help-36

[sacred night]

0/0

[odd seal]

you dont have to

[sacred night]

wow this is confusing

[sacred night]

oh wait

[odd seal]

like I said, 0=0 is already an equality

[sacred night]

i can just factorise

[odd seal]

you dont have to do anything at all

[sacred night]

o

aell

well

how can i do the next q please:)?

[odd seal]

if its 0, then you check if it satisfies the equation, and if it does, then its a solution

[odd seal]

just use the basic trig relations, your sohcahtoh or whatever

[sacred night]

i tried

part b

[odd seal]

yea

say the foot of the perpendicular is H

[sacred night]

isnt that y

[odd seal]

then you have split the big triangle into two smaller ones, BCH and ACH

[odd seal]

thats the length of the perpendicular

[sacred night]

oh the FOOT

okay

got it

[sacred night]

how do i find the area from there

1/2aysinalpha?

[odd seal]

you can calculate the area for the two small triangles, which are as you might realize, right angled at H

[sick pine]

hi

[odd seal]

so 1/2 * base * height

[sick pine]

i am new here

[odd seal]

!redir

[final saddle]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[sacred night]

wait

i have this

[sick pine]

so can anybody explain me vectors

[sacred night]

!redir

[final saddle]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[sick pine]

i really am struglling on that thing

[odd seal]

if you have a specific question, go open your own channel

#❓how-to-get-help

if you want a small discussion on topic, go to #geometry-and-trigonometry

[sacred night]

,, , A_{ABC} = \frac12ab\sin(\alpha+\beta) \ \frac12 ay \sin\alpha + \frac12by \sin \beta = \frac 12 ab \sin(\alpha + \beta)

[soft zealot]

1.732050807568877293527446341505

[sacred night]

from [A = \frac12ab\sin C]

[odd seal]

yea, this is good, you can sub in the results from part a

[timber scroll]

Put y = bcos(beta) and y = a cos(alpha)

[soft zealot]

1.732050807568877293527446341505

[sacred night]

which one do i sub each into

like there are two y’s

[odd seal]

you want the expression to have both a,b as well as alpha and beta together

[sacred night]

do i plug in both

[timber scroll]

First in first and second in second.
So that u can cancel out ab from both sides

[odd seal]

choose wisely

[sacred night]

hmm okay

[sacred night]

thank you gus

guys

i reached this before

but i plugged each one wrong

so i ended up with a^2 and b^2

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[rain sentinel]

can someone explain the bit in red

becus they've taken a quarter out

[lilac moat]

partial fractions

[spark swan]

partial fractions

wow

[lilac moat]

although you don't need to setup partial fractions and do the long way

[spark swan]

try adding 1/(2-y) and 1/(2+y) and compare it with 1/(4-y²)

[severe canyon]

Mmhh why?

[lilac moat]

$\frac{1}{4-x^{2}}=\frac{1}{4}\frac{\left(2-x\right)+\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}=\frac{1}{4}\left(\frac{1}{2+x}+\frac{1}{2-x}\right)$

[soft zealot]

Roy

[lilac moat]

you can do something similar to $\frac{1}{\left(x+a\right)\left(x+b\right)}$
$$=\frac{1}{b-a}\frac{b-a}{\left(x+a\right)\left(x+b\right)}$$
$$=\frac{1}{b-a}\frac{\left(x+b\right)-\left(x+a\right)}{\left(x+a\right)\left(x+b\right)}$$
$$=\frac{1}{b-a}\left(\frac{1}{x+a}-\frac{1}{x-b}\right)$$

[soft zealot]

Roy

[rain sentinel]

why are we doing partial fractions here and not..

[spark swan]

that gives you 1/(-4y²)

[rain sentinel]

oh

[severe canyon]

(and that's not about integration)

[rain sentinel]

thanks

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[ivory salmon]

How can I make a function $f(x, y)$ for $x, y \in \mathbb{R}$ that is equivalent to $\max{x, y}$, but \textbf{only} using the following functions:
\begin{enumerate}
\item Standard binary operations: $+$, $-$, $\div$, $\times$
\item Roots and exponentiation $\sqrt[n]{}$, $a^n$
\item Trigonometric functions: $\sin$, $\cos$, $\tan$
\item Inverse trigonometric functions: $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$
\item Absolute value: $|a|$
\item Factorial (defined only for positive integers): $!$
\item The constants $e$ and $\pi$ may also be used.
\end{enumerate}
For example, $f(-1, -2) = -1, f(4, 2) = 4, f(1, 1) = 1$

Is this even possible? If not, why?

[soft zealot]

polished andesite

[desert mantle]

you in fact only need +,-,/2 and absolute value

[spark swan]

think about the midpoint of x and y

[ivory salmon]

okay

oh wait

do you add the distance to the midpoint of x and y to both x and y and take the average again?

[barren hound]

does that work? try it

[ivory salmon]

i dont think it works for x, y < 0

[barren hound]

try finding it for an easier case. what if x and y were in {-1, 1}?

[final saddle]

@ivory salmon Has your question been resolved?

[final saddle]

@ivory salmon Has your question been resolved?

[ivory salmon]

it's just $\frac{x+y}{2} + \frac{|x - y|}{2}$ isn't it?

[soft zealot]

polished andesite

[ivory salmon]

i had the right idea here but for some reason i got a little confused and thought it wouldnt work if one of x or y < 0

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[subtle path]

how to solve 2.b)

[worldly spruce]

!status

[final saddle]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[subtle path]

I can send you what I've done

[worldly spruce]

[subtle path]

@worldly spruce

[worldly spruce]

Sorry!

I can't help you with this sorry 🥰

someone else will

[subtle path]

<@&286206848099549185>

[orchid osprey]

Sorry I don’t understand the language

[molten oak]

i really have no idea about any of this it looks like some transformation of the gamma function though lmao

[muted olive]

The derivative of $(1-x)^n$ is $-n(1-x)^n-1$, not $+n(1-x)^n-1$

[soft zealot]

techspresso

[charred comet]

Pour tout x appartenant à l’intervalle zéro un,
on a un moins x puissance n supérieur ou égal à zéro,
e puissance x est strictement positif
et n factorielle est strictement positif.

Donc, pour tout x de zéro à un,
l’expression
un moins x puissance n sur n factorielle multipliée par e puissance x
est positive.

