precalculus

[sick steppe]

what's a convergent in this context...?

[mystic cloak]

Its talking about Sequence

[willow bear]

i'm like 3 hours late but that's literally the least helpful reply possible @mystic cloak

[patent patio]

where u find this exercises?

[sharp lagoon]

its for my college pre calc class

o-o

[sick steppe]

it's a standard question to get.. idk why they asked where it was from

[cerulean salmon]

How can p/q be a sequence? Am I missing something?

[viscid thistle]

It's probably two sequences p_n and q_n that aren't shown

Although that would make the second part a bit nonsensical

[past meadow]

maybe talking about the sequence of cut-offs of the infinite fraction of sqrt(D)? https://en.wikipedia.org/wiki/Continued_fraction#Infinite_continued_fractions_and_convergents

[somber sigil]

In case ur still wondering after 4 days... to find the area of a circle u need to complete the square... in the form (a±b)²=a²±2ab+b², in this case b=2 for x and b=1 for y.

[olive pollen]

Hi geeks, I am a homeschooler who is currently aspiring to learn Maths and Physics and I know arithmetic, Basic Algebra and Basic Geometry. So I was wondering if I should start directly with a Precalculus text or study Algebra, Geometry and Trig in detail. My main aim is to get on to calculus so that I can understand Physics

Thank You

[sick steppe]

you need trig to do calculus and physics... so if you know nothing about trig then you're screwed, since you're neglecting an entire family of functions

[fleet yew]

Precalc texts will cover basic trig

Unit circle and the definition of the sinus and cosinus

[olive pollen]

Ya, exactly

and 3blue1brown is there for rescue as always

so I guess I should start with Precalc text

[patent patio]

guys

which book or course of precalculus is good? i think in learn in Openstax, but maybe have a better options

[jagged sun]

cricket noises

[steel venture]

this should be a good starting point https://www.youtube.com/watch?v=eI4an8aSsgw

it should be a good foundation for calc 1 if you practice and do mock exams along with the course

[wary mulch]

idk how to approach this sum.can anyone help me

[peak sandal]

You can evaluate it directly

[wary mulch]

no 0*infinity form

[steel venture]

👀

ok had to make sure ann wasn't watching

have you tried l'hopital

[wary mulch]

no

i forgot abt tat

@steel venture

is this crct?

[steel venture]

nope

[wary mulch]

wat error did i do

[steel venture]

to use lhop you need 0/0 or infinity/infinity

[wary mulch]

ohhh ok

[mild kestrel]

So I need to find the range of this and give it in interval notation... Having said that, I'm not sure if the answer in interval notation would be this (-4, inf) or [4, inf), since I was only given the graph and no equation

[nocturne jacinth]

what does range represent?

[mild kestrel]

The lowest and highest y points of the slope

[nocturne jacinth]

what would f(2) be?

f(1)*

[mild kestrel]

I believe (1,-4)

[nocturne jacinth]

correct which shows that -4 is a possible value of the function

do u see any y value lower?

[mild kestrel]

No, so then answer should be [-4, inf)

[nocturne jacinth]

u can also look at it like this, what function does this look like?

[mild kestrel]

f(x)=(x^2-1)-4

[nocturne jacinth]

and x^2 is centered at (0,0)

u can plug in 0 and still get 0 and the range of x^2=[0,inf)

[alpine hedge]

@wary mulch Just rewrite tan(x) in terms of sin(x)/cos(x) then use the product rule for limits s.t you get log(sin(x)/ cos(x) and can use l'Hopital as the expression is of the form 0/0.

[nocturne jacinth]

this graph looks like f(x)=(x-1)^(2)-4

[mild kestrel]

Got it, thank you!

[copper vigil]

[] means inclusion

() means exclusion

[gusty relic]

how can I graph a fast sketch of a fucntion like f(x) = 2/(3x-1)

[steel venture]

desmos

unless you mean by hand

[gusty relic]

yes I mean by hand

and without calculating it point per point if possible, just a general idea of what the function looks like

[steel venture]

function transformations

this is just 1/x, with various transformations applied to it

you can rewrite this as $f(x)=\frac{2}{3}\cdot\frac{1}{x-\frac{1}{3}}$

[obsidian monolith]

maximo

[steel venture]

so you're scaling it by a factor of 2/3, and shifting it 1/3 units to the right

[gusty relic]

by scaling you mean it goes up by 2/3 units?

[steel venture]

no, literal scaling

[wet surge]

WSP PPL

i need help.

[steel venture]

what do you need help with?

[wet surge]

the directions are making me confuse.

how do i solve?

[steel venture]

do you know what a sequence is?

[wet surge]

direction says "for each prob. write the first-four term"

yea

i know how to solve etc etc. but the directions are making me confuse.

[steel venture]

i'm assuming they want you to write the first four terms of each sequence

[wet surge]

okay how.

[steel venture]

i know how to solve etc etc

[devout trail]

[wet surge]

and nth term for 10th term is 10. and 15th term is 15.

[steel venture]

hm?

do you know how to solve these or not

let's start with that

[wet surge]

slight yes. definetly not.

cannot*

[steel venture]

alright so let's say no

[wet surge]

i have a book right here. but the book confuse me also. i need help with the person

[steel venture]

i'll give you an example problem

and an example solution

[wet surge]

so i assume replacing the n of 10 or 15?

[steel venture]

no

[wet surge]

but the nth term is 10 and 15.

[steel venture]

$a_n$ is the nth term in the sequence a

[obsidian monolith]

maximo

[steel venture]

no

let me ask again

do you know what a sequence is?

[wet surge]

slight.

honest answer. slight.

[steel venture]

what's your native language?

[wet surge]

sequence have a consecutive terms having a commong diff?

filipino.

[steel venture]

ang pagkakasunud-sunod ay isang order na listahan ng mga numero na sumusunod sa isang karaniwang panuntunan

[wet surge]

im guessing you used a translator?

[steel venture]

yes

[wet surge]

it doesnt translate better.

[steel venture]

"a sequence is an ordered list of numbers that follow a common rule"
"ang pagkakasunud-sunod ay isang order na listahan ng mga numero na sumusunod sa isang karaniwang panuntunan"

from either of these

do you understand what a sequence is?

[wet surge]

yes. the sequence is ordered by the list of common that follows the common difference?

[steel venture]

not a common difference necessarily

just a common rule

[wet surge]

term is following the common diff, solving for the next upcoming terms?

[steel venture]

no common difference

i'll give you an example

$a_n = n$

[obsidian monolith]

maximo

[steel venture]

this is a sequence $a$, the nth term is $a_n$, which means the 1st term is $a_1$ the 2nd is $a_2$...

and so on

[obsidian monolith]

maximo

[steel venture]

so for the case of n=1, the 1st term, we have $a_n = n \rightarrow a_1=1$

[obsidian monolith]

maximo

[steel venture]

does that make sense?

