precalculus

[distant pagoda]

I know bro

Idk where I was going with that

[quartz pagoda]

Could this equation be done without moving a sqrt to the other side?

Just squaring it as it is?

[willow skiff]

yes you have to square twice anyways

but also you should solve for equality first, cause the inequality gets messed up by the squaring

then make a sign diagram to figure out the inequalities

[willow skiff]

for the second squaring you have to move a sqrt though

[quartz pagoda]

So I squared it but when I divide by 2 to get rid of the 2 next to sqrt, and when setting up a new condition my interval becomes [-1,1/2] when it should br [-1,1/4]

And then it becomes >-1 +2x and when I divide by 2 its - 1/2+1x

[cloud steeple]

I cant understand inverse functions

[viscid thistle]

In essence

Make the x y and the y x

Then solve for y

[cloud steeple]

Yeah i get this

But

How do i find if a function is the inverse of the other

f(x) and g(x)

[viscid thistle]

Oh

Identify function

F(g(x)) = g(f(x)) = X

[cloud steeple]

Ok so, both composite functions should be equal to x?

[viscid thistle]

Yes

[cloud steeple]

Thank you NomNom

[viscid thistle]

@cloud steeple

[cloud steeple]

Oh yeah! That makes sense

I used to just power it to minus -1

[viscid thistle]

😵‍💫😵‍💫😵‍💫

That’s just the notation

Don’t mix it up and use the exponent rule

[cloud steeple]

Yeah i just read about iit

[viscid thistle]

That’s why we use f^-1(x)

Not f^-1

[cloud steeple]

I see

[uncut mantle]

Yeah I understood that already, but I don't understand how you would get the right answer instead

[honest moat]

Hi anyone here?

[echo sinew]

There are plenty of people here lol

As you can see from the chat

[honest moat]

Ah

[weary fjord]

Hi

[hoary iris]

@honest moathi

😁

[quartz pagoda]

Could someone explain what the hell was done here hahaha

[frank dome]

just factoring

[short orbit]

everything was taken common

[viscid thistle]

Factoring by grouping

[merry tide]

are u in high school?

[viscid thistle]

Yeah

[merry tide]

cool, what in math are you learning rn?

[viscid thistle]

Calc BC

[merry tide]

nice

[viscid thistle]

You were just curious? 😵‍💫

[merry tide]

yea i guess cause it reminds me when i was in highschool enjoying math

[viscid thistle]

I think it’s fine

I don’t like the way we learn math in school

[merry tide]

yeah it gets a lot better in university

[viscid thistle]

I get so much homework and it’s annoying

My GPA is like garbage

But I still think that all that really matters is if I’m learning the content

[merry tide]

Yes true

[viscid thistle]

Like I do really well on my exams so I think as long as I’m not wasting my time I’ll be fine

[merry tide]

is there any part of math that u are looking forward to learning rn?

[viscid thistle]

I’m not exactly sure

I think statistics

I’ve taken Alg 1, 2, Geometry and Pre Calc

My statistics is very very very weak

[merry tide]

nice

[viscid thistle]

And honestly so is my geometry

[merry tide]

i dont think i ever formally learned "geometry"

[viscid thistle]

I think my main strengths is algebra

I did

[merry tide]

im in canada we dont have a course like that

[viscid thistle]

It was whack

[merry tide]

that is my favourite branch of math

[viscid thistle]

Same

It’s fun

[merry tide]

you will really like abstract algebra

thats when I really started liking math

[winter comet]

[merry tide]

indeed

[viscid thistle]

We’ll see

I’m not exactly sure what I will study in college

It will depend on which university I go to

[merry tide]

yea if u dont major in math you most likely will not take a course in abstract algebra

[viscid thistle]

But I am 80% sure I will either do chemical engineering or finance

[merry tide]

rip

[viscid thistle]

I can always self learn

[merry tide]

true

[viscid thistle]

Or I can still take the class

They wouldn’t stop me

[merry tide]

they want your $

[viscid thistle]

It would maybe not be necessary but I still like math

[winter comet]

if you minor in math you'd prob need it tho ri?

[viscid thistle]

Don’t worry

[quartz pagoda]

Thanks, but maan I aint smart enough for this shit🤣

[merry tide]

depends on the school

[viscid thistle]

I’m not spending money on college

[merry tide]

in fact, the school i went to didnt even require it for a major in math

although that is unusual

[winter comet]

what was the "highest" required math?

[merry tide]

well at my school they basically just gave u free choice to pick any math courses u want

the requirements were lax

but usually schools require at least a first course in abstract algebra and real analysis

[winter comet]

how many did you need to take then 💀

[merry tide]

for a math major

[merry tide]

i did a math major so it was mostly math courses

[winter comet]

no complex analysis or topology?

[merry tide]

abstract algebra and real analysis usually would take priority over those ones

but yea many schools do require all that as requirements

im not a fan of analysis courses

[atomic linden]

@everyone SOMEONE HELP, I HAVE AN EXAM ON SUNDAY.

I NEED TO DO MATRICES AND DETERMINANTS AND VECTORS

I SUCK AT IT.

[hoary meadow]

Nah sorry we dont have this at school rn(

[plain marten]

you like math?

[weary fjord]

math

[plain marten]

math

mathematicas

cos(5pi/8)sin(5pi/8)

[dull raven]

u can ask questions!!! here in #linear-algebra

[mental wyvern]

man i miss

easy algebra

now its just like

sin(a+b) sin cos + cos sin

cos(a+B) cos cos - sin sin

tan(a+b) = (tana +tanb )/ 1-tanAtanB

[plain marten]

yup

whats tan(a/2)?

how bout sin(6a)

[tame pike]

Can some1 tell me how to find the equation for a tan function?

[willow skiff]

there's no horizontal shift cause it passes through the origin

for the same reason there's also no vertical shift

[tame pike]

Ok

[willow skiff]

so you just have y = p tan(qx)

[willow skiff]

what's the period of the graph

[tame pike]

I’m assuming it’s π

[willow skiff]

no, that's the period of tan(x)

[tame pike]

Ok

[willow skiff]

just look at the graph

which values repeat?

[tame pike]

U mean the y?