Par conséquent,
I n est supérieur ou égal à zéro.

De plus, pour tout x appartenant à l’intervalle zéro un,
on a e puissance x inférieur ou égal à e.

Ainsi,

I n est égal à l’intégrale de zéro à un de
un moins x puissance n sur n factorielle multipliée par e puissance x dx
et est inférieur ou égal à
l’intégrale de zéro à un de
un moins x puissance n sur n factorielle multipliée par e dx.

On obtient alors

I n inférieur ou égal à
e sur n factorielle multiplié par l’intégrale de zéro à un de
un moins x puissance n dx.

Or, d’après la question deux a,
l’intégrale de zéro à un de un moins x puissance n dx
est égale à un sur n plus un.

Donc,

I n est inférieur ou égal à
e sur n factorielle multiplié par n plus un
c’est à dire
e sur n plus un factorielle.

Ainsi,

zéro est inférieur ou égal à I n
et I n est inférieur ou égal à e sur n plus un.

i hope it works, i mean i used google translate to convert english to french. ignore the grammer

[subtle path]

hmm

[muted olive]

h'(x) is incorrect.
Applying product rule to h(x) gives you: $h'(x)=e^x/n![-n(1-x)^n-1 + (1-x)^n] = e^x(1-x)^n-1/n! (1-n-x)$
For n>=1, this is -ve on [0,1] = h is decreasing. You did get the monotonicity correct somehow, but your upper bound of 1/n! derived from h(0) isn't what question is asking (I suppose- IDK french but I know math, and I think I know what the question is asking).

[soft zealot]

techspresso

[subtle path]

well what i wanna prove is that ((1-x)^(n)/n!) decreases on [0;1]

[muted olive]

Understood it, this is how you do it...

[subtle path]

then i use the mean value theorem

but i get stuck when I use the mean value theorem cause the superior terminal is then e/n! and not e/n+1

is my reasoning understandable @muted olive

like with the mean value inequality

cause when I integrate the ineaquality it's just mulitplied by (1-0) which is 1 and doesnt help

the question is proving that for all n >=0, 0<=In<=e/n+1

2.b)

[muted olive]

I think you're being too generous with your approximations. By replacing the entire function with its max on [0,1]- you're essentially treating the area under the curve as a simple rectangle, which neglects the diminishing nature of the (1-x)^n term.

I'd suggest you first isolate the variable term. I'll need to solve it thoroughly to get you the next steps...

[subtle path]

so how can I get a more precise inequality ?

am i supposed to find a smaller upper terminal ?

if so, how ?

[muted olive]

Instead of replacing the entire function with its maximum value (which creates a massive, inaccurate rectangle), only replace the exponential...since e and n! are constants, you can pull them out and use your result from 2(a).

[subtle path]

but its e^x tho

I mean it isnt a constant right ?

ohhh mb

wait

sigma

[muted olive]

You get it?

[subtle path]

so it gives 1/(n+1)!

and I can say 1/(n+1)! < 1/n+1 with ease

[muted olive]

Yep!

[subtle path]

so like when I integrate the inequality in this case I got to integrate the function right ?

[muted olive]

Yes

[subtle path]

like i dont just multibly by (b-a) (1-0 in this case)

alrighttt I see

[subtle path]

is it common practice to do this ?

because I stumbled across another exercice where I found the similar problem

[subtle path]

the 2, where you have to prove the inequality

[muted olive]

No, I would never solve this in this manner...I would bound the integrand, get the equality expression, integrate it and substitute the answer from 2(a).

[subtle path]

bound the integrant ?

[subtle path]

where it would keep the functions on the two terminals like this :
1/2x < f(x) < sqrt(3)/2x then integrated the two functions

[muted olive]

For all x belongs to [0,1]; we know that $1 <=e^x<=e$. Since $(1-x)^n/n! >=0$ on this interval, we can write $0 <= (1-x)^n/n! . e^x <=(1-x)^n/n!$

[soft zealot]

techspresso

[subtle path]

instead of using max and min

[subtle path]

oh yeah ok I see but then you do the thing where you got a function that you integrate on one of the terminals

(well on the other exercice I sent it is on both)

[muted olive]

So here, cos x would be bound and 1/x will be the function you integrate. You'll integrate over [pi/6, pi/3]

[subtle path]

yeah that's the reasoning

is it a common practice to keep a function instead of using max and min then ?

for the terminals

and then integrating the functions

[muted olive]

IDK...I would say no, but then again I can't help you with knowing what's commonly used.

[subtle path]

okok

[muted olive]

But certainly, not the method I'll use when I first see such a question

[subtle path]

well you do use it on both the exercices I sent right ?

or I understood wrong

[muted olive]

Yes when you gave your reasoning, you made me think like you

[subtle path]

oh yeah ok haha

[muted olive]

Right about here...

[subtle path]

ok yeah

but I'm curious how would you solve it in another way ?

[muted olive]

Using integral comparison theorem. So you get $0 <= (1-x)^n/n! . e^x <=(1-x)^n/n!$ after bounding it, then you integrate across [0,1] and substitute what you got in 2(a).

[soft zealot]

techspresso

[subtle path]

mhmmm ok ok I see

that's much simpler

[muted olive]

Yep...

[subtle path]

well I thank you very very much for taking your time for helping me, I appreciate it !

very nice of you thanks

[muted olive]

Anytime 😄

[subtle path]

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[wraith glen]

is sin(theta + pi/2) = cos(theta)

[worldly spruce]

Yes

[loud sundial]

https://mathematica.stackexchange.com/questions/175791/why-is-mathematica-converting-sinx-pi-2-to-cosx

google is your friend

[worldly spruce]

Helper in Need!

[loud sundial]

!done

[final saddle]

If you are done with this channel, please mark your problem as solved by typing .close

[wraith glen]

naw sorry

u see the -2root(2) sin ((pi)t/3 - 3(pi)/4) term

i thought since sin(theta + pi/2) = cos, I could add pi/2 in the paranthesis to get -2root(2) cos(pi(t)/3 - pi/4)

nvm

.solved

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[unreal scaffold]

How do you find a non linear function if you have two sets of coordinates?

[steep nest]

"sets of coordinates"?

like (2,3)?