[wet surge]

yea i know nth term is based on a1 or opposite

[steel venture]

so what's the 20th term of this sequence?

[wet surge]

lemme see it.

actually let me try to solve it.

[steel venture]

$a_n = n$, what's the 20th term?

[obsidian monolith]

maximo

[wet surge]

okay brb

wait no

the problem is incomplete tho

wait the nth is 1right?

@steel venture

[steel venture]

the nth term is n

[wet surge]

yea

and whats the value on n?

give me problem with numbers on n.

[steel venture]

what's the 20th term

what do you think the value of n is?

[wet surge]

oh

oof

my bad

so sorry.

[steel venture]

no need to be sorry

so:

[wet surge]

wait can we do like a real problem on arithmetic like theres an common diff

[steel venture]

these problems don't have a "common difference"

[wet surge]

oki.

[steel venture]

so one more time, if we have $a_n = n$, and $a_1=1$ is the first term, what's the 20th term?

[wet surge]

okay so whats the formula. is it an=n or an=(n-1)

[obsidian monolith]

maximo

[wet surge]

oh

okay

wait

lmao

im really so sorry

[steel venture]

no need to be sorry

what's the 20th term?

let me know when you have an answer, it doesn't matter if its right or wrong

i just want to see where your head is at

[wet surge]

okay back

im confused at the formila

formula*

not sure if this is correct or not.

im sorry if its wrong.

[steel venture]

why are you using common difference again

[wet surge]

atleast i tried tho

wdym

i alr remove the d on my formula

is it allowed?

[steel venture]

that's one way to do it i guess

but it's much simpler

[wet surge]

okay. show me the real answer

[steel venture]

the nth term is equal to n

so the 3rd term is equal to 3

and the 15th term is equal to 15

and so on

do you see what im saying?

[wet surge]

yea

and theres a gap on your given term

[steel venture]

wdym?

[wet surge]

nvm

go on

[steel venture]

ok so in the sequence $a_n = n$, what's the 50th term?

[obsidian monolith]

maximo

[steel venture]

assuming the first term is 1

[wet surge]

hm

wait

whats the answer on the previous problem

[steel venture]

you got it right, it was 20

[wet surge]

50?

49?

is around 49-50

[steel venture]

it is 50

for the nth term in a sequence, you just need to plug in the n value into the formula

so for example

$a_n = n$ for the nth term

[obsidian monolith]

maximo

[steel venture]

the 50th term is the term where $n=50$ so $a_{50} = 50$

do you see what i did?

[obsidian monolith]

maximo

[wet surge]

yea

okay

[steel venture]

so if i ask for the 100th term, what is the value of n?

[wet surge]

100

[steel venture]

ok good

now lets do a more difficult one

$a_n = n^2$

[obsidian monolith]

maximo

[steel venture]

what's the 1st term?

[wet surge]

1?

[steel venture]

yup

[wet surge]

its up to you

[steel venture]

no

its not up to us

[wet surge]

okay

[steel venture]

it's based on the formula

[wet surge]

okay

so give me another prob

[steel venture]

so what would the 2nd term be in that sequence?

[wet surge]

get another value term?

[steel venture]

remember

[wet surge]

oof

[steel venture]

$a_n = n^2$

[wet surge]

uh

[obsidian monolith]

maximo

[wet surge]

2?

cause there id already a 1

therefore there qould be 2

[steel venture]

we're looking for the 2nd term

what is the value of n then? (if we're looking for the nth term

[wet surge]

oof sorry having problems in wifi

5?

wait sorry,can we continue this later. i gotta eat.

[steel venture]

sure

[wet surge]

OKAY BACK

so

whats the second term?

[steel venture]

4

[wet surge]

why

[steel venture]

because the 2nd term is the nth term when n = 2

[sick steppe]

2^2=4...

[steel venture]

and our nth term is defined as n^2

hence n^2 for n=2, 2^2 =4

do you get that?

[wet surge]

mhm

so 4^2

[steel venture]

what is that?

[wet surge]

nvm

ok got it

[steel venture]

so, what's the 5th term of the sequence $a_n = n^2$?

[obsidian monolith]

maximo

[wet surge]

a15=n2
a15=15^2
15th term = 225

[steel venture]

why did you do the 15th term

[wet surge]

OOF

I FORGOT ITS 5

[steel venture]

so what's the 5th term

[wet surge]

a5=n2
a5=5^2
5th term = 25

[steel venture]

alright

25 is right

[wet surge]

is it correct??

[steel venture]

i think you're starting to get it

so now if i give you

$a_n = \frac{1}{n}$

[obsidian monolith]

maximo

[steel venture]

what are the first 3 terms?

[wet surge]

wait brb

gotta do dishes

sorry

[steel venture]

np

[wet surge]

OKAY BACK

LETS CONTINUE

yea i got it

give me a problem

[steel venture]

$a_n = \frac{1}{n}$

[obsidian monolith]

maximo

[steel venture]

what are the first 3 terms?

[wet surge]

1,2,3?

[steel venture]

no

look at the sequence

[wet surge]

i see one nth term and another nth term but half with 1

[steel venture]

what?

[wet surge]

what

[steel venture]

let's go back to the other sequence

$a_n = n^2$

[obsidian monolith]

maximo

[steel venture]

what are the first 3 terms here

[wet surge]

OH

1,4,8

[steel venture]

no

1 and 4 are correct

[wet surge]

10?

[steel venture]

how did you get 1 and 4?

[wet surge]

cause you said earlier

wait no

2^2=4

and the 1 is the main nth term value

[steel venture]

the 1 is the 1st term, because 1^2 = 1

the 2nd term is 4, like you said, because 2^2=4, like you wrote

now what would be the 3rd term?

[wet surge]

if 2^2 is 4. the 2^4 is 8?

wait no

[steel venture]

remember how you found the 5th term?

[wet surge]

32?

[steel venture]

and the 15th term?

this is the same thing

[wet surge]

OH 15TH TERM?

[steel venture]

i want the 3rd term

[wet surge]

wait no

[steel venture]

but you find it the same way you found the 15th term

[wet surge]

1,4,3?

[steel venture]

remember, the sequence goes 1^2, 2^2, 3^2, 4^2, ... n^2

[viscid thistle]

What's $3^2$

[obsidian monolith]

cpl

[wet surge]

oh got it

wait no.

can you just tell me whats the third term

[viscid thistle]

Answer my question

[wet surge]

18

[viscid thistle]

That's wrong, but how did you arrive at 18?

[wet surge]

...

okay you know what

im just gonna have to fail my math this sem

[steel venture]

how did you arrive at 18?

how did you get that 3^2=18?