[willow skiff]

yeah

how far apart are the different branches of the tan function, in other words

[tame pike]

π/2 apart

[willow skiff]

yes now that's correct

also the pi is written super small on the graph, annoyingly

[willow skiff]

great so the original period of pi units gets compressed to pi/2

so that means q = 2

you can actually just write y = tan(2x)

the p doesn't matter for this question as they just want an equation

[tame pike]

Ok

Also, I need the domain & range

[willow skiff]

range is correct

for the domain, now think about which values of x are impossible

you can think of it like this, one asymptote of tan(x) is x = pi/2
so one asymptote of tan(2x) is x = (pi/2)/2 = pi/4

[tame pike]

Ok

[willow skiff]

okay so can you find an expression for all the asymptotes?

you don't need to wait for me to finish typing to start working, btw

[tame pike]

Ok

[willow skiff]

jeez

[safe basin]

Ok

[viscid thistle]

Ok

[willow skiff]

0k

[safe basin]

0 °K

[viscid thistle]

0 Kelvin

[safe basin]

O(k)

[cloud steeple]

On kay

[placid panther]

K(o)

[viscid thistle]

()K

[merry tide]

It gets better again

I thought that stuff was kind of boring

[cloud steeple]

Any tips for studying pre-calc faster?

[viscid thistle]

Just do questions and learn that way

[cloud steeple]

Ok

[viscid thistle]

You guys broke the chain 😞

[raven tundra]

Lul

[supple kiln]

I need help with numbers 15,17,19,21,25,27,45

I’m genuinely so lost

[cloud steeple]

What are you suposed to do exactly?

[supple kiln]

Partial fraction decomposition

[cloud steeple]

Oh

I havent studied this yet. Sorry

[supple kiln]

😞 Ty for responding

[cloud steeple]

No problem

[winter comet]

do you know how to start?

[supple kiln]

Kind of

I have these notes

But when I try applying it to the questions I get stuck and can’t progress

This is all I did for 17 and idek if it’s right

[hidden hollow]

Ok you’re just trying to simplify right

What grade lvl is this?

[hidden hollow]

Oh wait I see nvm

Oh fk idk how to do this

[winter comet]

so you had the right idea for plugging in 2, but it doesn't give you a value of C because it cancels out that term. it should give you a value of B...

and same with the others

[winter comet]

what grade level is anything breh 💀

[cloud steeple]

Is there a way to write aproaching

Like, as x is aproaching y

[supple kiln]

11

[supple kiln]

Oh ok 😞

[hidden hollow]

Ok

I get that this is a prerequisite to breaking down integration

Does this have any integration involved or could I figure it out without a knowledge of intergrals

[winter comet]

its a technique used for integration and inverse laplace transforms, but no you don't need to know integration or nothin to learn partial fraction decomposition

its not that hard to learn just a bit of algebra :>

more or less

[hidden hollow]

Okie dokie

[long shore]

its about stretching and shortening, took me a long time to solve it i just want to make sure its right.

[chrome barn]

calculus ii

[winter comet]

its used in calc 2 but sometimes they teach in precalc

[safe basin]

yeah definitely like precalc or algebra 2 or whatever

what Aestusy is thinking of is that often in calculus you apply it to decompose an integrand into something easier when solving an integral

it doesn't have anything uniquely to do with calc though it's just an algebraic manipulation

[viscid thistle]

Can someone help me to understand 5. quastion

Calcus 1

a 2 / b 2 / c -3 / d 2 , my answers

[shadow ether]

c is not -3

[shadow ether]

a and b are not 2

[viscid thistle]

ty for feedbacks but I got help from someone

thank you again

[glossy bane]

is there an operation like factorial that adds the previous numbers

like 4(this operation) is 4+3+2+1

[naive rock]

cant you just use summation

or just do n(n+1)/2

[willow skiff]

exactly ^

[quartz pagoda]

Anyone knows any good vids for factoring by grouping for those kinda big equations

[willow skiff]

sub u = x + 1

ah then you can definitely group $3u^3 - 5u^2 + 5u - 3 = 0$ and factor

[obsidian monolith]

south, just south

[quartz pagoda]

Huh ok will try

[willow skiff]

recall what u^3 - 1 factorises as

then yes you can find all 3 solutions this way (2 are complex)

[quartz pagoda]

Well I'm fckd hahahaha

[cloud steeple]

How to find a slant asymptotes

[cloud steeple]

In rational functions

[latent reef]

do synthetic division and ignore the remainder

well, ok you cant always do synthetic division, but in that case just do long div

[cloud steeple]

Thank you

[plain marten]

bro its so annoying to find those assemtotes

its the only rhing i hate about conic section

[jagged patrol]

Given:
f(x)=log8(x)
g(x)=16^x
h(x)=f(g(x))
Find:
h(x)=w in terms of w for some constant w
My solution:
x=3/4w
h(3/4w)=w

somone plug in value and check pls its 5am rn and i might have been too tired when doing htis

[willow skiff]

it should be x = (4/3)w, interesting

yes, change of base gives you $x \frac{\log 16}{\log 8} = x \frac{4 \log 2}{3 \log 2}$

[obsidian monolith]

south, just south

[quartz pagoda]

Does anyone maybe know any good resources for rational inequalities with absolute values

[shut willow]

I'll send you some questions

[quartz pagoda]

Aah thank you

[plain marten]

i have a trig test monday

yay

on solving trig equations

actually pretty boring

identities were better

[viscid thistle]

I agree

[silent prism]

Identities drove me insane

[unborn brook]

Can I apply rational zeros theorem to this polynomial?

[viscid thistle]

sure

[unborn brook]

Doesn't it state that 0 is the only possible rational root

[viscid thistle]

it states that the root must be a factor of 0, which tells you nothing as all numbers are factors of 0

you would, ideally, factor the -x^2 out of the equation before using it

[unborn brook]

Oh it was quite clear once I rederived it, thanks

The thing is I've read that all rational zeros would be in the final list of RZT, in this case there's only 0

[unborn brook]

Or is this not a great way to put it in the first place?

