help-19

[pseudo mesa]

New here

[odd edge]

@modest kite Has your question been resolved?

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[fading nimbus]

I got to the part where

[fading nimbus]

But idk what to do after

[odd edge]

@fading nimbus Has your question been resolved?

[agile sinew]

hmm where'd you get this from?

did you expand the 10cos(x-30)?

[agile sinew]

$R\sin(\theta+\alpha) \equiv (R\sin\alpha)\cos\theta + (R\cos\alpha)\sin\theta$

[clever fjord]

neshi

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[fading nimbus]

Yes

[agile sinew]

kay so

[agile sinew]

we need LHS (left hand side) in terms of sines and cosines right?

we have

[fading nimbus]

Just sin no?

[agile sinew]

Mm no we can have cosines

[agile sinew]

because of this

and then we simplify it into our one phase shifted sin

[fading nimbus]

Okay

[agile sinew]

yup so

[fading nimbus]

And then compare coeffiecients or what

[agile sinew]

mhm!

[fading nimbus]

Okay ty

Someone else i asked told me to use aome formulas

[agile sinew]

wellll you should convert the $\sin2\theta$

[clever fjord]

neshi

[fading nimbus]

I did

[agile sinew]

you can use these formulas!

[fading nimbus]

I did all of that and got to 11 sin

[fading nimbus]

This

[agile sinew]

righttt so then we have

$R\sin\alpha = 11$

[clever fjord]

neshi

[agile sinew]

$R\cos\alpha = 5\sqrt3$

[clever fjord]

neshi

[agile sinew]

again you can use these formulas! but ill show you how to solve them via comparing coeffs

and both will lead you to the same answer 💛

sorry btw slow wifi & on mobile- bear with me 🙂

[agile sinew]

dividing first expression by second expression:

$\frac{R\sin\alpha}{R\cos\alpha} = \frac{\sin\alpha}{\cos\alpha} = \frac{11}{5\sqrt3}$

and the left hand side turns into (tan a) right?

Soo we can solve that by taking the inverse tan on both sides

$\alpha = \arctan\frac{11}{5\sqrt3}$

[clever fjord]

neshi

[agile sinew]

note that this is the same thing we would've gotten via the formula!

[fading nimbus]

Tyyy so much

[clever fjord]

neshi

[agile sinew]

np np

for R we do

$(R\sin\alpha)^2 + (R\cos\alpha)^2 = R^2(\sin^2\alpha + \cos^2\alpha) = R^2(1) = R^2$

[clever fjord]

neshi

[agile sinew]

and we replace Rsin(a) and Rcos(a) with the values we found for them

$(5\sqrt3)^2+(11)^2=R^2$

[clever fjord]

neshi

[agile sinew]

$R=\sqrt{(5\sqrt3)^2+(11)^2}$

[clever fjord]

neshi

[agile sinew]

again this is identical to the formula :)

you can use your calculator to find the values for both! (i don't have a calculator on me rn 😝)

[agile sinew]

$R=14\text{ i think!}$

[clever fjord]

neshi

[odd edge]

@fading nimbus Has your question been resolved?

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[olive kernel]

What is this question even asking for

[amber veldt]

you found the average velocity for the time interval [4, 4+0.5]

it's asking for a general formula for [4,4+h]

[olive kernel]

shoot nvm I figured it out.

Ive been subtracting wrong this whole time

.close

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[glass swallow]

I have a question about synthetic division. Say you have a polynomial being divided by (x - a) my textbook says to use -a as the value but it gave me the wrong answer for this specific question, (I got a, the right answer is b) so how is my textbook wrong? Do I actually use x = a?

[wooden python]

that... looks like a mistake in the book?

[tidal matrix]

-3 + -2 = 1

[wooden python]

you should be using x=a

their example is bad bc it has both 2 and -2 as roots

[tidal matrix]

yeah, you're dividing by the root, which is a

[late sinew]

Congrats on active haseeb

[glass swallow]

So even this example is wrong?

[wooden python]

oh they're... subtracting at every step

dear lord

it might be correct but it's bloody confusing is what it is

[late sinew]

[glass swallow]

Oh so that’s great 😭 going forward should I use x = a?

[wooden python]

yes

[late sinew]

Unless you're specifically asked to show your working with long divvy

Also, welcome to the server!

[glass swallow]

Thank you!!!

.close

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[tidal matrix]

hehe ty ^^ happy to be a help channel addict

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[shrewd trellis]

woudnt it be a point of inflexion if f"(x)=0?

[shrewd trellis]

using the second derivate test

[wooden cypress]

Not necessarily

Think about f(x) = x^4 at x=0

[shrewd trellis]

wdym

[wooden cypress]

What is f''(0)?

[shrewd trellis]

well i need a function first

[wooden cypress]

.

f(x) = x^4

[shrewd trellis]

its 0

[wooden cypress]

Good

Now draw the graph

Is it a point of inflection?