[severe canyon]

!xyp

[unreal scaffold]

Yes

[final saddle]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[severe canyon]

That's not a "set of coordinates". It's just one single point (with two coordinates, of course)

And there are infinite functions going through one given point

[steep nest]

well do you realy wanna find "a" non linear function

as there are infinite

[severe canyon]

And also the linear ones would be, actually

Unless you have other points given

[severe canyon]

Lol

[unreal scaffold]

This was the original question (redrawn)

[severe canyon]

That's totally different information

[unreal scaffold]

In this example, how do I find the slope of the tangent?

[vital crag]

Can you just do this

[severe canyon]

Without some written information we can't do anything, you should realize this...

[unreal scaffold]

That is the entire problem

[vital crag]

Doubt

[severe canyon]

@unreal scaffold

[unreal scaffold]

Specifically: “find the slope at point 0,p”

[bronze void]

is there any extra info on what kind of function/shape it is?

[severe canyon]

(0, p)

[unreal scaffold]

Quadratic function

[vital crag]

Amazing

[severe canyon]

Never drop off brackets

[severe canyon]

Why didn't you tell it before??

[vital crag]

So this was wrong

[bronze void]

good, that's essential

[unreal scaffold]

Ok

[severe canyon]

Besides helpers, we always have to be detectives as well

[bronze void]

cut them a bit of slack, you guys have to deal with a bunch of people doing it but for this user it's only been this one time

and with the context of the type of problem, it might not be immediately clear to them what all is essential to the problem

[severe canyon]

Yeah sure, it's not at all the first time, just that we still have to do it

[bronze void]

just sayin for this one user

also i dont remember algebra much so i cant help any

[vital crag]

It's amazing how specific this was yet OP refused repeatedly

[unreal scaffold]

Bruh it was like 10 seconds

[bronze void]

i doubt that they would knowingly choose to self-sabotage from getting help with the problem

[ornate spade]

not the first time, i have witnessed some other time as well

[unreal scaffold]

I apologize 1 million times

I’d appreciate some help with the problem?

[severe canyon]

Do you know what's the geometric (a.k.a. graphical) meaning of the derivative at some point?

Let's say you have the function f(x).
What is f'(a)? It is the .... of the .... line to the function at the point x = f(...)

[unreal scaffold]

Well, you typically replace a with x

[severe canyon]

Don't try to change the question, and don't overthink please

[unreal scaffold]

So f’(a) would be the derivative of x, ie the slope at a specific point

[severe canyon]

The slope of what? Both of the function and a tangent line to it, right?

But yes, you have it correct

[unreal scaffold]

Yes

[severe canyon]

Have you already found the equation of that parabola?

[unreal scaffold]

No, and I am unsure how to do so

[severe canyon]

This should be something you've covered way before doing derivatives

[unreal scaffold]

Ok

The issue specifically being the equation, even in part, is not given, and i have an incomplete set of coordinates given j only have (0,p)

[worldly spruce]

In high school already

[severe canyon]

By the way, I gotta go now
If nobody helps you within 15 mins, you can tag helpers if you want

[unreal scaffold]

Ok

[bronze void]

a parabola with roots $r_1$ and $r_2$ is defined by the function $f(x)=a(x-r_1)(x-r_2)$ where $a$ determines width and direction. if you know a point on the parabola, like $(x,y),$ plug in $y=a(x-r_1)(x-r_2)$ and solve for $a$ to find the general equation of the parabola

[soft zealot]

JamR_71111

[bronze void]

then if you're familiar with standard derivative rules, it shouldn't be too bad to go from there

[unreal scaffold]

The r wording is confusing to me

[bronze void]

for example, the roots of this parabola would be $r_1=2, r_2=4$ and solve for $a$ in $1=a(0-2)(0-4)$

[soft zealot]

JamR_71111

[unreal scaffold]

Ok, that’s what r is

[bronze void]

the r's just represent the x-values at which the parabola crosses the x-axis

yeah

[unreal scaffold]

Hang on

[bronze void]

can you solve for $a$ and find the equation of the parabola?

[soft zealot]

JamR_71111

[bronze void]

sent that before i saw "hang on"

[unreal scaffold]

If r1 is 2, is x also 2?

[worldly spruce]

that's p, not 1

[bronze void]

ah my bad

[steep nest]

could I say my way of thinking about it

or do you got this

[bronze void]

you can go for it if you want

[unreal scaffold]

Surely there is only one way of finding the equation of a parabola?

[steep nest]

well so we have this right

so we want a line which crosses through the x-axis twice and twice ONLY right?

is the question Im guessing

[unreal scaffold]

We want the slope of the tangent

It is also a quadratic function

[bronze void]

given roots (2,0), (4,0), and the point (0,p) of a parabola

[steep nest]

well then we must have that if the parabola crosses the x-axis, the y coordinate must be 0 right

[unreal scaffold]

Yes, that’s obvious

[steep nest]

we know it crosses the x-axis at (2,0) and (4,0). so we must have a function such that at those points of x, the y value is 0

we get this by doing the following f(x) = (x-2)(x-4)

notice how at x = 2 and 4 the y value is 0?

[unreal scaffold]

Ok

[steep nest]

and we want to find p im guessing?

[bronze void]

they want the slope at (0,p)

with derivatives

[unreal scaffold]

So if it’s (x-2)(x-4)

It’s 8

[worldly spruce]

ScarlettRose

!xy

[final saddle]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[unreal scaffold]

P=8

[unreal scaffold]

Scroll up

[steep nest]

is that all that was given?

[bronze void]

they gave all info earlier; roots, that y-intercept, it's quadratic, and want slope

[steep nest]

well then we have an infinite amount of p's

[worldly spruce]

I see no original problem

[unreal scaffold]

So the slope is -4

[worldly spruce]

Your answer will result on the slope being dependant on p

[unreal scaffold]

.

[steep nest]

that's 1 solution

true

[worldly spruce]

I seriously doubt this was written in the book, or whever you got this problem from.

