help-27

[alpine python]

it just changes how you plot it

[orchid umbra]

omg

i was never taught that

omfg

[alpine python]

[devout snow]

@orchid umbra Has your question been resolved?

[orchid umbra]

wait is this accurate

[alpine python]

yeah seems right

[orchid umbra]

ty

[tame palm]

It's easier to understand if you see it in a graph.

[devout snow]

@orchid umbra Has your question been resolved?

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

@austere sluice Has your question been resolved?

[lean anchor]

What have you tried?

[austere sluice]

Resolving vectors etc

[devout snow]

@austere sluice Has your question been resolved?

[devout snow]

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[vestal ferry]

When to prove a sequence is Cauchy why its fine to suppose n>m or m>n

[devout snow]

@vestal ferry Has your question been resolved?

[vestal ferry]

Ass helpers

[lapis quarry]

Look - I can just speak for myself but: If I wanted help, insulting the people who are voluntarily going to help you isn't going to make them help you faster. You can ping the Helper role after 15 minutes, but that comment is just unnecessary.

[vestal ferry]

Sorry i love you bro

[lapis quarry]

also I would say the answer to your question lies in symmetry.

|am-an| = |an-am|

[devout snow]

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[merry ridge]

Can someone help with quadratic eqn 2x^2-3x+9

[pseudo basin]

this is not an equation. an equation always has an equals sign.

did you mean: 2x^2 - 3x + 9 = 69 ?

@merry ridge

[merry ridge]

I mean 2x^2-3x+9=0

[pseudo basin]

ok

and what do you need to do with that? just solve it? or find its sum of roots or something?

and also, tell us how much you've progressed on your own thus far. if you haven't, then just say you have no progress (no shame at all)

[merry ridge]

yes, solving it

[merry ridge]

Using factorisation didn't work

Completing square too hard

[pseudo basin]

do you have access to the quadratic formula

[merry ridge]

Forgot it

[pseudo basin]

do you have it in your notes and/or a textbook on hand

[vestal ferry]

You used chatgpt

[pseudo basin]

if not then ig i will have to be a stand-in for google since you didnt think to look it up

for the equation $ax^2+bx+c=0$ the two roots are given by $$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

[woven radish]

Ann

[pseudo basin]

@merry ridge

@vestal ferry @lapis quarry figure out yalls shit in #discussion or #chill please

idc whos right and whos wrong but we got a help channel going rn and i would like it uninterrupted, capisce?

[lapis quarry]

Nope. Covered that topic myself a few weeks ago so still know it ;-) But please can we move on to the current question of this channel? Thanks.

[merry ridge]

Si

[vestal ferry]

No problem bro i love you

[merry ridge]

Pls go or i'll close the chanel

[pseudo basin]

was not addressed to you

@vestal ferry @lapis quarry please leave this channel at once. go to #discussion, #chill, or DMs.

@merry ridge do you still need help here or can you continue on your own with the formula?

[lapis quarry]

@merry ridge For any quadratic equation, you can always use this formula. You'll surely learn the small formula too, but from my experience, it's practical to remember this formula by heart, as it helps you to solve all quadratic equations.

So a normal quadratic equation is ax^2+bx+c = 0. What would you put in for a, b and c in your case?

[pseudo basin]

op gone again

[devout snow]

@merry ridge Has your question been resolved?

[devout snow]

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[devout snow]

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[paper prism]

Start with part B

A I have solved

[inner sorrel]

Oh yeah your original post got archived

[paper prism]

I solved that

In that question,lambda=1/k eqn turns to x²-4kx+9k=0 leading coefficient is 1 so D must be perfect square integer

[inner sorrel]

My brain is lwk malfunctioning rn

[paper prism]

No problem someone else will help

[celest frigate]

@paper prism what have you done in parts B and further.

[paper prism]

Lemme tell

When u simply the first inequality in part B u get (x-k-1)(x-3k+1)>=0

[warm karma]

B doesn't seem too bad?

[paper prism]

Inequalities aren't too bad but when I jumped in into that question,I felt stuck

Second inequality in part B is (x-k)²+k>=0

when k>=0 second set in B is R,so whatever be the soln to first inequality u will get it

But I am stuck at k<0 case

[warm karma]

So just to be clear, In the first one we have x^2 + O(x) >= 0, so solution set is everything not between the roots, yeah?

same for the second

[celest frigate]

Why do you need to make separate cases for polynomials?

[warm karma]

Idk man, he has the most cursed questions.

Sometimes he really does need to make separate cases, but either way it doesn't hurt

[celest frigate]

fair

[warm karma]

You just need to show that both roots lie in the interval between k+1 and 3k-1, given that k<0, we can have 3k-1 < k+1 that range becomes [3k-1, k+1]

see?

[paper prism]

(-inf,3k-1]U[k+1,inf)

[warm karma]

No

I'm saying roots of the second one must be in the range [3k-1, k+1]

[celest frigate]

That is the part that is excluded except the endpoints which you have included, which are included.

[paper prism]

I didnt get it

[celest frigate]

You find the solution set of the second one and then solve the inequality where it lies in the bounds specified by CyclicTree

[paper prism]

Now I get it

So,just f(k+1)>=0 and f(3k-1)>=0 where f(x)= first one

[warm karma]

x^2 - 2kx + (2k + k^2) = 0
x = (2k +- sqrt(4k^2 - 8k - 4k^2))/2 = k +- sqrt(-2k)

oh?

am I right?

oh yeah k is negative

[paper prism]

I think I did wrong

I did opposite maybe
Lemme check

I think I got it

Tell me where I am wrong if I am

k<0 case

Call the second inequality g(x)>=0

g(3k-1)>=0 and g(k+1)>=0

I think I am right

That would be sufficient condition

Isn't it?

[warm karma]

yeah

wait, no?

[paper prism]

Yaa

I missed one thing

[warm karma]

it could have a root to the right of k+1

[paper prism]

That vertex thng

-b/2a lies in between 3k-1 and k+1

[warm karma]

ok, if it does

that' correct

[paper prism]

Thanks a lot

[warm karma]

I'm not gonna check if it does or not, but I'm assuming you are correct about that. How do you actually solve the ineqs?

[paper prism]

I mean we have to apply that inequality and find intersection with other 2 inequalites

[warm karma]

How do you find the set of k, for which g(3k-1) >= 0?

[paper prism]

just plug it in

U get quadratic in k

[warm karma]

ok, sure, I'm too lazy for that

but you are right that this would work

[paper prism]

Ill solve them(Bcd parts) and dm you if I get any trouble

Thanks a lot

[warm karma]

idk how much I helped tbh

It seems you are doing rubber duck debbuging here

[paper prism]

What's rubber duck debugging btw?

[warm karma]

https://en.wikipedia.org/wiki/Rubber_duck_debugging

[paper prism]

I mean The questions I am solving are slightly good level.sometimes,I get stuck and ask here.i reveal my thoughts and further get help

Anyways,thanks

[devout snow]

@paper prism Has your question been resolved?