[shrewd trellis]

let me see

its

a

minimum turning point

wait

@wooden cypress

minimum turing point

[wooden cypress]

Good

It's a minimum point

But it's not a point of inflection

So to answer your original question, not always

[shrewd trellis]

but wait i didnt input 0

in the second derivative

12x^2

the answer is giving me 0

what does that mean

12(0)^2

[wooden cypress]

that just means $12\cdot0^2$

[clever fjord]

depression

[shrewd trellis]

oh wait

so point of inflexion is when

u let f"(x)=0?

and solve for x

[wooden cypress]

Not exactly

All points of inflection have a second derivative of 0

But not all points with a second derivative of 0 are points of inflection

You can find them all by solving that equation, yes, but you still have to verify that there aren't any false positives

Because $f(x)=x^4$ has no points of inflection

[clever fjord]

depression

[late sinew]

You can use the wavy curve method for f''(x)=0 to verify

[shrewd trellis]

im so confused rn wait

[shrewd trellis]

I dont nderstand from here

understand*

[wooden cypress]

As in $\text{"x is a point of inflection"}\Rightarrow f''(x)=0$

But $f''(x)=0\not\Rightarrow\text{"x is a point of inflection"}$

Wait that's not right, sorry mb it's the other way round

[clever fjord]

depression

depression

[wooden cypress]

That's better

[wooden python]

f''(a)=0 is a necessary but not sufficient condition for being a point of inflection @shrewd trellis

[shrewd trellis]

wdym by neccessary

[wooden python]

do you know what the word "necessary" means in ordinary English?

[shrewd trellis]

yes

[wooden python]

it means basically the same thing in math only it's ironclad and absolute

a point cannot possibly be an inflection point if f''(a)=0 is not satisfied,

but that condition alone is not sufficient

https://en.wikipedia.org/wiki/Necessity_and_sufficiency?wprov=sfla1

[shrewd trellis]

@wooden python I understand but

what is the second point saying

[wooden python]

if the second derivative is a bitch to calculate then just analyze the first for sign and make a slope table.

that's what it says.

[shrewd trellis]

how would it be hard to find

[wooden python]

well if you read the thing

it gives you one example of what might be "hard"

you can do this without me regurgitating the text at you, right?

[shrewd trellis]

I know that but I dont see connections

why would it be hard

like show me an example

My other question is

In the table, why are they using first derivative?

and in another example, they were using y"

[late sinew]

Because stationary and inflection points are different things

Stationary points are points where f'(x)=0, and can be local maxima or minima

Inflection points are points where f''(x)=0, and the concavity of the curve may change

[shrewd trellis]

I understand what u are saying u but im not forming any connections

[shrewd trellis]

It could be the same if inflextion point is y'=0

[late sinew]

?

Inflection point is not where y' = 0

Oh do you mean that y'' = 0, and y' also is 0 at that pt?

[shrewd trellis]

yea

i think so

[odd veldt]

They can happen at the same place, but they're different things

Like catching a cold and breaking your leg. They can happen at the same time without being the same sort of problem

[late sinew]

You can take y= x^3 for an example

This has a potential for (0,0) to be both inflection and stationary

Can you find out if that's true

[shrewd trellis]

@late sinew yea sure but i may have

confused my self with point of inflexion

like this

this is a horizontal point of inflexion but where is the point of inflexion?

or are they two different things

[warped glacier]

it's at x = 0

because if you look at this other table, when x = 0, y'' = 0

so they mean a point which is both:

1. horizontal gradient, so y' = 0
   and 2) an inflection point, so y'' = 0

[shrewd trellis]

wait so

horizontal point of inflexion

is the same as

point of inflexion?

[warped glacier]

no

not all points of inflection have horizontal gradient

for example

[shrewd trellis]

oh i see

makes sense

[warped glacier]

yeah

[shrewd trellis]

so what does the first point mean

[warped glacier]

the 2nd derivative at x = a equals zero

[shrewd trellis]

yea

but what does that mean?

[warped glacier]

you don't know what the 2nd derivative is?

[shrewd trellis]

It means the concavity

of the graph

[warped glacier]

that's related but that's not quite correct

2nd derivative positive --> concave up
2nd derivative negative --> concave down

[shrewd trellis]

so when its 0?

its neautral?

[warped glacier]

no, it's inconclusive

take a look at these three graphs (y = x^3, -x^4, x^4)

at x = 0, the second derivatives of all of these are 0

[odd edge]

@shrewd trellis Has your question been resolved?

[shrewd trellis]

ok makes sense

.close

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[warped glacier]

no worries!

[odd edge]

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[sour terrace]

a three-dimensional shape with n faces can only produce cross-sections with a maximum of n sides when intersected by a plane?

Is this true? I want some rule for my students for cross sections questions due to imagining all the cross sections (or the cross section with max number of sides) being difficult.

I tried it with octahedron, but I couldn't get a shape with 8 sides.

[wooden cypress]

That works as an upper bound yes

The "why" isn't too difficult either

*assuming the faces don't intersect each other or do anything else strange

Actually I take it back, this highly depends on your definition of a 3d shape

[odd edge]

@sour terrace Has your question been resolved?