[bronze void]

scarlettrose this question ticket is cursed apparently

[steep nest]

we have the parabola f(x) = a(x-2)(x-4)

where a is some number

this keeps our roots of 2 and 4 right?

y = 0 at x = 2 or 4

now we solve for y when x = 0

[unreal scaffold]

Then perish

[steep nest]

there are infinitely many different tangent lines scarlettrose

[worldly spruce]

"die, especially in a violent or sudden way."

thanks 👍

[steep nest]

as we can change a as much as we want

[steep nest]

in this

[unreal scaffold]

Hey man maybe you should leave the thread

[worldly spruce]

I will, don't worry. I don't realize why you are so reluctant to post the original problem

[unreal scaffold]

Right

[bronze void]

you can find $a$ by plugging in the point (0, p): $p = a(0 - 2)(0 - 4)$ then solving for $a$

[soft zealot]

JamR_71111

[unreal scaffold]

So we get -8a?

[bronze void]

there are two negatives there, so $p = 8a$ or $a=\frac{p}{8}$

[soft zealot]

JamR_71111

[unreal scaffold]

Oh, right, double negative

8

[bronze void]

then the parabola function becomes $f(x)=\frac{p}{8}(x-2)(x-4)$

[soft zealot]

JamR_71111

[unreal scaffold]

Well I got it here anyways

[bronze void]

so, depending on p and our roots, we get that function. treat p like a constant and you can take the derivative

it might be easier to multiply it all out IMO

[unreal scaffold]

Right so we replace a with p/8

[bronze void]

ye

have you learned the derivative tricks yet or is it still using the limit definition?

[unreal scaffold]

The limit definition? You mean h->0 and the like? That is what I’ve been taught

[bronze void]

yeah that limit thing. is that what you guys are still using to compute derivatives?

[unreal scaffold]

Yes

[bronze void]

ok, can you try finding the derivative of this using that method?

[unreal scaffold]

p/8 * (2x - 6)?

[bronze void]

exactly

[unreal scaffold]

p/8 = 1/8 * p also

[bronze void]

right

since we want the derivative at point (0, p), just plug in x = 0 into that derivative and see what you get

[unreal scaffold]

p/8(-6)

[bronze void]

what does that simplify to

[unreal scaffold]

-6p/8?

[bronze void]

ye it can be reduced just a bit more

$\frac{-6p}{8}=\frac{-3p}{4}$

[soft zealot]

JamR_71111

[unreal scaffold]

-3p/4

Ye that

[bronze void]

I dont know how much your teachers focus on that, but usually they want most simplified

and that's the derivative at the point (0, p) of the parabola with roots (2, 0) and (4, 0)

[unreal scaffold]

Ok

[bronze void]

just using the $f(x)=a(x-r_1)(x-r_2)$ method and derivatives

[soft zealot]

JamR_71111

[bronze void]

hope that helped

[unreal scaffold]

Uh

I still don’t know the slope?

[bronze void]

it's exactly -3p/4

it's based on whatever p is

[unreal scaffold]

But p isn’t defined?

So it’s a trick question ig?

[bronze void]

here, p is just an unknown variable kinda

yeah

so it remains a function of p

[unreal scaffold]

That’s dumb but thanks for the help

[bronze void]

np 😁

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[green olive]

..

[finite sinew]

Hello

[formal trail]

Hello @finite sinew, do you have a math question?

[finite sinew]

No Im new

I have to start first

[formal trail]

this channel is for asking math questions in. if you just want to chat you can use #discussion

[finite sinew]

ok

[formal trail]

if you don't have any further questions, close the channel by typing .close

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[stable viper]

ive went through this like 4 times and i swear everythings clean

[tawny barn]

what are we looking for

[stable viper]

Answers say this

so im like lost

[tired walrus]

!xy

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Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[stable viper]

b)

the method in q3 is just

(a^2 - b^2) + 2abi = 24 + 7i

so then u equate

1.. (a^2 - b^2)= 24

2.. 2ab = 7

[tired walrus]

ok the problem is now clear, let me look through the thing

u = 2a^2

[tired walrus]

so 2a^2=49, not a^2=49

[stable viper]

ohhh no way

[tired walrus]

also your handwriting has some issues:

• 7 looks like 1
• a and 9 are mildly confusable

[stable viper]

this the first tmie i use 2a^2 = u

damn

[stable viper]

okay

thanks

ill try improe on that

.close

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[uncut topaz]

Help

[final saddle]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[uncut topaz]

Help me with the eggs

How should j analyze this?

[tired walrus]

write out a table of how many eggs she had left before and after every sale

[final saddle]

@uncut topaz Has your question been resolved?

[meager fiber]

if you're struggling to get started, try calculating how many eggs she must have had left before the third sale (in order for the third sale to leave her with 3 eggs)

alternatively, you can try every starting value that is given as a possible answer and find which one leaves you with 3 eggs at the end (though this only works because this is a multiple choice question)

[uncut topaz]

I've got it now

I just did back solve

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[onyx peak]

Prove that the red tangents are parallel to the sides of the yellow triangle

What I know is that the centers of orange circle lie on the the circmucircle of yellow triangle and they lie exact in the middle of the relevant arcs

That implies that they also lie on the angle bisectors of the yellow triangle

[final saddle]

@onyx peak Has your question been resolved?

[final saddle]

@onyx peak Has your question been resolved?

[slender basin]

hint: in my image you can prove that DG || AC

[onyx peak]

hmm, can i get a slightly bigger hint?

What I see is that AD is an angle bisector and so E is in the middle of arc BC

[slender basin]

on how to prove this or how its useful?

[onyx peak]

how to prove it

should i try some angle chasing on it maybe?

[slender basin]

you can find eq triangles

dont forget about incenter-excenter lemma

[onyx peak]

oh

[onyx peak]

oh there is one more cyclic quadrilateral

nvm there isnt

if i dont add it manually

oh shit that orange is a bisector

if i can prove that it bisects DEC, im done (B D orange E will be cyclic and thus GDE will be alpha, proving parallellity)

rn i know that i bisects FEG

Clearly, DEF = 2beta + 2gamma - (180 - 2alpha) = 0 (sounds like sth went wrong..)

oh yeah, my 2gamma is wrong

ig im lost

@slender basin can I get another hint?

I just need some angle near the tangent

I know that that B D orange E will be cyclic but idk how to prove it

it would suffice to know sth about the angle FEG

or about angles around that point

[slender basin]

look angle chasing is good, sometimes you just need to check from another end, it can help

for example in this case its enough to prove that EDG = alpha
and since BFE=DGE=90 you have candidates for eq triangles

[onyx peak]

wait equilateral triangles?

I only found 4 isosceles

[slender basin]

equal

[onyx peak]

oh

congruent?

