help-33

[brave marsh]

Do you have any question?

[marsh citrus]

@iron ocean Has your question been resolved?

[iron ocean]

hi guys how will I know its graph manually with plotting points

also

I have to get the reason why the opening is like that

Idk why can someone explain to me..

[gritty pumice]

Why the parabola opens up?

What do you mean?

[marsh citrus]

@iron ocean Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

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

[pallid tapir]

continuity of f requires its limit as x tends to 0 to equal b

f is otherwise continuous at all nonzero x

[buoyant jetty]

can u write in latex

which definition are you talking about?

elaborate

@pallid tapir

[pallid tapir]

writing $g(x)=\frac{e^{5x^2-x}-1-x}x$, we require that $\lim_{x\to 0}f(x)=\lim_{x\to 0}g(x)=f(0)=b$

[elfin berry]

Flip

[pallid tapir]

the first equality is forced by the nature of f

[pallid tapir]

you had pinged me while I was typing lol

[buoyant jetty]

what definition is that

from a book or ?

[pallid tapir]

a function is continuous at a point if its limit agrees with its evaluation

i.e. $f:D\to C$ is continuous at $x_0$ if $\lim_{x\to x_0}f(x)=f(x_0)$

[elfin berry]

Flip

[buoyant jetty]

okay is that definition from a pdf?

[pallid tapir]

the limit obviously must exist to say such a thing

[buoyant jetty]

is that a theorem sir

[pallid tapir]

it's a definition

[buoyant jetty]

which one

[pallid tapir]

the one?

[buoyant jetty]

did you have it in pdf or

[pallid tapir]

no lol

[buoyant jetty]

okay

is it publicly available or is common knowledge

or

[pallid tapir]

if you have any doubts you should check with yourself that that's the definition

it's common knowledge

[buoyant jetty]

okay

so everyone knows that

[pallid tapir]

yeah

is this calculus?

[buoyant jetty]

except me I'm the exception

[pallid tapir]

this is calculus yes?

[buoyant jetty]

intro to anal

[pallid tapir]

that's the definition for continuity at a point in real analysis

[wise musk]

[pallid tapir]

general continuity for a function between topological spaces is less recognizable

but still another context for continuity

just not at a point

[buoyant jetty]

a what

lets stick with non high caliber machinery for the job

[pallid tapir]

brother I'm just sharing what I know

[wise musk]

When we apply limit
We get the 0/0 form
I think we should apply l hospital's

[buoyant jetty]

okay let's do lhopital

no fuck no

lets use the definition of derivative at a point

[pallid tapir]

what does that do for you?

[buoyant jetty]

I mean that's for the second part the derivative BIT

when it says find f'(0) right?

@pallid tapir

for finding continuity, I'm unsure if there is a theorem for continuity at a point or a definition we can use

[wise musk]

There we can use (v'u - u'v)/v² as well I think

[buoyant jetty]

you mean quotient rule

[wise musk]

But I suppose this should also get you the same answer

[buoyant jetty]

<@&286206848099549185>

@fluid mica

[pallid tapir]

you still want f(0), which f'(0) does not give

[slate yarrow]

hello ren,whats the question?

[buoyant jetty]

if we found differentiability at a point then continuity comes bundled

hello convergence

[pallid tapir]

that's correct, but not insightful

[slate yarrow]

oh i see

[buoyant jetty]

I am not sure how to start flip

[pallid tapir]

why is lhopital off the table?

[buoyant jetty]

which are the definitions we have on the toolbox?

[wise musk]

What I'm saying

[buoyant jetty]

we can use it I thought it wasn't the way to go

sorry

[mellow sable]

use taylor series

[buoyant jetty]

@wise musk

[pallid tapir]

you seemed really passionate about disregarding it

was just curious why

[wise musk]

No problem 😌😌

[buoyant jetty]

I think lhopital is the only way to go, I'm not sure if we can do Taylor expansion here

please let me know what u guys think

[slate yarrow]

you can use taylor also

[pallid tapir]

the negative 1 kills the constant term in the taylor expansion

and the x in the denominator is killed off as well

[buoyant jetty]

kills? can you show a latex?

[pallid tapir]

can I choose not to?

I'm on mobile

it should be kinda clear how it plays out, not to be dismissive

[buoyant jetty]

hmmkay

I just want to be sure I don't miss any info

and to have the definitions and theorems on my toolbox 🧰🛠

[pallid tapir]

kills, in the algebraic sense, like "cancels out" but more dramatic

[slate yarrow]

$\frac{\left(1+\frac{5x^2-x}{1}+\frac{(5x^2-x)^2}{2!}+\dots \right)-1-x}{x}$

[elfin berry]

convergence

[slate yarrow]

we get

$\frac{x\left((\frac{5x-1}{1}+\frac{x(5x-1)^2}{2!}+\dots \right)-1)}{x}$

[pallid tapir]

missing a pair of brackets

[slate yarrow]

oh yeah

[elfin berry]

convergence

[buoyant jetty]

ye

[slate yarrow]

then we cancel the x

[pallid tapir]

actually we kill the x

[slate yarrow]

kill cancel byebye x are all the same

[cold pecan]

byebye x

[buoyant jetty]

now what(?)

I got lost

[pallid tapir]

it now comes down to evaluating the series at x=0

[buoyant jetty]

= b

(?)...