[sour terrace]

Right

That's why it's difficult for me to give a rule to my students.

[wooden cypress]

I think if all the faces of the "shape" are convex and don't intersect each other then your rule is true

And implicit in that is that they have to be flat faces

Otherwise you can get weirdness with one face corresponding to multiple sides in the cross section

For example, if your 3d shape is just a flat annulus, then you can get a cross section that's just two disjoint lines

i.e. a 2 sided shape

But annuli aren't convex

[odd edge]

@sour terrace Has your question been resolved?

[odd edge]

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[hard violet]

I need to find the primitive of (2x+6)*x^-2.

Using the inverse chain rule I find: (2x+6) * 1/2 * -1 * x^-1 + C

The awnser is: 2 * ln|2| - 6 * x^-1 + C

why cant I apply the chainrule here?

[hard violet]

I meant the inverse chainrule**

sorry for the confusion

[narrow crypt]

[\int (2x+6)x^{-2} = \int \frac{2}{x} + \frac{6}{x^2}dx ]

this is what u're finding, right?

[hard violet]

yes!

[narrow crypt]

how did u apply chain rule here

[late sinew]

What is an inverse chain rule

[narrow crypt]

u sub

[late sinew]

O

[clever fjord]

k

[hard violet]

So what I did is:

(2x+6) * x^-1 * 1/-1 * 1/2

So letting the (2x+6) stand,
taking the primitive of x^-2,
and powering the dirivative of (2x+6) to -1

idk why but thats what I learned

If possible I would like why this doesnt work

[narrow crypt]

im quite confused by the notation

[ \int (2x+6)x^{-1} \cdot \frac{1}{-1} \cdot \frac{1}{2}dx]

[clever fjord]

k

[narrow crypt]

also

[ \int f(x)g(x) dx \neq \int f(x) dx \cdot \int g(x) dx]

[clever fjord]

k

[wooden python]

2 * ln|2|
sus

[modern sundial]

They are trying to say that they want to use integration by parts by taking the integral of x^-2 and differentiating (2x+6), but it is written in a way that is kind of incomprehensible.

[hard violet]

oh, what do you do then?

[late sinew]

So you want to do integration by parts

And not u sub

?

[hard violet]

idk what sub is (._.)

[late sinew]

Inverse chain rule

[hard violet]

like writing it in one equation?

[narrow crypt]

reverse chain rule is u sub

[hard violet]

I dont get taught in English

oh oke

yes I want to do that

[narrow crypt]

Let ${u = g(x)}$
[ \int f'(g(x)) \cdot g'(x) dx = \int f(u) du ]

[clever fjord]

k

[whole lava]

yo wsg

[narrow crypt]

is this what u tryna do?

[hard violet]

If I understand the equation, no

Im just trying to intergrate the formula

[late sinew]

Hello, this is an occupied help channel, #❓how-to-get-help if you want to open your own and #360643390594875392 for off- topic

Welcome to the server

[graceful viper]

could you please use brackets when writing down your work

and place \ behind * so it doesnt italicise your work

[hard violet]

sure, im very inexperienced. Is there any tutorial?

[graceful viper]

no like (a*b)^(bc) rather than (a*b)^bc

[hard violet]

ahh

oke

[late sinew]

What you're saying is
$\int {\frac{2x+6}{x^2}} = (2x+6)\int {x^{-2}} - \int{2 \cdot \int{x^{-2}}$, yes or no?

[clever fjord]

Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[narrow crypt]

u can take a picture of the paper u've been working on

thats one way..

[hard violet]

so im trying to intergrate the formula (2x+6)*(x^-2)

Ill try to take a picture

[narrow crypt]

wonderful

[late sinew]

Bruv

[narrow crypt]

-2 is gone

[late sinew]

This is not the same thing at all

[narrow crypt]

💔

[late sinew]

Ok so

[late sinew]

What r u trying to do with this now

[late sinew]

And you changed the question again

[hard violet]

my bad I have some dyslectia I meant this 😅

[late sinew]

Ok what do you plan to do with this

[hard violet]

solve

[late sinew]

Yes but

How

[hard violet]

by integrating it

[late sinew]

Like this?

[hard violet]

I dont get the - sign

at this point just any way lol

[late sinew]

Do this then

[hard violet]

sure

that went well

but idk why subbing doesnt work

ill try searching for it on youtube

thanks for the help!

[late sinew]

What are u subbing tho

[hard violet]

idk tbh what it means

[late sinew]

u = x^2?

[hard violet]

no?

I want to integrate it without rewriting the function

because you can rewrite it to the function of k

and that works

[late sinew]

What is k now

[hard violet]

I tried it and I got the right answer

the name of that person

not a variable 🙂

[late sinew]

Show

Wait not by k's method

[hard violet]

this

[late sinew]

Yh that ik

[hard violet]

is the right method but I want to find another

to get a better understanding of the question

[late sinew]

Why

[hard violet]

because I dont understand why another way I had

doesnt work

and if I know why I figuered I get a way better understanding of the concept

[late sinew]

Show it bro

[hard violet]

the original answer?

