domain of f(x)

that is where I was confused

[willow bear]

uh nope

-pi/2 < arctan(x+1) < pi/2

[timid plank]

o

i am stupid

thx

[hidden linden]

Could someone help me with solving an equation? It's basically just x^2/3 - 4x^1/3 - 5 = 0

I got to the point of x^2 - 4x = 125

I assumed I'd want to get all terms on the left to factor, but I can't factor that, can I?\

[viscid thistle]

@hidden linden Actually yes, I think its technically called super imposing where a*b=0, either a or b is zero

We find the roots this way

so, we set it equal to zero

so we factor/quadformula/newton-rasphone/(my discovered on my way to get coffee way to solve for polynomials), etc

(x-n)(x-m)=0

so, we can take it over and we get x^2 -4x -125 = 0

[hidden linden]

wut

[viscid thistle]

In short, you can take it to the other side.

[river crow]

@hidden linden use quadratic formula after

[viscid thistle]

or you can complete the square

heh

[clever inlet]

I think you made an error converting to a quadratic?

If you are trying to cube everything

It wouldn't work like that

[viscid thistle]

let me check actually

[clever inlet]

Its still a question reducible to a quadratic

But not like that

[river crow]

anyone know how to solve for x here? (1/8)^(2x+1) = 32^(x-3)

[viscid thistle]

@clever inlet I mean, he should have got x^2 -4x -15 = 0

@river crow I can _walk you through it _

[river crow]

@viscid thistle ok that would be great

[clever inlet]

I don't think that works either

[viscid thistle]

@river crow well, someone who is good with the mathbotshould actually do it ~~ because Im very bad at it...because I never use it ~~

@clever inlet yeah, by the property of a/b-c/b = (a-c)/b

okay, so first off, you know what an exponent of 1/8 turns into, right?

[clever inlet]

x^2/3 - 4x^1/3 - 5 = 0

(x^1/3)^2 - 4x^1/3 - 5 = 0

let x^1/3 = t

[viscid thistle]

but, you multiple both sides by 3 though

[clever inlet]

oo

i assumed the 1/3 was the exponent

cause he got 125

so i assumed he cubed everything

in an attempt to eliminate the fractional exponents

[viscid thistle]

Yeah, okay, I was think, _where did you get the idea we should (x/3)^2

nvm

[river crow]

=calc (1/8)^(2x+1) = 32^(x-3)

[granite stirrup]

Failed to parse equation: Invalid syntax at position 8

(1/8)^(2x+1) = 32^(x-3)
        ^

[river crow]

wth

[viscid thistle]

** go to bot testing to experiment, would you kindly? **

@clever inlet yeah, I'm guessing he just put it as x^1

[river crow]

yeah sure sorry

[viscid thistle]

no problem, sorry if you thought it was being harsh

My monster of an equation

made with extremes in mind

https://www.desmos.com/calculator/h8pwgjjc20

[hidden linden]

=tex 2^{-3}^{(2x+1)} = 2^{5}^{(x-3)}

[granite stirrup]

Rendering failed. Check your code. You can edit your existing message if needed.

[hidden linden]

Fuck

Isn't that what you do for your problem though @river crow

So it ends up being -6x-3 = 5x-15 ?

[clever inlet]

yeah, seems like it

[hidden linden]

And just solve for x at that point

I did that all in my head so correct me if I messed something up

[clever inlet]

yeah

have you done that question above?

[hidden linden]

Which one?

The one I asked about?

[clever inlet]

yeah

[hidden linden]

Yes I did

[heavy sonnet]

o ya

[clever inlet]

nice

how'd you end up doing it?

[hidden linden]

Substitute in u² and u, giving you u² - 4u - 5 = 0

[clever inlet]

yeah

[hidden linden]

Factor giving you (x-5)(x+1)=0

U I meant

Not x

[clever inlet]

yeah

looks good

[hidden linden]

Then u ressubed in valves for u as x^1/3, cube to get rid of cube roots

Gives you -1 and 125

[clever inlet]

👍

[hidden linden]

Sweet shit bois

I think this is how you do it @river crow

But double check with someone smarter

Should be 12/11 :/

[clever inlet]

looks right to me

[river crow]

yeash its correct

thanks

[hidden linden]

Yeah those problems are basically all laws of exponents I believe, so if it doesn't make sense Google that shit

[viscid thistle]

Alright I need some help with pre-calculus problems

So I'm still trying to figure out using summation notation for finding the area under the rectangle.

https://gyazo.com/80539e886b79d022abfb8d3efe84b033

This is supposed to equal 24.8

However, I put it in for my calculator as 4^∑vx=1 (2(2x-1))

And I get 32

Why is that?

