help-36

[muted olive]

.close

Use that

[formal trail]

that channel is already closed

[final saddle]

@solar tapir Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[final saddle]

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.
• 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.

[modern estuary]

what are reasons to drop lower order terms in a polynomial model?

[modern estuary]

as in im looking for certain reasons to justify dropping them in my polynomial models of certain data

[opal plinth]

Context?

Are you asking about asymptotic behavior?

[modern estuary]

I have a certain set of data which I'm modelling using a degree 7 polynomial. its very accurate when i remove the terms below x^3

so i just wanna understand why is this, so i can justify removing the lower order terms to my teacher

[opal plinth]

Still missing context. I would just assume you were using the wrong polynomial to model your data

[modern estuary]

what context do you need?

[opal plinth]

How you chose the polynomial would be a good start

What the data is, along with what polynomials you came up with

[modern estuary]

this is the regression equation im using. Whenever I add lower order terms (x^3, x^2, ...), it just doesnt fit as well

[opal plinth]

What the data represents

[modern estuary]

to be honest, randomly

im just looking to find polynomials that fit my data

[opal plinth]

...

[modern estuary]

they dont have to be great

i have a project where i have to evaluate models and how well they fit

[desert mantle]

all those coefficients are basically zero already

how much of this is just noise?

scale up your data or something

[modern estuary]

it seems to be good enough, even at scale

the red dots are my data

[desert mantle]

I dont know what regression solver desmos uses

but more options should not give worse results

at worse you should still recover the old ones by setting the relevant coeffs=0

[opal plinth]

Are you the one who asked something similar and tried the logistic function and a bunch of other things?

[modern estuary]

indeed

logistic function worked great

[opal plinth]

So... what's your goal exactly?

[modern estuary]

yup, i could do that. i'd have to justify it though, hence my question

[desert mantle]

no, I mean the solver should do it on its own

if those actually give better results

if it doesnt then there is a mistake somewhere

[modern estuary]

my data is time vs frequency of popcorn pops. I have a project where I'm modelling this data using various models, to try to find an optimum time of roasting

[stone flint]

are you trying to find an instance of each model that can fit your data to a good extent? otherwise, I do not see the point of trying so many models when you have quite a few good ones already.

[modern estuary]

they don't need to fit well, i have some that are atrocious, and thats fine

but this case has confused me a bit. i can't understand why excluding certain terms would make it have a worse fit. it's something i'd like to discuss in my project, but i don't know the reason

[opal plinth]

This was probably already discussed in your earlier discussion, but a Gaussian distribution would just make sense for it, and the function to fit your data would be its CDF

[desert mantle]

you would need to look at the solver desmos uses

if it prohibits that you can just set one of the coeffs as zero, then you are working in a different space

[modern estuary]

yup, thats already been done

[desert mantle]

hence you can get a different optimum

[opal plinth]

So why are you even bothering with a polynomial?

[modern estuary]

the point is to use different models.

[desert mantle]

and again, might just be noise due to floating point accuracy

how much worse is the fit with lower terms

show

[modern estuary]

Ah, I see. So could this be a case of just some weirdness with the solver desmos uses?

[opal plinth]

Could you just link a Desmos snapshot?

[modern estuary]

can't sadly. not really supposed to share my data publicly

[opal plinth]

Can't share data about roasting popcorn?

[modern estuary]

academic honesty 😭

[desert mantle]

you basically gave us your data in that screenshot

if we wanted to extract it

[opal plinth]

Plus we're not going to do the work for you anyway

[desert mantle]

btw you can also take a look at google sheets

it also does regression

see if it gives different results

[modern estuary]

interesting, i will try this out as well. thanks!

[modern estuary]

it's not that yall are wrong, it's just that it's not something i can do 🤷 .

the coefficients are basically 0 anyways so i might just write a section about polynomial models being bad fits

[opal plinth]

You do you, but this isn't about academic honesty

[stone flint]

I think at some point there's a difference between academic honesty and diagnosing the underlying data to see why certain models don't fit as well instead of having helpers grasp at straws about the potential nature(s) of your data.

[modern estuary]

thats fair

yeah then i guess this is unfortunate. thanks for all the help regardless guys

.close

[final saddle]

Channel closed

Closed by @modern estuary

Use .reopen if this was a mistake.

[final saddle]

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.
• 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.

[autumn lodge]

How csa of cylinder 4πr²

[civic obsidian]

bcuz the height of the cylinder is 2r

[autumn lodge]

Ohhhhhhh

I'm stupid

K ty

[final tangle]

also make your r look less like pi

[civic obsidian]

no worries, it happens lol

[autumn lodge]

Ty for the advice but this is a yt vid

.close

[final saddle]

Channel closed

Closed by @autumn lodge

Use .reopen if this was a mistake.

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.
• 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.

[snow verge]

I need help I joined a class a week and a half after it started and I missed the explaining period I have no idea how to do this

[worldly spruce]

This means everything?

[snow verge]

pretty much but im fine if we can just focus on a few questions till I get the hang of it

[worldly spruce]

Well to begin with the 1.

[civic dome]

g(😃) is crazy

😭😭🥀

[snow verge]

I have no idea what it means

[worldly spruce]

Think of functions as some sort of math machines

[snow verge]

we never covered this stuff in pre calc 11

[worldly spruce]

You input a value (the input), the function does some stuff to it, and outputs a value (the output)

[fast pendant]

Yeah as u put wheat in a mill and it gives u flour

[worldly spruce]

So, by this, f(x) is a function of x. It tells us how it "modifies" x.