[devout snow]

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[devout snow]

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[plucky siren]

i have to solve this limit

[plucky siren]

am i doing it right ?

[untold ravine]

is that $\lim_{x\to0}\sin(8x)\cot(3x)$

[thorny bloom]

there is easier way

try putting all in sinx/x form

[woven radish]

CherryMan

[plucky siren]

yessir

[plucky siren]

thas what am tryna do

i have solved it already

[thorny bloom]

sinx^2?

[plucky siren]

but i wanna know if we can reach the solution from the circled part

[plucky siren]

?

[untold ravine]

should be 3/8 btw

[thorny bloom]

8/3 is correct tho

[untold ravine]

or am i tripping

[plucky siren]

nops

[plucky siren]

yes you are :p

[thorny bloom]

where did that come from?

sin8x . cos3x/sin3x right?

just divide by 8x and multiply by 8x

[untold ravine]

mb

[thorny bloom]

same with 3x

[plucky siren]

i took out the 8 and then multiplied and dividen the numerator and denominator by x

[untold ravine]

you cant take the 8 out of sin (8x)

[thorny bloom]

huh

[untold ravine]

and also you multiplied x inside the sine and the other x outside the sine in the denominator, right?

[plucky siren]

new to calc T~T. Dont they take out the real part ?

[plucky siren]

sin x times x upon 1 times x

[plucky siren]

pardon my foolish ideas

[thorny bloom]

x can also be 8x

[plucky siren]

how ?

[thorny bloom]

its not complusory for it to be x

[smoky slate]

Where are you from?

[plucky siren]

how did you end up on the conclusion that I 'seem' like an Indian ?

[plucky siren]

ahhh

[smoky slate]

Nothing brother

[smoky slate]

Where are you from? ×2

[plucky siren]

earth

[smoky slate]

😹 crazy but i meant from which country

[smoky slate]

Would you

[plucky siren]

india brotha

[smoky slate]

I knew it 😎

[plucky siren]

anyways, help with limits ?

[plucky siren]

no shit sherlock

[smoky slate]

Tell me what does "randi" mean?

💀

[plucky siren]

so technically everything i did just went in vain ? 😫

[plucky siren]

tf ? brother grow up (seriously)

[thorny bloom]

<@&268886789983436800> really bad word damn

[smoky slate]

Bro it's means "come"

[plucky siren]

fr

[plucky siren]

we all know what you meant

[smoky slate]

Bro it's not i can tell you

[thorny bloom]

im indian

[thorny bloom]

yes it was wrong

[plucky siren]

nahhhhhhhhhhhhhhh (peak math ragebait)

[untold ravine]

sin(ax) is not a sin(x)

[plucky siren]

thanks for help tho !!!

[plucky siren]

ahhh i see

so when they talk about taking the real part out what does that refer to ?

[copper harbor]

either way, not an appropriate convo topic for a help chanenl, and seemingly none of your interactions in help channels so far @smoky slate have been appropriate so far so take a day to review #rules

[untold ravine]

it would be helpful if you could specify the context they said that in

[rustic heron]

why havent you done the taylor series of sin in 0 ?

[plucky siren]

24hrs mute ?

[copper harbor]

We do not discuss a user's moderation actions with other users

[plucky siren]

ouuu okok ic

[thorny bloom]

but the property applies sin(ax)/ax = 1

[plucky siren]

da hail is tha ???

[untold ravine]

thats for the limit

im talking about general properties

[plucky siren]

like whenever dealing with limits we take the real part out, ig they meant like 2 sin(2x) then we take out 2 and it becomes 2 lim sin(2x)

[plucky siren]

idk none of that bro sorry

i only know who taylor swift is ✊😔(not a swiftie tho)

[untold ravine]

you can say that $\lim_{x\to 0} \frac{\sin(ax)}{x} = a$

wait

[rustic heron]

oh, it's very useful to calculate hard limit, especially in 0

[plucky siren]

oo

[plucky siren]

mhm

[woven radish]

CherryMan

[untold ravine]

is that what you are referring to?

i saw that vie

[thorny bloom]

i thought the a was in denominator

[plucky siren]

is this correct ?

[untold ravine]

yes

[plucky siren]

how ?

[untold ravine]

$\lim_{x\to 0} \frac{\sin(ax)}{x} = \lim_{x\to 0} \frac{\sin(ax)}{ax} \cdot a = 1 \cdot a$

uh

[woven radish]

CherryMan

[untold ravine]

also another mistake you did

was writing sin x as sin(x^2)/x

which is incorrect. you can write sin x as sin(x) *x /x but not the former

[plucky siren]

i see

my bad

but ig the other method is correct ?

[untold ravine]

yes

[plucky siren]

ok

thanks for the assistance !!!

.close

[devout snow]

Channel closed

Closed by @plucky siren

Use .reopen if this was a mistake.

[thorny bloom]

cherryman

[untold ravine]

yea?

[thorny bloom]

i needed help with real analysis

[untold ravine]

#math-discussion or a help channel?

[thorny bloom]

here?

[untold ravine]

this is a help channel that will end soon, lets at least go to #math-discussion

[thorny bloom]

oh it didnt end damn ok

[plucky siren]

.reopen

[devout snow]

✅ Original question: #help-27 message

[plucky siren]

no nvm

.close

[devout snow]

Channel closed

Closed by @plucky siren

Use .reopen if this was a mistake.

[plucky siren]

ill have to open a new channel to post a new qn ?

[untold ravine]

yea

[plucky siren]

ok

[devout snow]

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[devout snow]

@spring oasis Has your question been resolved?

[devout snow]

@spring oasis Has your question been resolved?

[vital sedge]

Do you know how to construct a lagrangian?

Im guessing you need to find the extremas

[spring oasis]

yes and yes

[devout snow]

@spring oasis Has your question been resolved?

[hot steeple]

Can't you just go through the cases?

[spring oasis]

kind of yes

but to be honest would like to start from scratch

because now I have more of a solid understanding of what I need to do

like, let me explain, it seems like lagrange multipliers only apply when the constraints are equalities

having said that, the region that is given here is different, is made out of the intersection of an elliptic disk and a plane. if we were about to use lagrange multipliers the constraints needs to be an ellipse and a line here

having said that, if we were about to use lagrange multipliers for the frontiers of this region (aka when the inequalities are equalities), the intuition is that the maximum and minimums if they exist will lie in the frontiers themselves, not in the interior of the region, that is why we need to find the critical points of f and evaluate the critical points of f and see if they lie inside the interior of the region, in the frontier or outside the region, if they fall outside the region we discard them, if they fall in the frontier of the region we discard them because we will find them when we are using lagrange multipliers on the frontiers, and if they fall in the interior of the region then we save them, and that's how we analyze all the maximum and minimums in this region

[dusk dawn]

the whole point of lagrange is to convert these inequality problems into equality ones

[spring oasis]

care to elaborate?