[willow skiff]

you've misunderstood the statement of the rational root theorem

the constant term is 0

so p is a factor of 0

but that means p could be anything as micabo stated

like 5 is a factor of 0, cause 0/5 = 0, which is an integer

it's like how 4 is a factor of 12, cause 12/4 = 3, which is an integer

[willow skiff]

so 0/(any non-zero number) = 0

[willow skiff]

this reads perfectly clear to me but if you want a summary

the rational root theorem gives you a list of all possible candidates for the roots

you still have to check every single candidate to see if they are actually a root

[willow skiff]

this part is what they didn't say

like if you want to solve sqrt(x) = 2, the candidates from x^2 = 4 are x = -2, 2

you still have to sub x = -2 and x = 2 into the original equation

check if both sides are actually equal

[viscid thistle]

the first part of your statement is correct: all rational zeroes should be in the list found using RZT
the second part: the list contains all factors of 0, which, as stated before, is every number

[solid shuttle]

usually RZT stated with the conditions that a_n and a_0 are not zeros. So, if you want to apply RZT you better first factor out x^2. and then possible rational roots are integer divisors of 4 (and 0 from x^2 of course).

[uneven nexus]

Any ideas?
The q is to prove this inequality

[nimble comet]

what is this? is this part of the inequality? no, right?

[willow skiff]

for $x \ge 1$

[obsidian monolith]

south, just south

[nimble comet]

how about shifting (2(x-1))/(x+1) to the rhs, and taking both sides as e^lnx and e^rhs

[willow skiff]

https://math.stackexchange.com/questions/2992839/prove-fracx1x-1-ln-x-geq-2?noredirect=1

this may be useful

a solution without any calculus would be pretty challenging

[willow skiff]

hopefully you accept this solution then

you want to show that $(\ln x) \frac{x + 1}{x - 1} \le \frac{(x - 1)^2 (x + 1)}{4} + 2$

[obsidian monolith]

south, just south

[willow skiff]

by finding the Taylor series, $LHS \le 2 + \frac{1}{6} (x - 1)^2$

[obsidian monolith]

south, just south

[willow skiff]

but $\frac{(x - 1)^2 \cdot 2}{4} + 2 \le RHS$, hence proven

[obsidian monolith]

south, just south

[willow skiff]

ah there's a particularly nice argument I think here, cause if LHS = f(x), f(x) = f(1/x)

and also note that $2 + \frac{1}{6} (1/x - 1)^2 = 2 + \frac{1}{6} \frac{(x - 1)^2}{x^2}$

[obsidian monolith]

south, just south

[shut willow]

Guys beware of what ur getting into with calc

https://www.youtube.com/watch?v=NzK11DrRdks

[dull raven]

common physics trash notation wtf is that

LOL

[brittle ruin]

Hey Im new to this channel, I am looking into going to my local college to get the prerequisites needed to work towards a bachelors in mechanical engineering. Unfortunately I am behind of where I would want to be at this time and I am going to need to take Gen Chem, and Pre Calc 1 and 2 at the same time for Winter quarter. Any recommendations as for the Pre calc? I've messed around with Calc before, but never have actually sat down and seriously studied it. Just kinda wanna stick my toes out there and see what people have to say about it. Thank you very much guys!!

[night yew]

Anyone need help?

[willow skiff]

aside from Khan Academy there's also Paul's Math Notes

[willow skiff]

https://tutorial.math.lamar.edu/Extras/AlgebraTrigReview/AlgebraTrigIntro.aspx

this is an overview of precalculus (even though it's called algebra)

and the calc 1, 2, 3 stuff is also very good

[brittle ruin]

I'll have to go through it when I am a little more alive tomorrow 😂 Thanks for the help, it is much appreciated, I get the sense Imma be in this channel alot more in the future probably 😂

[willow skiff]

cool no worries!

[winged peak]

Could anyone please explain to me? How does pi over 4 = square root 2 over 2?

I understand the part where u simply look at the unit circle but... im gonna have an exam on this and im assuming they want me to have it memorized

Is there any math involved?

I tried getting chat to explain it to me, but ion get it. Is it more of a visual thing to grasp or what?

[winter comet]

pi/4 is not equal to sqrt(2)/2
however, if you plug theta = pi/4 into sin(theta), sin(pi/4) = sqrt(2)/2. Similarly, cos(pi/4) = sqrt(2)/2

[winged peak]

Im only allowed a scientific caluclator on this ajsufhsjdb. How do i write this down? 😭

Im sorry, I totally get where ur coming from, but its just how to get there is throwing me off

X is cos and y is sin. I look at the unit circle and theyre right there, but im not so sure what to do from there

[winter comet]

the x and y depends on the angle right?

[willow skiff]

$\sin(-7 \pi/4) = \sin(-7 \pi/4 + 2 \pi) = \sin(\pi/4)$ by periodicity

[obsidian monolith]

south, just south

[willow skiff]

now $\sin(\pi/4) = \frac{\sqrt 2}{2}$

[obsidian monolith]

south, just south

[winged peak]

My class is just using right triangles?

[winged peak]

Im sorry, Im a very slow learner. I have... this section and 5 other units after. Imma die fr 😭

Ill be gettinf tutored on this tmr. I saw a yt video and while it kind of made sense, I just dont feel like I got it completely.

Fr i appreciate the help.

[willow skiff]

it's okay

sometimes the solution to not understanding is to repeat the same small steps a million times

repetition legitimises - Adam Neely

[unborn brook]

Which one should I go with?

A, D, & B all >1

[willow skiff]

<@&268886789983436800>

[long shore]

can someone help me stretch or shorten this graph?

im having problems with this topic

[willow skiff]

f(2x + 1) - 1

so that is a:

1. horizontal compression by factor 2
2. horizontal translation by 1/2 unit left
3. vertical translation, 1 unit down

[long shore]

im having issues to compress or stretch the values

to compress i must divide by the factor?

[willow skiff]

divide the x coordinate by 2 yes

it would be easier to do the transformation on specific points

(-4, 4) on the original graph, for example

where does it go?

[long shore]

ok first i would compress so (-2,4)

but my professor said compress first

is that wrong?

[willow skiff]

oh right both ways work, okay I'll update it

[long shore]

ok so why 1/2 to the left shouldnt it be 1?