[tired walrus]

https://youtu.be/7JZ0IfCQ488?si=P4z53JpEiQXq4jzn

consider giving this a watch

[worldly spruce]

And for f (a), there is really nothing to understand about it so I will just say it. f is just the name (label) of the function. We can call it f, g, h or even something more literal like height

[snow verge]

should I watch the video or do you wanna explain this to me?

[tired walrus]

the intended answer there is "f represents the name of the function" ig

[worldly spruce]

Whatever you wish. I would suggest watching the video though because it's from a high quality channel which probably put hours of thought into it

[snow verge]

alrighty ill watch the video and if I need further help ill message here

[worldly spruce]

Yes of course, if you don't understand anything come here

<@&268886789983436800>

Can we like get rid of people's computers who use these bots

🙁

[wispy grove]

well there are way better channels than this one

[tired walrus]

https://youtu.be/E9YEUQR9NAU?si=f3MsHp8uRLbR1uCf

there is also this, which is more targeted

[worldly spruce]

I think something as simple as the basics of functions is difficult to miss

by a channel of at least a decent following

[wispy grove]

well if u refer to channels of named institutes

they dont

[worldly spruce]

oh <@&268886789983436800> I am sorry

[tired walrus]

these are the ochem tutor and khan academy

khan academy is quite good

[worldly spruce]

Are you kidding me

<@&268886789983436800>

😭

[opaque ember]

mfw math help channel has more modpings than math help

[worldly spruce]

I mean, I disagree as I don't need a master chef tell me that water boils when it gets to a certain temperature.

[snow verge]

ok I tried the first one after watching the video

[worldly spruce]

For more complex topics, of course though

[worldly spruce]

Yes, but that is the second

[snow verge]

sweet ill watch that one soon

[snow verge]

oops

im gonna ignore that first question cause really its just notes

what do i do for the smiley face question

[worldly spruce]

I mean, it's not really just notes

It's important to understand the meaning of symbols

Otherwise they are, well, meaningless

[snow verge]

ah then can you help me 🥲

[worldly spruce]

Right

So from before

You understand that f is just the "label"

Or the name of the function

That we picked

It could have been g(x), h(x), or circumference(x)

Just a name.

That should be understandable

[snow verge]

yep

[worldly spruce]

Good

What would you say x represents then

In f(x)

[snow verge]

great question ngl I dont entirely know

[worldly spruce]

It's just the input to the function

[snow verge]

ohhh

[worldly spruce]

The value we feed into it

We can pick say x = 40

And we can evaluate what the function "does" to the value 40

We would write it as f(40)

We just replace x with the value of x

[snow verge]

makes sense

what does f(x) represent then

[worldly spruce]

The output

[snow verge]

ah oki

[worldly spruce]

Let's take the example you solved

f(x) = -3x - 2

So again

f is the name of the function

could have been p(x) = -3x -2

And then we are given a rule for the function

-3x - 2

In words

This tells us

That f takes an input x

Multiplies it by (-3), and subtracts 2

And returns the value

And you were asked what f(-3) is

So as I said above, we "picked x = -3"

And we want to see what the function outputs

So we literally just substitute x = -3 into the rule, -3x - 2

And we get the output

So f(-3) = -3 * (-3) - 2 = 9 - 2 = 7

Therefore the output, f(-3), is 7

Get it?

[snow verge]

yeah i think I do

[hoary tiger]

f(x) is essentially the output for some arbritrary value of x

[worldly spruce]

Great

So the key idea is

We have f(x) = -3x -2

We want to find f(-3)

So we literally just replace every x with -3

And that's all

<@&268886789983436800> ...

oh they are here

[snow verge]

alrighty now how do I do the (½)

this is where ive gotten i just wanna make sure im on track

[worldly spruce]

Yes, exactly, we replace x with 1/2

Though I don't know what you are doing in the second line

We are multiplying two fractions

Same denominators are not required

[snow verge]

alrighty ill erase that

[worldly spruce]

I mean, it's not incorrect

[snow verge]

I just think my teacher didnt want non whole numbers

[worldly spruce]

Just doing something you will have to undo later

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

[soft zealot]

USS-Enterprise

[tired walrus]

not really

in this sort of question fractional answers will be quite unremarkable and not a sign by themselves that you would be doing sth wrong

[snow verge]

so is this right?

[worldly spruce]

Yes

[snow verge]

sweet

so how do I do the little smiley face

[worldly spruce]

The same way

Remember, earlier we said x is an input

I didn't explicitly say it's a number

[snow verge]

oh jeez

[worldly spruce]

For highschool, it will pretty much always be a number yes

But it doesn't have to be

It can be a smiley

And we do it the same way

It multiplies the smiley by (-3) and subtracts 2

Doesn't make sense to us

But doesn't mean it can't exist

[snow verge]

so is it just this?

[worldly spruce]

Yes

[snow verge]

weird

lol

ok ima try 3 a)

so are these 2 right?