[viscid thistle]

Depending on the contents of your math class, not knowing how to square numbers is not the end of the world

[versed arrow]

[steel venture]

[versed arrow]

[sharp lagoon]

what can i do about the square root when it is subtraction

[trim hemlock]

nothing, you just subtract it like normally how you would subtract two functions

[sharp lagoon]

is (f o g)(x) which is also f(g(x))the same as (fg)

nvm it isnt

or iss ittt

[stuck lark]

@sharp lagoon depends what you mean by fg

[sharp lagoon]

(fg)(x) srry i wasnt clear

would that just be multiplying both functions

and (f o g)(x) is like f(g(x))

[stuck lark]

i ask what YOU mean by fg

some books mean composition, others mean product

[willow bear]

why is the answer to (f+g)(x) highlighted in green despite being wrong?

[sharp lagoon]

EMM composition? am not sue

[sharp lagoon]

i have no idea

thats college for me

https://tenor.com/view/dog-sad-funny-animals-funny-turning-gif-10670068

[patent patio]

guys

anyone know the Prof Leonard precalculus class? where i can find the exercises to practice function, trigonometry and all?

[uncut mulch]

you can get some from khan, or use a generic math textbook

[gusty relic]

is there a direct way to check if a function is bijective other than check for inyective+surjective?

[uncut hollow]

you can sometimes just give an explicit inverse

[bleak relic]

how can I solve that?

[night wyvern]

Does anyone know where I can find the curriculum for Pre-Calc honors? I wanna see if I got the basics down

[thorny geyser]

uhh probably ka precalc?

some pcalc h classes have like an introduction to limits or something similar in addition to whats in KA pcalc

[night wyvern]

Khan Academy?

[patent patio]

if i did the check of James Stewart, i'm ready to start learn calculus 1? or will gonna be much difficult?

i need learn more something first? or i can keep going? what u suggest?

[steel venture]

did you read the entire thing?

[sharp lagoon]

How can I get rid of the 1/3

[viscid thistle]

Take the log

[trim hemlock]

theres no need for taking logarithms

just isolate the x^(1/3) and raise both sides to the power of 3

[sharp lagoon]

OO THANK YOU idk why i was stuck for so long, that was all i needed!

[viscid thistle]

I kept looking at it trying to figure out how the answer was arrived at. Already assuming it’s right due to the green color.

[viscid thistle]

i was wondering that too LOL

[trim hemlock]

just wanted some clarification.
Translation: Given f(x) as such. Assume F is the antiderivative of f on R satisfying F(0)=2. The value of F(-1) + 2F(2) equals?
So what im confused about is are they making us assume that the antiderivative of the upper branch of the function also satisfies F(0)=2? But wouldnt the antiderivative of the upper branch of f(x) have the domain of x>=1?

basically, what im saying is that the antiderivative for 2x+5 is undefined for x=0 isnt it?

or am i missing something

[fleet yew]

Is F meant to be continuous

[trim hemlock]

its not listed in the question

so i cant confirm

the phrase "is antiderivative of f on R" kind of throws me off

but then again plugging x=0 into the antiderivative of the upper branch of f(x) seems kinda illogical to me. However, this is from a past national exam paper, so they must have been checked through very thoroughly, i.e its very unlikely a mistake can be found here.

[fleet yew]

Typically in this type of problem they should give you integration bounds

[trim hemlock]

i suppose so too

[stuck lark]

thinking of 'branches' that way isn't sensical

[trim hemlock]

oh, i didnt quite know how i would say/call it

[stuck lark]

"is antiderivative of f on R"
all this means is F'=f

[trim hemlock]

like the portion of f(x) where it is defined as 2x + 5

hmmm

[stuck lark]

how to approach this without getting tripped up by the piecewise def of f is to think of how best to explicitly define F

[trim hemlock]

should i define it as a piecewise function also?

i mean, it must be right?

[stuck lark]

for all $x\in\bR$ we have $F(x)-F(0)=\int_0^xf$

[obsidian monolith]

Cryptic Dance of the Fireflies

[stuck lark]

if x<1 then the rhs is easy to compute

if x>=1 then not so easy

[trim hemlock]

hmmm interesting

never looked at it that way

oh so basically we just have to split the integral into two different bounds?

[stuck lark]

but we can be clever if $x\ge1$ and write
$$F(x)-F(0)=\int_0^xf=\int_0^1f+\int_1^xf$$

[trim hemlock]

right

[obsidian monolith]

Cryptic Dance of the Fireflies

[trim hemlock]

alright, thats something new for me

[stuck lark]

now both integrals on the rhs are easy to compute

so F is best explicitly defined piecewise as such

[trim hemlock]

wait for the integral from 0 to 1, with the upper bound, shall i take the left hand side limit of x tending to 1?

since that 1 is not in the domain of 3x^2+4 right?

[fleet yew]

The problem occurs when you let x=0 because F(0)=2

And the integral from a to a is necessarily 0

[trim hemlock]

strange

[stuck lark]

an alternative version of ftc lets us swap evaluations of F by limits of F

[trim hemlock]

erm theres a website that has quite a different approach but comes up with the same results as yours

so many ways that we could define antiderivative of piecewise functions

[stuck lark]

by this alternative version, $\lim_{x\to1^-}F(x)$ exists and
$$\int_0^1f=\lim_{x\to1^-}F(x)-F(0)$$

[obsidian monolith]

Cryptic Dance of the Fireflies

[trim hemlock]

right

[stuck lark]

this ensures we need not worry over computing int[0,1]f (even without this alternate version of ftc, one can intuit this is how the computation 'should' work out anyway)

[trim hemlock]

alright

thank you very much for these information

this question is interesting

[stuck lark]

you're welcome

[hollow flint]

by any chance could somone pin the half angle formulas

[sick steppe]

#geometry-and-trigonometry

[hollow flint]

k

do i have to memorize the sum to product and visa versa for a pre calc test?

[fading rapids]

in the equation $x^x=x$, the solution is $x=1$. I've been trying to solve $cx^x=x$ for $c \in \mathbb{R}$ but I can't get very far. any ideas/tips?

[obsidian monolith]

gmod

[toxic flame]

let me rephrase, break it into cases when c is in [0,1] and when c is not in [0,1]

[fading rapids]

how does that help? I still have to deal with the number

[violet vale]

When is it a good time to define 0^0 =1 or 0^0 = undefined?

[trim hemlock]

0^0 is undefined

[violet vale]

I think its depends on context?

[trim hemlock]

well idk, that seems like a discussion-wise topic

[violet vale]

Should I even need to worry about this for use in calculus?

[trim hemlock]

but 0^0 is undefined on its on

however x^x converges to 1 as x goes to 0+

[violet vale]

I agree with that limit

[patent beacon]

So for example, Taylor series does use 0^0 = 1 in order to make the formula prettier. This is something you'll see in a calc course. 0^0 is almost never an important issue though.