[twilit zodiac]

depends what f is

4^∑n=1, (2(2n-1)) <=> 2 * (2 * 1 -1) + 2 * (2 * 2-1) + 2 * (2 * 3 -1) + 2 * (2 * 4-1)

[willow bear]

that <=> should be =

also

if you want to do sums in plaintext

sum[n=1..5] n^2

[primal crater]

if h(x) is g(f(x)) would the restrictions of g(x) (domain and range wise) carry over to h(x) as well as f(x)?

[late haven]

ping <@&286206848099549185>

[severe verge]

yes the restrictions carry over @primal crater

[viscid thistle]

@severe verge are you good at algerbra

[severe verge]

why whats up

[viscid thistle]

look at general 1

@severe verge

[severe verge]

ya i posted in questions 2 lol

woops

[viscid thistle]

lol

[calm thicket]

I don't know how to do e-h

[viscid thistle]

e) does that refer to end behavior ?

as in is it increasing as x goes to positive infinity and so on

[calm thicket]

Basically, solve as x goes to infinity

[viscid thistle]

ye

ok so its that

right

[calm thicket]

Mhm

Preferably in limit notation

[viscid thistle]

oh, so without the odd/even and degree of leading coefficient info?

since u posted in precalculus i figured that's what you'd be expected to do it

lol

[calm thicket]

If you have another, I can translate it 😛

[viscid thistle]

so u mean

u want to evaluate it with limits as x goes to infinity

right

[calm thicket]

Generally yea

[viscid thistle]

so you want $$\lim_{x \to \infty} f(x)$$

[calm thicket]

Yea

[viscid thistle]

are you comfortable with limits to infinity ?

[calm thicket]

Not overly

[viscid thistle]

well, and keep in mind that its 11pm here so i may not be at my best, let me try to figure this out on paper really quick

so i dont waste ur time

[calm thicket]

But I get some concepts, for instance, I'm 95% sure this is 1

because the highest degree is the same

[viscid thistle]

its better if we go thru the evaluating limits at infinity

it will come at more use with integrals and shit, at least im using it in calc now

gimme a sec

ok got it

it is 1 as far as i got

i can walk you through the limit process if you want

I got 1 too

actually, lets do it

[calm thicket]

Yea because not all of 'em have the same degree in num and den

[viscid thistle]

=tex \lim_{x \to 0} \frac{x^2-9}{x^2-x-6}

[viscid thistle]

the way i went about solving this

and the way my book did it too (calc book tho)

is, i divided all terms by the term largest exponent of the denominator (in this case x^2)

[late haven]

\infty ?

[viscid thistle]

shit right lmao

[calm thicket]

:p

[late haven]

lol

[viscid thistle]

11pm intensifies

anyways xd

what the next step is doing

[calm thicket]

that'll help with my x-ints but I think just plugging in 0 would be plenty 😛

[viscid thistle]

is dividing each term by x^2

[calm thicket]

Okay

[viscid thistle]

so then

cough l'hopitals rule cough

try to divide all terms by x^2

[calm thicket]

cough what's that? cough

Mhm

[viscid thistle]

You'll cover it in calculus

tell me what your fraction looks like

once you divide all terms by x^2

[calm thicket]

=tex \frac{1 - \frac{9}{x^2}}{1 - \frac{1}{x} - \frac{6}{x^2}}

[viscid thistle]

yep

so then

our limit now looks like

[calm thicket]

The stuff on the right would get to 0 as x gets larger it seems

[viscid thistle]

=tex \lim_{x \to \infty} \frac{1 - \frac{9}{x^2}}{1-\frac{1}{x} - \frac{6}{x^2}}

[viscid thistle]

agreed

?

[calm thicket]

So 1

mk

makes sense

[viscid thistle]

right

now remember the property about limits to infinity

being that

[calm thicket]

Does the same method work on this one? Dividing all by x^3?