[wise musk]

Yea

[slate yarrow]

$\lim_{x \to 0^{+}\lor 0^{-}}\left((\frac{5x-1}{1}+\frac{x(5x-1)^2}{2!}+\dots \right)-1)=b$

[elfin berry]

convergence

[wise musk]

We get b = -2 right?

[slate yarrow]

yes

[pallid tapir]

interesting notation

[slate yarrow]

im too lazy to tupe both those limits

[pallid tapir]

valid

[buoyant jetty]

how is b two

[slate yarrow]

-2

[buoyant jetty]

-2

[pallid tapir]

the series is -1 + (5x-1) + x(junk)

[slate yarrow]

indeed

[pallid tapir]

x goes to 0, killing the junk

and what remains is -1 + 5(0) - 1 = -2

from a real analysis perspective I can appreciate that this might not be a complete argument

[buoyant jetty]

can someone share the latex

[pallid tapir]

this is more like what a calculus student would do

but it gets the point across

[buoyant jetty]

oh yeah b is -2 I did it mentally

[pallid tapir]

also might not be a bad idea to tex it yourself

[buoyant jetty]

no need to show me the latex is easy

[cold pecan]

that was really clean

[buoyant jetty]

so how do I find the derivative and find if it's continuous?

<@&286206848099549185>

[pallid tapir]

why do you ping the helpers every time lol

[pallid tapir]

on all nonzero x, f is defined as a mishmashing of other continuous functions, and is continuous

the limit as x approaches 0 of f is shown to exist, so we simply define f(0) as this limit, and now f is defined on all of R and is continuous

[buoyant jetty]

in human what does that mean

[pallid tapir]

continuity is no longer a problem

[buoyant jetty]

mishmashing?

[pallid tapir]

mishmashing felt pretty human

[cold pecan]

peak comedy

[pallid tapir]

it's the sum, product, composition of various continuous functions

[cold pecan]

@buoyant jetty try understanding by this

[slate yarrow]

filp has one of the most interesting terminology..

[buoyant jetty]

I dont understand

can you show in latex

[cold pecan]

you just found that b = -2 which means it is a subset of R and according to question f is continuous on x = 0

read the question carefully

[buoyant jetty]

about the derivative what

the differentiability at a point what

[cold pecan]

it is already continuous, that's not what is asked in the question

[buoyant jetty]

hey takezo

we found b in R such that f is continuous on x = 0

about derivative at a point....

how to compute f'(0)

[slate yarrow]

two things to take note of

• f(x) for all values of x \neq 0 is a continuos function
• second its is stated in the question that f(x) is continous at x=0
  from these two points we can say that f(x) is continous for all real numbers

[wise musk]

f(x) is given
First find f'(x)
Then put x= 0

[pallid tapir]

the derivative of a function at a point where it is not continuous doesn't exist. so we need b to be found in order for f to be continuous at 0. now that f is continuous at 0, the task makes sense

[buoyant jetty]

is a homographic function

[pallid tapir]

homographic?

you mean like, homeomorphic?

[cold pecan]

answer is zero maybe

[pallid tapir]

or holomorphic?

[cold pecan]

nah just say heterotrophs

[slate yarrow]

why did abstract algebra come here...

[pallid tapir]

it's topology and he said a word I'd never read before to be fair

[pallid tapir]

huh

[buoyant jetty]

sorry maybe is a term in spanish

[pallid tapir]

neat

[buoyant jetty]

sorry I'm sorry

[pallid tapir]

that's very specific lol

[wise musk]

Did anyone calculate f'(x)?

[pallid tapir]

no

[buoyant jetty]

I am trying

but I don't know how

[slate yarrow]

oh btw for peice wise functions you have to check the RHD/LHD also to see if the derivative exist

[cold pecan]

I am maybe wrong but I think that at x = 0 we get b. then f(0) = b = -2. So f'(0) = 0

[wise musk]

I tried (v'u - u'v)/v²

[pallid tapir]

that logic would apply for all functions

[wise musk]

No I think that's wrong

[buoyant jetty]

if Saitama is not finding f' what is it expected from. me

[wise musk]

Bro I swear I'm trying but I'm not getting the final answer 😂

[pallid tapir]

since f(x) isn't just -2 for all x in any neighborhood of 0, we don't have information on f'(0) just from that

[pallid tapir]

so f is still a mishmashing of differentiable functions right. we know how to differentiate sums, products, and compositions of differentiable functions, so we can attain a closed form for f'(x) for all nonzero x

[pallid tapir]

as convergence said, however, we need to still verify that f'(0) exists, i.e. that lim_(x->0) f'(x) exists

for this to be the case, left- and right-side limits must exist and agree

but based on how the question is worded

it probably does exist

[wise musk]

Ok just a min

Are we sure if this is differentiable at zero?

[pallid tapir]

it'll be another limit

you see there's an x^2 in the denominator

[wise musk]

I tried differentiating the function f(x)
But the problem is that there is x² in the denominatoe

[wise musk]

Exactly

[pallid tapir]

we can probably still kill it

[wise musk]

I've been stuck here for near about 5-10 mins now

[pallid tapir]

taylor series expansion might work again

[wise musk]

How tho?

[wise musk]

Oh yea

[cold pecan]

I am myself very curious to know the answer and how it came

waiting for you guys

[slate yarrow]

damn this was quite long...