[hot steeple]

i found someone's example for this actually

https://users.wpi.edu/~pwdavis/Courses/MA1024B10/1024_Lagrange_multipliers.pdf

[hot steeple]

apparently you use something called slack variables to turn inequality constraints to equality constraints

generally these variables are positive and so they are written as $s^2$

[woven radish]

Katharine

[hot steeple]

instead of just s

to indicate positiveness

[devout snow]

@spring oasis Has your question been resolved?

[spring oasis]

wait. let me check this

[spring oasis]

whaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaat?

[young spade]

tbh i dont know how to do lagrangian but i found the extrema pretty easily just by following the same process as the other time

still requires for you to parametrize correctly tho

[spring oasis]

all my classmates are using lagrange multipliers

well is not too hard though the issue is

[young spade]

yeah, probably its quite easy, but i have personally never done it

[spring oasis]

we have inequality constraints not equalities

[young spade]

esto sigue aplicando:
f'(x) = 0
∄f'(x)
La frontera de la region A

Bueno, casi, en realidad usas el gradiente

pero si, es lo mismo

[granite venture]

either the inequality is satisfied with slackness or it is tight

if it is tight, then that is an equality

[young spade]

you get quite a lot of points to go through tho in this problem

[granite venture]

if it is slacked, then it is as if it was never there

[spring oasis]

why is that

[young spade]

The specific shape of the surface you get

[spring oasis]

care to elaborate?

[spring oasis]

what?

[young spade]

looks something like this, you can see one of the sides is almost flat, so you get a lot of slope 0 points at the edge.

[spring oasis]

what is this

[young spade]

the surface you get out of this f(x,y) and restrictions

[spring oasis]

what?

the f(x,y) is a quadric surface

the A is a region of points

[young spade]

This is the surface made by f(x,y) = x^3 - y^2

The restriction of A basically makes a cutout of it

[spring oasis]

okay

[young spade]

the red surface is x^3 - y^2 without the restriction, the blue is with the restrictions

Remember, A is basically a projection from xy to the surface.

So you get only the points that xy coordinates satisfy the restrictions of A

[spring oasis]

what?

[young spade]

This is the curve that determines the frontier of A

right?

Is basically the half of an ellipse.

Well, this is the red and blue surfaces seen from above alongside that curve

[spring oasis]

how?

[young spade]

using this

[coral dragon]

amogus

[spring oasis]

how

[young spade]

Same thing we did last time

You get two inequations

$$y\geq\sqrt2 x$$
$$\frac{x^2}{2}+\frac{y^2}{4}\leq1$$

[young spade]

The first is everything above the line y = sqrt2 x

and the second is an ellipse with sqrt2 width and 2 height

[spring oasis]

my point is

we dont need the geometrical intuition

if you would have a picture of the situation you can see exactly where the maximums and minimums are

[young spade]

The picture is just for reference, i can do the algebra of it using the curves purely algebraically.

you get
c1=(t,sqrt2 t) [-1,1]
c2=(sqrt2cos t,2sin t) [arccos(1/sqrt2), arccos(1/sqrt2)+pi]
replace these back into f(x,y) = x^3 - y^2, to make it into F(t) and take a derivative with respect to t
solve for F'(t) = 0

[fiery marten]

The method of Lagrange multipliers also works with inequalities. In this case, instead of φ_k(x, y)=0 we have λ_k φ_k(x, y) = 0 with λ_k >= 0

For reference, it's often called Karush–Kuhn–Tucker theorem

[devout snow]

@spring oasis Has your question been resolved?

[spring oasis]

yes but this is killing a fly with a bazuka

[spring oasis]

in order to remove the case work and have an easier time with the system solving, we can do the following, use lagrange multipliers to look for possible extreme values in the frontier of the region and in the interior of the region use regular old way of finding extrema, by finding when the gradient is zero and using determinant of the hessy, that way the case work of the kurash kuhn tucker problem reduces to just finding critical points of f(x,y), and those critical points we evaluate them to see if they either fall inside the interior of the region, outside the region, or in the frontier of the interior, if it falls outside the region we discard them, if it falls in the frontier of the region then we ignore them because we are guaranteed to find them again when we use lagrange multipliers to analyze the frontier of the region, and if they fall inside the interior of the region then we save them because they can be used for later, what I am saying is, is we do this, then the problem becomes a regular problem of finding extrema for f + lagrange multiplier problem for two equality constraints insteaad of a kurash kuhn tucker problem when we have to do multiple casework for the multipliers, also, there is guaranteed that the maximum and minimum both lie inside the region because the region is a compact set, and f is a continous function, so using weierstrass theorem we know this global maxima and global minima will be attained either in the interior of the region or on the boundary

@fiery marten

[fiery marten]

This is exactly the same amount of case work though

KKT is almost literally what you've just described

When both λ1=λ2=0 we're essentially in the situation where we just find extrema of f without any constraints, and then we check whether they are inside the region or not

When some λ is not zero, it corresponds to the situation when we investigate the boundary

• Extrema in the interior <=> λ1=λ2=0
• On bounding curve № 1 <=> λ1>0, λ2=0
• On bounding curve № 2 <=> λ1=0, λ2>0
• On the intersection of bounding curves <=> λ1>0, λ2>0

But of course you can choose the most comfortable approach for you

I'm just explaining that KKT is essentially the same thing you described

[spring oasis]

we can also find the intersection outside lagrange by finding the intersection between the boundaries

[fiery marten]

Yes, and this is what KKT leads to when λ1>0, λ2>0

In that case our system of equations contains φ1(x,y)=0, φ2(x,y)=0

Which is precisely the intersection of 2 curves

[spring oasis]

sure, but instead of solving for multiple use cases we reduce the case work is what I am saying, just setting the boundaries of the region equal to each other outside lagrange instead of analyzing the cases λ1>0, λ2>0 and solving the system which is arguibly harder

[fiery marten]

How is it harder if it's the same thing?

Like, in both approaches you just solve a system of 2 eqs in 2 vars

[spring oasis]

well in one case we have lambdas lying around, and if we find the intersection between the boundaries of the region outside lagrange then we have no lambdas lying around

life is already hard by itself we dont have to make it harder is what I am saying

[fiery marten]

Yes, but then you have to manually check whether each intersection point attains an extrema or not
This is the same amount of work as to find lambdas and check their signs

[fiery marten]

I get it, you're convinced that KKT is hard, but what I'm saying is that it's the same exact thing as the one you want to do

It's just that previously you didn't see the geometry behind the case work

And now you saw it, and realized that life isn't hard afterall

[spring oasis]

life is hard choosing whatever method we choose

all I am saying is, messing with lagrange multipliers is already hard by default

using kkt is just the cherry on top

but we can do in whatever way you like, I dont really mind anymore, I just feel this problem is being overcomplicated from the start

if we do the interior + boundary + intersection way, then we separate the problem into three smaller problems

if we use kkt evertything is smashed together into one big and hard system to solve