[willow skiff]

it's a different transformation order that I originally did

you have to be careful cause 2x + 1 = 2(x + 1/2)

[long shore]

then the (-3,4) 1 to the left

[willow skiff]

so it's actually a translation by 1/2 units left, not 1 unit

[willow skiff]

nope

[long shore]

lastly (-3,5)

but

[willow skiff]

https://www.desmos.com/calculator/63ign6oojf

[long shore]

ok so it would be a vertical stretch of 2

[willow skiff]

no

[long shore]

thats a stretch

[willow skiff]

x^2 was a bad example then cause it also can be a stretch

but do you see how it's also a compression?

[long shore]

yea i tried it on desmos for minutes, and saw no difference

why is this so confusing

[willow skiff]

cause you don't know the information

[long shore]

but it says there it is a vertical stretch

thats what im saying

[willow skiff]

no

you have f(2x + 1) - 1

compare this with f(kx)

so it's a horizontal compression

[jagged patrol]

Could someone help me with how to make log functions given 2 points. I keep going the wrong way and not knowing how to get to the correct solution.

I did eventually solve with inverses but I'd like to know how to do it correctly

[willow skiff]

$3 = a \log_4 (1) + b$ and $8 = a \log_4 4 + b$

[obsidian monolith]

south, just south

[willow skiff]

that gives you $3 = b$ and $8 = a + b$ directly

[obsidian monolith]

south, just south

[willow skiff]

you can choose log base 2 to make the calculations easier

[jagged patrol]

Where does the 3 come from?

[willow skiff]

y-coordinate for (1, 3)

[jagged patrol]

So you just made it an exponential?

[willow skiff]

no?

you literally sub in x and f(x) for each point you are given

[jagged patrol]

The point is 3,1 not 1,3

[willow skiff]

nah it says x = 1 and f(x) = 3 right?

your table is cropped out

[jagged patrol]

Where?

[willow skiff]

oh I see

I wasn't doing your question

[jagged patrol]

Ahh

Yes the other one

[obsidian monolith]

south, just south

[willow skiff]

this is the one you were doing then

yeah it's confusing

you can choose log base 3 to make the calculations easier

or log base 9 works

[jagged patrol]

So a log 3+b is just arbitrary right

Is that just the easier skeleton

[willow skiff]

it depends on the base of the logarithm yes

[jagged patrol]

Is b the base?

Or just a variable

I was using f(x)=alog(x-h) because that was one of the problems from last class

[willow skiff]

yeah don't use that

you should just use $y = a \log x + b$

[obsidian monolith]

south, just south

[jagged patrol]

Ok, I guess but what if I need to

As I said the problem my teacher solved last class was in that form as a given

[willow skiff]

well like you know how an exponential function passing through (1, 3) and (2, 9) must also pass through (0, 1) right

so for the logarithmic function

it must pass through (1, 0)

[jagged patrol]

Does it?

[willow skiff]

which corresponds to no horizontal shift

[willow skiff]

common ratio

[jagged patrol]

What about 3 times 2^x -3

That was one of the ones I got

[willow skiff]

that's different

I'm just saying for your specific question

also like the 2nd question gives you an indication that you do not need to consider the horizontal shift

it just works out

[jagged patrol]

Ok I'll try that then

[willow skiff]

and that's an exponential function not a logarithmic function

[jagged patrol]

Yeah I just took the inverse

[willow skiff]

basically the full form would be y = a log(b(x - h)) + c

but no one would want you to use that

[willow skiff]

oh actually just a log(x - h) + c is enough by log rules

[jagged patrol]

Yeah I guess you could just set c to 0 or h to 0 if you needed to answer in that form

[willow skiff]

yeah

[willow skiff]

then you could just start off here

[jagged patrol]

Ok ty

[willow skiff]

no worries!

[jagged patrol]

I got f(x)=2.096logx for standard log and it worked. Ty again

[quartz pagoda]

How did he get an x in the table

[vague basalt]

isn’t it the x multiplying the 3x+2

[quartz pagoda]

I mean there is x^2+x-2 and then x and then 3x-2

Dunno how he got the solo x lne

Oooooh

1/x?

[vague basalt]

i think it’s the x multiplying the 3x+2

[quartz pagoda]

From 1/x?

[quartz pagoda]

Ooooooh got it thanks

[vague basalt]

you’re welcome

[uneven nexus]

How did they know that xn>xn+1 from the equation?

[dull raven]

are there conditions on $x_n$?

[obsidian monolith]

rain

[dull raven]

$\sum_{i=1}^n x_n^i = \sum_{i=1}^{n+1} x_{n+1}^i = \left(\sum_{i=1}^{n} x_{n+1}^i\right) + x_{n+1}^{n+1}$

[obsidian monolith]

rain

[dull raven]

you have the extra term on the right

[cloud steeple]

How do i simplify this?

[vague basalt]

i may be wrong but the inverse of that would be ln(x)=ln(1.06), x=1.06 so e^ln(1.06)=1.06

[cloud steeple]

Ok thanks

[dull raven]

$e^{\ln(x)} = x$ for all positive $x$ by definition

[obsidian monolith]

rain

[dull raven]

@cloud steeple

[cloud steeple]

Yes i know that, i wanna know why 😞

[cloud steeple]

This girl awnsered pretty well

[dull raven]

$e^x$ and $\ln(x)$ are inverse functions by definition

[obsidian monolith]

rain

[dull raven]

e^x and ln(x) are reflections of each other across y = x

[terse aspen]

Are csc and cot the only graphs that start at pi/2

[winter comet]

wdym 'starts at'

[terse aspen]

Like the parent function

Mybe im mixing it or sum

[winter comet]

i don't know what a graph 'starting at' pi/2 could mean

[terse aspen]

Im just trying to wrap my head around graphing 1 period of csc vs sec since i what I am seeing is that for sec we graph a. Half a whole and than a half while csc is 2 wholes

[terse aspen]

Like rhe minimum

Right?

[winter comet]

csc has a minimum at pi/2, cot doesn't tho

[terse aspen]

Yea but does cot cross there

And tan crosses 0

[terse aspen]

..

[winter comet]

are you just graphing in the first quadrant or smthn?

[winter comet]

it keeps goin

they just didnt draw it

[terse aspen]

Yea

Ok

[winter comet]

goes on infinitely

[terse aspen]

Ok

[terse aspen]

Hey how do I know the start of y=tan(3x)+2 is -pi/2

[winter comet]

you mean how do you know there is an asymptote at x=-pi/2?