[slender basin]

in this case they are actually equal

[onyx peak]

yeah, BFE and EFC are congruent

ohh

[slender basin]

no i mean to prove EDG = alpha there is one more trianle

[onyx peak]

and i think ur point is that DGE is also gonna be congruent

[slender basin]

yes

[slender basin]

this can help

[onyx peak]

EF = EG is obvious

90° is also obvious

then I know BE = ED

oh damn

That was quite simple

im blind

now that I look at, i just see it

[onyx peak]

Thank you so much for the hints, i really appreciate your help

[slender basin]

np

[onyx peak]

This proves that if a tangent of that circle passes through incenter, then its parallel to the side

so last step is proving that the common tangents do pass through the incenter ig

[slender basin]

you can just make proof using incenter (proving that tangents from incenter to said two circles are actually one line and actually common tangent)

in this case using incenter is just simplier than trying to do it straightforward because fact that we proved involves only one of the circles

[onyx peak]

Oh, both are parallel to the same line and both pass through the same point

so they are equal

Nice, thanks again!

.close

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[signal mica]

is that the Redmi pad pro

My case looks so similar

[uncut topaz]

Yep

[signal mica]

twinssss

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[velvet pivot]

How do I find the midpoint of triangle ABC?

[brisk lion]

midpoint of a triangle?

do you mean one of its centers?

[severe canyon]

!xyp

[final saddle]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[velvet pivot]

If I drew a perpendicular bisector of all the sides of triangle ABC, would I get a midpoint for them?

[narrow dirge]

no, that was orthogonal point iirc

[velvet pivot]

Sorry this isn't a homework problem or anything, I was playing around with the desmos geometry tool and I got curious.

[severe canyon]

Alright don't worry 👍

[brisk lion]

so you are looking for the circumcenter of the triangle?

[velvet pivot]

I think so

[narrow dirge]

there's a huge ass formula for that or you find equations of perpendicular lins and find poiny of intersection

[brisk lion]

then yes, the three perpendicular bisectors from each vertex to its opposite side will always coincide at a single point.

this point is known as the circumcenter of the triangle.

[velvet pivot]

If I do this, does this help? I believe I made a perpendicular bisector for the unlabeled point for both of the 90 degree angles

[velvet pivot]

Yea!

[thin jasper]

what do you mean by from each vertex? , it is the intersection of perpendicular bisectors of each side

wdym bisector from a vertex,
the median from a vertex is not always perpendicular to the opposite side

[brisk lion]

right, I think I mistook that part. I'll cross that part out.

though you could have said it just once and I'll change it anyway.

[velvet pivot]

Using the tool highlighted in red, I find the angle to be 90 degrees

[brisk lion]

this is a special result, by the way.

[thin jasper]

was providing more clarity, that's all

[brisk lion]

the circumcenter of a right triangle is always the midpoint of its hypotenuse.

[velvet pivot]

Also, this is how I constructed the unknown point, it's not very clear how I did this but I hope you can somehow see. You can ask me a lot if you'd like

the unknown point as in the midpoint of AB

[brisk lion]

but that aside, do you have any other questions?

[velvet pivot]

Or at least I think it's supposed to be the midpoint based on the construction

[velvet pivot]

how do i do the midpoint here?

[brisk lion]

which midpoint?

[velvet pivot]

sorry i mean circumcircle

of the whole triangle

[brisk lion]

the circumcircle of a triangle is the circle that passes through each vertex and has a center on the circumcenter of the triangle.

so the way you can do it by hand is to use a pair of compasses, with the sharp tip on the circumcenter and the drawing end on a vertex, and then draw the circle.

whereas with software, plot the circumcenter, find its radius (distance to one of the vertices), and use the software's circle command to draw a circle of that radius and center.

[velvet pivot]

Like this? I made a circle for every vertex i think?

[brisk lion]

I'm not sure what you're trying to do with a circle for every vertex here?

[velvet pivot]

idk how to make a circumcenter

[brisk lion]

then first, draw all perpendicular bisectors.

[severe canyon]

Probably there's a command

[brisk lion]

the point at which they meet is the circumcenter.

[severe canyon]

Or this yeah

[brisk lion]

which, technically, you have found here.

see that intersection on the hypotenuse? that's the circumcenter of the triangle.

[velvet pivot]

using the angle bisector tool i get this

[brisk lion]

um... I mentioned perpendicular bisectors.

[thin jasper]

those are the medians....

[velvet pivot]

oh my bad

[brisk lion]

if you use angle bisectors, you are getting the incenter of the triangle.

not the circumcenter.

[velvet pivot]

I see

[brisk lion]

anyway once you've done that, find the radius of the circumcircle. this would be the distance from the circumcenter to any one vertex.

[thin jasper]

oh he drew angle bisectors - my bad too lol

[brisk lion]

then plot a circle with that radius centered at the circumcenter.

[velvet pivot]

So 3/2 is the length of the radius of the circumcircle?

[brisk lion]

I did not do any calculation, but if you want, you can share your working here and I'll check it for you.

[thin jasper]

you're right

[velvet pivot]

i did it now?

[velvet pivot]

Ty

[brisk lion]

correct.

[thin jasper]

in this case, as this is a right angle triangle, the midpoint of the hypotenuse gives the circumcenter

[velvet pivot]

sorry i think i confused you then, this doesn't look like what i was hoping to find, i wanted to see the midpoint of ABC inside of the triangle

[velvet pivot]

Gotcha!!

[brisk lion]

a triangle does not have a midpoint.

do you mean a center?

[velvet pivot]

Yes

[brisk lion]

and if you meant a center, which center?

[velvet pivot]

of vertices ABC

[brisk lion]

a triangle has four different centers. please pick the one that you want to find.

[severe canyon]

Ohhh so that's indeed the intersection point of the medians
It's called barycentre or centroid

[brisk lion]

ah I see, then I apologize for the misunderstanding OP.

[velvet pivot]

I'm not sure which one i meant initially

i like all of those

[severe canyon]

Don't worry, I also misunderstood from the beginning 😅

[brisk lion]

well, the sheet shows you how to construct each center.

[velvet pivot]

I think i meant the circumcenter because it's the perpendicular to the midpoints of each side

[brisk lion]

if you meant the circumcenter, it's point 3 in this image.

[velvet pivot]

Fascinating

Is this a special triangle?