[wise musk]

Even after the expansion it's something like this

Ohh wait

O missed a bracket

[pallid tapir]

found my own mistake as well thankfully

[cold pecan]

x^2 remains

[wise musk]

x remains*

[cold pecan]

it will cause the same prob ig coz we have to put x=0 right ?

[wise musk]

Yes

[cold pecan]

[pallid tapir]

legible?

my mistake is I need one more term from the taylor series than I took lol

so this is at the very least on its way to a solution

[wise musk]

Ohh

[pallid tapir]

the glyphs below the last "junk(x)" on the bottom right is "lol"

[wise musk]

I think that should be the solution right?
Put x=0 and that's it
You get 6

[mellow sable]

I did it a little differntly

[pallid tapir]

the first term of the taylor series doesn't contribute a 0 so it must be heard

[mellow sable]

i just applied taylor to original f(x)

[pallid tapir]

lhopital?

ah

[mellow sable]

cancelled the x in denominator

[pallid tapir]

and then differentiate

[mellow sable]

then differentiated

[pallid tapir]

that's much cleaner

[mellow sable]

i was getting 5.5

[cold pecan]

🫡

[pallid tapir]

makes sense

looks like it

my series gives a -1/2 to the total, plus a 6 yields 5.5 as well

[mellow sable]

Oh ok

[pallid tapir]

from the whiteboard that is

[cold pecan]

so the answer is 6

[pallid tapir]

i.e. junk(0) = -1/2

[wise musk]

Bro I got lost in brackets😩
Finally found mistake🤦‍♂️

[pallid tapir]

I agree with 5.5 I think

[wise musk]

Okk👍

[cold pecan]

@buoyant jetty do you have an answer to check ?

[mellow sable]

also the differentiation of f(x) should be undefined

as at exactly x = 0, f'(0) is 0

cause b is constant

[slate yarrow]

Yeah it's not defined at 0

[pallid tapir]

that's not how that works, it's a singleton point

[cold pecan]

well that's what I though few minutes ago

[pallid tapir]

you can have a differentiable function g on all of R, remove a point, then define f as basically just g but specify f(a) = g(a) or whatever

[cold pecan]

undergaduate helped the pre university understand it

[pallid tapir]

f is just g, but treating the point as its own thing as you claim asserts f'(a) = 0

which need not be true

[mellow sable]

yeah, ik all that, its a removable discontinuity of derivative function

[pallid tapir]

like so

[mellow sable]

i might be wrong but

[wise musk]

@pallid tapir I think the junk will have the value as zero

[pallid tapir]

it will not, sadly

[wise musk]

I'm not sure
Pls correct me if I'm wrong

[wise musk]

How tho?

[pallid tapir]

the first term is half of (5x-1)^2, evaluating to 1/2

the poly in front contributes a negative

[wise musk]

No but it is also multiplied by an extra x beside it

[pallid tapir]

sorry, it isn't

[wise musk]

Wait let me check again🥲

[pallid tapir]

np

from how I decided to write it, the indexing starts at 0

so the first term in the series corresponds to n=0 and gives (5x-1)^2 / 2

afterwards there's x's from the (5x^2-x)^n

[cold pecan]

I love maths fr it's so interesting

[wise musk]

Got it thanks

[pallid tapir]

this junk(x) thus contributes (10x^2 - x - 1)(5x-1)^2/2 evaluated at x=0

hurray

[wise musk]

Hurray😂

How long has it been?

[pallid tapir]

joy is ours

idk

for me, 1h45m

[wise musk]

Oh my god

Worth it tho

[cold pecan]

1 hr 46 min

joining this server was worth it

time well invested

[wise musk]

Yea😌

[cold pecan]

but having a whiteboard is cool

[pallid tapir]

it is

speaking of worth things

whiteboards

[cold pecan]

I want to have it too once my college starts

[pallid tapir]

hell yeah get one for your dorm

[cold pecan]

[pallid tapir]

I brought mine to my dorm and a local REU it was hype

[cold pecan]

it just gives the feel of solving things

[pallid tapir]

ye

virtual whiteboards are fine as well, like I have a tablet or whatever

but it's not quite the same

[cold pecan]

nothing compares to the OG Physical

[pallid tapir]

pros and cons to both though

ye

also

chalkboards are sick

hagoromo chalk is lovely

[slate yarrow]

this

[pallid tapir]

made me like chalkboards

[cold pecan]

yeah it's better I saw it on yt

[pallid tapir]

the feel of it though

much much nicer than the cheap stuff

literally never gave me shivers writing with hagoromo

[cold pecan]

it's smooth

[slate yarrow]

damn i onw day hope to use hogoromo

[pallid tapir]

you gotta try homoromo

my uni had it

[cold pecan]

at the end the ans was zero

[pallid tapir]

it was -2 and 5.5 respectively though

[slate yarrow]

isnt it -2 and DNE?