@fiery marten is what I am saying

[fiery marten]

4*
We have 2 different curves forming the boundary

[fiery marten]

Okay, feel free to use the approach you're most comfortable with

[spring oasis]

i still need handholding

[spring oasis]

I have spent long time trying to understand this problem but never really solved it in the first place

[fiery marten]

Okay, start with finding the extrema of f without any constraints

[spring oasis]

fx = 3x^2
fy = -2y

the gradient is zero at (x,y) = (0,0)

and this point is inside the compact region

furthermore, is it a maxima or a minima? @fiery marten

fxx = 6x
fyy = -2
fxy = 0
det(H(x,y)) = fxx * fyy - (fxy)^2

[fiery marten]

The task doesn't ask us to specify the type of the extrema

[spring oasis]

ok

[fiery marten]

Although I don't know the exact translation of the task

[spring oasis]

det(H(x,y)) = (6x)(-2) - 0^2

[woven radish]

Renato

[spring oasis]

@fiery marten

[fiery marten]

Thanks

Then we don't need hessians

[spring oasis]

why not

[fiery marten]

(0,0) is an extremum anyways
Maybe local max, maybe local min, maybe saddle

But it's an extremum either way

[spring oasis]

we only care about evaluating that on f(x,y) and comparing it with the other extremas

[fiery marten]

Ah, so "absolute extrema" means global min and global max?

Yes, then it's sufficient to compare f at every critical point against every other critical point

[spring oasis]

i think so, yes

[spring oasis]

how to proceed with the other shit

[fiery marten]

If I understood correctly, you wanted to consider different boundary curves separately

Let's see whether we have any extrema on the ellipse

Being on an ellipse means 1/2 x² + 1/4 y² - 1 = 0

Now do Lagrange multipliers with one equality constraint

L = x³ - y² + λ(1/2 x² + 1/4 y² - 1)

Solve dL/dx=0, dL/dy=0, dL/dλ=0

[spring oasis]

is doing lagrange multipliers with one equality constrained any simpler than iwth 2 constraints? and what about the intersection points of the boundaries of the region

[spring oasis]

i thought we needed a lambda0 for the lagrangian

[fiery marten]

Depends on which constraints you mean
If you want both of them to be equality constraints, like φ(x,y)=0, ψ(x,y)=0, then it corresponds to corners of the region, and we have only 2 corners so it's easier to find them right away
If you mean inequality constraints, then we're back at KKT

[spring oasis]

equalitiy constrains i mean

forget about inequality constraints now that we analyzed the interior of the region as of now with the first step we did

[fiery marten]

If λ0=0, then we don't get anything meaningful in this particular case, so we can assume λ0≠0 and immediately assume λ0=1 (we can skip λ0=-1 because it doesn't affect the equations; we can just multiply all equations by -1; also here λ is any real number, we don't have non-negativity)

[spring oasis]

ok lets continue

[fiery marten]

If both equality constraints hold, then we're in one of the corners
At this point just calculate the corners, there're just 2 of them

[spring oasis]

help please

[fiery marten]

I think you found them the other day

(1, sqrt(2)), (-1, -sqrt(2))

[spring oasis]

how do I do this

[fiery marten]

If one curve is defined by φ(x, y)=0, and the other one is defined by ψ(x, y)=0, how do you find the intersection points of these curves?

[spring oasis]

well

we jusst set them eaualq to eachj otjer

correct ? @fiery marten

[fiery marten]

Yes, and we set them equal to zero

[spring oasis]

2x^2 + y^2 -4 = sqrt(2)x - y = 0

[fiery marten]

Yes, so what are the boundary curves?

[spring oasis]

@fiery marten

[fiery marten]

1/2 y² but yes

This is a system of 2 equations with 2 vars

[spring oasis]

2x^2 + y^2 -4 = sqrt(2)x - y = 0

[fiery marten]

Correct

[spring oasis]

,, \begin{cases} 2x^2 + y^2 - 4 = 0 \ \sqrt{2}x - y = 0 \end{cases}

[woven radish]

Renato

[spring oasis]

@fiery marten help

[fiery marten]

Express y in terms of x, sub into first equation, solve for x as a standard quadratic equation

[spring oasis]

2x^2 + 2x^2 - 4 = 0

4x^2 - 4 = 0

x^2 - 1 = 0

(x-1)(x+1) = 0

@fiery marten

[fiery marten]

Amazing

[spring oasis]

now wat

[fiery marten]

Write down the corners

(x, y)

[spring oasis]

how

[fiery marten]

You wrote (x-1)(x+1)=0 and y=sqrt(2)x

[spring oasis]

ok

[fiery marten]

2 solutions then

[spring oasis]

x1 = 1
x2 = -1
y1 = sqrt(2)
y2 = -sqrt(2)

[fiery marten]

Correct

[spring oasis]

(x1,y1) = (1, sqrt(2))
(x2, y2) = (-1, -sqrt(2))

@fiery marten

this?

[fiery marten]

Yes, so now we have 3 candidates

The third one is (0,0)

[spring oasis]

candidates = {(0,0), (1, sqrt(2)), (-1, -sqrt(2))}

[fiery marten]

We care only about a point where f has the greatest value among 3 and the lowest value

[spring oasis]

we still need to do lagrangian magic

[fiery marten]

Yes, but we can already discard one candidate

[spring oasis]

care to elaborate on that?

also, maybe shouldnt we do the discarding at the end? lets try to be organized

[fiery marten]

f(0, 0)=0
f(1, sqrt(2))=-1
f(-1, -sqrt(2)) =-3

[fiery marten]

Okay

I was trying to be optimized

[spring oasis]

this one is discardede right away
f(1, sqrt(2))=-1

[fiery marten]

Yep

Anyways, let's do some Lagrange multipliers

[spring oasis]

help

[fiery marten]

.

[spring oasis]

shouldnt we make the lagrangian have two equality constraints?

[fiery marten]

We have a function of 3 variables, we want to find where all 3 partial derivatives are zero

[fiery marten]

No, why?

[spring oasis]

theres. two constraints in the region

[fiery marten]

These?

You've just finished dealing with them

[spring oasis]

care to elaborate before we start with lagrange

[fiery marten]

Okay, so let's see what are these two constraints you're talking about

1. 1/2 x² + 1/4 y² - 1 = 0

2. y - sqrt(2)x = 0

I didn't write inequalities because we collectively decided to avoid them

Because they'd lead to KKT

Instead we decided to explicitly separate the problem into 4 cases

[spring oasis]

I mean the two equalities

[fiery marten]

1. Interior extrema [done]
2. Boundary extrema, curve № 1 (ellipse)
3. Boundary extrema, curve № 2 (line)
4. Boundary extrema, corners [done]

[spring oasis]

forget about inequalities alltogether

[fiery marten]

Yes, so we're here:

Do you see that if these 2 equalities hold simultaneously, we're sitting in the corners?