[late sinew]

This way

Show

[hard violet]

ohhhh

give me 1 min

[late sinew]

Waiting

[hard violet]

Im trying to apply the reverse chain rule on this equation.

When you have the equation [\int 2(4x-3)^{2}dx]

It becomes [(2\cdot \frac{1}{3} \cdot \frac{1}{4} \cdot (4x-3)^{3} + C]

oh how do I get the bot to draw the function?

[graceful viper]

$[(2\cdot \frac{1}{3} \cdot \frac{1}{4} \cdot (4x-3)^{3} + C]$

[hard violet]

[\int 2(4x-3)^{2}dx]

[clever fjord]

dj

[hard violet]

ahh

$[\int 2(4x-3)^{2}dx]

[graceful viper]

okay so the reason you are wrong is because

x^-2 itself is a function

so you cant really integrate products of functions like that

[late sinew]

[graceful viper]

but compositions of functions on the other hand... thats where you use the reverse chain rule

let (4x-3) = f(x)
x^2 = g(x)
then g(f(x)) is going to be the composition of g and f

[graceful viper]

in cases of compositions, we can often use a method called u-substitution or as you call it, reverse chain rule

[hard violet]

ahh

so I cant just apply the u-substitution because the x^-2 is a kind of chain by itself?

[graceful viper]

it is because you wont find a way to separate the product of (ax+b)^n and x^-2 as a composition of functions

you could do it if it were just (ax+b)^n

[hard violet]

ahhh

so if I have the chance to put them together

I must do that before applying the reverse chain rule?

so I got to make the most dense composition of inner functions possible, before intergrating the whole function?

[graceful viper]

dont think of integration as an algorithimic process where you "must" do something

you will benefit more if you watch videos on u-sub than me explaining it to you

https://youtu.be/fNhW-3Hvy7k?si=5N2MwYq2H2-m36p3

you can start with this

[hard violet]

alright, I will do that

thanks for all the help!

[odd edge]

@hard violet Has your question been resolved?

[hard violet]

.close

[odd edge]

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[obsidian mauve]

by using rational root theorem I got ± 1, ±2, ±4, ±8
then by using synthetic division I got 2 zeros.. -1 and 2

but I can't seem to figure out how to find the other 3

[sand horizon]

have you write it as a product of (x-r) where r are the roots you already have?

[obsidian mauve]

I dont understand what u mean here

[sand horizon]

you know that -1 and 2 are roots

you can write this quintic as (x+1)(x-2)P(x) where deg(P) = 3

[obsidian mauve]

u mean (x+1) (x-2) (x^5+x^3-2x^2-12x-8) ?

[sand horizon]

no

[swift lake]

I usually do horner

[swift lake]

well if you goot other zeros

you can divide it by

[sand horizon]

works aswell

[swift lake]

$\frac{x^5+x^3-2x^2-12x-8}{(x+1)(x-2)}$

[clever fjord]

Newt

[sand horizon]

(x+1) (x-2) (ax^3 + bx² + cx + d)

[swift lake]

,w \frac{x^5+x^3-2x^2-12x-8}{(x+1)(x-2)}

just checking

i see

those zero maker are corret just missing another 1

[obsidian mauve]

can I use long division for this?

[swift lake]

sure why not

or use horner

just use whatever the teacher teach you:)

[obsidian mauve]

yeah idk what that is yet

[obsidian mauve]

well apparently the answer says its -1
so its a double root at -1

how would I know if its double root while solving tho?

[swift lake]

,w factorize x^5+x^3-2x^2-12x-8

[clever fjord]

Results provided by WolframAlpha

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

Requested by .newts_

[obsidian mauve]

I gotta factorize 5 terms

[swift lake]

there is two -1, one 2 and -2i and 2i

bad english lol:)

and andd:

[late sinew]

Do RRT

[obsidian mauve]

whats that

[swift lake]

what is rrt?

[late sinew]

Rational root theorem

It says that for any polynomial, the probable rational roots are (the factors of the terminating coefficient)/ (factors of the leading coefficient)

So here the terminating coeff is -8

And the leading coeff is 1

What are the factors of -8?

[obsidian mauve]

+- 1, 2, 4, 8

[late sinew]

Okay

What about 1

[obsidian mauve]

1

[late sinew]

and -1

[obsidian mauve]

+- 1

[late sinew]

So now you can try possible combinations of p/q where p belongs to the factors of 8 and q belongs to the factors of 1

[swift lake]

$\pm$ lol she meant

i think:)

so 1 and -1 is included

[late sinew]

she, but prolly yh

[clever fjord]

Newt

[obsidian mauve]

😭 yes idk how to do the +- using keyboard

[late sinew]

(I recommend starting out with the smaller ones first)

[obsidian mauve]

yeah I did that and used synthetic div and -1 and 2 are zeros

[low locust]

I mean the channel started with RRT immediately and found two roots

so thats not the problem here

[late sinew]

O

did not see

well then you can just do long divvy and try

[obsidian mauve]

yeah😭

[swift lake]

yeah:)

[low locust]

after you do polynomial division (in whichever way you want) you need to guess the numbers again

until eventually you get a quadratic

[late sinew]

the two roots uve found, multiply them and use them as the divisor

[obsidian mauve]

"you need to guess the numbers again" wdym? like do the RRT thing all over again for the new function?