[worldly spruce]

,rccw

[worldly spruce]

Yes

[snow verge]

now is this one right aswell it was a littlw confusing but I think i got kt

[worldly spruce]

Yes, end result is okay

I would take a second to improve notation though

[snow verge]

the way I went about doing it looks weird I know lol

[worldly spruce]

You've got side calculations in the middle of everything

It's best to just lead = all the way till the end result

So that the left does actually equal the right

[snow verge]

yeah i was just trying to work with the limited space

[worldly spruce]

And do little side calculations, if necessary, on the side

or below or above if there's no space

[snow verge]

alrighty

[worldly spruce]

Because right now it isn't that big of a problem

But later on you might encounter much bigger problems

Where you will first have to do tons of other little stuff

And it's extremely good for clarity to separate things

[snow verge]

ok next is these and they seem really confusing

[worldly spruce]

,rccw

[worldly spruce]

well

Why

I assume by now you know f(2) and g(3)

And then, I assume you know how to do 3 * f(2)

[snow verge]

well there's now 2 but its also the fact there's a 3 in front of the f

[worldly spruce]

And then 3 * f(2) + g(3)

[worldly spruce]

Yes

You are just multiplying f(2) by 3

It's 3 times that output

First, calculate f(2) and g(3)

[snow verge]

when do I multiply after everything is done?

[worldly spruce]

Right I see

Let's forget about functions for a minute

Say we had $3 \cdot 5 - 5$

[soft zealot]

USS-Enterprise

[worldly spruce]

How would we compute this

Don't give me just the answer

But step-by-step

[snow verge]

3×5 = 15 then 15-5 = 10

[worldly spruce]

Correct

Order of operations right

We multiply first and then subtract

Okay

What if we had

$3 \cdot m - p$

[soft zealot]

USS-Enterprise

[worldly spruce]

Where m and p are some number we don't know

Does the way we compute this change?

[snow verge]

3m-p

no not really

[worldly spruce]

Exactly

We still first multiply 3 * m

and then subtract p

So then

How would we compute

$3 \cdot f(2) + g(3)$

[soft zealot]

USS-Enterprise

[worldly spruce]

Think of f(2) as m and g(3) as p

They are some number we don't know yet

[snow verge]

so 3f(2) + g(3)

but how does that play in when solving the rest of the equation?

[worldly spruce]

Let's first find f(2) and g(3)

It should be clear then

What is f(2) and g(3)

[snow verge]

so do I now multiply the 5 by 3? then add then add the other 5?

[worldly spruce]

Yes

But again

This notation is pretty bad now

[snow verge]

really? 😭🥲

[worldly spruce]

You are just equaling things that aren't equal

[snow verge]

im so confused on how I should be writing it out

[worldly spruce]

3f(2) = 2*2 + 1 + 9(3) = 3^2 + 3 -1

Does not make sense

Right

On the side

Compute f(2) and g(3)

Always divide a problem into small pieces

$f(2) = 2 \cdot 2 + 1 = 5$

[soft zealot]

USS-Enterprise

[worldly spruce]

$g(3) = 3^2 - 3 - 1 = 5$

[soft zealot]

USS-Enterprise

[snow verge]

does this look a little better?

[worldly spruce]

Yes a bit

but on the left side

You have 3*f(2) written

But on the right of = you are computing only f(2), not 3 * f(2)

And at the end write 3 * 5 = 15

get rid of the 3f(2) and just write f(2)

and it's okay

and then write 3f(2) = 3 * 5 = 15

And then as you did yes, since we know f(2) = 5 and 3*f(2) = 15, and g(3) = 5. then 3*f(2) + g(3) = 15 + 5 = 20

[snow verge]

so kinda like this?

[worldly spruce]

yes exactly

but it's important you understand why we are writing it like this

not just because I said so

We are trying to compute 3f(2) + g(3)

so 3 times f(2) plus g(3)

We can separate this problem into smaller problems

[snow verge]

alrightyy

[worldly spruce]

first we compute f(2) and g(3)

then we need 3 times f(2)

because of order of operations

and then we must add 3 times f(2) and g(3)

Just think of it like that in your head

and write it in chunks I guess

[snow verge]

kk ima try the next question

[worldly spruce]

👍

[snow verge]

oops wait

I made a mistake

there

[worldly spruce]

Yes exactly

Look at how much cleaner it looks

[snow verge]

yeah lol

[worldly spruce]

I instantly see what you were doing at each step, and at each side

[snow verge]

I also started the next one

[worldly spruce]

Correct!

[snow verge]

sweet ill move on to the next one it looks a littlw tricky

is this right?

[worldly spruce]

Yeah

exactly

[snow verge]

sweet

[worldly spruce]

see it doesn't matter if we are finding f(2), 2f(3), or f(x+1), or 4f(x^2)

The process is always the same

[snow verge]

alrighty ima try the next 2

[worldly spruce]

👍

[snow verge]

ok these were hard lol

[worldly spruce]

,rccw

[worldly spruce]

Yep, correct

Great job 🙂

I assume you understand the point of these now

Judging by the much better writing

[snow verge]

I think so but now ive flipped the page and im confused all over again lol

[worldly spruce]

What's the new page

[worldly spruce]

Well for the first

[snow verge]

mainly the fact that there equal and then the find part

[worldly spruce]

We do the exact same thing

[snow verge]

kk

[worldly spruce]

We find f(x+1) and g(x-1)

[snow verge]

ill start rn

[worldly spruce]

And then we are saying they equal

So we just set them = to each other

And solve the equation we get

[snow verge]

ok so im here rn

[worldly spruce]

Yep

And now you just set them equal to each other

And solve

[snow verge]

now how do I do that again its been a while

[worldly spruce]

Well we have

f(x+1) = 2x - 1

and

g(x-1) = -4x + 2

And we want to solve f(x+1) = g(x-1)