[copper swallow]

<@&268886789983436800>

[viscid thistle]

@fading rapids do you still need help?

[torn parcel]

hello can someone pls explain to me why the graph of the semi circle in this case is |x|≤2 and not |x|≥2

[strong marlin]

The semi circle is contained between -2 and 2

So you want to display the semi circle when x <= 2 and x >= -2

which is the same as saying |x|<= 2

[torn parcel]

OHHH I SEE

idk why i wasnt able to grasp it while our teacher was explaining tysm

[strong marlin]

np

[viscid thistle]

I'm having a hard time understanding this

Why is this a thing that you can do?

Wait nevermind I got it

Feelsgoodman

[gusty relic]

any hint on how to approach this limit?

[toxic flame]

hmm. @gusty relic
\begin{align*}\left(\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}\right)^2&=x+\sqrt{x}-2\sqrt{(x+\sqrt{x})(x-\sqrt{x})}+x-\sqrt{x}\&=2(x-\sqrt{x^2-x})\end{align*} and
$$(x-\sqrt{x^2-x})=(x-\sqrt{x^2-x})\frac{x+\sqrt{x^2-x}}{x+\sqrt{x^2-x}}=\frac{x}{x+\sqrt{x^2-x}}=\frac{1}{1+\sqrt{1-\frac{1}{x}}}$$

[obsidian monolith]

c squared

[toxic flame]

so your original limit approaches ||1||

i suppose alternatively you could do
\begin{align*}
\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)&=\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)\frac{\left(\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}\right)}{\left(\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}\right)}\&=
\frac{x+\sqrt{x}-x+\sqrt{x}}{\left(\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}\right)}\
&=\frac{2\sqrt{x}}{\left(\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}\right)}\&=
\frac{2}{\left(\sqrt{1+\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{\sqrt{x}}}\right)}
\end{align*}
and see what happens as $x\to\infty$

[quick bluff]

Rationalizing and dividing both numerator and denominator by sqrt(x) might be simpler

[obsidian monolith]

c squared

[quick bluff]

Ye

[obsidian monolith]

T0lgi01

[shut totem]

I think you can do it a bit easier:
$$\lim_{x \to \infty}\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} =$$ $$ \lim_{x \to \infty}\sqrt{x^2+x}-\sqrt{x^2-x} = \lim_{x \to \infty}(x+ 1/2) - (x- 1/2) = 1$$
The first equation is a substitution $x \mapsto x^2$, the second one is the general formula
$$\lim_{x \to \infty} \sqrt{x^2+ax} -x = a/2$$
You can prove this by turning this into a fraction and multiply numerator and denominator by the conjugate

[obsidian monolith]

T0lgi01

[shut totem]

Ah I should still append the proof
$$\lim_{x \to \infty} \sqrt{x^2+ax} -x = \lim_{x \to \infty}\frac{(\sqrt{x^2+ax} -x)(\sqrt{x^2+ax} +x)}{\sqrt{x^2+ax} +x} $$ $$= \lim_{x \to \infty}\frac{ax}{\sqrt{x^2+ax} +x} = \lim_{x \to \infty}\frac{a}{\sqrt{1+a/x} +1} = a/2$$

[obsidian monolith]

T0lgi01

[fading rapids]

sorry for the late response but I would appreciate help

[bleak relic]

$c=\frac{x}{x^x}\Rightarrow{c=({\frac{x}{x})}^{1-x}}\Rightarrow{c=1^{1-x}$ if $x=1$ then $c=0$

[obsidian monolith]