[low locust]

so you would divide out (x+1) and (x-2) (either individually or at once) and then you get your cubic

and then you guess again

when you guess -1 again you'll find that it works again

so its a double root

divide by x+1 again, get quadratic, use quadratic formula

alternatively (although I wouldnt necessarily recommend it), you can also check whether -1 is a double root by checking whether it is a root of the derivative

[obsidian mauve]

didnt learn derivative yet

[low locust]

then ignore it

[obsidian mauve]

well I get it now! imma do these step first before closing the channel tho coz i might get it wrong or smth

[late sinew]

ok

[obsidian mauve]

omg got it thanks everyone who helped 🩷

yall better than my teachers

.close

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[low locust]

<@&268886789983436800>

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[rich iris]

I’m doing double derivatives and I think I did something wrong? Can someone check it out for me

[wooden python]

yeah your signs for y" are the opposite of what they should be, all throughout

4x^2-1 is negative between -1/2 and 1/2

[rich iris]

aint no way 😭

[sand horizon]

ax² + bx + c is of the sign of a on the exterior of the roots

[rich iris]

yea i think i know that but i just had a brain fart lmao

wait another question

how do you know where the arrow starts and points?

[sand horizon]

Sign of derivative and second derivative

[rich iris]

yea but like i forgot what those mean

[sand horizon]

y' >= 0 means increasing (else decreasing)

y" > 0 means convex i.e the tangents are below the curve

[rich iris]

that sounds confusing but ok

[sand horizon]

Convex == concave up

[rich iris]

So like this?

[sand horizon]

seems good

[rich iris]

nice! Thanks

.close

[odd edge]

Channel closed

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[sand horizon]

,w graph exp(-2x^2)

[clever fjord]

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[solar peak]

ive gotten f(x) = x

how do i find that limit?

preferable without using l'hopital thrice

[late sinew]

ok

[bitter folio]

thought of hamel basis then saw differentiable

[solar peak]

what is hamel basis 😭

[late sinew]

same bro

[waxen talon]

wow

the fraction simplifies to

[solar peak]

yes?

[brittle plinth]

????

[solar peak]

what does it simplify to

[waxen talon]

srry

on other channel

but since f(x) = x

then probably you can simplify it

[solar peak]

...

[late sinew]

kekw

[solar peak]

can anyone help

[late sinew]

Whats the problem with lhopital thrice

[solar peak]

not a big problem its just im not too good w differentiation yet and i thought there was a better way

[opaque eagle]

@solar peak do you still need help

[solar peak]

yes

[icy egret]

you want A?

[opaque eagle]

Did you figure out that f is an exponential

[solar peak]

huh

i thought it was linaer

linear

[wooden python]

$\lim_{x \to 0} \frac{2^{\tan(x)} - 2^{\sin(x)}}{x^3}$ is what y'all are looking at right

[clever fjord]

Ann

[opaque eagle]

oh mb

yeah

linear

[solar peak]

yes

[wooden python]

multiply and divide by tan(x)-sin(x) for a start

then $\frac{2^{\tan(x)}-2^{\sin(x)}}{\tan(x)-\sin(x)} \to \dv{x} 2^x \bigg|_{x=0}$

[opaque eagle]

factor out 2^(sin x) first

[icy egret]

f(x) must linear

[clever fjord]

Ann

[opaque eagle]

it'll become a standard limit

[wooden python]

we've already established f(x) = x actually \

[icy egret]

k

[opaque eagle]

why exactly is this true

even if you do the small angle approximations

[solar peak]

i dont understand this

[opaque eagle]

[ \lim_{x \to 0} \frac{a^x - 1}{x} = \ln a ]

[clever fjord]

dyxn

[wooden python]

it's the difference quotient of 2^x over the interval [sin(x), tan(x)]

[opaque eagle]

do you know this limit?

[wooden python]

said interval shrinks to 0

[solar peak]

yeah

[opaque eagle]

Oh I guess it ends up being the same thing

[opaque eagle]

multiply and divide by tan x - sin x as ann said

and factor out 2^sin x from the top

[opaque eagle]

becomes this

[solar peak]

oh right

[solar peak]

but where did the d/dx come from

[opaque eagle]

ann used the definition of the derivative

[opaque eagle]

here

[opaque eagle]

it's technically the same thing here except the point is fixed to 0

[wooden python]

yeah but it's not jeebro-accessible i guess 😔

[solar peak]

im getting
2^sin x (tan x - sin x) ln 2

now i put x=0?

[opaque eagle]

theres still /x^2

[solar peak]

oh mb

[opaque eagle]

2^sin x --> 1 so you don't have to worry about that

[icy egret]

A=ln2/2 ?