[viscid thistle]

=tex \lim_{x \to \infty} = \frac{c}{x^n}

[viscid thistle]

it wouldn't be x^3

but x^2

remember, its the denominator you evaluate

[calm thicket]

Oh

k

[viscid thistle]

but anyways

on the problem we were now

you should also remember that $$\lim_{x \to \infty} c = c$$

[calm thicket]

Yea

[viscid thistle]

c being a constant in both "rules" i just put up

so then ye

u get 1/1 = 1

[calm thicket]

A'ight

[viscid thistle]

this is from where the rule you were talking about comes from, from the evaluation (butchered english?) of the limit to infinity

also

if ur evaluating horizontal asymptotes with these

you should remember that some functions may have different horizontal asymptotes on one end of the x axis vs the other

u can verify that its the same on both by checking the limits as x approaches positive as well as negative infinity

[calm thicket]

A'ight

[viscid thistle]

are u seeing this in precalc tho ?

we did limits at infinity like a month or so ago in calc xd

[calm thicket]

I asked my teacher if instead of the usual method, I can use limits to infinity as practice for precalc

*calc

and she said it's fine

[viscid thistle]

good call

[calm thicket]

so I'm learning limits while everyone else is doing weird stuff like $$x\to\infty, f(x)\to1$$

[viscid thistle]

well

what they'd be saying is something like

as x -> infinity, f(x) -> infinity or so on

but ye

[calm thicket]

and correct me if I'm wrong, limits are a more formal (and accurate) way of doing it

[viscid thistle]

i dont qualify myself to say whats accurate/formal, but i am able to tell you that it is used in calc

[calm thicket]

k

[viscid thistle]

and it starts to be more and more used in integration too, so you can bet its going to be used on calc 2 and so on

[calm thicket]

But how do I evaluate the limit of this one?

[viscid thistle]

aight so

remember what i told u about

u want to divide every term by the term with the largest exponent of the denominator

[calm thicket]

Dividing by x^2, we get x + 1 - 2/x / 1 - 1/x^2

[patent beacon]

Limit going to infinity?

[calm thicket]

mhm

[viscid thistle]

=tex lim_{x \to \infty} \frac{x + 1 - \frac{2}{x}}{1-\frac{1}{x^2}}

[viscid thistle]

right ?

[calm thicket]

teacher said something about oblique asymptotes if the numerator's degree is larger than denom

And yea

[viscid thistle]

so then

limit as x -> infty of 1 = 1

as we saw before

[patent beacon]

Mineswell not bother. The numerator has a larger degree, so it will always overtake the denominator

[viscid thistle]

lim x-> infinity of x = infinity

[patent beacon]

Goes to infinity

[calm thicket]

So it's infinity

[viscid thistle]

therefore

[calm thicket]

Yea

[viscid thistle]

yes

infinity + 1 is still infinity

infinity / 1 = infinity

= infinity

[calm thicket]

=tex \lim_{x\to\infty}\frac{x^{n+a}}{x^n} = \infty \text{ for positive numbers } a

[calm thicket]

If I'm not wrong

[viscid thistle]

it would appear so

[patent beacon]

The only interesting rational function is the one where top and bottom are both the same degree

[viscid thistle]

then the limit you'll get

will evaluate to coefficient/coefficient

[calm thicket]

I concluded that if it's the same degree, you get leading coef/leading coef

[viscid thistle]

like you did with the previous one

[calm thicket]

Yea

[viscid thistle]

yep

[calm thicket]

And if it's a lower degree in numerator, it's 0

[viscid thistle]

and then there's the 3rd ruel

yep

lmao beat me to it

[calm thicket]

But higher, is there no asymptote because it goes to infinity?

[patent beacon]

There is a slant asymptote

[calm thicket]

Whoa

[viscid thistle]

only if the difference between the exponents is one

[calm thicket]

How to find?

[patent beacon]

You can get it with polynomial division
But it will go to infinity

[viscid thistle]

unless i remember it wrong

but the degree of the denom has to be at least 1

and the difference of degrees has to be 1 exactly

[calm thicket]

Mkay

[viscid thistle]

i plugged some values into desmos and it appears to be correct

but feel free to correct me xd

[patent beacon]

(x² + x + 1) / (x + 1) = x + 1 / (x + 1)

As x goes to infinity, 1 / (x + 1) starts to disappear, leaving x alone. So, (x² + x + 1) / (x + 1) will act like x for large values of x

[calm thicket]

So the asymptote is y=x?