[terse aspen]

Yes

[winter comet]

tan(3x) = sin(3x)/cos(3x)
when x=-pi/2, the denominator is cos(-3pi/2) = 0 and you cannot divide by 0, so there will be an asymptote at x=-pi/2

[terse aspen]

🤯

Yea im cooked

[old fern]

I have a doubt...

what about tan^2 3x

what do we call the line x = pi/2

still an asymptote??

[willow skiff]

yes, that is still a vertical asymptote of $y = \tan^2 3x$

[obsidian monolith]

south, just south

[willow skiff]

cause tan(3pi/2)^2 is still an infinite discontinuity

[old fern]

Got it...

Thank you south!!

[viscid thistle]

Isn't the answer B? My answer key says its D for q11

[willow skiff]

,rotate

[willow skiff]

(2x - 1)(x - 1) / (3(2x - 1)) = (x - 1)/3

yeah the answer should be D

[viscid thistle]

I used the derivative to find the answer here

[willow skiff]

wtf

just sub in x = 1/2 now and you want this to equal the other part of the piecewise function

[viscid thistle]

For q16 isnt this also C? Answer key once again says D

[willow skiff]

wait I can't do arithmetic

[viscid thistle]

Why not? Its the fastest shortcut for me it takes only a second

[willow skiff]

why are you even taking the derivative

[viscid thistle]

Cus i was told i can

[willow skiff]

that's not how you approach this question anyways

just because a function is continuous doesn't imply it is also differentiable

[willow skiff]

also then h(x) = -1 or whatever doesn't carry any slope information

[viscid thistle]

Well its a multiple choice question its not smthn where i am required to show my solving

[willow skiff]

what did you do after you found the derivative

[viscid thistle]

Plug what x= to

[willow skiff]

no!

you would be finding h'(1/2) then

not h(1/2)

[viscid thistle]

😃

Alright alright

[viscid thistle]

Is this one C?

[viscid thistle]

For q16

[willow skiff]

basically you have a removeable discontinuity for q11

you want the red dot to fill in the hole

[viscid thistle]

I was talking abt q16

[willow skiff]

okay then

[viscid thistle]

Oh

Nvm

Can u see if q16 is C? I got C but the answer key says D

[willow skiff]

can you rotate q16 then

[viscid thistle]

Oh okay give me a second

[willow skiff]

again it's the same thing, you should rationalise f(x) or something

the answer is again D

yep rationalising gives $f(x) = -\frac{2}{\sqrt{2-x}+\sqrt{x}}$

[obsidian monolith]

south, just south

[viscid thistle]

Thanks

[willow skiff]

no worries

[outer glacier]

Is anyone here doing matrixes for pre calc

[willow skiff]

do you want to ask a question or nah?

[outer glacier]

Yes I was wondering if anyone here did are they familiar with word problem matrixes

[willow skiff]

if you have a question you can just post it here

or you can also post it in an open help channel

#❓how-to-get-help

[outer glacier]

It's been posted

In a help channel

Just thought I'd ask in here too in case anyone knew it

[willow skiff]

oh lol

[magic meteor]

Whats the answer

[raven tundra]

wtf is that

thats not precalc

[slender zenith]

wrong channel mate

[night shore]

What is this even asking for

[narrow oxide]

pretty sure midline is the horizontal line that goes through the middle of the graph

like, for y = sin x, it would be y = 0, since sin x centres around y = 0

[night shore]

yeah I figured it out

didnt think it wanted y =

[narrow oxide]

ah

online math do be a pain

[night shore]

Couldnt figure out how -1 was wrong

[fading monolith]

Dont expect an answer without context and on this channel

[willow skiff]

also they're clearly trolling

[fading monolith]

Ik

[magic meteor]

The answer is 2 chill

[fading monolith]

Sure

[tawdry canopy]

bros, just asking, what are the range of sine and cosine functions? because some says it is (-∞, ∞)

[viscid thistle]

that would be the domain

range of sin and cos is [-1,1]

[tawdry canopy]

still confused bro, the answer is (-∞, ∞) for the reason that "a" and "d" were unknown

thats what they told me

from formula
asin(b(x - c)) + d

they say that "a" and "d" could be any number

[viscid thistle]

💀

i thought you only meant for "sinx" and "cosx"

[tawdry canopy]

i thought the same

[viscid thistle]

yes since a and d can be anything the range of the function asin(b(x - c)) + d is $(-\infty , \infty)$

[tawdry canopy]

lesson learned: read carefully 💀

[wary tree]

Not gonna be on my pc soon so I didn't want to open a math room for this, but when testing for horizontal/vertical symmetry on the polar plane, why do some combinations work and others don't?

Ex: both (r,-theta) and (-r,pi-theta) test for symmetry over "x-axis", but sometimes (r,-theta) doesn't work and (r, pi-theta) does. (The problem, if anyone was curious, was r=4cot(theta)

[dull raven]

when does $(r, -\theta)$ not work?

[obsidian monolith]

rain

[gray magnet]

7

[magic meteor]

It's 2

[gray magnet]

dam

close

[wary tree]

r=4cot(theta)

Should be symmetrical over x-axis, but plugging it in, 4sin(-theta)/cos(theta)= -4sin(theta)/cos(theta)

[dull raven]

it seems symmetric already?

like

that works

its symmetric over the x and y axis

[wary tree]

Well yeah it's supposed to be, but plugging -theta in to solve algebraically doesn't result in that

[dull raven]

wdym its supposed to eb

be

it literally is

[wary tree]

4sin(-theta)/cos(-theta)= -4sin(theta)/cos(theta) = -4cot(theta) ≠ 4cot(theta) though the graph suggests it should be symmetric

[dull raven]

polar functions dont work exactly like that

they're parametric so they can be symmetric without being equal pointwise

[wary tree]

Sorry, can you explain what you mean by that? That they can be symmetric without being equal pointwise?

Idk if there's more nuance to polar functions I haven't learned about yet, but I was just taught that if a polar function was symmetric over the x-axis, you can plug -theta in instead of theta and the r should be the same. And it intuitively makes sense too; they're being rotated an equal amount while extending out the same distance

And if there is more nuance, does that mean the "if you plug -theta in instead of theta and get the same output, then it's reflective over the horizontal axis" is wrong?