[solar peak]

yeah i can figure out the rest from expansions

[opaque eagle]

[ \ln 2 \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} ]

[clever fjord]

dyxn

[opaque eagle]

I think its ln(2)/6

[icy egret]

[opaque eagle]

oh wait no

this is weird

[solar peak]

i think its just ln 2

[opaque eagle]

1 - cos(x)

[late sinew]

ay everything is accessible

but just not

taught

[opaque eagle]

,w limit as x goes to 0 (tan(x) - sin(x))/x^3

[clever fjord]

Results provided by WolframAlpha

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Requested by dyxn06

[solar peak]

.

oh

[opaque eagle]

ln(2)/2 mb cro @icy egret

[icy egret]

The approximate value is 0.3466

ln2/2 is correct

[solar peak]

im getting this

[opaque eagle]

what did you do lol

[solar peak]

expansions

[opaque eagle]

[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} ]

[solar peak]

sin x as x- (x^3/3!)

tan x as x+x^3/3

[clever fjord]

dyxn

[wooden python]

that which is not taught must be presumed illegal to even speak of

[solar peak]

oh wait it gets added

[opaque eagle]

yeah

[solar peak]

ok yeah i get 1/2

ok got it

thanks everyone

.close

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[mystic saffron]

Integrate 5x^2 /(x2+4x+3)

[mystic saffron]

With upper limit 2 and lowerrr limit 1

[worn horizon]

Try splitting the denominator and do partial fractions

[mystic saffron]

I got 5/2(log3/2 - 9log5/4) as ans

But my book has 5/2 (9log5/4-log3/2)

[vivid wyvern]

Factorize

The denominator

[mystic saffron]

I did

[vivid wyvern]

Do partial

[mystic saffron]

Dude i even got the ans 😭

But its opposite

[vivid wyvern]

Then?

Again

Lemme try

[mystic saffron]

Im getting the same

Lemme try once again

Still gettin the same

Oh wait ill try division

[vivid wyvern]

Ay doonr yall get 4/5

[mystic saffron]

Since degree is same

[mystic saffron]

Wat

[modern sundial]

Honestly when you are practicing at home for something like this, you should just throw the function into wolfram alpha just to check if you got the partial fractions correct

[mystic saffron]

Ill divide wait

[modern sundial]

That way you at least know if you did the partial fractions wrong, or the integration step wrong

[mystic saffron]

Ok

Ye i got it now

I forgot to divide since the degrees are same

.close

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[devout schooner]

Can someone check my work?

[icy egret]

what?

0?

[late sinew]

How are u getting 0

if u cut everything ur left with 1

[devout schooner]

oh

brhu

[late sinew]

rest is correct

[devout schooner]

is the work right though? because i still have 2 more of these

oh thanks

[nimble blaze]

ensure you have a + there

[devout schooner]

okok

thanks guys

ill be back with the rest eventually maybe if i have doubts

.close

[odd edge]

Channel closed

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[devout schooner]

.reopen

[odd edge]

✅

[devout schooner]

I may be lost

Idk how to do this

[graceful viper]

where are you lost

you seem to be doing it perfectly fine

[devout schooner]

how do i multiply it?

or is that the answer?

[narrow sinew]

pls help

[narrow crypt]

!occupied

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[graceful viper]

cancel out 5

[jagged grail]

You forgot to remove the 5

[devout schooner]

i can do that? oh

[limpid adder]

multiply out 5(x^2+2) then multiply by that by (x-1) using factoring thing (i am blanking) then divide by 5x

or that

yeah

[devout schooner]

so ive got (x-1)(x^2+2)

is that it?

wait no

(x-1) / x

[jagged grail]

No

[jagged grail]

Yes

You can also

If you want

[devout schooner]

what do you reccomend?

[jagged grail]

Write 1 - x^-1 * (x^2+2)

If you want no fractions

[devout schooner]

how did you get that

[jagged grail]

So yk how if you have (x-1)/x it’s the same as x/x - 1/x?

[devout schooner]

i dont really understand this is a very new topic for me

not factoring but these types of questions

[jagged grail]

Ok so when you have a binomial (x-1)

Where there’s an operation

And one denominator

You can split the fraction in 2

Let’s say you have (10+5)/5

You can split it up as 10/5 + 5/5

= 2 + 1

=3

Same applies here

[devout schooner]

ok

[jagged grail]

X + 1/x is x/x + 1/x

Due to indeces laws, you can write 1/x as an exponent

x^-1

Wait what grade level is this btw

Because this may not even be necessary

[devout schooner]

im going into 11th

[jagged grail]

Ok then I’d recommend knowing this

[devout schooner]

this is the summer packet for pre calc

indeces law you said?