[fiery marten]

And here are these equalities, you wrote them yourself:

[spring oasis]

this is why we do lagrangian with just one constraint equality?

[fiery marten]

Yep

We manually investigated the case with 2 equalities, without lagrange

[spring oasis]

wait, but we are just going to do one lagrangian

[fiery marten]

Two, here's the second one:
x³ - y² + λ(sqrt(2)x - y)

[spring oasis]

or two separate lagrangians

[fiery marten]

Yes!

[spring oasis]

ok ok, lets proceed with the first one

[spring oasis]

.

L = x³ - y² + λ(1/2 x² + 1/4 y² - 1)

@fiery marten

[fiery marten]

Indeed

[spring oasis]

now what

[fiery marten]

As prescribed by the method, find where dL/dx=0 & dL/dy=0 & dL/dλ=0

[spring oasis]

Lx = 3x^2 + λx = 0

Ly = -2y + λ(1/2)y = 0

[spring oasis]

As prescribed by the method, find where dL/dx=0 & dL/dy=0 & dL/dλ=0
how do I even do the last one

Lλ = 1/2 x² + 1/4 y² - 1 = 0

[fiery marten]

Treat λ like a variable, just like you did with x and y

[spring oasis]

this?

[fiery marten]

Yes

Very nice

[spring oasis]

Lx = 3x^2 + λx = 0
Ly = -2y + λ(1/2)y = 0
Lλ = 1/2 x² + 1/4 y² - 1 = 0

now what

[fiery marten]

It's a system of 3 equations with 3 variables

Let's solve it

[spring oasis]

this is the part that sucks ass

[fiery marten]

Probably the easiest way is to express λ in terms of x using the first equation

Then sub to the second and express y in terms of x

Then third becomes a quadratic

[spring oasis]

,, \begin{cases} 3x^2 + \lambda x = 0 \\-2y + \frac{1}{2} \lambda y = 0 \\ \frac{1}{2}x^2 + \frac{1}{4} y^2 - 1 = 0 \end{cases}

[woven radish]

Renato

[spring oasis]

isnt this assuming x is not 0 and y is not zero

@fiery marten

ah, we already captured the point (x,y)=(0,0) so we dont give a damn

[fiery marten]

Correct, x=0 is a separate case

[spring oasis]

no because if x = 0 then y = 0 if I am not missing any shit

[fiery marten]

Yes, we don't care about x=y=0 anymore, but there're cases when x=0, y≠0 and vice versa

[fiery marten]

From the third equation y can't be 0 if x is 0

Otherwise the eq doesn't hold

[spring oasis]

ok

life sucks

[fiery marten]

Not too much

If x=0 then y=±2 and we can move on
If y=0, then x=±sqrt(2)

[spring oasis]

λx = -3x^2
λ = -3x

[spring oasis]

care to elaborate?

[fiery marten]

Sub x=0 into third eq and solve for y

[spring oasis]

y = 2 or y = -2

[spring oasis]

i think life is good now that you mention it

λx = -3x^2
λ = -3x

-4y + λy = 0

λy = 4y

λ = 4

4 = -3x

x = -4/3

@fiery marten

[fiery marten]

Yep that works

Let's find y though

Using the third eq

[spring oasis]

2x^2 + y^2 - 4 = 0

2(16/9) + y^2 - 4 = 0

y^2 = -(32/9) + 4

y^2 = -32/9 + 36/9

y^2 = 4/9

y = +- 2/3

@fiery marten

[fiery marten]

We now obtained a shit ton of new candidates

(0, ±2), (±sqrt(2), 0), (-4/3, ±2/3)

[spring oasis]

candidates = {(0,0), (1, sqrt(2)), (-1, -sqrt(2))}

[spring oasis]

how did this shit happen? and when

[fiery marten]

First let's discard those that don't belong to our region

[fiery marten]

We've solved the system

These are all the solutions (λ omitted)

[spring oasis]

help please

[fiery marten]

I really don't understand what's the issue rn

You probably lost track of events

[spring oasis]

yes, when did we captured this?

[spring oasis]

.

[fiery marten]

Brb

[spring oasis]

what about (±sqrt(2), 0)?

@fiery marten

[spring oasis]

here

[spring oasis]

yes

,w 2x^2 + y^2 <= 4 and y >= sqrt(2)x , for (x,y)=(0,2)

[woven radish]

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[spring oasis]

,w 2x^2 + y^2 <= 4 and y >= sqrt(2)x , for (x,y)=(0,-2)

[woven radish]

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[spring oasis]

,w 2x^2 + y^2 <= 4 and y >= sqrt(2)x , for (x,y)=(sqrt(2),0)

[woven radish]

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[spring oasis]

,w 2x^2 + y^2 <= 4 and y >= sqrt(2)x , for (x,y)=(-sqrt(2),0)

[woven radish]

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[spring oasis]

,w 2x^2 + y^2 <= 4 and y >= sqrt(2)x , for (x,y)=(-4/3, 2/3)

[woven radish]

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[spring oasis]

,w 2x^2 + y^2 <= 4 and y >= sqrt(2)x , for (x,y)=(-4/3, -2/3)

[woven radish]

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[spring oasis]

candidates = {(0,0), (1, sqrt(2)), (-1, -sqrt(2))}

,w 2x^2 + y^2 <= 4 and y >= sqrt(2)x , for (x,y)=(0,0)

[woven radish]

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[spring oasis]

,w 2x^2 + y^2 <= 4 and y >= sqrt(2)x , for (x,y)=(1, sqrt(2))

[woven radish]

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[spring oasis]

,w 2x^2 + y^2 <= 4 and y >= sqrt(2)x , for (x,y)=(-1, -sqrt(2))

[woven radish]

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[spring oasis]

candidates = {(0,0), (1, sqrt(2)), (-1, -sqrt(2)), (-4/3, -2/3), (-4/3, 2/3), (-sqrt(2),0), (0,2)}

f(x,y) = x^3 - y^2

,w evaluate x^3 - y^2 with (x,y)=(0,0)

[woven radish]

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[spring oasis]

,w x^3 - y^2 with (x,y)=(1, sqrt(2))

[woven radish]

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[spring oasis]

,w x^3 - y^2 with (x,y)=(-1, -sqrt(2))

[woven radish]

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[spring oasis]

,w x^3 - y^2 with (x,y)=(-4/3, -2/3)

[woven radish]

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[spring oasis]

,w x^3 - y^2 with (x,y)=(-4/3, 2/3)

[woven radish]

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[spring oasis]

,w x^3 - y^2 with (x,y)=(-sqrt(2),0)

[woven radish]

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[spring oasis]

,w x^3 - y^2 with (x,y)=(0,2)

[woven radish]

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[spring oasis]

candidates = {(0,0), (1, sqrt(2)), (-1, -sqrt(2)), (-4/3, -2/3), (-4/3, 2/3), (-sqrt(2),0), (0,2)}
f(x,y) = x^3 - y^2

f(0,0) = 0
f(1, sqrt(2)) = -1
f(-1, -sqrt(2)) = -3
f(-4/3, -2/3) = -76/27
f(-4/3, 2/3) = -76/27
f(-sqrt(2), 0) = -2sqrt(2)
f(0,2) = -4