[cold pecan]

and derivative of them is 0

[mellow sable]

i think so too

[slate yarrow]

since LHD/RHD \neq to derivate at 0

[cold pecan]

not over yet it seems

[pallid tapir]

I think I disagree because it worked out nicely to evaluating a polynomial

proof by computer approximation also says f(-small) = f(small) = 5.5

[wise musk]

It did

[pallid tapir]

charge your phone

[slate yarrow]

yeah i agree about that but for the derivative to exist RHD/LHD should have the same value as the derivative at that point

[mellow sable]

exactly my point

[pallid tapir]

ah fuck I see what you mean lol

ok the question is worded in such a way where we should expect its derivative to exist

like how a number b can be found so that f(0) being defined as b makes it continuous at 0, so too can c be found so that f'(0) being defined as c makes f' continuous

I guess that's what I'm really saying

but not necessarily arguing that's how we should be interpreting it

[wise musk]

Ok so what I'm trying to say is that s should not be 0.0000001
It should be 0.(infinite zeroes)01
If you understand what I mean
I'm pretty sure that then the lhd will be equal to rhd

[mellow sable]

yeah LHD = RHD but not equal to f'(0)

[pallid tapir]

actually giving it too many 0's makes the computer holler in pain

[cold pecan]

have to take a break

[wise musk]

Yea no prob

[wise musk]

But f'(0) came out to be 5.5 so i think I should be same no?

[wise musk]

Lol me too

[mellow sable]

f'(0) = 0

because b = -2

[pallid tapir]

f'(0) is definitely not 0

[wise musk]

Yea

[mellow sable]

why??

[pallid tapir]

knowing the value of f(a) does not give you any information on f'(a)

only knowing what f(x) is for x in arbitrary neighborhoods of a gives that

in arbitrary neighborhoods of 0, f acts precisely like the nicer function

the derivative is defined as a limit anyway, which is why this is the case intuitively

[mellow sable]

maybe you are right..

[wise musk]

Check here

[pallid tapir]

you reminded me to close desmos ty

[cold pecan]

graph explains a lot more than the numbers itself sometimes

[wise musk]

Yea

[mellow sable]

hover at x= 0 on the derivative graph

it will show undefined

i kno

[pallid tapir]

it's because "f" in the desmos screenshot is only defined on the nonzero reals

but inserting the missing point into f makes it defined

[mellow sable]

and we are talking about neighborhood of 0

[pallid tapir]

yeah, but it is still the case that derivatives are only defined at least where the function itself is defined

the two domains aren't exactly equal because e.g. sqrt(1-x^2) has closed domain, but its derivative's domain is open

I gotta take a drive, farewell for now

[marsh citrus]

@buoyant jetty Has your question been resolved?

[marsh citrus]

@buoyant jetty Has your question been resolved?

[buoyant jetty]

<@&286206848099549185>

[buoyant jetty]

<@&286206848099549185>

[red nimbus]

is this the question?

[buoyant jetty]

hi

im glad you are back

[red nimbus]

cool

so many people here yet unsolved lol

i can try to also help if you are willing

or rather you need to state questions

looks like your channel was conquered

by intruders

[buoyant jetty]

we got b

but I was trying to find the derivative at the point 0

you know what I mean

[red nimbus]

yea

[elfin berry]

bacc

[red nimbus]

x = 0 is undefined for the derivative

but you can figure out the limit at 0 and define a piece wise derivative

[elfin berry]

bacc

[red nimbus]

and c is here the limit of this

prob wasn't as helpful lol

[lost haven]

omfg what is this

[buoyant jetty]

H definition? @red nimbus

[red nimbus]

you can't because x = 0 is undefined

[buoyant jetty]

I need more guidance

I dont get it

you mean to use quotient rule or lhopital

[red nimbus]

L'hopital sounds good

and yes we would need quotient rule

[red nimbus]

no first we differentiate it

[buoyant jetty]

quotient rule?

[red nimbus]

,, \frac{\dd}{\dd x} \left ( \frac{e^{5x^2-x}-1-x}{x} \right ) = \left ( \frac{ \left [ e^{5x^2-x} \cdot (10x-1) - 1 \right ] \cdot x - (e^{5x^2-x}-1-x) \cdot 1}{x^2} \right )

[elfin berry]

bacc

[buoyant jetty]

yea

now wat

[red nimbus]

,, = \left ( \frac{10e^{5x^2-x}x^2-e^{5x^2-x}x-e^{5x^2-x}+1}{x^2} \right ) = \left ( \frac{ e^{5x^2-x}(10x^2-x-1)+1}{x^2} \right )

We can use L'hopital

but let me think if there is a simpler way

[elfin berry]

bacc

[red nimbus]

,, \lim_{x \to 0} \left ( \frac{ e^{5x^2-x}(10x^2-x-1)+1}{x^2} \right )

[elfin berry]

bacc

[red nimbus]

now we can use L'hopital

oh btw

are you still here

[buoyant jetty]

y

[red nimbus]

i dont wanna do work for nothing

[buoyant jetty]

?