[jagged grail]

Yeah search it up and you’ll see all of them

It’ll help when you start calc

Maybe pre calc

[devout schooner]

thanks

i will need it anyways

how do you recommend me practicing indeces laws

because i literally want to excel in all the topics before i start pre calc

[jagged grail]

Rn you have (x^2 + 2)(-x^-1 + 1)

Honestly

I think all that’s left is to foil

So do that

[devout schooner]

ok

[jagged grail]

And lmk what you get

Such a weird question though

[devout schooner]

fr

the one before it was easier

i got

x^2 -x -2x^-1 +2

[jagged grail]

Yup that’s correct

Yeah man idk what else you can do from here lol

I’ve never seen a question like this before until now

Just assume that’s the answer

If you really wanna simplify it more

Then you can even common factor out the x^-1

But I doubt that’s necessary

[devout schooner]

thanks lol

[jagged grail]

Np

[devout schooner]

i have one more if i have doubts ill be back

[jagged grail]

Sure

[devout schooner]

i might have doubts

just by looking at it

[jagged grail]

I’m reviewing your steps rn and everything seems fine

[devout schooner]

thats good

[jagged grail]

Might be an error with the question itself

Ask your teacher what’s up with it when you can

[devout schooner]

i sure will

if theres no factoring needed

can i just start cancelling out

[jagged grail]

Maybe he thought the 5x on the denominator would cancel out nicely. And if that was the case it’d make more sense since you’d get a quadratic and factor away

But as we can see it’s not that simple

[jagged grail]

Cancelling out what?

[devout schooner]

nope

like the numbers

this is the last question

[jagged grail]

😭

Ok let’s unpack this

[devout schooner]

is it bad...??

[jagged grail]

Nah

[devout schooner]

ok

i can get rid of the x+1 and the 4?

[jagged grail]

I think you should do the operations on both sides first

To remove the - and +

[vivid wyvern]

No

[jagged grail]

No

[vivid wyvern]

It's division first

Then multiplication

Badmas basically

[jagged grail]

But didn’t he already do it though

[vivid wyvern]

U can get rid of 4

[jagged grail]

Since the first thing he did was do the reciprocals

[vivid wyvern]

He didnt complete it

U can cancel out 4

[devout schooner]

so 4/x * x+1/4?

[vivid wyvern]

And then do rhe add and sub
Or u could do it altogether

[vivid wyvern]

Yeah 4 gets cancelled out here

[devout schooner]

ok

[vivid wyvern]

4/4 is what? 1 ryt?

[devout schooner]

i got x+1/x out of this

[jagged grail]

But it won’t cancel out nicely

Yes there’s 4 on both sides

But it’s not on all the fractions

[vivid wyvern]

It doesnt need to be does it?

[vivid wyvern]

Do the whole thing together now

[vivid wyvern]

Replied to this

[jagged grail]

I mean this works but like

I feel like it’s cleaner and easier to understand if he just adds the 2 fractions together

On each side

Then cancel things out after

Instead of having a bunch of operations flying around

You have a nice, clean fraction a * fraction b

And you go from there

[devout schooner]

it looks weird now

[jagged grail]

Yeah that’s exactly why I don’t like that method lol, easier to screw up

[devout schooner]

how else could i do it 😭

[jagged grail]

Try this

And see if you get something different

Don’t cross anytning out until you add the fractions together

Well, not add but

Add the right side fractions and subtract the left side fractions

Basically make it so that the only operation on the paper is multiplication

[devout schooner]

ok

[jagged grail]

Then you factor out the LCD and see what happens

[vivid wyvern]

Uh no

That's not how u do it

U can cancel out that way only when the terms r in multiplied form

[devout schooner]

oh, i forgot

Bruh

[devout schooner]

How can i do it then

[vivid wyvern]

See u hv denominators

X ,x and (x+1)

[vivid wyvern]

Let's continue from here

Sk among x and x ,both r same

That gives u x as an lcm

Wait b4 that do u k what's lcm or lcd,wthever u call it

[devout schooner]

yewah

least common multiple

[vivid wyvern]

So between x and (x-1)
We cant find a lcm
By that i mean common factor

Cuz think abt it

So u directly multiply them

If x =5
X-1 will be 4

They dont hv any common factors

So u multiply them

Similarly 2,3
1,2
3,4
6,7

That's y here u need to multiply them

To find lcm

So x,x,(x-l) has an lcm of?

[devout schooner]

x(x-1)?

[vivid wyvern]

Yep

Now u do the math

[devout schooner]

k

my pencil broke doing this

i found a new one

brb

bro my brain hurts

[vivid wyvern]

Help

[devout schooner]

pls tell me this is right

[vivid wyvern]

I mean according to badmas ir isnt
But let me confirm

[devout schooner]

bruh

[vivid wyvern]

Yeah

U r goon

Good

[jagged grail]

Crazy

Also I’m back

[vivid wyvern]

GOOD

Wc

[devout schooner]

peak

k so i learned both ways

@jagged grail was an easier route

but it doesnt hurt to learn both

thank god im finished for today

[jagged grail]

Use the one that’s more efficient

If you have a number in the denominator for both sides

Let’s say a/4 + b/2 = c/4 + d/7

You can’t just like

Cancel out the 4

[devout schooner]

Yah lol

[jagged grail]

If you divide both sides by 4, the stuff that doesn’t have a 4 on the denominator will need to be divided by 4 as well

But the other method he gave you apparently didn’t show that

[jagged grail]

Since it’s an equation

So if you have 4/3+ 2/7 = 5/4

You can’t just have 1/3 + 2/7 = 5

Because there was nothing done to the 2/7

The equation wouldn’t be balanced

You have to do the operation on every term

[devout schooner]

Sorry im back

I went to do the dishes

[vivid wyvern]

Was ur problem solved?