So can't we just replace f(x+1) with 2x - 1, and g(x-1) with -4x + 2

[snow verge]

also I realized -4x+2 could also be -2(2x-1)

[worldly spruce]

That's correct

[worldly spruce]

But try doing this

And I will explain why the -2(2x-1) can work later

[snow verge]

so wait im confused what do I write down

[worldly spruce]

We need to solve $f(x+1) = g(x-1)$

[soft zealot]

USS-Enterprise

[worldly spruce]

What is f(x+1)

[snow verge]

2x-1

[worldly spruce]

and what is g(x-1)

[snow verge]

-4x+2

[worldly spruce]

so can we not literally replace f(x+1) with 2x-1 and g(x-1) with -4x+2 here

since f(x+1) equals 2x-1 (and g(x-1) equals -4x+2)

So it doesn't matter if we write f(x+1) or 2x-1

[snow verge]

so we need to find the x value?

[worldly spruce]

yes

[snow verge]

ok I assumed that

[worldly spruce]

We are solving when f(x+1) equals g(x-1)

[snow verge]

but how do we find x its been a really long time since ive done that

[worldly spruce]

So when the output for f of x+1 equals the output of g of x-1

[worldly spruce]

Well first do what I said

[worldly spruce]

.

what do we get

[snow verge]

im confused I'm sorry

[worldly spruce]

Okay

Say we had

$m = n$

[soft zealot]

USS-Enterprise

[worldly spruce]

and we know m = 16, and n = 40

What can we write

[snow verge]

m ≠ n?

[worldly spruce]

exactly

[snow verge]

Ohhhh

[worldly spruce]

this is a pretty stupid example from me

because it doesn't make sense

but the idea is

we know m = 16

and n = 40

so we can replace m and n with what they know they are

so we get 16 = 40

Which is clearly not true

So if we were asked to solve m = n, we say it's false

In our real example though, we have to find x that satisfy this (if any)

same thing

We need to solve $f(x+1) = g(x-1)$

[soft zealot]

USS-Enterprise

[worldly spruce]

since f(x+1) equals 2x-1 (and g(x-1) equals -4x+2)

so we can replace f(x+1) with 2x-1 and g(x-1) with -4x + 2

To get

$2x - 1 = -4x + 2$

[soft zealot]

USS-Enterprise

[worldly spruce]

get it?

[snow verge]

yeah so in the end do I just say they're not equal or do I leave it at that

[worldly spruce]

no no

as I said that was a bad example

[snow verge]

ohh

[worldly spruce]

It had nothing to do with our example

it was just to show how can we replace m and n

with something we know they are equal to

like here

we know f(x+1) EQUALS 2x-1

so we can replace f(x+1) with 2x-1

Because they are the same

[worldly spruce]

And we get an equation which we have to solve

[snow verge]

ok but how do we solve it?

[worldly spruce]

um

You don't know how to solve linear equations

[snow verge]

fuck I feel so stupid

[worldly spruce]

😅

[snow verge]

NO I SAID THAT CAUSE IT'S BEEN A HOT MINUTE SINCE I'VE DONE ANY MATH

[worldly spruce]

oof

I am not making fun or anything

I just thought we were past that since we are doing functions

[snow verge]

ive said like 3 times ive gotten to this point but I dont remember how to solve jt

[worldly spruce]

I thought you didn't know how to set it up

[snow verge]

its semester 2 of grade 12 I have done math since semester 1 of grade 11

[worldly spruce]

sorry

[snow verge]

no i knew how to set it up just not how to solve

[worldly spruce]

Well okay I'll try to explain it.

$2x - 1 = -4x + 2$

[soft zealot]

USS-Enterprise

[worldly spruce]

For you, enough is to know the idea is to get the variable alone

Here, the variable is x

We are solving for x

So we want x = something right

[snow verge]

ah wait I think im starting to remember

this is how you do it right

[worldly spruce]

ah so you do know

😂

[snow verge]

yeah its just been like over a year

[worldly spruce]

right

Well yeah

pretty much

for linear equations

(x degree 1)

all we have to know is multiplying, dividing, adding and subtracting on both sides

So here we add 1 on both sides to get $2x = -4x + 3$

[soft zealot]

USS-Enterprise

[worldly spruce]

And then add 4x on both sides to get x alone, so $6x = 3$

[soft zealot]

USS-Enterprise

[worldly spruce]

And then divide both sides by 6 to get x = ...

That's all

[worldly spruce]

for linear equations

[snow verge]

yeah so what i did?

[worldly spruce]

||in one variable||

Yes, exactly

[snow verge]

ok so is this the final question or do I have to right it our a specific way

[worldly spruce]

what do you mean

[snow verge]

like can I leave this as this

[worldly spruce]

The answer is $x = \frac{1}{2}$

[soft zealot]

USS-Enterprise

[snow verge]

ah ok

now what does find f(g(x)) mean

[worldly spruce]

well remember what we were talking about earlier

f(smiley)

we just replace every x with the smiley

same thing here

g(x) = 1 - 4x

so we just replace every x with g(x), or (1-4x)

[snow verge]

so it kinda becomes f(1-4x)?

to start

[worldly spruce]

it does become f(1-4x)

g(x) = 1-4x

so you can replace g(x) with 1-4x

and then think of the 1-4x as smiley

and do the same thing

[snow verge]

so wait im confused

wait I think i get it

[worldly spruce]

yeah?