I know it is but

[brisk lion]

in a way, yes.

[velvet pivot]

in this case for circumcenters?

[thin jasper]

a right angle triangle, yes

[brisk lion]

there are three different cases for circumcenters and their relative position to the triangle.

[velvet pivot]

the special right triangles are 30-60-90 and 45-45-90 (which is this one)

[velvet pivot]

inside, outside, and in the middle of the hypotenuse like this one?

[brisk lion]

if the triangle is:

• an acute triangle - circumcenter lies within the triangle.
• a right triangle (ANY right triangle) - circumcenter lies on the midpoint of the hypotenuse.
• an obtuse triangle - circumcenter lies outside the triangle.

[velvet pivot]

Cool!!!!!

Tysmmm for helping me everyone

[brisk lion]

anything else you would like to ask?

[velvet pivot]

Well i was just wondering if i have 2 points on the cartesian plane that have the same x coordinate, their slope is 0 but their distance isn't 0 right?

that's why i was opening the desmos geometry tool because i'm learning the distance formula

[brisk lion]

well, that would depend on their y-coordinates, wouldn't it?

[velvet pivot]

my initial question didn't have much to do with this but i was wondering it because i was curious

[velvet pivot]

Yea

[brisk lion]

if the y-coordinates are also the same, then you really have two coincident points and their distance is also 0

[velvet pivot]

Yep

[brisk lion]

but if not, then this turns into a 1D distance problem, and the distance between them = |y_1 - y_2|.

[velvet pivot]

what's directed distance called?

[brisk lion]

directed distance is called a vector.

[velvet pivot]

Cool!!!!!

[velvet pivot]

divided by 2 right?

[brisk lion]

but do note that vectors have more use than being quantities with distance.

[brisk lion]

no, just that.

[velvet pivot]

linear algebra?

[velvet pivot]

ah ok!

but

[brisk lion]

vectors are prominent in linear algebra and vector calculus but you'll start seeing vectors in a lot of other places as well.

[velvet pivot]

its slope is y2-y1/x2-x1 right? and if delta x or the change in x is 0 then it's undefined right?

[velvet pivot]

ok!!!

[brisk lion]

delta x being 0 means that they have the same x-coordinate, not y.

[velvet pivot]

Yep

[brisk lion]

in that case, the slope is undefined and you get a vertical line, yes.

[velvet pivot]

so the distance between the y's if they're not coincided is a constant

[brisk lion]

what do you mean, constant?

[velvet pivot]

and a horizontal one is always 0

[velvet pivot]

a number

[brisk lion]

well yeah, distance is always a number.

[velvet pivot]

Tysmmmmmmmmmmmm ❤️

[brisk lion]

no problem! anything else?

[velvet pivot]

that's all!

i'm very joyful and thankful!

[brisk lion]

alright, have a nice one!

[velvet pivot]

Bye have a goodday everyone! im gonna close this then

.close

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[small stump]

this is the correction of an exercice
im trying to understand why they chose to keep sqrt(x) instead of x^5

this is the explanation i came up with:
for a good upper bound, you want it to be as small as possible
so the upper bounded function should be bigger then the original function, but the least possible bigger
so the denominator of the upper bounded function sould be smaller, but the least smaller possible
x^5 makes it smaller more than sqrt(x), so we chose to keep sqrt(x)

is that basically it?

[onyx peak]

Yeah, thats basically it

x^5 wouldn'tve worked, because it would be too small already

[small stump]

but it would still be a valid upper bound right? just not the best one

[onyx peak]

It'd be a valid upper bound, but the later comparison test would be inconclusive

integral of 1/x^5 wouldnt converge around 0

[small stump]

ah yeah that makes sense

youre basically saying it's smaller than a divergent function

[onyx peak]

Yeah, which is useless

we need the upper bound to result in something that'll converge

[small stump]

perfect thanks

[onyx peak]

and you have the best chances for that when you choose a small upper bound

[small stump]

.close

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[orchid crystal]

in the third part, I was hoping for a map with kernel = U_2 so something like log|x| but then this has a kernel which contains U_2 but is not U_2 itself, any clues?

[orchid crystal]

I mean if it were something like R* I can show they're isomorphic but idk how to show its isomorphism with this

[scarlet sequoia]

I also think it should be R*

the quotient is easily group-isomorphic to (R*, x)

and (R*,x) and (R,+) are not group-isomorphic

[orchid crystal]

that settles it then

makes sense

thanks!

.close

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[celest marsh]

Hello if anyone is avaliable i would really love it if someone could help me understand algebra on a deeper level I have a 35 in the class i understand basic concepts like pre-algebra and variables but when it comes to trying to build on those concepts I tend to struggle. If anyone could help I would really appreciate,, it .

[celest marsh]

<@&286206848099549185>

[crystal pawn]

maybe ask your questions in #study-discussion #book-recommendations #prealg-and-algebra

this video could help you too

https://youtu.be/grnP3mduZkM?si=FwG-Xnzpo4_KJ4wN

[celest marsh]

oh im sorry

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[peak tapir]

Whats the direction of this vector im confused

[charred comet]

well, the thing you did wrong is that

You drew delta v as if both components were positive.
But the x component must be negative, because velocity changed from east to zero, not the other way around.

[peak tapir]

What

Its to the east tho

[charred comet]

ur asking for the direction of vector, i cant tell you direct answer but the steps towards it

[peak tapir]

Ye ik

[charred comet]

but u drew the delta v. assuming both are positive which isnt

[peak tapir]

Wait is vf supposed to be the opposite direction

[charred comet]

yep

[peak tapir]

O right

[charred comet]

god i hate it sometimes

$delta_v = v_f − v_i = (0,11) − (14,0) = (−14,11)$

[soft zealot]