[willow skiff]

all 3 of them are the same for your function

the period of cot(theta) is the same as tan(theta), pi radians

so cot(-theta) = cot(-theta + pi)

$(-r, \pi - \theta)$ will be the same as $(r, (\pi - \theta) - \pi) = (r, -\theta)$ for any polar function

[obsidian monolith]

southlander!

[willow skiff]

the reasoning is because $r \mapsto -r$ is a rotation of $180$ degrees or $\pi$ radians

[obsidian monolith]

southlander!

[willow skiff]

for the cot example, it's because of the periodicity that $(r, -\theta)$ happens to be the same as $(r, \pi - \theta)$

[obsidian monolith]

southlander!

[quartz pagoda]

How did she get (1+nx)

[willow skiff]

induction hypothesis

you can assume that $(1 + x)^m \ge 1 + mx$ cause it is given to you

[obsidian monolith]

southlander!

[willow skiff]

what you can't assume is $(1 + x)^{m + 1} \ge 1 + (m + 1)x$ though

[obsidian monolith]

southlander!

[willow skiff]

that's what you want to prove!

[quartz pagoda]

What if it was less or equal to?

[willow skiff]

that would be a false statement

,calc (1 + 0.01)^2

[obsidian monolith]

Result:

1.0201

[willow skiff]

which is not less than 1.02

[quartz pagoda]

Huuuh

I think I kinda get it when you used a real number hahaha

[willow skiff]

basically I used x = 0.01 and m = 2

now compare both sides

[quartz pagoda]

Huh its correct

[willow skiff]

wdym?

yeah like it's actually true

there are some conditions which you can search up

Bernoulli inequality

[quartz pagoda]

Yea

[quartz pagoda]

Thank youu

[gusty crypt]

is khan academy good to learn pre calc?

[low yoke]

Hello, I'm struggling to understand the factor theorem, why a zero is a factor of a polynomial?

[quartz pagoda]

Probably a dumb question but how did she get the end, 1+(n+1)x+nx^2

[summer ruin]

it's not? if it was, then the entrie polynomial would be zero

perhaps you meant to ask why 0 is a root of a polynomial, but it's not always the case either

[viscid thistle]

please

the whole base is 50

[low yoke]

I mean, what im seeing so far:

[summer ruin]

well it doesn't say that 0 is a factor of a polynomial

it's x - c that is a factor, provided P(c) = 0

[low yoke]

But I don't understand it

If P(c) = 0 then x - c is a factor?

Because c - c = 0?

[summer ruin]

what you said above is correct, but that's not the reason

the theorem says that if P(c) = 0, then x - c is a factor of P(x)

and also if x - c is a factor of P(x), then P(c) = 0

this is what "if and only if" refers to

[low yoke]

yeah, but why, why P(c) = 0 implies that x - c is a factor?

[summer ruin]

because the theorem says so - the factor theorem

in order to really understand how you can prove that you need to investigate the proof

[low yoke]

I also have the proof, but I don't really understand it:

what has to do Q(x) in the function

[summer ruin]

the first line says what P(x) having a factor of x - c means

it means P(x) is a product of some polynomial and x - c

and we then show that c is a root of P

then it proves the second direction which assumes that c is a root of P

[low yoke]

but why is Q(x) in here?

[summer ruin]

it is some arbitrary polynomial ☝️

[low yoke]

oh

okay

I though Q(x) was the quotient

Or something

[summer ruin]

it is in the second line

they're not the same polynomials

[low yoke]

Yeah, but in the first line it just means a "random" polynomial, right?

[viscid thistle]

is this solvable? the whole base is 50m

[summer ruin]

you can still treat it as a quotient with remainder 0

[low yoke]

okay

now I get it

that was breaking my brain

Thank you very much !!

[solid shuttle]

yep

[scenic dome]

Correct?

That 3 rd line didn't add +f(x) on the other side but i added it already

[viscid thistle]

how do you solve it?

[solid shuttle]

find the other side b adjacent to the angle 68 with the law of sines, then x=b*sin(68).

[viscid thistle]

but you cant get the adjacent ??

[solid shuttle]

why not?

[viscid thistle]

sinv is a/c

[solid shuttle]

Law of sines. You know all the angles alpha, beta, and one side a. So you can find any other side b.

[viscid thistle]

alright can you solve it? since ive just started learning (this) trigonometry and not in english so i got no idea what youre talking about 😭

i need to see how you solve it

[solid shuttle]

Then it is better to take some textbook or webpage in your language showing how to solve a triangle by law of sines. Something like this https://youtu.be/qOjfBYdS_eM only in your language.

[viscid thistle]

i know what you mean but if i see you solve it ill understand

[solid shuttle]

nope I dont want to solve instead of you

[viscid thistle]

wait so for a right? 50* (sin81/sin31) right?

95.8

[solid shuttle]

yep

[viscid thistle]

now what

[solid shuttle]

but it is for the left

[viscid thistle]

yeah the one on the left

how do i get the height after that

[solid shuttle]

do you know what sin(x) is?

[viscid thistle]

yes

[solid shuttle]

so apply it for the right triangle with height

[viscid thistle]

the problkem is that ik sin tan and cos but i havent learned the "Laws of sines" and all that stuff or whatever you call it

[viscid thistle]

wym? for the right riangle with height

[solid shuttle]

this one. The dotted line is the height

[viscid thistle]

is the answer 94.6?

[viscid thistle]

yeah but you need the base though

for sin

[solid shuttle]

nope, you have a hypotenuse and the angle 68. That is enough if you remember what sin(68) is.

[viscid thistle]

sin68=x/95.8??

[solid shuttle]

yes

[viscid thistle]

yeah this is the only part i can actually do

idk what my teacher is smoking

so = about 89m

[solid shuttle]

well it can be solved without those sine laws

[viscid thistle]

how??

im guessing thats how my teacher wants me to solve it

i got no idea tho

[solid shuttle]

x * cot(68)+x * cot(81) = 50.

[viscid thistle]

wtf

[solid shuttle]

x* cot(68) is a small horizontal segment on the left (a projection of the left side)
and x* cot(81) is a horizontal segment on the right

do you know what cot(alpha) is?