[snow verge]

yeah ima do these 2 questions then stop for now cause its lunch and ima run out of time before my next class

[worldly spruce]

👍

[snow verge]

ok here

[worldly spruce]

Yep

Correct

[snow verge]

actually I still have about 20 minutes till the end of lunch can we do the next question

ive already figured out which are and aren't functions because of the vertical** line test but I really don't know how to explain in words and I need a little refresher on how to write out the domain and range

[worldly spruce]

Well that is incorrect

[snow verge]

shoot really

😭

[worldly spruce]

We test whether a function is injective with the horizontal line test

[snow verge]

ohhh

[worldly spruce]

To test whether it's a function, we use the vertical line test

[snow verge]

OH I MEAN VERTICAL

I meant to say vertical not horizontal

[worldly spruce]

oh right

[snow verge]

big brain moments over here rn

[worldly spruce]

I didn't even look at what you marked with x and checkmarks

yeah you're correct

[snow verge]

lolll

[worldly spruce]

Anyway

Quick definitions

Domain: all possible input values (x) the function can accept (is defined for)
Range: all possible output values (y) the function can produce

So for the domain we look at the x-axis, for the range at the y-axis

[snow verge]

yeah i know that i just dont remember how to right it out depending on what it is

[worldly spruce]

right give me a second

[snow verge]

like the weird E and R lookin things

[worldly spruce]

This is a graph of f(x) = x^2

[snow verge]

like for a is it this or am I missing something

[worldly spruce]

Yes exactly

[snow verge]

ohh ok sweet

[worldly spruce]

If each square is 1 unit

[snow verge]

yeah

ok so now that that id out of the way how do I explain in words if its a function or not

[worldly spruce]

Well, the definition is, a function from a set X to a set Y assigns to each element of X exactly one element of Y

In other words, each input (each value from the domain) must represent exactly one output value (from the range).

On graphs, you just explain the vertical line test: the vertical line must intersect the graph of the function at most once

This means, if you have for example f(x), any x in the domain you pick you always must get one y value, not more

[worldly spruce]

That's why for example b isn't a function

[snow verge]

ok I figured now do I have to write the range and domain for the non functions?

[worldly spruce]

If we pick x = 1, we get both y = 2 and y = about -1.2 right

[snow verge]

and im kinda also stumped on questions f

[worldly spruce]

Yes, relations which aren't functions still have domain and ranges

[worldly spruce]

Well what would be the domain and range

your idea

[snow verge]

x ≠ 0
y ≠ 0

[worldly spruce]

I mean

you listed values which aren't in the domain and aren't in the range

so everything else is right

[snow verge]

ngl I just realized that there's an answer key for this part of the work book and thats all it says lol

[worldly spruce]

uh

well

it's not really correct

but not really incorrect

It's just that it's asking us the domain and range

and we are saying what the domain and range aren't

so I guess we can then deduce what they are

but still

[snow verge]

lol isnt math fun

[worldly spruce]

much cleaner to write domain = $\mbb{R} - {0}$

[soft zealot]

USS-Enterprise

[worldly spruce]

and range the same

[snow verge]

ah okie

ive got 3 more pages of questions but my p3 class is about to start so im gonna have to call quits till im off work tonight

[worldly spruce]

yeah of course

best of luck!

hope I've been of help

[snow verge]

you have thank you so much

[worldly spruce]

👍

👋

!done

[final saddle]

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

[snow verge]

.close

[final saddle]

Channel closed

Closed by @snow verge

Use .reopen if this was a mistake.

[final saddle]

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.
• 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.

[earnest cobalt]

can someone help me with derivatives? im trying to learn the complex ones

is this solved correctly?

[next thorn]

looks good to me.

wait a minute 😟

[upbeat creek]

Yeah, that's correct, but I would convert 5throot(1/x) to x^{-1/5}. It doesn't change anything, but makes derivatives a bit clearer.

[next thorn]

nevermind, yeah I was having a big
1/5 - 1 = 4/5
Moment

[earnest cobalt]

i also got this one solved, looks good but if i can double check

i know its not equal to each other just wrote it this way

[next thorn]

when you get the antiderivative of a differentiated function it should throw you back to the original function. if you have some sort of software that can antiderivative anything then you can by simply plugging the f'(x) in, get the antiderivative.

[earnest cobalt]

hmm you know any website like this?

[next thorn]

there is a whole bot for this

[next thorn]

wolfram alpha

you can visit its website but you can also use it from here

starting with a ,w followed up with "integrate" then said function.

[earnest cobalt]

lemme try

[next thorn]

,w integrate -sin( (1/x)^(1/5)) * 1/5 (1/x)^(-4/5) * (-1/x^2)

[soft zealot]

Results provided by WolframAlpha

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

Requested by acarvroomvroom

[earnest cobalt]

nice

[next thorn]

oh sorry

i didnt see that you wanted to try

[earnest cobalt]

ill try the other one

[next thorn]

yeah sure.

[upbeat creek]

cool, never knew something like this existed.

[earnest cobalt]

,w integrate (5/(sqrt(1-25x^2))*tg(2x)-(arcsin(5x)(2/cos^2(2x))))/(tg^2(2x))

[soft zealot]

Results provided by WolframAlpha

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

Requested by adamborygo

[earnest cobalt]

dam i suck

[next thorn]

is this an integral?