leonardomoura

$c=\frac{x}{x^x}\Rightarrow{c=({\frac{x}{x})}^{1-x}}\Rightarrow{c=1^{1-x}$ if $x=1$ then $c=0$
```Compilation error:```! Missing } inserted.
<inserted text> 
                }
l.55 ...\frac{x}{x})}^{1-x}}\Rightarrow{c=1^{1-x}$
                                                   if $x=1$ then $c=0$
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.```

[fading rapids]

@bleak relic I don't think you can go from x/x^x to (x/x)^(1-x)

[shut totem]

by that you would get x^{1-x}

[bleak relic]

i made an error

[obsidian monolith]

leonardomoura

[bleak relic]

which is $1^{b-c}$

[obsidian monolith]

leonardomoura

[bleak relic]

you're right

[fading rapids]

so do you have any other ideas?

[shut totem]

So I don't think this has any analytical solution like $x^x = x$. When I write the equation like
$$x^x = cx \Leftrightarrow e^{x\ln x}-cx = 0,$$
this doesn't seem solvable in general. I know that $e^x -x = c$ in general is like this, too.

[obsidian monolith]

T0lgi01

[shut totem]

but I'm pretty sure you can approximate this very quickly somehow I feel like

[strong marlin]

You said you solved x^x = x?

surely cx^x = x -> cx(x^x-1 -1)=0, thus c=0,x=0, x^x = x ?

[steel venture]

x^x = x
x^(x-1) = 1

[shut totem]

I think c=1 is the only c for which this is easy

[steel venture]

as long as x does not equal 0

[ornate island]

im stuck on finding the coordinates of each side of this inscribed rectangle

i know the zeros are 6 / -6

but idk how to find the actual coordinates of x

[strong marlin]

nvm lol#

[ornate island]

can someone ehlp

[steel venture]

i don't get your question

which part are you on?

[ornate island]

im trying to figure out the length of the sides of the rectangle

by trying to find out what coordinate the corners are on

[steel venture]

which part are you on?

[ornate island]

part a

lol

[steel venture]

look at how the 4 coordinates are shown in the picture

those will be the 4 vertices for any value of x between -6 and 6

so now pay attention to the bottom side of the rectangle

what are the 2 vertices that form that side?

[ornate island]

wouldnt it be 0

[steel venture]

?

[strong marlin]

I think it has infinite solutions -> x^x-1 = 1/c. So any solutions is dependent on c

[steel venture]

wdym

[ornate island]

like

(x, 0) and (-x, 0) are the vertices

that form the bottom side

[steel venture]

yes

so what is the length from (x,0) to (-x,0)

[ornate island]

uh

[steel venture]

try using the distance formula

[ornate island]

sqrt((-x-x)^2 + 0)?

[steel venture]

not quite

oh wait

thats it yeah

what does that simplify to?

[ornate island]

-2x

?

[steel venture]

2x, but yeah

[ornate island]

oh wait i forgot

to square both

mb

so the distance is 2x

[steel venture]

yes

that's for the bottom and top sides

now look at either of the sides on the left and right

[fading rapids]

yeah so what would the expression be

[steel venture]

and apply the same idea

[ornate island]

if the y coord is f(x)

we just plug the equation in for that

[shut totem]

pretty sure c>=1 allows for exactly 2 solutions

in fact a bit lower should also work

[ornate island]

not sure what to put in for f(x) using the distance formula

[strong marlin]

We restricting x?

[steel venture]

what's our function?

[ornate island]

36 - x^2

so we plug that in

?

[steel venture]

yeah

[ornate island]

ok

[steel venture]

do you understand why?

[ornate island]

yeah

[shut totem]

I thought, given c, we should find all x that satisfy cx^x = x (or x^x = cx)

[steel venture]

x^(x-1) = 1/c is essentially x^(x-1) = k

so for any k>0 it should have 1 solution (i think)

[strong marlin]

Plotting yx^{x}-x=0 produces something otherwise

[steel venture]

no that's not the case (to what i said)

[strong marlin]

if you plot yx^{x}-x=0 you get a range of values for y and x

[ornate island]

wait the height is just going to be 36-x^2 right @steel venture

[steel venture]

yes

[ornate island]

i plugged it in the distance formula and yeah

so the area of the entire rectangle is just 2x(36-x^2)

[steel venture]

exactly

[ornate island]

aight ty

[shut totem]

c = 0.01 shouldnt give any answers x>0, going for x<0 is weird because then we have complex numbers

,w x^x

[strong marlin]

are we saying x has to be an integer or something?

[ornate island]

how do we express the area as a function?

[shut totem]

no, just a real positive number

,w plot x^x

[ornate island]

is it just A(x) = 2x(36-x^2)

[obsidian monolith]

Results provided by WolframAlpha

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Requested by T0lgi01#6506

[steel venture]

yup

[ornate island]

a ok

[strong marlin]

quite strange, because plotting some yx^x -x = 0 shows a graph.

[shut totem]

If we plot cx, then if we choose c small enough, they don't intersect

[steel venture]

i think you guys are talking about different graphs

[strong marlin]

Intuitively we can rewrite this as x^X = 1/c x = ax. So if a is large enough (as in x^x-1), it works. So there should be plenty of solutions

And like i said before, the graph has infinitely many solutions for y = 0

which *I do * like the idea of

y=0 *

^ actually very strange

[steel venture]

x^(x-1) = 0 only has 1 solution

[strong marlin]

Incorrect actually, its just the grapher was getting too close to 0

[steel venture]

actually i don't think it has a solution at all

[strong marlin]

Yea, my bad. The demos grapher thing was saying 0 when it meant really close to 0

[shut totem]

x = 1 is a solution

nvm I misread

[steel venture]

ln(x) = 0/(x-1), nvm you can't even do this

[shut totem]

wait ICSF could you explain what exactly you want to solve?

[steel venture]

so there's no solution to the equation

[strong marlin]

Yes there is

x = 2, c=0.5 for example

0.5 * 2 ^2 = 2

[steel venture]

to the equation x^(x-1) = 0

[strong marlin]

This is just a big miscommunication :/

(from me)

[shut totem]

let's just say f(x) = x^x, yes? We want to find, given c, solutions to f(x) = cx (I just put c on the other side because I feel it's a bit easier)

if c = 1, then we have exactly one solution x=1

if c=2 then we have two solutions x=2 and x = 0.3463233622785809

[strong marlin]

Right then we have y=cx, which we can change to think of it as tan(theta)x=y. So rotating this straight line intersects y=x^x infinitely many times.

[shut totem]

I dont think it does

[strong marlin]

x^x is continuous? y = tan(theta)x. And theta can get infinitely fine. We can intersect any angle between 90 and 0

[shut totem]

since the derivative of x^x after a certain point is always bigger than c, there can be at most only one intersection after that

[strong marlin]

Yes but we can find a c which equals 1/x^x-1

giving by definition cx^x = x

as x^(1-x) * x^x = x = x

[shut totem]

So you want to fix x and find possible values for c?

[strong marlin]

no

Im saying we can always produce a c which intersects with x^X at any point

[shut totem]

sounds trivial to me

[strong marlin]

Ah, i see what you mean

there is only 1 solution per value of c

[shut totem]

you can prove that for any continuous function

[strong marlin]

thats what you are saying?

[shut totem]

actually what I'm trying to say:
If c<1, there exists no x, such that x^x = cx
If c=1, there exists one x, such that x^x = cx (x=1)
If c>1, there exists two x, such that x^x = cx

for example c = 2 admits x = 2 (as you said) and x = 0.3463233622785809 as solutions

[strong marlin]

I see what you mean, yes

[bleak relic]

@fading rapids you got your problem solved?

[fading rapids]

not yet

oh actually I didn't realize people were having a conversation about my problem

lemme read these

[bleak relic]

ok haha

lemme know if you got an answer

[fading rapids]

thanks for providing this! but how would I go about finding such solutions without using wolframalpha