V

$delta_v = v_f − v_i = (0,11) − (14,0) = (−14,11)$
```Compilation error:```! LaTeX Error: Unicode character − (U+2212)
               not set up for use with LaTeX.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.49 $delta_v = v_f −
                        v_i = (0,11) − (14,0) = (−14,11)$
You may provide a definition with
\DeclareUnicodeCharacter```

[peak tapir]

Huh

[peak tapir]

Never learnt this

[charred comet]

in whih grade are u

[peak tapir]

11

[charred comet]

ahh

okk

gotta tell u the 11th way

[peak tapir]

Its the grade where u learn kinematic equations or smth

[charred comet]

i know, but you wont be understanding delta.

[peak tapir]

Delta is change of?

Is that right or am i dumb

[charred comet]

here, like if say according the question delta would be , final velo - initial velo.

[peak tapir]

Ohhh

Wait i get it

Oops

[charred comet]

nopee

not corect

i mean you are close just, If you redraw Δv from the end of vᵢ to the end of v𝒇, your diagram will be fully correct

[peak tapir]

Wait delta is vf - vi

Which is vf+ (-vi)

So u flip the direction of vi?

[charred comet]

yea that's what i told you earlier

godd im handling 5 channels at once

ouchh my brain

[peak tapir]

Pls tell me this right

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[fresh bobcat]

could someone explain why they have written 2x * dx ?

[fresh bobcat]

dy = 2x * dx
dy/dx = 2x ?

[arctic plank]

Do you know the formula for differentiation y=x^n then dy/dx=n(x^(n-1))

[fresh bobcat]

yep
but why do they write it with the dx there

[arctic plank]

It's just a modification , multiply by dx to both sides for the equation in my expression and see what you get

Basically the same thing ,but this is called a differential change i think rather than a derivative

[fresh bobcat]

mm ok

cus they were calling it a derivative hence the confusion

[arctic plank]

Uhmm derivative is when it is expressed as dy/dx

[fresh bobcat]

yh

same story here right

[arctic plank]

What are the symbols representing? Like i think g and f are the upper and lower functions respectively but what is' a '?

[fresh bobcat]

my bad, a is the initial point. the thing we are "solving for"
and then dx is the slight nudge away from a

[arctic plank]

Oh you mean g(a) and f(a) ? Is that what it is given?

[fresh bobcat]

yeah

[arctic plank]

It's the same concept as the earlier one, we are finding out the derivative at the point a

Instead of a general point x

[fresh bobcat]

yeah i mean i just don't really understand this specific part

why is it written as that

"the value of f at that point, is approximately its derivative, df/dx evaluated at a times dx"

df/dx evaluated at a ✅ ok fine with that
but why times dx

[arctic plank]

I wouldn't say the value of f at that point is approximately it's derivative times dx
It's more precisely the approximate change in value of f

Which is df

[fresh bobcat]

$$dy = \frac{df}{dx}\left(a\right)\cdot dx$$
$$\frac{dy}{dx} = \frac{df}{dx}\left(a\right)?$$

[soft zealot]

SimpleGamer14

[fresh bobcat]

is it just this again

but he writes it in this form just to.... cancel them in the next step

them = dx

[arctic plank]

Yes

That is what it is

The change in a function is approximately the derivative of the function times it dx

[fresh bobcat]

yeah, aka change in y aka dy for this case. okay thank you for clearing it up

[arctic plank]

Yup

[fresh bobcat]

❤️

.close

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[rustic wedge]

consider two ellipse E1 and E2 as follows: $x^2 + 5y^2 = 8$, $x^2 + 3y^2 = 4$. If a tangent at E2 intersects E1 in two points P and Q, what is the angle between the tangents at P and Q to E1?

[rustic wedge]

eg figure

whats the best approach here?

note: i changed the data so that i could understand a general approach

in the original question the locus of the point of intersection of the tangents is the director circle of the outer ellipse

hence the answer is pi/2

[soft zealot]

rak³en

[rustic wedge]

wait ok

o

i see

.close

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[rain sentinel]

Can someone explain why we do -P[1-P(z<-1.5)

[tiny gorge]

!original

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Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[tiny gorge]

this notation makes no sense

probability of a probability?

[lyric obsidian]

think the outer P was unecessary

[tiny gorge]

even if you remove it, it's still wrong

should be 1 - P(Z >= -1.5)

which by symmetry of the unit normal distribution equals 1 - P(Z <= 1.5)

[final saddle]

@rain sentinel Has your question been resolved?

[rain sentinel]

I see

here is the original question

part c

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[lucid nymph]

!da2a

[final saddle]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[severe canyon]

Are you doing a test?

😂

<@&268886789983436800>

Yes, no comment!!! But because you shouldn't be here!!!!!!!!!

[opal plinth]

No comment apart from the date

[tulip coyote]

Why exactly do you need answers asap

[quasi rapids]

british maths olympiad?

yeah mate youre dead

[severe canyon]

What about this?? @fleet bough

[lucid nymph]

you joined the server like just now

to get answers

[tiny gorge]

accurate statement, cheating is not smart

[tulip coyote]

Yeah, no

[vital crag]

Can't be bothered to do math but wants answers anyway

[tulip coyote]

.close

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[severe canyon]

Do you know what happens when someone enters here trying to cheat in exams/tests? @fleet bough We call the ...

[opaque ember]

https://ukmt.org.uk/intermediate-challenges/intermediate-mathematical-challenge

[charred trail]

ghostbusters!

[lucid nymph]

dammit i should have screenshotted so i could try it lol

[opaque ember]

looks like an established math contest. they ded

[severe canyon]

Yeah I was also curious about the the questions lol

[tiny gorge]

maybe in your browser cache?

[lucid nymph]

WAIT YES I DID SCREENSHOT IT i wanted this moment saved

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[sage zenith]

I started this problem by figuring out where sqrt(x) and e^-3x meet (via desmos graphing calculator, but I could have done it with a different one) and got 0.239 (rounded), so I took the integral from 0.239 to 1 of the top line (sqrt(x)) minus the bottom line (e^-3x) and got my answer of 0.443
this is good and all for part A, but I have no clue how to get part B done. How does one solve a problem like this?
I am new to this server and if there's any more information I can provide or something else I should say, please direct me to do so. I read the rules and tried my best to make a good question.

[green ibex]

https://tutorial.math.lamar.edu/classes/calci/volumewithcylinder.aspx

https://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithRings.aspx

[sage zenith]