[viscid thistle]

uh no

just looked it up so ig

[solid shuttle]

cotangent

[viscid thistle]

a/b?

[solid shuttle]

yep

[viscid thistle]

yeah what the fuck

[solid shuttle]

if a and b are legs

[viscid thistle]

is there NO other way to solve this?

cause i havent learned any of these

[solid shuttle]

thats the easiest. and it needs just the definition of sin, cos.

just cot(a)=cos(a)/sin(a)

but you dont need that.

[viscid thistle]

oh

thank you

[solid shuttle]

so the answer can be written also as x=50/(cot(68)+cot(81))

[viscid thistle]

wow

alright thank you so much

got no idea what any of this is but atleast i got the answer

[stiff sluice]

what

is the difference between

inflection point and critical point

[solid shuttle]

just refer to your textbook on what cotangent is.

[viscid thistle]

alkright but im sure that im not supposed to know any of this

[solid shuttle]

this subject always goes with the definition of sin and cos. So, I suppose you should know

[viscid thistle]

yeah but like rn im learning math 1 as we say

[viscid thistle]

ok so just searched it up, you learn cot in math 4 and lawes of sines in 3

im not supposed to know that lmao

[solid shuttle]

there are different levels of trigonometry. You are probably looking at more advanced part. The definition of cot( ) is somtimes given with the triangle, same with sin and cos.

[stiff sluice]

what would be the difference between

critical point and inflection point

[low yoke]

Hey there, can anyone give me a hand understanding the upper and lower bound theorem?

I do understand the theorem, but I don't see what relation does dividing a polynomial and having the quotient > 0 has to do with the upper bound?

the same for the lower bound

How the sign of quotient can represent a lower and upper bound?

[plain marten]

im doing precal homework right now and its eazy

[solid shuttle]

You have P(x)=Q(x)(x-b)+r.
If Q(x) has only non-negative coeffs. and r>=0 and b>0 then plug in any x>b and you get P(x)>0. So there are no roots greater than b.

[cloud pivot]

chat

im having so many issues with trig identities

i cant even do them without having to look at solutions

and at that point whats the point

i get to the middle then i just get lost after a jump to another step\

like for example where did the 1+ go

i know 1=sin^2x+cos^2x

but then what happened to the ^2

[winter comet]

you understand first step to second step, but not second step to third step right?

[cloud pivot]

yeah

this jump confused me

[winter comet]

they just rewrote 1 to have a common denominator, then combined the numerators

[cloud pivot]

ohhhh

that makes a lot more sense i thought like the 1 magically turned into a pythagorean identity

[winter comet]

naw 💀

[cloud pivot]

thats what confuses me about trig identities its so over

ty though

[winter comet]

np

ngl if i were them i would have started from right side

doing that

like just expanding sin(x+y) first

easier to see imo

[cloud pivot]

yeah tbh

but when doing that im able to take out cotx tany easy

but idk how to get the 1

im left with (sinx)(cosy) after expanding and converting to cotx tany

[winter comet]

you're left with (sin(x) cos(y))/( sin(x) cos(y))

which is 1

[cloud pivot]

so the denominator still remains even when its used for the cot and tan?

[winter comet]

yes
(a+b)/a = a/a + b/a = 1 + b/a

[cloud pivot]

ah i see that makes more sense a lot faster than doing LS too lmao tyvm

[edgy tusk]

anybody know what function this is?

anybody online?

[winter comet]

honestly pretty hard to tell, why?

the x-intercept seems to be not quite at x=-1

[viscid thistle]

f(x) = 6x + 7

[winter comet]

i would say the close is probably closer to 7

but it doesn't look like its exactly 7

[viscid thistle]

(-2, -5) and (0, 7) are on the line

(0, 7) is the y intercept

[winter comet]

oh

yea you're right that seems fair

ignore me :>

[tall forum]

What to do with the term x^n?

[glossy cosmos]

What's the name of that technique when you add something that equals to zero to your equation?

[summer ruin]

smart zero

[glossy cosmos]

[solid shuttle]

the answer does not depend on n.

You have 1012=kn+253m, where k,m>=0 and k+m<=10. Since 1012=253 * 4, we have kn is divisible by 253. But kn<=10 * 22=220. So, kn=0 and thus k=0 and m=4.

[tall forum]

@solid shuttle ....thank you so much

[solid shuttle]

tbh they could have eliminated this n<=22. The answer is always the same if n is not divisible by 253 as 253=11*23 and 11>10, 23>10 and k<=10.

[proud spindle]

how do we solve this

[summer ruin]

applying log to both sides

[mossy wolf]

try to solve this

insanely hard

good luck!

[stiff sluice]

how to find the range of a exponential function

[arctic willow]

yo

[jolly pebble]

Trivial + well known

[arctic willow]

can anyone help me

i have precalc final in about 2 weeks or 1 week and im so overwhelmed because i barely know anything abt precalc and idk what to do

[jolly pebble]

you should study and do problems

[arctic willow]

idk where to start

or what to start with

[jolly pebble]

https://www.khanacademy.org/math/precalculus

[forest blaze]

This is not Precalculus

[forest blaze]

KhanAcademy + prof leonard + paul notes online

[forest blaze]

just do it

[arctic willow]

these are the topics im taking

[jolly pebble]

all of these topics are on khanacademy

and a billion other sources

[arctic willow]

okay

im so overwhelmed whenever i try to study i just panic and give up

i cant imagine myself understanding this

[winter comet]

it's really not as hard as you think it is 😅

[arctic willow]

i know it isnt

its "pre"calc

its not even the hard part yet

but for some reason i just cant understand it

[winter comet]

understand what

[arctic willow]

everything im taking in precalc

i tried keeping up with my teacher from the start

we started with functions

i couldnt even get past that

i just cant understand it

are there tutors in this server

[stiff sluice]

how to find the range of a exponential function

[forest blaze]

If you work everyday you will make it

[arctic willow]

i plan on doing that

[forest blaze]

You must learn to focus

[arctic willow]

but do you know how i can shake this feeling

[forest blaze]

What feeling

[arctic willow]

i feel this sick feeling in my stomach and i feel horribly depressed

[forest blaze]

Whyv

?