@fiery marten

,calc 76/27

[woven radish]

Result:

2.8148148148148

[spring oasis]

,calc 2sqrt(2)

[woven radish]

Result:

2.8284271247462

[spring oasis]

x³ - y² + λ(sqrt(2)x - y)

candidates = {(0,0), (0,2)}

x³ - y² + λ(sqrt(2)x - y)

Lx = 3x^2 + λsqrt(2)
Ly = -y - λ

Lλ = sqrt(2)x - y

,, \begin{cases} 3x^2 + \lambda\sqrt{2} = 0 \ -y - \lambda = 0 \ \sqrt{2}x - y = 0 \end{cases}

[woven radish]

Renato

[spring oasis]

if x = 0 then y = 0 I think

if y = 0 then x =0. I think

λsqrt(2) = -3x^2
λ = (-3x^2) /sqrt(2)

-y = λ

(-3x^2) /sqrt(2) = -y

3x^2 / sqrt(2) = y

sqrt(2)x - (3x^2 / sqrt(2) ) = 0

2x - 3x^2 = 0

x(2-3x) = 0

x1 = 0
2-3x = 0
2 = 3x
x2 = 2/3

,w (-3x^2) /sqrt(2) = -y where x = 2/3

[woven radish]

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[spring oasis]

(x1,y1) = (0, 0)
(x2, y2) = (2/3, 2sqrt(2)/3)

,w x^3 - y^2 where (x,y)=(2/3, 2sqrt(2)/3)

[woven radish]

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[spring oasis]

,calc -16/27

[woven radish]

Result:

-0.59259259259259

[spring oasis]

candidates = {(0,0), (0,2)}

,w x^3 - y^2 where (x,y)=(0,0)

[woven radish]

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[spring oasis]

,w x^3 - y^2 where (x,y)=(0,2)

[woven radish]

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[spring oasis]

@fiery marten

[spring oasis]

https://www.geogebra.org/3d/nhnz6sus

@fiery marten

check this simulation I did for the situation

[spring oasis]

hey friend, i really appreciate your help....

you had helped me with some hard ass elementary linear algebra questions aswell last year

like some change of basis for linear transformation shenanigans

this year with lagrange multipliers aswell, you really helped me a ton today

IMHO, separating this KKT problem into simpler subproblems that we can solve easier is the best approach instead of using kkt theorem

you say its the same, but in the other scenario we had a bunch of λk laying around

it really made everything simpler when we do it like this, it becames a normal lagrange multipliers problem instead of a KKT problem

other than that, I have been struggling with this kind of problems for such a long time sir, I really appreciate the time you took for helping me out here

just like I was having troubles with change of basis for linear transformations last year in pre university, after someone explains it to you once clearly I think everything becomes clear and everything becomes simple aswell, but when you are on the other side of the wall things can really look dark, if you know what I mean

[fiery marten]

You probably meant -2y - λ = 0

Other than that it looks like you really have some grasp of the method now, congrats

I will have a look at your illustration in a minute

[fiery marten]

Of course, learning any topic for the first time can be challenging and confusing at times. It can get even worse if one has certain gaps in required preliminary knowledge

Which seems to he the case with you, tbh. I suggest revising the geometric meaning of systems of equations

[spring oasis]

Lx = 3x^2 + λsqrt(2)
Ly = -2y - λ
Lλ = sqrt(2)x - y

[spring oasis]

wdym?

,, \begin{cases} 3x^2 + \lambda\sqrt{2} = 0 \ -2y - \lambda = 0 \ \sqrt{2}x - y = 0 \end{cases}

[woven radish]

Renato

[zinc flame]

jesus christ what sort of forbidden conjecture is being proved here? this channel has been open for 3 days lol

[spring oasis]

-3x^2 / sqrt(2) = λ

λ = -2y

-3x^2 / sqrt(2) = -2y

3x^2 / 2sqrt(2) = y

[fiery marten]

Renato is learning optimization with constraints

[spring oasis]

sqrt(2)x - 3x^2 / 2sqrt(2) = 0

2x - 3x^2 = 0

x(2 - 3x) = 0

x1 = 0

2-3x = 0

2/3 = x2

[fiery marten]

Remember we discussed today that intersection of 2 curves = solutions of a system of 2 equations? I'm not sure you understood that fully

If that's the case, my suggestion holds: revise/learn what systems of equaitons do gemoetrically

That'll be very helpful in the future

[fiery marten]

very nice

[spring oasis]

geometrically no I dont follow, but for example if you find the points that constitute the intersection of two curves, then those points satisfy those two equations, which is equivalent to finding the solution set of a system of two equations, no? maybe I am tripping . . .

[spring oasis]

(x1,y1)=(0,0)

,w 3x^2 / 2sqrt(2) = y where x = 2/3

[woven radish]

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[spring oasis]

pretty much the same I got earlier

strannge

[fiery marten]

You're not tripping, this is correct intuition for 2 curves on a plane

y = sqrt(2) x
lambda = -2 sqrt(2) x
3 x^2 - 4 x = 0
Hence x=0 or x=4/3

If x=0, then y=0 which we already considered
If x=4/3, then y = 4 sqrt(2) / 3

[spring oasis]

strange

wait a second

,w 2x^2 + y^2 <= 4 and y >= sqrt(2)x , for (x,y)=(4/3,4 sqrt(2) / 3)

[woven radish]

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[spring oasis]

luckily is discarded

hehehehehehehehehhe

because it is not inside the region, no?

anyways this should be it @fiery marten no more solutions exists for x,y in this system

its either (0,0) or (4/3, 4sqrt(2)/3) for this system

so thats it

[fiery marten]

correct

[fiery marten]

seems so, yeah

[spring oasis]

I mean, the only thing that is still dangling around for me is why we do two separate langrangians instead of one lagrangian with two constraints

[fiery marten]

You should ask yourself what lagrange method does in that particular case

When we introduce two constraints like g1(x, y)=0, g2(x, y)=0, and apply lagrange blindly, what are we doing, essentially?