[final saddle]

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

[earnest cobalt]

first one is the function and after the = is the derivative of it

,w integrate (5tan(2x)/sqrt(1-25x^2) - 2arcsin(5x)/cos(2x)^2)/tan(2*x)^2

[soft zealot]

Results provided by WolframAlpha

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

Requested by adamborygo

[earnest cobalt]

hm

[next thorn]

so- you either integrate the differentiated or differentiate the function

[earnest cobalt]

so i did write it correctly

[next thorn]

you can just differentiate the left side directly.

[earnest cobalt]

i might be misunderstanding that terms english is not my main lang 1s

[next thorn]

what is tg2x?

[next thorn]

please show the original problem.

[earnest cobalt]

tangent (2x)

[next thorn]

espiecally if its in arabic, i speak arabic.

[earnest cobalt]

this is the function i need to get derivative of

[next thorn]

,w differentiate arcsin(5x)/tan(2x)

[soft zealot]

Results provided by WolframAlpha

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

Requested by acarvroomvroom

[next thorn]

is that your right-side?

[earnest cobalt]

no its this

[next thorn]

maybe they are the same

[earnest cobalt]

ye this is the problem

[next thorn]

ok ill try to do this

[earnest cobalt]

idk what csc is for example

from trygonometry i only know
sin, cos. tg, ctg, arcsin, arccos, arctg, arcctg

[next thorn]

,w integrate ( (5/sqrt(7-25x^2) * tan(2x) -(arcsin(5x) * (2/cos(2x)^2))/(tan(2x)^2)

[soft zealot]

Results provided by WolframAlpha

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

Requested by acarvroomvroom

[next thorn]

ok...

[earnest cobalt]

thats not a big deal tho this one im pretty sure is correct

i also got a problem with derivatives in other execrise

may i ask here?

[next thorn]

csc is cosecant, reciprocal of sine

csc(x) = 1/sin(x)

you know it.

[next thorn]

of course.

[earnest cobalt]

this is the task in english: Examine the monotonicity, extrema, limits, continuity, concavity, and inflection points.

and where i got:

[next thorn]

im not very sure what monotonicity is...

[earnest cobalt]

this one doesnt matter tbh

last three i dont understand the most

so i did count the derivative twice

and dont really know what to do from there

from the last line i got:
so the only root is 0 cause this quadratic has no roots

[next thorn]

so wiat wait

what are you doing here

[earnest cobalt]

and dont really know how to get to concavity and inflection points.

[next thorn]

what are you exactly finding

[earnest cobalt]

from the task, im calculating first derivative

and then second derivative

[next thorn]

I'm not ery sure if your second derivative is correct but thats fine

[next thorn]

i cant confirm the second derivative

but ill assume its correct... hopefully it is

so what is this?

[earnest cobalt]

so this is the numerator of the second derivative

[next thorn]

aha

[earnest cobalt]

i assumed the denominator doesnt change the sign cause its always >0

and compared the numerator to 0

and solving this i got x=1 or x=0

but x=1 cannot be

[next thorn]

the denominator is (x-1)^4

[earnest cobalt]

ye

and x=1 is excluded from the domain

so it cannot be 0

[next thorn]

mhm i see its not a part of hte domain

[earnest cobalt]

and everything to power 4 is positive right

[next thorn]

you're right, I just didnt pay attention to the original function

[earnest cobalt]

so yeah im left with x=0 as the root to the nominator of the second derivative

[next thorn]

ok lets tackle the requirments one by one

[earnest cobalt]

so basically the second derivative is 0 only when x=0

right

[next thorn]

well x=0 makes f''(x)=0

so its def a root

but im not sure if theres any other roots

and im very certain that im not willing to- solve that equation

but

inflection points are like the last thing they're asking for

lets focus on the simpler stuff

like extrema and minima

[earnest cobalt]

alr

so for this we need first derivative right

[next thorn]

indeed, what do you think we can do

[earnest cobalt]

alr let me try

we need to solve when its 0

let me try

[next thorn]

well

we are really looking for** critical points**

which are points that occur when f'(x) is

0

undefined

or doesnt exist

doesnt exist only really applies to piecewise functions

this is not a piecewise function

so we only have to consider the** two** cases of 0 and undefined

[next thorn]

ok yes

very well

but still, you didnt check for the **f'(x) is undefiend **case

[earnest cobalt]

i dont really know how to

i only learned this

[next thorn]

oh its basically when f'(x) has has a 0 in the denominator

[earnest cobalt]

so x=1?

[next thorn]

yes.

altho its not a part of the domain, its still worth to get a habit of checking the sign of f'(x) around it.

for increase and decrease.

alright!

seeems like we're done with the extrema and minima

great job

next is... continuity

[earnest cobalt]

yea i think i know it well

[next thorn]

and limit?

limit as x approaches what?

[earnest cobalt]

actually wait

cause i think i translated the task wrongly

lemme doublecheck

[next thorn]

take your time.

[earnest cobalt]

task says: examine monotonicity, local extrema of functions, convexity, concavity, intersection points

so monotonicity we can skip its basically saying that function grows from -infinity to 0 and from 3/2 to infinity etc

extrema we did

and the hard part left

[next thorn]

well yeah

to search for concavity

and convecity ig

we need to find possible inflection points

[earnest cobalt]

all i know is that we need second derivative for that and we also equal it to 0

[next thorn]

which indeed occur when f''(x)=0

[earnest cobalt]

and roots are inflection points i think

[next thorn]

alright

thats very true

[earnest cobalt]

assuming my solution is correct that would be x=0

[next thorn]

i think the root to the f''(x) =0 is indeed x=0 (and only that)

[earnest cobalt]

cause we remove x=1 solution cause its not in the domain

[next thorn]

so yeah

its worth to mention

that there are multiple requirments for a point to be considered and inflection point

its only one of hte requirements that f''(said point)=0

there is a requirment that the point has to be defined within f'(x)

[next thorn]

which it fortuanetly is! considering that when you plug in x=0 in f'(x) you just get- a defiend value

alright so

now that you're almost sure that x=0 is an inflection point

your next step is...

actually- to avoid this being a yak shave

do you know what our next step should be and why?