thank you, i'll read these and see if I can figure it out

[green ibex]

they're structured very nicely and even have some examples :3

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[loud sundial]

<@&268886789983436800>

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[crimson ermine]

how to prove the triangle inequality

[muted prairie]

For what?

Actual triangles?

[crimson ermine]

yes

[muted prairie]

ah ok

[plucky rover]

Oh god I don't even remember the geometric proof of this anymore

[muted prairie]

well we could prove it from the distance formula

idek how you can have a geometric proof of this

[crimson ermine]

😭 i only know about the law of sines proof and distance formula proof

[muted prairie]

maybe project the two legs onto the third

then use Pythagorean theorem?

[hidden marlin]

wait I think out book has this in a cut section

di you want me to check?

[crimson ermine]

is there a more classical proof?

[crimson ermine]

whats out book

[hidden marlin]

our*

like I meant my book

[muted prairie]

I'm getting deja vu

[crimson ermine]

oh

[crimson ermine]

yes

[hidden marlin]

it's like in a cut section so lemme double check

Ok so uh

in a triangle ABC. Produce the line AB to D such that AD = AC. Then,

angle ACD = angle ADC

b isosceles

which means that Angle BCD > angle BDC.

which means that BD > BC

(side opposite to the largest angle is bigger)

BD = AB + AD and AD = AC (by construction)

you can do this for all three sides

@crimson ermine

[crimson ermine]

how to prove this

is it an axiom

[hidden marlin]

God damn I don't remember that 😭

[hidden marlin]

no

wait wait no I do remember

I DONT HAVE TIME

[muted prairie]

law of sines

[hidden marlin]

sorry

😭

[crimson ermine]

i want a classical proof

[muted prairie]

maybe you can draw an isosceles triangle inside your triangle

by shortening the biggest angle

here blue is my big triangle

i want to show the base is longer than g

so i draw an isosceles triangle like so

now we see that part of the base is equal to g

so the base is longer than g

[crimson ermine]

thats not very rigorous

[muted prairie]

its a sketch

[crimson ermine]

i saw a video proof which made sense

[muted prairie]

maybe we start with a circle centered at one vertex passing through the closer vertex

[crimson ermine]

.close

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[muted prairie]

then we mark the point where it intersects the other side

that's our isosceles triangle

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[ivory vessel]

why is P(C) wrong here?

why should it be this instead?
[
\binom32(P(A)P(B\mid A)^2P(\overline B\mid A) + P(\overline A)P(B\mid \overline A)^2P(\overline B\mid\overline A))
]

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[thick depot]

hii

DAMN.. IM A DUMBASS AND I DECIDED TO JOIN THIS

[plucky rover]

Hi, please claim an available help channel such as #help-2

[thick depot]

nahh its fine

[plucky rover]

Suit yourself ¯_(ツ)_/¯

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[wraith glen]

i get that the limit is 1/4 and how to show work to get there

but how would you describe the limit in words? for example in calc 1, you just say the left limit = right limit and this is the limit. what would you say here?

lmk if the question doesn't make sense

[brisk lion]

f(x, y) approaches the limit from all possible directions? that's the closest I can think of at the moment.

but you may want to get a second opinion.

[supple jolt]

yes

[wraith glen]

ok so f(x,y) approaches (2,0) goes to 1/4 with all paths

ok

is there a good graph showing where a limit at a point DNE?

i can't picutre it in my head

professor said the limit at (0,0) DNE

but im confused, it looks like at 0,0 it all meets up at one point on the z axis

<@&286206848099549185>

[supple jolt]

so f(x)= x^2 / x^2 + y^2

?

[wraith glen]

f(x,y)

yes

[supple jolt]

yeah ok

yeah it shoudn't exist

at (0,0)

if you approach from f(x,0)

you get x^2/x^2 =1

[wraith glen]

ah

[supple jolt]

if you approach along x=0

you get something else

[wraith glen]

thank you bro

so is the graph messed up or something

wait

yes

why does it look like that then, it looks like it meets at a point

is it because x wants to stay at 1, and y wants to stay at 0? that's making it messed up

[opal plinth]

What software is that?

[final saddle]

@wraith glen Has your question been resolved?

[wraith glen]

math3d.org

[opal plinth]

Increase the x and y samples

You'll see it doesn't actually look like that

(Desmos 3D seems to handle that automatically: https://www.desmos.com/3d/xi0ekcx6yq)

[wraith glen]

i see thank u bro

[wraith glen]

yeah tried it thanks

looked way better

.solved

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[candid pulsar]

wait a min how are any of these correct

[candid pulsar]

given answer is a, d

[tired walrus]

this is the graph of f

its range is a medium-ugliness subset of R

[scarlet sequoia]

Since A is made up of a whole lot of intervals, each isolated from the others

To show f^-1 is continuous, all is needed is to show that it's continuous on each interval making up A

$A = \bigcup_{k\in \bZ}[2k,2k+1)$

[soft zealot]

Rafilouyear2026

[candid pulsar]

what the helly

how does that work

is there a definition for continuity in a set

[scarlet sequoia]

A is a subset of R like any other

[candid pulsar]

mhm

[scarlet sequoia]

it's just that when you look at delta small enough

when you take $x\in A\cap (x_0-\delta,x_0+\delta)$

[soft zealot]

Rafilouyear2026

[scarlet sequoia]

the only values of x that really matter are the ones in the same interval as x_0

otherwise it's so far away that we can just pick delta small enough

like delta < 1 is enough here

[candid pulsar]

ok nvm makes sense

but the points in the domain are still in ascending order

and y = x is just chopped and spread in this new set of numbers

in the inverse

[scarlet sequoia]

that's a weird way to interpret f^-1 but ok

[candid pulsar]

i think thats it yeah

just a moment

ah yeah was about to ask if there was any other civil way to find this limit other than inhumane lhopital abuse

[scarlet sequoia]

what was your way of doing it?

[tired walrus]

take logs of that shit

[scarlet sequoia]

that and ln(x+1) ~ x

[tranquil pine]

:resisting my urge to say every jee q can be solved by lhopital

[candid pulsar]

lhopital abuse ❤️‍🔥

actually it doesn't look that bad nvm

ill just check if theres any weird sums left

[old quarry]

my way was ln y = ln (x+1) ln x
x=0
ln y=0 , y=1 😄

(ik its wrong)

[scarlet sequoia]

0 * ln(0) is 0, for shure

[old quarry]

🫠

if it works it works

[tranquil pine]

which mostly works

[scarlet sequoia]

getting rid of my gif, but "right answer with wrong method" isn't gonna work anywhere 💀

[old quarry]

:(

[candid pulsar]

ok i dont think theres any prickly qns for now

thank yo ufor the halp