We're asking "dear Lagrange, we're trying to find min and max of f(x, y) on a set of points carved by these two equations: g1(x, y)=0, g2(x, y)=0"

Since our equations define an ellipse and a line, I'm pretty sure Lagrange would give us a confused look and say something like "bro, these two curves intersect at only 2 points, just find them"

And he would be damn right

Because equations 1/2 x^2 + 1/4 y^4 = 1 and y = sqrt(2) x yield precisely 2 points

So lagrange with 2 lambdas is an overkill

[spring oasis]

I see

[spring oasis]

yes

we can still do it though, is just that doing lagrange with more than one constraint yields a harder linear system because we have more equations and more unknowns

[fiery marten]

Technially yes

[spring oasis]

i mean we still have x and y, but now we have λ1 and λ2

[fiery marten]

yep

Although lambda1 and lambda2 don't have any constraints (like being non-negative), and the equations including them will be linear in lambda1, lambda2. And this linear system will be non-degenerate and hence have a solution

So we can just forget about lambda1 and lambda2

And lagrange with 2 constraints just becomes a task of finding intersections of curves

So once again we see how different methods do the same thing

A common thing in math

¯_(ツ)_/¯

[spring oasis]

what if my region has 3 constraints, the intersection of the three might be empty but there could be intersection between two of them and such

[fiery marten]

As long as you're in 2d you can avoid lagrange multipliers with 2+ constraints, because you can always just calculate all intersection points instead

However, if you're in 3d+, then dealing with 2 constraints using lagrange method is reasonable

[spring oasis]

the pairwise intersections?

[fiery marten]

yes

[spring oasis]

it is possible that the triple intersection is empty

[fiery marten]

It's almost always the case

[spring oasis]

but the pairwise intersections are not

[fiery marten]

Of course, consider a circle and 2 lines that intersect it

As long as these lines do not meet on the circle the triple intersection is empty

[spring oasis]

your point?

[spring oasis]

the thing is, is strange as to why we look for intersection between the two constraints when they are equal to zero when we have two constraints but when we have three or more we look for the pairwise intersections between the constraints when they are equal to zero

any logical explanation to that?

[fiery marten]

In dimension 2, two curves typically intersect at a finite number of points. If you intersect more than 2 curves, you will end up at a smaller finite set. Because adding a third curve is basically like "which of these points from the finite set belong to the 3rd curve?"
Basically what I'm saying is that in 2d intersections of 3 curves don't yield new points. They only yield subsets of intersections of 2 curves

Amazingly, this generalises to higher dimensions

In n-dimensional case, intersection of n hypersurfaces typically is a finite set of points. Adding more surfaces to the intersection will just reduce this finite set (maybe even make it empty) and won't give us any new points

[spring oasis]

it will be empty if and only if the pairwise intersections between the curves are empty

[fiery marten]

no

Not if and only if

If pairwise intersections are empty, then of course triple intersection is empty

But not vice versa

[spring oasis]

i see

[spring oasis]

I will look for exercises with 3 or more constraints to practice this topic

but most likely in the exam two constraints will appear

.solved

[devout snow]

Channel closed

Closed by @spring oasis

Use .reopen if this was a mistake.

[spring oasis]

I appreciate it 🙏

[devout snow]

Available help channel!

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[drifting relic]

Find a number, 𝑎, such that 𝑎^{2} ≡ 2 mod 5, or show that no such number exist.

I just want to make sure I'm doing this correctly? You should go through all the possible remainders and check to see if which one fits, right? So r = {0, 1, 2, 3, 4}? Thank you.

[topaz beacon]

yup, nothing wild in this

[drifting relic]

Ok, thank you, sorry, I'm really bad at math, so I just want to make sure I'm not doing this incorrectly.

I'm also wanting to make sure about this, am I correct in saying that these are congruent and therefore you can do what I'm doing in the second picture? Because if r divides by 14 and the remainder's 5, it has to be the same as 5 mod 14 because that's also 5.

[late scaffold]

if r leaves a remainder of 5 on division by 14, r is indeed congruent to 5 mod 14, yes.

[drifting relic]

Ok, thank you. For some reason it's just hard for me to wrap my head around what congruence is.

[topaz beacon]

congruence is a fancy way of saying remainders

it might help if you think $r \equiv 5 (\mod 14)$ is the same thing as $r=14k+5$ for some integer $k$

[drifting relic]

And finally:

Find the lcm and gcd of the following numbers by the Euclidean Algorithm

I think I got this right, unless I screwed up how the Euclidean algorithm works.

[woven radish]

726=|4 step 3D lattice trails|

[drifting relic]

Thank you, let me paste that in a note.

[topaz beacon]

seems like they all look fine

[drifting relic]

Ok, thank you so much.

[topaz beacon]

,w gcd(355,603)

[woven radish]

Results provided by WolframAlpha

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

Requested by garlicbredfries

[drifting relic]

Thank you for checking too

[topaz beacon]

ok just wanted to make sure with that one

[drifting relic]

Yea, I think I got it wrong beforehand, when I was looking back at it to double check I realised I wrote 365 for some reason and did the Algorithm. 🤣

Ok, thank you very much you two. Have a good day/night.

.close

[devout snow]

Channel closed

Closed by @drifting relic

Use .reopen if this was a mistake.

[topaz beacon]

you too

[devout snow]

Available help channel!

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[sand knot]

hey, ive just got a few specific questions on my study guide for my upcoming exam. Can someone hop in vc to answer a few questions?

[midnight echo]

We don't do VCs. If you have a specific question, send a clearer photo and someone will help you

[sand knot]

ok

this is the clearest photo

since its large

[old seal]

Which question

[midnight echo]

No one's not gonna walk you through 12 multi-part questions. If you have a specific question, zoom in on it, explain what you've tried so far and where you got stuck, and what you need help with.

[sand knot]

uh

can we do #5 ?

[old seal]

write out f(x) i can't see it

[midnight echo]

Yes, now "zoom in on it, explain what you've tried so far and where you got stuck, and what you need help with."

[sand knot]

ok one sec

[old seal]

$\frac{x}{x^2 + 1}$

[woven radish]

Miyamoto

[old seal]

do you know how to find the derivative

[sand knot]

yea

[old seal]

do you know how to find the critical points

[sand knot]

i think i just dont know what to do after

uhmm

not with the derivative i have]

can we do whiteboard?

[old seal]

what is whiteboard

[sand knot]

oh nvm

its disabled

it just shows shared writing

wait ill send a pic of my derivative

[old seal]

show us all the derivative

show us what you got

[old seal]

okay did you do the quotient rule to get that

[sand knot]

yes

[old seal]

nice

[sand knot]

uh

[old seal]

u made a mistake in your algebra skills

[sand knot]

oh

[old seal]

it is a serious mistake

if you want to pass you must correct this mistake

[sand knot]

well this is my full workout

idk what i did wrong

waittt

[old seal]

around the middle you have 2x*x = 2x

[sand knot]

yea i just realized that

[old seal]

yes

[sand knot]

it should be x^2

2x^2

[old seal]

yeah you know

[sand knot]

k one sec

there we go

oh

i shoudl simplify

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

[old seal]

yeah

you have good algebra skills now

you must have had a good teacher

[sand knot]

i guess

how do i set this all to 0???

or do i just do the numerator

[old seal]

The denominator cannot be negative

[sand knot]

it isnt

[old seal]

yes it isnt

for this function, you would solve the numerator for 0

[sand knot]

oh

why do we not touch the denomintor?