[shut totem]

that final answer would be numerics. I can assure that one solution is in the interval (0,1) and the other in (1,inf) (open intervals). Maybe you could get an better upper estimate like c (for c>=2). Simply because no exact solution should exist, only for special cases

[fading rapids]

oh alright, thank you for the insight

[viscid thistle]

what is this? @cyan cape

and you might have to take a closer pic if you need help with any

[cyan cape]

Alright @viscid thistle

[cyan cape]

Calculus,, limits

[ornate island]

can someone help me clarify what it means to "State the number of nonreal complex zeros"

[copper vigil]

it means to state the number of nonreal complex zeros

there is no simpler way to state it

find the nonreal solutions to the polynomial

how many of them are there?

[hard onyx]

what are the prerequisites, i.e. what do I need before learning this book?
https://www.amazon.com/Calculus-James-Stewart/dp/1285740629/ref=sr_1_3?dchild=1&keywords=calculus&qid=1629069326&sr=8-3

[viscid thistle]

Dude

I don't know how to describe what I'm feeling but it's the first time it was explained to me why it is like this

Not sure if this is the correct channel for this

[stuck lark]

not really precalc material

each intermediate step is justified by a field axiom or a corollary of them

[patent patio]

where i can find some exercises to practice precalculus?
i'm studyng with this course
https://www.youtube.com/watch?v=eI4an8aSsgw&t=9064s

[steel venture]

most precalculus textbooks should have problems to go along with them

https://cec-code-lab.aps.edu/downloads/precalculus-text.pdf

that's one you can use

[willow bear]

@molten garnet no advertising.

[molten garnet]

oh sorry

is this not allowed in other servers as well?

[grizzled orchid]

Yeah don't post advertisements here

Or really in most servers... you should read #rules

[molten garnet]

oh

my bad sorry guys

[dusk edge]

Send invite bro

[viscid thistle]

no

[dusk edge]

https://tenor.com/view/send-it-bro-send-it-helms-deep-theoden-gif-20755446

[molten garnet]

Lol I don’t wanna get banned

Or removed from the server

[toxic flame]

to late

[molten garnet]

They removed my invite so no

[glacial lantern]

how do i solvew 2(x+1)(x-3)^2 - 3(x+1)^2(x-3)

[trim hemlock]

you cant

[uncut mulch]

wdym by solve?

[dusky estuary]

i presume he wants to find x

[glacial lantern]

factor

[glacial lantern]

factor

[glacial lantern]

factor

[glacial lantern]

factor

[dusky estuary]

it's already factored??

oh wait

nvm

[trim hemlock]

Instead of saying factor 4 times

Just say, the problem asks to factor the expression

Also factor does not mean solve

(Just saying)

Anyway, look for the common things both terms have

And then factor it out

[willow bear]

you are forgetting that 90% of ppl here have zero communication skills

[hollow flint]

i just took a pre calc test

and 1/4 of the test was logs

i didnt do much logs in the text book...

(im just mad cause i failed the test after putting 150+ hours and wasting my summer break)

[viscid thistle]

how do i find the domain definition of this one? Ik its basic but i am not sure

[uncut mulch]

try not to overthink this

do you know the definition of domain?

[viscid thistle]

It's Df = - infinity to + Infinity

[viscid thistle]

That's not the definition of domain

But it is the right answer in this case

[ornate island]

can someone help explain this?

because i dont think there are any zeros

[willow bear]

there are, they're just complex.

[ornate island]

how can i factor the function knowing that 1+2i is a zero?

idgi

[willow bear]

do you know that for a polynomial with real coefficients, the complex zeros come in conjugate pairs?

[ornate island]

i know that they come in pairs

didnt know they were conjugates

so is (1+2i) & (1-2i) the pairs?

,w roots x^4-x^3-14x^2+24x+5

[obsidian monolith]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by grizzly bear#8060

[willow bear]

that's one pair

you can construct a quadratic with those as roots, and divide your polynomial by it

[ornate island]

ah ic

so i multiply x -1 -2i with x -1 +2i since they have to = 0 right

[willow bear]

...yes

[ornate island]

,w roots x^4-8x^3+27x^2-50x+50

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by grizzly bear#8060

[unique heath]

Please help me Solving this

[uncut mulch]

what have you tried?

[ornate island]

Can someone help me with part a of this question

I need to find how to combine the volume of a sphere and a cylinder together

[steel venture]

do you still need help?

[steel venture]

volume of shape = volume of hemisphere + volume of hemisphere + volume of cylinder

[ornate island]

is the height of the hemisphere also x?

[steel venture]

do you know what a hemisphere is?

[vagrant ingot]

ive tried multiple times i just dont know what to do

[willow bear]

@vagrant ingot so you don't know the defining formula for average rate of change?

oh, sorry, i'm 50 minutes late. do you still need help with this?

[vagrant ingot]

no i do not probably because im kinda getting ahead of thte class

and school started less than a week ago

[willow bear]

so you're doing assignments ahead of the class, are you?

okay, so the average rate of change is simply the slope of the line connecting the two relevant points on the graph of your function

to put that more symbolically, the average rate of change of a function f(x) from x=a to x=b is the slope of the line connecting (a, f(a)) and (b, f(b))

or to put that even more formulaically to relieve you of any need for an INT skill check, it's (f(b)-f(a))/(b-a)

[limpid schooner]

$ sin^4 (x) + a sin^2 (x) + 1 = 0. $ For what value of 'a' equation has solution? How to proceed. I tried substituting t = sin^2(x) but didn't reach to any result.

[willow bear]

you will get the equation t^2 + at + 1 = 0

you want this equation to have a solution in [0,1]

do you understand why? @limpid schooner

[limpid schooner]

No.

[willow bear]

the range of sin^2(x) is [0,1]

[limpid schooner]

Yes.

[willow bear]

if t^2 + at + 1 = 0 turns out not to have solutions in that range, then the original trig equation will fail to have any solutions

[limpid schooner]

Oh.

So, i should satisfy these value to obtain 'a'?

[willow bear]

i don't know what you mean by 'satisfy these value'

[limpid schooner]

What to do next?

[willow bear]

well now you want the equation to have at least one solution in [0,1]

there's a nice shortcut way to determine what a can be

and it may sound a little odd, but you can do so by isolating a in the equation.

[limpid schooner]

isolating? I didn't understand the term.

[willow bear]

solve the equation for a (in terms of t)

make a the subject

whatever you call it

[limpid schooner]

Ohh. I see. I will try it

[hallow nimbus]

Hi everyone. I need help with a matter for which I can't seem to find a solution.

On Desmos, I am trying to do a cubic regression to find a cubic function of best-fit for a particular series of data points. You can see this on the left in the picture I have posted below. The regression works out well, as it displays a cubic graph that goes through 94% of the points, and the specific parameters of a,b,c, and d that go with it.

However, when I try to place these parameters into the general form of the equation y = ax^3 + bx^2 + cx + d, I actually do not get the same graph (see: the black graph is displaced quite far above the blue graph even though they both seem to take the same appearence)! I copied down the parameters correctly, so I have no clue what the problem is. A YouTube tutorial also suggests that you can simply copy the parameters and place them into the general form of the equation to get the same graph, but as you can see, the black graph (the equation) is completely different from what I get in the cubic regression in the blue graph.