[earnest cobalt]

i think instead of doing =0 we need to do >0 and <0

[next thorn]

we already know that x=0 is an inflection point. what do we do next and why

[earnest cobalt]

wait so

we just draw the polynomial function

read where its positive and where negative

and we basically got it

am i right?

[next thorn]

not quite. "read where its positive and where negative" is lowkey gonna tell you when the function decreases and increases

but wait! we dont even need to draw

[earnest cobalt]

wouldnt that be true for first derivative?

[next thorn]

do you know what an inflection point is?

[earnest cobalt]

where it changes from one to another

[next thorn]

indeed!

so wej sut found our inflection point to be x=0

all you have to do now

is know

when is the function concave up

and when is the function concave down

theres a very special property about the inflection point, **the sign of ** f''(x) changes around it

from negative to positive or from positive to negative

if the signs do not change around uit, it is not an inflection point.

but here's the thing

if f''(x) is positive then the function was concave down (convex up)
if f''(x) is negative then the function was concave up (convex down

so, search the sign of f''(x) around the inflection point (left and right)

and see if its positive or negative

this way, you can find convexity and and concavity.

[earnest cobalt]

so should i just put 1/2 and -1/2 for the x and see the sign
or
solve for >0 and <0

[next thorn]

yes!!!

any numbers within the domain will work.

[earnest cobalt]

yes for which option

[next thorn]

.

[next thorn]

you can graph or do this algebriaclly.

i offered you the algebriac approach

[next thorn]

solving for >0 and <0 is not very necessary

if a value after the inflection point is positive for example

then ALL of the values after the inflection point is positive

you dont have to solve an inequality

just solve for one value and others should apply. (with respect to where they are at relative to the inflection point)

it also kinda depends on what you've been taught... if you've been taught to do this algebriaclly then you should do what i said. if you were isntructed to solve this with a graph then you can try and see how the graph looks like.

after you're done checking, i will clean up the algebriac method for you to understand better.

[earnest cobalt]

i got to this

[next thorn]

that is correct

wait a minute, what did you solve the inequality for

whats
2x(x^2-3x+3)

[earnest cobalt]

this

[next thorn]

is that your second deritaive simplified?

[earnest cobalt]

shortened version of numerator

yes

but im concerned that getting rid of that (x-1) early is a mistake

[next thorn]

ok but you lowkey need your denominator as well. but lucky for you, the denominator is laways positive and in the denominator and it cant be 0 thus it wouldnt affect the sign or the inequality.

but dont rely on that luck every time 😔

[next thorn]

yit worked in this very specific situation.

[earnest cobalt]

i got rid of it under the condition that its always positive hence it doesnt change the sign of the second deriv

[next thorn]

you also needed to make sure its not 0

multiplying an inequality by 0 forces it to change its parameters.

[earnest cobalt]

it cant be zero when i got domain x not equal 1

[next thorn]

2 > -3
multiply both sides by 0
0 > 0

[next thorn]

amazing

so now we got that

the second derivative is positive from -inf to 0

what does that translate to in terms of concavity

[earnest cobalt]

that its convexity at this range

and concavity at 0 to inf

but the problem is, its not correct

[next thorn]

you need to specify

is it convex up or down

[final saddle]

Channel closed

The owner is missing!

[next thorn]

what????

the fuck?

[earnest cobalt]

im not missing

[next thorn]

ignore the bot

[earnest cobalt]

isnt that the case?

[next thorn]

wel the phrase are inaccurate

there is a difference between concave up and cave down and convex up and convex down

[earnest cobalt]

either my language doesnt specify that or i just dont have this much detail yet?

[trail mango]

[trail mango]

yes this is correct

there is no "convex up or down"

concave up and concave down are terms that calculus education have tried to push on everyone

[next thorn]

تقعر و تحدب لاسفل و لاعلي

it really depends on your teacher

[trail mango]

well nobody says convex up and convex down

[next thorn]

how is this not correct

its idneed "convex" from -inf to 0 and concave from 0 to inf

[trail mango]

but i just say convex and concave

[next thorn]

well my education uses the 4 terms.

[earnest cobalt]

when i checked it says:
convexity: (-inf to 0) and (1 to inf)
concavity: (0 to 1)

this 1 is important somehow then

[next thorn]

isnt the original function x^3/(x-1)

[earnest cobalt]

it is

[next thorn]

the function doesnt have inflection points at x=-1 and x=1.

it doesnt go fronm concave to convex anywhere execpt for x=0

[earnest cobalt]

when you plot this function you can see this is correct tho

[next thorn]

where?

[next thorn]

what the fuck

[earnest cobalt]

there is the asymptote at x=1 and it changes things i guess

[next thorn]

can you show me the function you inputted?

[next thorn]

blud what

[final saddle]

@earnest cobalt Has your question been resolved?

[earnest cobalt]

no

[next thorn]

oh...

wait

okay i was just

too zoomed in

yes.