[old seal]

because the denominator is always positive for any value of x

[sand knot]

o

ok so for my sign chart i have x<1 is negative and x>1 is positive

[old seal]

The interval is from negative infinity to positive infinity for this question

[sand knot]

mhm

[old seal]

so in the case of x^2 = 1
You would square root both sides to remove the square on x

but the issue is that you need to include all values which make this true

if x^2 = 1 then inputting -1 into this x value is still true

because -1 * -1 = positive 1

[sand knot]

oh

so +- 1

[old seal]

yeah

[sand knot]

i forgor

[old seal]

it would be correct if the interval was not including -1

[sand knot]

so how owuld i put that on a sign chart if i have -inf and inf

i cant plug any number in

[old seal]

tbh idk what a sign chart is . Do you want to test for values between these and show if the result is negative or positive y value?

[sand knot]

yes

[old seal]

you have the critical points you could chose arbitrary values between (-inf, -1) and (-1,1) and (1,inf)

for example -2, 0, and 2

[sand knot]

ok

[old seal]

this will tell you the slope of f(x) at those points

[sand knot]

uhm

oh

[old seal]

what is it

[sand knot]

hm

when i plug in -2 i have 3/25

then that is pos?

[old seal]

yes because any value above 0 is positive

but you calculated wrong it should be -3/25

[sand knot]

oh

[old seal]

does it make sense that the derivative of f(x) results in a function that gives you the slope values of f(x)

[sand knot]

yes

i think i understand now

thanks

[old seal]

your welcome

[sand knot]

i think i can try the rest

[old seal]

okay nice

[sand knot]

ok cya

[old seal]

cya

[sand knot]

.close

[devout snow]

Channel closed

Closed by @sand knot

Use .reopen if this was a mistake.

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[torn loom]

Need help with this calculus problem

Problem 5

I know that e^x doesn’t have an absolute min or max, but I think I need to prove why it doesn’t.

Where do I start?

[trail eagle]

Well have you taken the derivative?

[midnight echo]

Start by listing the critical points of g, as the question states.

[torn loom]

Not yet

[trail eagle]

That would be a start then

Then find critical points and classify them using sign tables or the second derivative.

[torn loom]

the derivative is e^x - 1 correct?

[trail eagle]

Yes

[torn loom]

Ok

Then I set that = to 0?

I forgot how this works. I have a test on integrals next week but these are the review questions assigned

is it x = 0?

[fossil locust]

great, so to find whether that is a local minimum, local maximum, or neither, you could find the 2nd derivative and then check its sign

[torn loom]

W helpers, yall are on it today hahaha

[torn loom]

alright so I derive e^x-1?

[fossil locust]

yep

[torn loom]

isn't that just e^x

[fossil locust]

yeah

and you also need to sub in x = 0

[torn loom]

then set that to 0?

[fossil locust]

no

[torn loom]

Hmm

Ok so its just 1?

[fossil locust]

yes

[uneven coral]

(How did I never find this🤌😭)

[fickle chasm]

omg i remember ts from 7th grade

it pmo sm

[torn loom]

give me a minute

[torn loom]

So its concave up?

So its a min?

[fossil locust]

yep!

[torn loom]

at 0,1?

[fossil locust]

okay, but note that it's asking you whether there is an absolute max and min

[fossil locust]

yes

[torn loom]

Well I don't think its an absolute min, because its just a change in direction

as I see on the calculator

[fossil locust]

what do you mean by that?

[torn loom]

like both sides go up to infinity but the "min" is the only a change in direction

idk if that does anything

like there is only one change in direction

The graph is kinda like this

[fossil locust]

yeah! okay so you've plotted the graph

whether that point is an absolute minimum is asking you if there's any lower y-value

[torn loom]

Nah I used a calculator, because our professor allows calculators on the the test

[torn loom]

well no, so it is an absolute min

[fossil locust]

yep

[fossil locust]

ah that's good

[torn loom]

but it also says explain why they fail to exist

which confuses me

[fossil locust]

so the red point isn't an absolute minimum, just to clarify

[torn loom]

ye its a local min

[fossil locust]

if there is an absolute minimum, find it

so you've found that already

[torn loom]

Oh or give reasons why it doesn't exist

LOL

[fossil locust]

but as for the absolute maximum, that doesn't exist (it's +infinity)

yeah so you need to give a brief reason for why

[torn loom]

wait so do I explain why it exists?

or do I just give the answer

the wording on this problem is confusing me

Ok I got it

thank you!

.close

[devout snow]

Channel closed

Closed by @torn loom

Use .reopen if this was a mistake.

[torn loom]

I'll be here

[fossil locust]

explain why the absolute maximum doesn't exist

[torn loom]

I see u typing

[torn loom]

Oh okok

I said

[fossil locust]

something like $\lim_{x \to \infty} e^x - 1 \to + \infty$ (the derivative goes to $+\infty$), or '$e^x$ grows much faster than $x$', should do

[torn loom]

The absolute max doesn't exist because both sides of the function rise to infinite and never change direction where the derivative is = to 0

[fossil locust]

that actually doesn't answer the question

if you zoom out enough on the graph, the function might not actually be increasing

it might have some turning point really far away

[late scaffold]

(hey, I think the channel's not reopened yet, fyi)

[fossil locust]

.reopen

[devout snow]

✅ Original question: #help-27 message

[woven radish]

south

[torn loom]

.close

[devout snow]

Channel closed

Closed by @torn loom

Use .reopen if this was a mistake.

[soft umbra]

👀

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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[scenic hollow]

I really just don’t understand how to do these

[keen sonnet]

for q5 and 6, you are given two pieces of info. an xy coordinate of your graph and a derivative (slope of tangent line of graph) at a point

your task is to approximate two values of the function around the given point

have you learned what linearization is?

[rotund umbra]

Q.5 and Q.6 are telling you aproximately which line (y = mx + c) the function looks like

[scenic hollow]

nope

[keen sonnet]

are you familiar with the equation of a line?

hint: shuba mentioned it

[scenic hollow]

y=mx+c?

[keen sonnet]

yes, and do you know what m and c stand for?

[scenic hollow]

M = slope and c I don't remember what it stands for, but I remember what it does, kinda

but how do i solve?

[fickle chasm]

i used mx+b lol

[scenic hollow]

?

[rotund umbra]

1. use the information given to aproximate the straight line the function looks like
2. use that line to aproximate the answer

[scenic hollow]

i dont understand

sorry

[uneven coral]

You've been given y and m at a certain point, you can find the equation of a line throught that

[rotund umbra]

Consider f(x) = x^2, but we're only given f(1) = 1 and f'(1) = 2. I can say the function looks like g(x) = 2x - 1 near x = 1, so f(1.1) is aproximately g(1.1) = 2*1.1 - 1 = 1.2, which it indeed is f(1.1) = 1.1^2 = 1.21.

[uneven coral]

You can also use $\frac {y2 - y1}{x2 - x1} = m$ to find f(1.9) and f(2.1)

[woven radish]

doctorstrangejr

[scenic hollow]

.close

[devout snow]

Channel closed

Closed by @scenic hollow

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.