[patent beacon]

= pup graph (x² + x + 1) / (x + 1)

[granite stirrup]

Query made by @patent beacon
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=graph+(x²+%2B+x+%2B+1)+%2F+(x+%2B+1)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
This command is in development. Suggest improvements on the MathBot server (type =about for the link).

[opaque bay]

hm

intresting

[calm thicket]

How does one divide polynomials?

[patent beacon]

The graph is far from good looking, but that's what you get, lol

[viscid thistle]

long division

[opaque bay]

nah

synthetic is faster

[viscid thistle]

or synthetic (if degree of term in the denom is 1)

[calm thicket]

(assuming it's quick, I still have 4 more questions that'll be on the test)

[viscid thistle]

synthetic is faster, but there's the condition of the denominator's leading term exponent being of degree 1

[opaque bay]

lol

yea

[viscid thistle]

in other words

[opaque bay]

welp

[viscid thistle]

if u got sth like x^2 + whatever in denom = synth doesnt work

[opaque bay]

gn ya'll

[viscid thistle]

so then long division

[patent beacon]

I avoid division processes wherever possible. It's typically possible:
(x² + x + 1) / (x + 1)
= x(x² + x + 1) / (x² + x)
= x(x² + x) / (x² + x) + x / (x² + x)
= x + 1 / (x + 1)

[calm thicket]

Nice.

[opaque bay]

op

[calm thicket]

So how do I find vertical asymptotes?

[patent beacon]

When the denominator is zero, and the numerator is NOT zero

[calm thicket]

Okay

[patent beacon]

My function has a vertical asymptote at x = -1

[calm thicket]

Referring to this, it factors to (x+3)(x-3)/(x+2)(x-3) so the asymptote is at x=-2, but NOT x=3?

[viscid thistle]

va = zeroes of the denom that do not make the num 0

if i remember it right

[patent beacon]

Indeed, no asymptote at x = 3. You get a "hole" there instead

[viscid thistle]

removable discontinuity if ur profs will take points off for saying hole

hides in B-

[calm thicket]

They won't

In fact, i is determine the coords of any holes, I assume limits are necessary there

[patent beacon]

And you can use a limit to find where this hole is

[viscid thistle]

i fail to see why you would in a case like this though

you see that RD's are factors that cancel out

[calm thicket]

Yea

[patent beacon]

To keep the function the same, you'd need to consider the restriction that x ≠ 3. But you can be like EEHH

[calm thicket]

Oh, can you just remove the x-3 from both sides and plug in x and ignore the x!=3 to get coordinates?

[patent beacon]

You can use a limit to turn "I don't really care" into "I don't really care but I'm rigorous about it"

[calm thicket]

I like this quote.

[viscid thistle]

well by saying that function cant be = 3

arent u saying that "for any value other than 3, this function is defined"

[calm thicket]

(other than 3 and -2)

[viscid thistle]

but x = -2 is a VA

[patent beacon]

Yeah. We can then go on to say that the function can be defined at x = 3 and remain continuous.

But x = -2 is a no-go

[viscid thistle]

ye

i didnt say x = -2 since we came to the conclusion that it was a VA based on the zeroes of denom that don't have the numerator as 0

also keep in mind that you should say x = -2 since its a line, dont want u to lose points for it

hides in B- again

[patent beacon]

= wolf graph (x^2 - 9) / (x^2 - x - 6)

[granite stirrup]

Query made by @patent beacon
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=graph+(x^2+-+9)+%2F+(x^2+-+x+-+6)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
🐺 Try out the new =pup command! It's much more concise.

[viscid thistle]

there xd

[patent beacon]

Again, not a great graph to show those features off :/

But you can kinda tell there's an asymptote there

[viscid thistle]

the zoomed in version up top lets us see the VA

[patent beacon]

Ahh good point

[viscid thistle]

sadly no point out of the RD at x = 3 : (

its even removed from the graph itself

unwanted : (

[patent beacon]

Unloved.