[earnest cobalt]

haha

i just dont know what to do to get to this solution with 1

[next thorn]

yes i might have forgotten that one of the "can-be" for an inflection point is "f''(x)=0 or undefined"

we forgot to check for hte undefined case.

[earnest cobalt]

yep, and how to do that?

[next thorn]

the second derivative has (x-1)^4 in the denominator

f''(x) will be undefined if the denominator =0

like any value

1/0

2/0

x/0

they're all undefined because the denominator = 0

so you're gonna check for when (x-1)^4=0

[earnest cobalt]

x=1

[next thorn]

which indeed outputs x=1 as an option

yes.

so

we check the signs from - inf to 0 then from 0 to 1 then from 1 to inf

wait a minute...

f'(1) is not defined.

we cant call this an inflection point

[next thorn]

but we still have to do this

[earnest cobalt]

cause its not an inflection point

[next thorn]

i will make sure to clean up everything in the end

[next thorn]

one of the requirments for a point to be an inflection point is that the first derivative at that point is defined

f'(1) is not defined

and thus we cannot CALL it an inflection point

but we still have to earch for f''(x) sign around it

because as you can see, the function jumped after x=1 and that gives proposes a possibility of f''(x) changing signs.

is the reasoning clear?

we just cannot call it an inflection point, but we should still consider that f''(x)'s sign might change around it.

[earnest cobalt]

reasoning is clear but i dont know how to get to answer still tho

[next thorn]

what answer?

[earnest cobalt]

convexity: (-inf to 0) and (1 to inf)
concavity: (0 to 1)

[next thorn]

well

lets ignore that you have the answer

and go back from the step of "find inflection points"

we jsut found that x=0 is an inflection point

and because the function might change its behavior around x=1

[earnest cobalt]

so solving the second derivative when its =0 got us 2 roots

x=0 and x=1

since x=1 is not in the domain we removed that root but it still impacts the convexity and concavity and i dont know how to see the impact

[next thorn]

we check the sign of f''(x)

before 0

after 0 and before 1

and after 1

[next thorn]

you dont need to see it. you just assume that it might have an impact hten you test if there is actually one

[earnest cobalt]

okay so the way to test if it impacts it is to calculate value arround this undefined point

[next thorn]

function behaviour gets nanoying when it comes to the amount of things you have to check.

[earnest cobalt]

?

[next thorn]

indeed!!!!

[earnest cobalt]

okay let me try

[next thorn]

yes

remember, you have x=0 and x=1

search in every area between them

and after them

and before them

[earnest cobalt]

so ill try f''(x) for
x=-1
x=1/2
x=2

this fine?

actuallt this

[next thorn]

that is indeed fine-

thats still fine

[earnest cobalt]

alr let me calc that

[next thorn]

x=-1 is before 0
x=1/2 is between 0 and 1

x=2 is after 1

which is exactly the areas around your inflection point + sus point

[earnest cobalt]

can i input those values to that simplified version aka this:

actually no cause it doesnt have that (x-1)

lets not overcomplicate it

[next thorn]

welll

welkl

well

you know (x-1)^4 is always positive

and thus it wont affect the sign of whatever you're gonna input

[earnest cobalt]

yea but im not talking about the denominator

[next thorn]

so you can indeed exclsue it

[earnest cobalt]

rather this:

we had an extra x-1 at the start

its not included in the simplified version

[next thorn]

well, it didnt go well when we exluded the possibility of undefined values did it?

so yes. keep it

it affects the sign. pretty hard.

oh yeah by the way

nevermind

[earnest cobalt]

x=-1 is 28 = positive
x=1/2 is (-0.5*1,75) = negative
x=2 is 4 = positive

[next thorn]

so convex --> concave -->convex

[next thorn]

which correct!

is

[earnest cobalt]

so x=-1 corresponds to -inf to 0 and its positive hence its convex

x=1/2 corresponds to 0 to 1 and its negative hence its concave

and convex for 1 to inf

alr we got it

[next thorn]

indeeddd

[earnest cobalt]

took only 1hour

[next thorn]

is there anything else on the list

[next thorn]

its okay, we both learned from our mistakes

ill give you a summarization for the logic we've used

[earnest cobalt]

there is actually one more exercise type i dont understand at all

[next thorn]

i can try to explain if i udnerstnad myself

so do send it.

[earnest cobalt]

exercise is just to calculatge

and from the requirements for the test im pretty sure i have to use de hospitals rule

problem is i have no idea how

[next thorn]

what is

[earnest cobalt]

cotangent(x)

[next thorn]

well

isnt cotan(x) = 1/(tan(x))

[earnest cobalt]

dam

this hits

[next thorn]

😭

its okay

yeah so

if you turn cotan to 1/(tan)

and have your limit be

[earnest cobalt]

alr let me write this

[next thorn]

okay sure do

[earnest cobalt]

alr and when we put 0 in

top is 0

and tg is undefined at 0?

[next thorn]

im very sorry but i dont like the way you type tan

i dont think anyone does it that way

[earnest cobalt]

i belive its only in poland this way

[next thorn]

you might want to get a calculator for this

[earnest cobalt]

its 0 i just checked mb

[next thorn]

or say that tan(0)=sin(0)/(cos(0))

[next thorn]

ok so

polugging 0 in

outputs

0/0

[earnest cobalt]

so its 0/0 and we can use de hospitals then

[next thorn]

do you know what this means

[next thorn]

exactlyy!!!!!