Any help is appreciated. Thank you so much

[glacial lantern]

how do i solve [-x^3 (1-x^2)^-1/2 - 2x(1-x^2)^1/2)]/x^4

[steel venture]

$\frac{-x^3 \times (1-x^2)^{ - \frac{1}{2}} - 2x \times (1-x^2)^{\frac{1}{2}}}{x^4}$

[steel venture]

does that look about right?

[obsidian monolith]

maximo

[glacial lantern]

nvm i got it

[wooden reef]

meth

[trim hemlock]

doesnt look like meth

wasted a police report

[wooden reef]

pls help me

[fleet yew]

@wooden reef
what can you tell me about the slopes of lines that are perpendicular and parallel to another line?

[wooden reef]

nvm I got ty

[jovial stratus]

Is this the right place where I can ask about finding domain and range?

[uncut mulch]

yes

[jovial stratus]

y = 3/x+4

The domain is all real numbers am i correct?

And x ≠ 4?

[uncut mulch]

do you mean
$$y=\frac3x + 4$$
or
$$y = \frac{3}{x+4}$$

[obsidian monolith]

ℝamonov

[jovial stratus]

-4*

[jovial stratus]

The latter one

[uncut mulch]

set of reals excluding -4, yes

[jovial stratus]

Ok, and I can't quite understand how to get the range.

Is it all real numbers except 0?

[uncut mulch]

yes

[jovial stratus]

Ok, thank you!

[ornate island]

,w roots x^3 - 1234567800x^2 + 1358022057x + 323703700758

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by grizzly bear#8060

[uncut hollow]

ouch

[ornate island]

i typed it wrong

[thorny geyser]

a good way is to solve for x in terms of y and see what y values are possible

since generally, finding the domain is much easier than finding the range of rational funcs

[obsidian monolith]

TheToadSage

TheToadSage

[thorny geyser]

but you also have to make sure that you cant have a y value that obtains a $x$ value that is not in the domain of the func

[obsidian monolith]

TheToadSage

[thorny geyser]

so you cant have $\frac{3-4y}{y}=-4$

[obsidian monolith]

TheToadSage

[thorny geyser]

this equations has no solutions, so the only real number that is not in the range of the function is $0$

[obsidian monolith]

TheToadSage

[untold otter]

hii

[jovial stratus]

Can anyone help me find the range of this equation? y=4x²+3x-1

Any help would be appreciated

[viscid thistle]

okay.

have you tried anything?

[jovial stratus]

Yeah but I cant seem to get to the answer

[viscid thistle]

can i please see it?

[jovial stratus]

I know the domain is all real numbers since it is quadratic

[viscid thistle]

yeah

[jovial stratus]

I tried the formula -b²+4ac/4a

And found that the limitation is -25/16

[jovial stratus]

But the thing is our teacher haven't taught us about this formula, so I am looking for any other way of finding the range

[toxic flame]

-b/2a is the vertex of the parabola, which is the minimum value of f, since your function is a parabola that opens upwards

[viscid thistle]

yeah, with this we can figure out that the range will be from the minimum up to +infinity.

[random root]

You can also try making the quadratic a perfect square

[toxic flame]

why tho

[random root]

The square part would be zero and the remaining constant would be the minimum value I guess

[ornate island]

f(5-x) = f(-x-5)

can someone explain why this is the same

[random root]

It's not same for every function but for some unique group of functions

[velvet jungle]

if f(x)=x,
then
f(5-x)=5-x=-x+5
but f(-x-5)=-x-5, which is not -x+5

[jovial stratus]

Hi i need hep in writing my answer in range

{y element R : y>=-25/16}

How can i write this in word form? like set of non-negative real numbers such that y ≥ 1

Wil i write set of all real numbers or set of non-negative real numbers?

[steel venture]

${y \in R : y\geq \frac{-25}{16}}$

[obsidian monolith]

maximo

[steel venture]

is this what you have @jovial stratus ?

if so, this is saying that y belongs to the set of real numbers, where y is greater than or equal to -25/16

[viscid thistle]

Does anyone get why did we have to plug in x1=6 into equation 2 x1+3x2=12 instead of x1+4x2=24? even though B is the point of intersection between x1=6 and x1+4x2=24 and not x1+3x2=12

[viscid thistle]

It might have been an error, but your right, it should have been x1+4x2=24 for finding point B.

[jovial stratus]

Yes, sry for late repy

[wooden reef]

how do you do this

[random root]

Solve the inequality x^2-4x>=0

[copper vigil]

as with any upwards quadratic there will be two intervals

you must take their intersection

[unborn nimbus]

how do we do this question?

[willow bear]

do you know what f(2) = 1 means in terms of graphs?

@unborn nimbus

[unborn nimbus]

nope, i haven’t learned that

[willow bear]

you haven't learned anything about graphing functions at all?

[unborn nimbus]

ive only self studied graphing quadratic functions

[willow bear]

but surely you must know the basic principle of function graphs?

surely it's not just an empty ritual to you?

like

okay, let's put it this way:

if the graph of a function f(x) passes through the point (1, 19)... it signifies that f(1) = 19.

is this familiar to you?

[unborn nimbus]

ahh yeah i understand that but i was never taught lol

[willow bear]

it's really hard to believe you were never taught that.

not being taught that means not being taught how to read graphs.

[unborn nimbus]

i think were gonna learn it in a few weeks

[willow bear]

it's an inexcusable omission on your teacher's side, if indeed they never so much as mentioned it.

you're going to learn it in a few weeks?

then why have you been assigned this exercise?

[unborn nimbus]

i think soo, since were on composite functions rn

[unborn nimbus]

ahh not sure :(

[willow bear]

you're on composite functions??

[unborn nimbus]

yeah

[willow bear]

then you MUST have been taught things about graphs

like cmon, this is the basics of the basics

[unborn nimbus]

i think we’ve only went over quadratics and thats it… i self studied the rest

[willow bear]

then your teacher's a sham, or you didn't pay attention at the critical moment.

in any case, ok, one way or another you have the required knowledge here.

if the graph of a function f(x) passes through the point (1, 19)... it signifies that f(1) = 19.
is this familiar to you?
freya — Today at 09:27
ahh yeah i understand that but i was never taught lol

based on this... refer back to your question, and observe that you are told f(2) = 1 and f(5) = 3.

this means that the graphs of your functions must all pass through two points. can you name the points?

[unborn nimbus]

(2,1) and (5,3)

[willow bear]

great.

so your job now is to draw any three function graphs which pass through these points.

as in, you should get a piece of paper, draw a set of coordinate axes, mark those two points, and then sketch three different curves going through them.

[unborn nimbus]

alrightt

so any curve?

[willow bear]

yes, so long as it passes the vertical line test

[unborn nimbus]

okay got it, thanks!!

[viscid thistle]

How do I deriviate this function? I don't really know what to do with the 7 neither do I know what to do with the 2 above the ()

it's 7(-4x^3 + x^2 - 6x)^2

[willow bear]

,rccw

[viscid thistle]

<@&268886789983436800> you should probably delete this one too

[slender river]

banned but why is it not deleting

hmmcat

[viscid thistle]

And check the other channels for more 😄

[slender river]

ok there we go

[dry mauve]

Evaluate

[peak sandal]

no you

[ornate island]

does anyone know how to rotate a graph 90 degrees on desmos

[copper vigil]

say you have the point (1,1)

to rotate it by 90 degrees you turn the point into (-1,1)

so if you have y=f(x) just turn it into f(-x)

[formal walrus]

I would probably use (0,1) as another example and that gets transformed into (-1,0) because then you know that (x,y) maps to (-y,x)

(in the (1,1) case, x = y = 1)

[copper vigil]

oh yeah nvm then i'm wrong

right because the perpendicular line is the negative reciprocal