Will always have a hole where its definition should be

[viscid thistle]

fucking hell 11:30 and im not sleeping

[patent beacon]

Same^

[viscid thistle]

back to math discord instead of regular sleep schedule

agony intensifies

[calm thicket]

Same tbh

This place makes me want to be a math major

and I end up staying up until like 1 discussing with the smart people about this concept where they're *probably like "Goddamnit i it's obvious"

[patent beacon]

I love math, but I'm glad I didn't go to be a math major. Just not for me.

[viscid thistle]

i actually slapped a math major on top of my game programming one

i was fine with it since i am getting rid of all petty shit like composition 2 and stuff this semester lol

[calm thicket]

I plan on dual majoring compsci and math.

[viscid thistle]

no offence to english majors but damn lol

[patent beacon]

That's a pretty good idea, you'll be bounds ahead of other compsci majors, who don't typically know enough math

[calm thicket]

you just offended them by spelling offense like offence

😛

[viscid thistle]

hahaha

good point

guess that's why i'm not one of them 😛

and ye

since i'm into game programming specifically

i'm expected to be strong at 3d math

agony intensifies

[calm thicket]

Game programming isn't my tea tbh, I would rather write software to solve problems 😛

[viscid thistle]

have u looked at reverse engineering?

[patent beacon]

I see too many people:
"I'm trying to program this game in 3D space, can someone help me with this right angled-triangle"?

[viscid thistle]

hahaha

true that

[calm thicket]

noice

[viscid thistle]

you should look up reversing in games

[calm thicket]

sohcahtoa intensifies

[viscid thistle]

thats why i had the 2 weeks of not touching this discord server

i was legit obsessed with it

like a new toy i was into it 24/7

then i was like "i cant leave math now halfway lol"

[calm thicket]

Mk I get f-j until problem 3 D:

[viscid thistle]

so big return

[patent beacon]

Freakin' rotation matrix intensifies. You need some pretty intense math to be able to do what they want to do and they haven't left grade 10 smh

[calm thicket]

*f-i

and nice

I like my 2d games thanks 😛

[viscid thistle]

well f-i are like

just stating ur asymptotes

i is stating ur RD's

[calm thicket]

Oh sorry

g-i

[viscid thistle]

oh

well horizontal asymptotes u understood in the limits thingy didnt u

[calm thicket]

mhm

[viscid thistle]

it was the difference in the degrees and stuff

[calm thicket]

but you end up with +- infinity

(~x/1 for large values)

[viscid thistle]

which one are u doing 3 or 4

[calm thicket]

3

already did 4 but 3 has me stuck

[viscid thistle]

what do u get as ur lim -> positive infinity

[calm thicket]

infinity

[viscid thistle]

and same for lim -> negative infinity

?

[calm thicket]

-infinity

[viscid thistle]

right

so that means

[calm thicket]

No asymptote?

[viscid thistle]

that the function is unobstructedly (rip english majors x2) making its way to positive/negative infinity

[calm thicket]

Mk

[viscid thistle]

i'd check with kayne just to be sure

[calm thicket]

x(x+2)(x-1)/(x+1)(x-1)

[viscid thistle]

since its 11:40pm after all xd

is that for the same function

[calm thicket]

Yea, factored

[viscid thistle]

its not rightly factored in the numerator

its not a diff of squares

theres a plus

[calm thicket]

So i is (1,3/2)

Numerator?

[viscid thistle]

denom

sorry

[calm thicket]

No, x^2-1

[viscid thistle]

it is ?

[calm thicket]

Mhm

[viscid thistle]

then i copied it wrong

LMAO

so ye

[calm thicket]

😛

[viscid thistle]

u have the vertical asymptote x = -1

[calm thicket]

Ye

[viscid thistle]

and a removable discontinuity at x = -1 which is the point (-1, f(-1))

[calm thicket]

g and h though

Do I say no horizontal asymptote?