[red nimbus]

what

i am just making sure if you are here and if everythings clear so far

[buoyant jetty]

is not clear

[red nimbus]

you can interrupt anytime it's not clear

otherwise what am i doing bro

[buoyant jetty]

is not

clear

bro

[red nimbus]

can you tell what's not clear

at which step

[buoyant jetty]

https://media.discordapp.net/attachments/907022703230345237/1281227324200845392/587715738484211754.png?ex=66daf36e&is=66d9a1ee&hm=e09c9509d5f9cd6e368784fb3212e6b53b83b4f87c036a4dc7a519672d0cce63&

how to continue after this

[red nimbus]

...

i was about to continue

[red nimbus]

so here you can notice for x = 0

you get 0/0

we can use l'hopital

[buoyant jetty]

l.h

[red nimbus]

ok good

[buoyant jetty]

product rule

[red nimbus]

,, \lim_{x \to 0} \left ( \frac{ e^{5x^2-x} (10x^2-1) \cdot (10x^2-x-1) + e^{5x^2-x} \cdot (20x-1)}{2x} \right )

[elfin berry]

bacc

[buoyant jetty]

ye

[red nimbus]

f'g + fg'

Now I will again factor

the exp term

and group somethings

[buoyant jetty]

okay

I'm eating a crossaint sorry if I don't speak much but I am following

[red nimbus]

,, \lim_{x \to 0} \left ( \frac{ e^{5x^2-x} \left [ (10x^2-1) \cdot (10x^2-x-1) + (20x-1) \right ] }{2x} \right )

[elfin berry]

bacc

[buoyant jetty]

exactly

[red nimbus]

,, \lim_{x \to 0} \left ( \frac{ e^{5x^2-x} \left [100x^4-10x^3-20x^2+21x \right ] }{2x} \right )

[elfin berry]

bacc

[red nimbus]

I use a calculator btw for these multiplications

[buoyant jetty]

factor 10x

from denomi

[red nimbus]

we can factor x

[buoyant jetty]

photomath

[red nimbus]

ye

,, \lim_{x \to 0} \left ( x \cdot \frac{ e^{5x^2-x} \left [100x^3-10x^2-20x^1+21 \right ] }{2x} \right )

[elfin berry]

bacc

[red nimbus]

,, \lim_{x \to 0} \left ( \frac{ e^{5x^2-x} \left [100x^3-10x^2-20x^1+21 \right ] }{2} \right )

[elfin berry]

bacc

[buoyant jetty]

21/2

[red nimbus]

wait

[buoyant jetty]

🚬

[red nimbus]

Oh I wrote accidentally 10x²-1 instead of 10x-1

,, \lim_{x \to 0} \left ( \frac{ e^{5x^2-x} \left [100x^3-20x^2+11x \right ] }{2x} \right )

,, \lim_{x \to 0} \left ( \frac{ e^{5x^2-x} \left [100x^2-20x^1+11 \right ] }{2} \right ) = \frac{11}{2}

[red nimbus]

the rest is fine then

[buoyant jetty]

-11/2

[elfin berry]

bacc

[buoyant jetty]

right

[elfin berry]

bacc

[red nimbus]

So the limit is 11/2

but f'(0) is undefined still

so the only way would be to patch it

by defining for x = 0 the value c = 11/2

and then you would have f'(0) = 11/2

[buoyant jetty]

ye

[red nimbus]

,, f'(x) = \begin{cases} \frac{ e^{5x^2-x}(10x^2-x-1)+1}{x^2} & x \neq 0 \ \frac{11}{2} & x = 0 \end{cases}

[elfin berry]

bacc

[buoyant jetty]

okay

[red nimbus]

yea that's it

[buoyant jetty]

. close

.close

[marsh citrus]

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[near elbow]

guys, this one is compound angle formula. and the person on yt says -B is on quadrant 4, i dont understand why is it quadrant 4 and not quadrant 3?

[raw hawk]

Do you count quadrants clockwisely or anticlockwisely

@near elbow

[near elbow]

anticlockwisely

[raw hawk]

Yes

[near elbow]

uhmm

[near elbow]

so how do i know that -B is in quadrant 4?

[raw hawk]

, rccw

[near elbow]

yess

[raw hawk]

Trigonometry

All trig ratios are positive in 1

Right?

[near elbow]

yess

[raw hawk]

In fact this is unnecessary

Just do this

Ok so 90 degrees will be in what quadrant

[near elbow]

in quadrant 1

[raw hawk]

Because you move 90 degrees in the anticlockwise direction

[near elbow]

yes!

[raw hawk]

So -90 degrees will be rotating 90 degrees in the opposite direction

So where will an angle like -40⁰ be

[near elbow]

yes

[near elbow]

in quadrant 4

ohhhhh

[raw hawk]

🎉

[near elbow]

and B must be less than 90 right

[raw hawk]

Yes

[near elbow]

is it the rule itself

[night mica]

,rotate

[elfin berry]

Couldn't find an attached image in the last 10 messages.

Sep 5, 2024, 1:05 PM

[raw hawk]

If B is acute -B will be in quadrant 4

[near elbow]

ohhh

will the formula work for obtuse angle?

[raw hawk]

Yes

[near elbow]

ohh so it works for any angle?

[raw hawk]

sin(-B) =-sinB
Is a property of sine

[raw hawk]

Yes

[near elbow]

oooooh

[near elbow]

is it different with cos and tan?

[raw hawk]

It's different with cos

Cos-B = CosB

[near elbow]

ohhhh

[lucid turret]

cos(-x) = cosx

[raw hawk]

Tan( -B) = sin(-B)/cos(-B)

[lucid turret]

sin(-x) = -sinx

[near elbow]

okayy thank you soo muchh

[lucid turret]

tanx = sinx/cosx

[near elbow]

thank youuu!!

[raw hawk]

So can you tell what tan(-B) = ...

[near elbow]

-sinB/cosB !