[arctic willow]

it feels like i just constantly keep failing whenever i try to learn

not to mention i have a quiz on thursday and a finals in a week or 2

[forest blaze]

Yeah thats normal

Because you lack of confidence

Because you think maths is not for you

Because you are over stimulated

[arctic willow]

how could i confident when im down this bad

[forest blaze]

Not easy to work

You will not feel confident tomorrow

Or the next day

You will feel confident when you will have good grade

Because you worked a lot without counting time

Because you can’t

You don’t have the choice

You must win

Or you will like shit for ever

you know what to do

[arctic willow]

i really want to

[forest blaze]

Stop watching yt/instagram etc…

Work first

Every day

Be efficient

If you strugle use pomodoro technique

20m work - 5m pause

Don’t use your phone

Do a lot of sessions

It will be easy with 20m of focus

Do a lot of exercices

If you want to be good you need to exercice

A lot

[arctic willow]

will do

thank u

[forest blaze]

Good luck

[arctic willow]

can u check dms

[vague basalt]

hey i’m currently taking pre calc as well, if you need help studying let me know i’ll try and help

[winter harness]

guyss rq question

in the logistic growht model

p(t)=carrying cap/1+ae^-kt

what does a stand forrrr

[tawdry canopy]

guys, i've been working for my reviewer for precalculus. Is my cotangent graph visualization correct?

because i haven't seen any visualization of cotangent function

guys???

[jolly pebble]

this shit is cotangential

[tawdry canopy]

i tried that, but it doesn't work

[tawdry canopy]

but guys, what do you think?

[jolly pebble]

wdym it doesn't work

[tawdry canopy]

the right triangle thingy, if i use it, it's so hard to visualize

[upbeat ivy]

wassup

[stiff sluice]

Find the surface area of the surface of revolution obtained by rotating about the $x$-axis the strip of the circle $x^2 + y^2 = 4$ for $0 \leq x \leq 1$.

[obsidian monolith]

Renato Chavez

[indigo ruin]

hey guys, lowk im not in pre calc but my school offers a program where we can take classes over the summer, I took alg 2 and a mix of pre calc this past summer and want to finish pre calc throughout this summer so i can get into calc AB my sophmore year. Lowk my school is sweaty and im jsut tryna catch up. Would you guys think this would be a good enough course to learn a good chunk of pre calc or is it a waste of my time.

[fading monolith]

To #calculus

[woven dragon]

-x^2+6x-12

complete the square

how do i do this?

[placid lotus]

Factor out the negative
-(x^2-6x+12)
-apply the “square”. Divide the middle term’s coefficient by 2 and square it. Since you are adding this value, you also have to subtract it so that there have been no changes to the equation
You get:
-((x^2-6x+9)+12-9)
-group the perfect square
-((x-3)^2+3)
distribute the negative
-(x-3)^2 - 3

[upbeat ivy]

that's just the surface area of a sphere with radius 1

4pi

[tall shoal]

no?

[upbeat ivy]

Yes

[tall shoal]

reread it

[upbeat ivy]

it's asking for the surface area of the surface created by rotating the circle of radius 1

[tall shoal]

its not radius 1

[upbeat ivy]

ohh

Shit

ok

got it

strip

[tall shoal]

yep

[upbeat ivy]

you first take a dx element on the x axis at a coordinate of x, then you find out the y coordinate here, in terms of x from the circle's equation. this is the radius of the circle of thickness dx. now you find out the circumference of this circle, this is 2piy (y will be in terms of x) then you multiply dx, which will give dA the differential area. now just integrate from x = 0 to 1

if you don't remember the formula, just u substitute x as 2sin(theta) or 2cos(theta) and solve

[stiff sluice]

let $f$ be a function with continuous derivative such that $\lim_{x \to 0} \frac{f(e^{3x})}{x\cos(x)} = 12$. what are the values of $f$ and its derivative at $x = 1$

[obsidian monolith]

Renato Chavez

[upbeat ivy]

use L'hospital

and find f'(1)

[slate birch]

hi

[long shore]

Help, how can I find the dimensions and the maximum?

Cuz I did 2 cases when 2(x + y) and 2(3x + 2y)

Which one is the correct?

[summer ruin]

can you post full problem?

[long shore]

It’s in Spanish but that’s about it I think

[summer ruin]

well it doesn't even contain a question, it just states something

[long shore]

Yea the question is find the dimensions and maximum let me show

I need to find the Area then with that the dimensions and maximum

[summer ruin]

have you done part a) and b)?

[long shore]

Yes I did two different options

[long shore]

This

[summer ruin]

the perimeter of a rectangle with sides x, y is 2(x+y)

[long shore]

Yea

But it’s tricky cuz it also talks about a division between the rectangle of 6

6 different slots for flowers

[summer ruin]

I don't think it mentions that in any of the questions

it only asks for total area

[long shore]

Hmm

I’ll have to ask the profesor then

Thanks tho

[dawn pond]

suhhhh

[dawn pond]

ask in prealg and algebra

that's algebra I

well someone already solved it so

[dawn pond]

Good job

[tawdry canopy]

guys, do this identity exist?
(tan x)(cos x) = sin x
(tan x)(sin x) = cos x

[winter comet]

yes

[tawdry canopy]

what do these identities called then?

[winter comet]

i don't believe they had a name

but they are true, granted that you don't divide by 0

[tawdry canopy]

whyy?

other identities have a name

like double-angle identity

or even-odd

[winter comet]

an identity is just an equation that is true for all x (sometimes with domain restrictions)

i think the amount of restrictions has to be finite tho

[winter comet]

yeah some common trig identities have a name

[tawdry canopy]

bruh

[winter comet]

like

5=5 is an identity lol

doesn't have a name tho 💀

[tawdry canopy]

ok

why is it defined at that point?

[winter comet]

wdym?

[tawdry canopy]

like tan x is undefined at x = π/2

[winter comet]

yeah

it isn't defined at x=pi/2

[tawdry canopy]

no

[winter comet]

theres a removeable discontinuity there

an open hole

[winter comet]

yes?

[tawdry canopy]

it is continuous

[winter comet]

tan(x) cos(x) isn't continuous, no

[tawdry canopy]

try input (tan x)(cos x)

[winter comet]

not at x=pi/2

[winter comet]

into desmos you mean?

there's a grey dot at each discontinuity

a grey dot means that its a special point 💀