[viscid thistle]

actually

look at the non-factored func again

can u repaste pls

[viscid thistle]

right

so notice the degree of the num's leading term

= 3

the one for the denom = 2

there's a diff of one

in them

and the denom degree > 1

so we have a slant asymptote

(which implies no horizontal asymptote)

[calm thicket]

So there's an oblique one

[viscid thistle]

thing is how to find that line (remember slant/oblique asymptote is a line)

if the degree on the denom were 1

then u could use synth division

but u cant since its 2

[calm thicket]

rip

[viscid thistle]

so long division, or the trick that kayne showed u (i dont understand it so i'd do long division)

[calm thicket]

Mhm

What do

[viscid thistle]

let me see if i got it righ

long ass time since i did this lol

ok think i got it

if its y = x - 1 (which it appears to be) then i can explain

if not then not xd

[calm thicket]

It appears to be y=x**+**1

[viscid thistle]

oh

[calm thicket]

via guess and check'

[viscid thistle]

im retarded i copied it wrong

yep

i got it on desmos too lmao

[patent beacon]

= (x³ + x² - 2x) / (x² - 1)
= x(x³ + x² - 2x) / (x³ - x)

= x(x³ - x + x² - x) / (x³ - x)
= x(x³ - x) / (x³ - x) + x(x² - x) / (x³ - x)
= x(x³ - x) / (x³ - x) + (x³ - x - x² + x) / (x³ - x)

= x + 1 + (-x² + x) / (x³ - x)

[viscid thistle]

essentially ^^

leave out the remainder

[patent beacon]

Holy hell that one ended up being harder than it needed to be

[viscid thistle]

well

i can put up the division

it wasnt that hard

xd

[patent beacon]

Yeah that's probably the better option at that point

[viscid thistle]

i dont know how to put it on latex tho

so ill post a pic

12pm almost

[calm thicket]

So

what did you do?

[viscid thistle]

that's long division

[calm thicket]

Started by multiplying by -1 yea

you got your negative thing without x^3 somehow 🤔

then yuo did it again with 0x and -1 whaaa?

[viscid thistle]

ok so

i think its easier if u do long division without poly's first

like just plain numbers

ill show u one

https://www.mathsisfun.com/long_division.html

after u understand the basic one, you just do the same but with polynomials

[calm thicket]

Was just reading through it :p

[viscid thistle]

xd

[calm thicket]

How do I turn this into polynomials?

[viscid thistle]

well

lets look at the pic i sent above

ill try to explain by steps

[calm thicket]

mhm

You're doing x^3... / -1?

[viscid thistle]

so first step is i write the denom in its own box

then write the num right by it

so it leaves me with

x^2 - 1 | x^3 + x^2 -2x

i put a line on top of the num to write out the result of the division

so then i start

i look at the first term

of the num

x^3

i think to myself

"what do i multiply x^2 by to get x^3"

the answer is x

[calm thicket]

mhm

[viscid thistle]

so then i write x on top since its part of the result

and now i have to keep going with the division

so

i set up the - and the parenthesis

so that i dont forget that they are being subtracted

you could just go straight ahead and put -x^3 -0x^2 +x

but i find it nicer with the - and parenthesis explicit

so then i go through with the multiplication

of the answer that i have so far

which is "x"

now we multiply this by the denominator

so first x * x^2 = x^3

notice how if you do this right then this first thing should always cancel itself out

if it doesn't, you did the division incorrectly

anyways, are u good up to this point ?

[calm thicket]

Mhm

[viscid thistle]

right

so now we need to subtract those terms

to come up with our next "iteration" of the division

if it helps to see it taht way

so we come up with x^2 -x

now we do the same thing

we look at x^2

and we say

"what can i multiply x^2 by to get x^2"

the answer is 1

so then i add 1 to the answer

answer so far is x + 1

so now

i do the same with this recent + 1

i multiply the denominator by it

so i add the - and the parenthesis

then i go 1 * x^2 = x^2

1 * (-1) = -1

so my thingy is looking like

-(x^2 - 1)

after that

we do the subtraction and end up with -x + 1

(its wrong on the picture)

so then we look at -x + 1

and we say hold up

-x has a smaller degree than x^2

so then we've reached the point where -x + 1 is our remainder

so our full answer is

-x + 1 + (-x+1)/(x^2-1)

but for the oblique asymptotes

we ignore the remainder

and so we get the line

y = -x + 1

tadaa

xd

[calm thicket]

*+x

[viscid thistle]

plus x where

[calm thicket]

asymptote is x+1

not -x

[viscid thistle]

oh right

lmao

effects of discord on 12am

anyways

[calm thicket]

nice

[viscid thistle]

you should practice some more of these slant asymptote ones

to get the hang of long division

[calm thicket]

you see ivan, when you do math like me, numbers are of confused because everything is absolute value!