😆😆

[raw hawk]

Yes

[near elbow]

oh wait is it also equals to -tanB

[raw hawk]

Yes

[near elbow]

ohhhhhhh

yay i understand it now

thankyouthankyou!!

😆😆

.close

[marsh citrus]

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[unkempt gull]

I am struggling to start this equation
(x - 2) / 5x = (x + 3) / 6x

Can I just use a LCD of x? or does it need to be something like 6x?

[unkempt gull]

.close

[marsh citrus]

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[marsh citrus]

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[dusky viper]

not really sure what to do

through some various means that im not entirely sure are correct, i ended up with |z|=1=|w| not sure if it helps

dw

.close

[marsh citrus]

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[nova pier]

Can anybody teach me standard form for 8th grade IGCSE

[marsh citrus]

Please don't occupy multiple help channels.

[eager tinsel]

standard form of what?

[nova pier]

Just standard form

It is a chapter in my text book

[late geode]

there are standard forms for multple things

be more specific

[eager tinsel]

do you have any specific problem

?

[lucid turret]

Ok

ICSE

[wheat rover]

like how you write 25 = 2*10 +5 ?

[nova pier]

No but I need to learn the topic

[lucid turret]

isn't IGCSE just ICSE?

[nova pier]

I think so

[eager tinsel]

no

[eager tinsel]

it is not

[lucid turret]

When 25 is divided by 10, it leaves a remainder of 5 and quotient 2.
So 25 = 2*10 + 5

[wheat rover]

looks like it is acc to the syllabus

[late geode]

Are the notes in your book insufficient?

[nova pier]

Yes

[lucid turret]

What is the question?

exactly

[late geode]

Are you able to provide a question in the book you're struggling with?
and/or notes in your book that you have issues with?

[nova pier]

I can’t but I can provide it for angles that I am struggling with

[lucid turret]

25 can be written in many ways:
25 = 10+15 = 5+20 = 1 + 24 = 12 * 2 + 1 = 6 * 4 + 1

So what exactly is your question?

[late geode]

Its really hard to help if we don't have a clear idea on what the topic is or what you're actually struggling with.

[lucid turret]

which question?

[nova pier]

4

[lucid turret]

Ok so

The triangle in which the 32 deg angle is marked has the other angles (180-32)/2 = 90 - 16 = 74 deg

Now the exterior angle of the angle consisting of the angle x is 74 deg which is equal to x + x = 2x
so x = 37 deg

[nova pier]

And if you can can you also help me in expressions

[marsh citrus]

@nova pier Has your question been resolved?

[nova pier]

Yes

[marsh citrus]

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[fiery walrus]

Hi

[fiery walrus]

Can someone help me with b

can someone help me pls

i really need help

[wheat rover]

yeah so first write the terms that contain x's together along with their signs

and then do the same with y

and group the remaining constants

like there is +2x and +5x, so write them together, then write +7y-y and finally +3 and -7

[fiery walrus]

what

so

i put 7x+6Y-4

that the answer right

@wheat rover

[wheat rover]

yes that's correct

[fiery walrus]

fr

how abt

d

[wheat rover]

first open the parantheses

[fiery walrus]

ok

[wheat rover]

what do you get?

[fiery walrus]

idk

[wheat rover]

you know how to open the parans?

[fiery walrus]

yea

This what I got

[wheat rover]

boy that's +2ab

because there are 2 (-)

so they will become +

[fiery walrus]

where

[wheat rover]

2nd step last term

[fiery walrus]

what.

i doont understand

can u show me

[wheat rover]

after -6b², you wrote -2ab

it should be +2ab

[fiery walrus]

how

[wheat rover]

what is -1 * -1 ?

[fiery walrus]

positive 1

[wheat rover]

yeah

same logic

[fiery walrus]

yea

[wheat rover]

-2b* -a = +2ab

[fiery walrus]

but what negetive make it positive

[wheat rover]

2 negative numbers when multiplied give the positive value of their product

clear?

[wheat rover]

yeah

now group the terms having ab

wait

how did you get 3b²

[fiery walrus]

on no is 6

[wheat rover]

yeah

now it will be 2a² -6ab +2ab -6b²

what do you get in the middle

[fiery walrus]

4ab

[wheat rover]

• or - ?

[fiery walrus]

idkn -

[wheat rover]

see

which number is bigger, 6 or 2 ?

[fiery walrus]

6

[wheat rover]

what is the sign beside 6?

to the left of 6

[fiery walrus]

negetive

[wheat rover]

so the resultant will be negative as well

so -4ab

[fiery walrus]

yea

how

oh

nvm

[wheat rover]

you got that?

[fiery walrus]

yea

[wheat rover]

so what is the final answer?

[fiery walrus]

so it 2a2 - 4ab -6 b2

[wheat rover]

yeah

you got it right

👏

[fiery walrus]

how did my friend got

-8ab in the middle

[wheat rover]

he messed up the signs

[fiery walrus]

ok

[wheat rover]

since the sign of 6 and 2 are different, you subtract them and if they have the same sign (either only + or only -) then you add them

and the resulting sign will be same as that of the larger number

[fiery walrus]

okl

[wheat rover]

if you're done then you may close

[fiery walrus]

no

im not done

[wheat rover]

ok then continue

what's next?

[fiery walrus]

p

[wheat rover]

all of them?

oh

open the parantheses as before

and take care of the signs

[fiery walrus]

wait

do number o first

[wheat rover]

the 1st step remains same

[wheat rover]

.