[steel venture]

there's 2 main things to know about problems like these

each of these functions have distinct shapes which you can sort of memorize

as you deal with them more and more it'll become second nature what functions belong to what depictions

the other thing is that you can always plot a few points to see the pattern

take the case of y=e^(-x)

f(0) = e^(-0) = 1

f(1) = e^(-1) = 1/e

f(2) = e^(-2) = 1/(e^2)

f(10) = e^(-10) = 1/(e^1)

this hints at the idea that as x becomes bigger and bigger, e^(-x) will become smaller and smaller

so the best graph in this case for e^(-x) will be the second graph on the right

you can also show that as x becomes bigger on the negative side (-1, -2, -3...)

then e^(-x) will become larger and larger in value

[ornate island]

does anyone know if i apply the bound theorem to [-5, 5] or [-200, 1000]

or both

[ornate island]

,w roots 6x^4 - 11x^3 - 7x^2 +8x -34

[obsidian monolith]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by grizzly bear#8060

[ornate island]

,w roots x^3+x^2-8x-6

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by grizzly bear#8060

[cinder reef]

Why is it that we can factor out an equation like 2(x+1)-x^2(x+1) by making it into (-x^2+2)(x+1) ?

I can group, I just don’t know why we can do that

[viscid thistle]

Distributivity

[cinder reef]

What do you mean by that ?

[viscid thistle]

The property that allows you to do that is distributivity

It is inherent to multiplication and addition

[full meteor]

If you know how distributivity works, you can see that (a + b)(c + d) = a(c + d) + b(c + d).
We are taking the equality the other way.

[cinder reef]

Alright I understand that now, thank you

[cloud olive]

How do you do this?
Find the values of θ between 0 <_ θ < π that form the portion of the graph r = 5sin (3θ) in Q4.

[cloud olive]

<@&286206848099549185>

[narrow niche]

im so lost on finding ordered pairs on the unit circle any help?

[frigid wagon]

Why can't logarithm have a negative base?

[autumn plover]

well it can

but the answer wont be real

[shut timber]

how can you solve one sided limits analytically?

[willow bear]

and what, pray tell, do you think the limit of f(x)g(x) should be?

$f(x) \lim_{x\to0} g(x) + g(x) \lim_{x \to 0} f(x)$ is not a valid answer, as the limit is a number and its value cannot depend on $x$

[obsidian monolith]

Ann

[willow bear]

really, i think you should familiarize yourself with how the derivative product rule is defined.

as-is, it sounds like you see it as nothing but a symbolic rule handed down from on high

[sudden patrol]

Anyway...

,help

[obsidian monolith]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[daring wolf]

I need help with this

Im new to pre calc and I'm having trouble

[torn acorn]

end behavior basically describes how the y value acts as it approaches either x=infinity or x=negative infinity

so you would write it in the format of X->infinity , y-> either positive or negative infinity depending on what the y value is doing

[daring wolf]

whats the y value on this

[torn acorn]

depends on which side youre looking at

[daring wolf]

Left

[torn acorn]

well, as the x value approaches the left, what number is the y value approaching @daring wolf

[daring wolf]

X -3

[torn acorn]

not quite. so, since the image goes off the graph and continues on forever, the x value is going to continue towards negative infinity. and since the y value is also going down, IT goes towards negative infinity as well. so on that side, the end behavior is, "as x approaches negative infinity, y approaches negative i infinity

@daring wolf

[daring wolf]

So x is -infinity

[torn acorn]

<@&286206848099549185> can one of you help me better explain this

[torpid grotto]

@daring wolf this is an example ... y=tanx , here at x=pi/2 y tends to infinity ... if you keep zooming out you will notice that the grpah near pi/2 will almost coinside and will look similar to x=pi/2 (so y is not stoping(as their is no finite value of y at x=pi/2) but x is const )

on the left side in ur question we can assume at a particular value of x , y will tend to - infinity

[icy spruce]

I need help with this problem f(x) = 7/1-lnx I need to find the domain in interval notation. I barely understand natural log except that it's the opposite I believe

[sick steppe]

dont crosspost

[icy spruce]

My bad

[dusk edge]

I think you forgot parentheses

But if you look at the graph of ln x you see it doesn’t take on the value of 0

Meaning x>0

Also the denominator can’t be 0

Meaning x can’t be e

Meaning 0<x<e , x>e

[sick steppe]

They got help already

[dusk edge]

Oh ok

[daring wolf]

Alright thanks

[willow bear]

$\int x^3 \cos(x/2) \sqrt{4-x^2} \dd{x} = 0$ by symmetry, so your integral reduces to $$\int_{-2}^2 \frac12 \sqrt{4-x^2} \dd{x}$$

[obsidian monolith]

Ann

[willow bear]

do you at least understand what i've said so far?

also i just realized i missed the bounds on the very first integral in my latex sorry

anyway here is the graph of y=sqrt(4-x^2)

it is a semicircle

the integral of an odd function over a symmetric interval is 0

you know what an odd function is, right?

this is a bit shoddy but it's the first google image search result

[drowsy helm]

can somone help me with my homework??? its about ellipse

determine the standard equation of an ellipse

nothing at all HUHUHUH

co vertices

nope

[drowsy helm]

i have to find the standard form using these

[crimson iron]

finding the standard equation a hyperbola with given asymptotes and one vertex

im stuck

how do i find b^2 here?

[prisma hearth]

Hi! i was wondering if anyone knew how to solve this?

[viscid thistle]

Just find the intercepts, asymptotes and holes of each function. Then compare it to find what is similar or different.

[prisma hearth]

Thank you!

[viscid thistle]

For example both function has the same y intercept. Np

[stuck lark]

nani

[distant valley]

i rememeber

x/x^2 + c

it literally took me

30 mins to figure out LMAO

ignore pls

[stuck lark]

f

[acoustic galleon]

I think this is true: If I have a ball and I slice the ball with two parallel planes, is it true that both hemispheres had the same volume "cut off" by the planes?

[willow bear]

no?

take the unit ball centered at the origin and take z=0.5 and z=-0.1 as your slicing planes

[ornate grove]

Could some one help me out

[uncut hollow]

what is [[x]] here, floor?

try going through conditions one by one e.g. checking f(sqrt(5)) for all of them to narrow down possibilities

[digital cloud]

what part of the question are you having trouble with?

[worldly badger]

So you can just test each function with x = sqrt(5).

1. f(sqrt(5)) = 5 so it doesn’t t match
2. f(sqrt(5)) = sqrt(5) so it doesn’t match too
3. f(sqrt(5)) = 5 - 3 = 2, it match + it s define in R (ez to prove if you need to do that) + but the range isnt Z because the range of x^2 is R, etc… (I let you end the demonstration). So it doesn’t work
4. f(sqrt(5)) = [[2.2… ]] = 2 match + domain is R + here the range is Z (we only have integer by taking the floor)

[spice wave]

can i get help

with 7 questiosn

[night cave]

Why is there no close bracket

[dapper hollow]

its a piece-wise function

thats just the way they are written

[night cave]

Yeah but why

There has to be a reason

[viscid thistle]

its just notation thats all

you might be able to look up the history of piecewise functions to find out

[uncut mulch]

though there should be a separator like a ; , , , if, for

[digital cloud]

it's not a bracket in that context

it represents it sorta being split up

into 2 different functions depending on what x is

[naive moss]

hi could i get some help on understanding this question? am a bit stuck with the absolutes. I tried to fit in x = -1 and x = 1 but am getting that f(x) = 1 in both cases

[willow bear]

no shit