[viscid thistle]

it always hurts my soul when synth division is not possible

[calm thicket]

😛

[viscid thistle]

i dont think i understand that meme lol

[calm thicket]

me neither tbh :/

[viscid thistle]

🤔

true math memeing

anyways i should get to bed

that should get u sorted for the questions f-i

so gn

and have fun with the rest of the problems xd

[calm thicket]

thanks :p

[viscid thistle]

Damn it, missed a good convo

whatever

@viscid thistle hey dude, haven't seen you in a while, whats up?

[calm thicket]

Out of curiosity, what's the value of your nick?

=tex \sum_{x=1}^{25}\sqrt[3]{2x^2}

[calm thicket]

=pup \sum_{x=1}^{25}\sqrt[3]{2x^2}

[granite stirrup]

Query made by @calm thicket
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=\sum_{x%3D1}^{25}\sqrt[3]{2x^2}
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
This command is in development. Suggest improvements on the MathBot server (type =about for the link).

[calm thicket]

=pup \sum_{x=1}^{25}\sqrt[3]{2}x^2

[granite stirrup]

Query made by @calm thicket
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=\sum_{x%3D1}^{25}\sqrt[3]{2}x^2
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
This command is in development. Suggest improvements on the MathBot server (type =about for the link).

[calm thicket]

🤔

=pup \sum_{x=1}^{25}(\sqrt[3]{2}x^2)

[granite stirrup]

Query made by @calm thicket
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=\sum_{x%3D1}^{25}(\sqrt[3]{2}x^2)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
This command is in development. Suggest improvements on the MathBot server (type =about for the link).

[viscid thistle]

@calm thicket ironic, 166 is one of my favorite numbers, I didn't intend to do that, atleast, conscience-wise

[calm thicket]

Nice

I'm a big fan of the number 2 :p

== 166/2

[granite stirrup]

83

[calm thicket]

83 is eh :/

[viscid thistle]

granted, its not exact 166 but yeah

the number 2 is great as well

2, when an exponent or degree of x forms my favorite shape, when its the leading term of a trinomial, it forms my favorite trinomial, if not not my favorite polynomial

I like 166 because I did a lot of physics work, that I used to prove some things and it appeared a lot in my calculations, I liked that experience, so I remember the number.

another another favorite number is .1666667 for the same reason.

[viscid thistle]

if anyone needs help in calculus/or math in general i can help(edited)
im on a lot these days
but please PM me tho, im in so many channels i forget to come here
just wanna help since my projects are finished 😃

but if someone can answer for them that works for you but if you see me online shoot me a question

[bright jetty]

can someone teach me precalculus

[viscid thistle]

I can

but granted,Im not going to be always here, a lot.

so yeah

[calm whale]

ask away if you have questions lol

[bright jetty]

@calm whale teach me the basics

this server inspired me to learn more about math

[viscid thistle]

The basics?

Are we talking the basics of pre-calc?

[bright jetty]

yes

[viscid thistle]

because that can be anything

I need to understand your level of mathematics

[bright jetty]

ibt as saying from yesterday i am like a political analyst/ mathematician

trying to uncover the Russia collaboration project through statistics and polls

[viscid thistle]

ah

[bright jetty]

ye

[viscid thistle]

There can even be more than just statistics involved

[bright jetty]

ik

polls

wanna help me?

[viscid thistle]

no

[bright jetty]

😮

[viscid thistle]

A lot more math canbe applied

[bright jetty]

true true

[viscid thistle]

Obyeag can tell you that logic can be applied, big important concepts

[bright jetty]

well duh

but logic is like 10% of the stuff

[viscid thistle]

its more than that

probability

[bright jetty]

ye

s

[viscid thistle]

counting

number theory

[quartz garnet]

=tex \sin^2 x = \cos^2 x - \cos x~(0\le x<2\pi) \
x=?

[quartz garnet]

Basic pre-cal stuff if you wanna.... test if you're good at precal? x'D

I guess?? :')

[viscid thistle]

He's not wrong

[bright jetty]

=tex \sin^99 x = \cos^2 x - \cos x~(0\le x<2\pi)
x=?