[fiery walrus]

wait

so the 2 infront has no sign so it positivw

right

[wheat rover]

yeah

you're doing O first right?

[fiery walrus]

yea

[wheat rover]

ok continue

[fiery walrus]

yea

swo

the answer

for o is

sfgsdugisdyfguiwsafgiugfousawyfduiswydfuised

[wheat rover]

tf

[fiery walrus]

-4w+11x+3

heh heh heh

answer

it midnihgt and me want sleep

[wheat rover]

well you're correct

[fiery walrus]

u srure

u sure sure

u sure sure sure

[wheat rover]

yeah

100%

[fiery walrus]

close

tysm

[wheat rover]

it is .close

[fiery walrus]

yea

did u heard abt smartschoolboy9

[wheat rover]

who's that

[fiery walrus]

it an account on tiktok i been up all night bec of it

it so weird

https://vt.tiktok.com/ZS2yMRhrF/

[wheat rover]

umm this is not the channel to post these

[fiery walrus]

ik

that why am always until 11

/close

close

.close

[marsh citrus]

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[marsh citrus]

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[mellow kraken]

Swapping any two rows inverts the sign.
The determinant of a matrix is the volume scaling factor of the unit volume. Each row in a matrix describes a plane for the transformed unit volume. Interchanging two rows flips the orientation of the volume described by these ordered planes.

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[night lion]

how to find the coeffs in ostrogradsky method any shortcuts or tips?

[tepid dust]

what do you mean? just set up a system of linear equations and solve

[night lion]

my book uses

comp of the coeffs

also u get

one equation right nr= A(dr)+ B(dr')+C

how to get the other equations?

[marsh citrus]

@night lion Has your question been resolved?

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[copper mist]

Hello! In Desmos how to sum a range of table cells? In screenshot I sum some manually. How to specify a range to sum?

Thank you for help!

[prime ivy]

you use summation

[copper mist]

thanks. I found out there is also a total() function:

total(T_c)

[prime ivy]

that seems simpler

[copper mist]

but it just totals all of course, whereas your summation has ability for bounds

thanks for the help!

.close

[marsh citrus]

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[vast glacier]

im tryna graph this

im on the wrong path

someone help

hello?

@stark trail

helpppp

[marsh citrus]

@vast glacier Has your question been resolved?

[marsh citrus]

@vast glacier Has your question been resolved?

[main idol]

show your attempt at a graph

[marsh citrus]

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[marsh citrus]

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[covert thorn]

Dumb question

[covert thorn]

How do I do derivative of sqrt (x^2 +1)

Is it u sub?

Am I just doing u sub wrong

[storm ember]

integrtion by parts is what first comes to mind

[hazy lion]

you dont usub for derivatives

[covert thorn]

Im in calc 3 trying to do partial derivatives

[storm ember]

omg my mind

[hazy lion]

you just use the chain rule

[covert thorn]

That’s what I thought

But then I’m blanking

Gimme a sec

[hazy lion]

do you know the chain rule?

[covert thorn]

Lemme work it out again

Yea

[hazy lion]

i mean have some kind of useful definition at hand

alright

[covert thorn]

Derivative of outside then mult by der of inside

[storm ember]

yes

derivative of inside

[covert thorn]

OH

1/2 turns to -1/2

That’s why it’s in the denominator

[storm ember]

$h'(x)=f(g(x))= f'(g(x))\cdot g'(x)$

[elfin berry]

Carter

[covert thorn]

Yeee

I was trying to do d u^1/2

But didn’t turn 1/2 to -1/2

in my head*

Thanks guys

This was embarrassing

/close

[hazy lion]

.close

[marsh citrus]

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[wise obsidian]

I'm so lost with D and E. how am I supposed to find f(-8)???

[hazy lion]

its odd

so, $f(-x) = -f(x)$

[elfin berry]

jan Niku

[wise obsidian]

so I guess -f(x) = -10 --> so 1/-10 is the answer?

[hazy lion]

no

start here

f(-8)

now what

[wise obsidian]

okay then that is equal to -f(8)

[hazy lion]

okay

what is $- \qty( f(8))$

[elfin berry]

jan Niku

[wise obsidian]

-10

[hazy lion]

okay

youre done

[wise obsidian]

what happened to the f^-1

[hazy lion]

oh i misread

[wise obsidian]

f^-1(x) = 1/f(x) tho right?

[hazy lion]

no

what makes you think that

[wise obsidian]

idk just felt right

[hazy lion]

its not a bad guess

[wise obsidian]

so what does f^-1 (x) = ?

[hazy lion]

its the inverse

[wise obsidian]

uh

reciprocal and inverse aren't the same?