[devout schooner]

It was

Eventually

Thanks guys

Have a good day

.close

[odd edge]

Channel closed

Closed by @devout schooner

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[devout schooner]

I will be back tomorrow

With a new topic

Probably

Bye!!

[odd edge]

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[random pumice]

Solve (b+c)^2 = 2011 + bc
With b,c positive integers

[ivory grove]

!show

[odd edge]

Show your work, and if possible, explain where you are stuck.

[random pumice]

sfft dont works

[ivory grove]

?

and do you have additional context? you have 2 variables here, but you only have 1 equation

[random pumice]

nope

[late sinew]

It's one of those qns where u have to see which sqrt is the closest

And factorise it

[ivory grove]

ah

[late sinew]

Can you find the closest square near 2011

[random pumice]

45^2

[late sinew]

Yeah got the ans too, qn is correct

Um actually might have messed up hollup kek

[graceful viper]

have you tried bounding b

[random pumice]

i think it doesnt work

Wlog b<=c
Then 2011<=3c^2

25<=c

[late sinew]

Scrap that mb i used that to find (bc)^2 = 2011 + b + c by mistake

[graceful viper]

I think you will inevitably have to check a bunch of values from 1 to 44 for b (unless you find better bounds)

[late sinew]

Yh

Is just guessing

Tookback the sully kekw

[tribal ingot]

this was a more fitting emoji

[late sinew]

Perfect use case for a program

[random pumice]

hm ok

[late sinew]

Dyk what the question intended

Where is it from like which chapter

[random pumice]

Aime 2011 p15

[late sinew]

https://math.stackexchange.com/questions/2162390/solving-bc2-2011bc here's an MSE post i found

Though the soln is talking about rings

So qn is prolly still guesswork

[odd edge]

@random pumice Has your question been resolved?

[odd edge]

Channel closed

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[vernal yacht]

.reopen

[odd edge]

✅

[graceful viper]

if (10, 39) is a satisfying ordered pair then so is (39, 10)

[graceful viper]

they bound it down to 19 checks using rings

and complex conjugates

pretty cool, their problem isnt for positive integers though

[odd edge]

@random pumice Has your question been resolved?

[odd edge]

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[tough remnant]

This is the answer

[rose coral]

,w differentiate ln(sqrt(x) + sqrt(x + 1))

[clever fjord]

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[rose coral]

I think you can try integration by parts

u = integrand
dv = 1

[cosmic wyvern]

can someone teach me math

currently stuck on substraction

[odd edge]

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[cosmic wyvern]

what is 15-10

[quasi sparrow]

use a calc

[solar peak]

<@&268886789983436800>

[narrow crypt]

(calc is short for calculator)

[blazing shuttle]

@cosmic wyvern dont abuse help channels

.close

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[odd edge]

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[narrow crypt]

what u wanna with it

[bitter folio]

What have you tried

[narrow falcon]

Oh x=7 and my answer didn't matched the book answer

[bitter folio]

Have you checked your solution?

!nogpt

[odd edge]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[bitter folio]

.

[turbid comet]

Do we need to simplify the expression in this ??

[solar peak]

what is the question

[odd edge]

@narrow falcon Has your question been resolved?

[odd edge]

@narrow falcon Has your question been resolved?

[low locust]

you havent written down what your actual question is

you just wrote down an expression

what are you supposed to do with that expression

[odd edge]

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[odd edge]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[timber spoke]

help

[timber spoke]

can someone explain the recursive formula to me

so f(1) = 0

term 1 of the sequence goes in front which is hence 0 + rate of change(n - 1)

which shows the difference is 1

is that right

waitt

[low locust]

before trying to think about it abstractly, do an example

whats f(2)

and f(3)

[timber spoke]

i thought it was supposed to be

denominator multiplied by the number outside the parantheses

which is 18

and then you add the 1 to the 18

so 19/3

i kinda forgot how to do fractions

i know you could just convert it into decimal form

im not rage baiting

[low locust]

ok what does that have to do with the other problem?

[timber spoke]

its irrelevant

let me ask chatgpt

[low locust]

well if you had 6 1/3 as a mixed fraction, then you would indeed convert that to 19/3 like you said

but you basically never encounter mixed fractions

and something like 6 (-1/3) is definitely not a mixed fraction, thats just a product

[timber spoke]

wait

i see what i did wrong

mjan

i keep making these silly mistakes

[odd edge]

@timber spoke Has your question been resolved?

[swift lake]

f(2) = f(1)+2 = 2
f(3) = f(2)+2=4
f(4) = 6
est...

[swift lake]

bruh what are you counting

[timber spoke]

PLS

THATS SO EMBARRASSING