[hazy lion]

they are not, unfortunately

[wise obsidian]

bruh, a-c is wrong then 😭

[storm ember]

the inverse for numbers and variables but not necessarily functions

[hazy lion]

the usual algebra way to think about inverses is to switch the subject

[elfin berry]

jan Niku

jan Niku

[hazy lion]

this gives you a way to start with the value of the function at some point, and back out the original input

[elfin berry]

jan Niku

[hazy lion]

at least, thats the inverse in this case

make sense?

at least, the primer on inverse functions

[wise obsidian]

kinda don't get it ngl

[hazy lion]

the inverse is a function such that $f ^{-1}\qty( f (x) ) = x$

it undoes f

[elfin berry]

jan Niku

[wise obsidian]

so what's the final answer for D. I think if I know the answer I can figure out how to do E and then we'll see if I rlly understand

[wise obsidian]

this kinda makes more sense

[hazy lion]

you have to convince yourself that if an inverse exists, then the inverse ||is also odd||

you should have no good reason to believe this is true at-its-face

but if you want to assume it to finish your assignment im sure thats fine

do you see what i mean?

like

$f^{-1} (-8) = -f^{-1}(8)$

[elfin berry]

jan Niku

[hazy lion]

and we know this is true because 1. f is odd and 2. the inverse is given to exist (and be defined for a reasonable domain, blah blah)

[wise obsidian]

I'm cooked

[hazy lion]

heres all u gotta do

im telling you f inverse is odd

[elfin berry]

jan Niku

[hazy lion]

we know the definition of f inverse being odd, yea?

[elfin berry]

jan Niku

jan Niku

[hazy lion]

that is, what is the x such that f(x) = 8?

[wise obsidian]

6?\

[hazy lion]

yea, 6 sounds good

[hazy lion]

so now use this

[elfin berry]

jan Niku

jan Niku

[wise obsidian]

f^-1(x) isn't the same as either of these?

[hazy lion]

we dont know f^-1 (x)

[wise obsidian]

so how would f^-1(8) = 6 help with f^-1(-8)

[hazy lion]

we only know things about it

one of the things we know is that f^-1 (-x) = - f^-1(x)

[wise obsidian]

okay

[hazy lion]

so connect two things:

[elfin berry]

jan Niku

jan Niku

[hazy lion]

use the first first, and the second second

[wise obsidian]

okay

-f^-1 (-8) = 6?

[hazy lion]

it does, yea

idk if youre guessing but thats actually true

can you use this to get your answer?

$-f^{-1} (-8) = 6$

[elfin berry]

jan Niku

[hazy lion]

youre super close to what you want

except for that pesky -

[wise obsidian]

okay so if -f^-1 (-8) = 6? is true

and we know

[hazy lion]

you dont need this

but it is true

i think we might be treading water

[wise obsidian]

I'm using 100% of my brain rn

[hazy lion]

\begin{align*}
f^{-1} (-8)
&= - f^{-1} (8) \
&= - 6
\end{align*}

[elfin berry]

jan Niku

[hazy lion]

first line because f^-1 is odd

second line because f(6) = 8

[wise obsidian]

uh

sort of get it

you down to hear me out with E?

just so I can make sure I actually get it

[hazy lion]

WEW thats a doozer

yea, what you got?

[wise obsidian]

I've gotten through literally 2 problems in the past like 3 hours lol

okay let me try E

[hazy lion]

sure

[wise obsidian]

f^-1 (-7) = -f^-1 (7)?

[hazy lion]

it does

[wise obsidian]

what is the x such that f(x) = ?

[hazy lion]

Can I state some properties you have I think will be helpful?

you're off to a good start

[wise obsidian]

okay

[elfin berry]

jan Niku

[wise obsidian]

my brain is drawing so many blanks

okay so

[elfin berry]

jan Niku

[hazy lion]

you have something that looks almost like the second thing, but there's a negative floating around. You'll need to use oddness to handle that.

sorry if thats too much spoiler

[wise obsidian]

wait

I think I see the path

okay so f^-1 (-7) = -f^-1 (7)?

[hazy lion]

it does

[wise obsidian]

how can I print this out neatly like u did so I can see it better

[elfin berry]

jan Niku

$f^{-1} (-7) = - f^{-1} (7)$

[hazy lion]

codes up above

[hazy lion]

I like your idea of starting at the very inside

i think you did a good job there, so maybe zoom out just a little

lets stare at $f \qty[ - f^{-1} (7) ]$

[elfin berry]

jan Niku

[wise obsidian]

$-f \qty[ f^{-1} (7) ]$

[elfin berry]

Paul04

[hazy lion]

yea, i think so

since f is odd

we should be able to simplify this further now

how do f and f inverse interact

if $f^{-1} (7)$ is the $x$ such that $f(x)=7$

[elfin berry]

jan Niku

[hazy lion]

you also said it here

[wise obsidian]

-f(7)

[hazy lion]

not quite

maybe it helps to define it? I hope this isnt gonna be symbol soup, tell me if its confusing, but it might help

[elfin berry]

jan Niku

$f^{-1}(7) = x_7$

[hazy lion]

lets just name it, its equal to some value, call it x_7

[elfin berry]

jan Niku

[hazy lion]

so, lets start here, and apply f to both sides

$f^{-1}(7) = x_7 \longrightarrow f\qty[f^{-1}(7)] = f[x_7]$

[elfin berry]

jan Niku

[hazy lion]

a lot of symbols floating around, but don't get too mixed up. All we did was start with an equation that's true, and apply f to both sides

whats f(x_7)?

[wise obsidian]

5?

[hazy lion]

.

[wise obsidian]

f(5) = 7 is what I was getting at

[hazy lion]

oh, sorry

yea, looks good

wait

well, okay

what you said is true

but its not relevant here

[wise obsidian]

but not related

bruh

[hazy lion]

the inverse and f eat each other

maybe you can't see it right now

but maybe you can believe it