precalculus

[stable knot]

sorry, again

[proven night]

By the way

2^x - 3^x = - 1

So can we say that lhs is monotonously decreasing function?

[hallow tide]

no

[proven night]

Why did you delete it?

how do i determine whether a function is monotonous?

[hallow tide]

with calculus

[proven night]

So cant i solve precalculus exponential equations without it?

[proven night]

I mean, hard ones

[summer ruin]

if you can show that x < y implies 2^x - 3^x < 2^y - 3^y or that x > y implies 2^x - 3^x > 2^y - 3^y, then you will prove that the function is monotone

[slender hedge]

f:g:h g is 60 percent more than h , f is a third of g , simplify f:g:h

[thorny prism]

Hello! Idk where I can ask it tbh, because it is a simple thing. Is there a general formula for derivative of f(x)=u(x)•v(x)•w(x)•…? I mean I can find the derivative of the ‘triple’ product, it isn’t that hard but is there a general way? Because if I would have to find the derivative of the of 6 functions, I will spend a lot of time counting a lot of default derivatives for the product of two functions

[paper tartan]

~~ see this: https://en.wikipedia.org/wiki/General_Leibniz_rule ~~ nevermind, I made a mistake

(also, this belongs in #calculus, but no big deal)

see this: https://en.wikipedia.org/wiki/Product_rule#Product_of_more_than_two_factors

I made a mistake, the first link shows you show to differentiate a product of two functions n times

the second link gives you the derivative for 3, and then k functions

just below that, you can also find out how to differentiate a product of k functions n times

there's also stuff for partial derivatives, but I'll let you explore all of that lol

[thorny prism]

Thank you so much!

I haven’t got calculus because, I think, I have only got pre-university group… my bad with it

[paper tartan]

give yourself the Advanced role

type ,iam advanced in #bots

that should let you type in #calculus, and the other early uni/adv math channels too

oh, you've taken the UG role

that works too!

[thorny prism]

Thank you!

[viscid thistle]

I just got the AoPS Precalc book and I'm stuck on trig unit circles :-: does anyone know the basics to it?

[chrome ether]

that's not nearly a specific enough question

try again

what don't you get about it?

[viscid thistle]

Hold on lemme find a picture

ok so for example you have graph sinx=sin(x+2π)

and it asks if it's periodic and if so find the period

how am I going to solve this question?

[bronze rock]

Having trouble understanding why multiplying the x value in a trig function, decreases the length of a period.
It just doesnt make logical sense to me. I feel like it would make more sense, for it to be reversed instead of 2pi/b.
Why not just 2pi * b, since you would be multiplying the input x value?

[hallow tide]

trig functions aren't special

[viscid thistle]

@bronze rock Thanks for the explanation! I think I got mixed up lol

[hallow tide]

f(ax) represents a horizontal dilation of f(x)

when |a| > 1, the function is compressed; when |a| < 1, the function is stretched

[bronze rock]

Yeah I just find it super odd that its a dilation I guess. It just seems reversed logically to me

[muted shore]

wat is

[bronze rock]

multiplying the x value of a function.
Its Inverse Variation which I find weird

[hushed sphinx]

Suppose f(x) is periodic with some period and consider g(x)=f(2x).
Then you only need half of the x before g starts repeating.

Since multiplying by 2 means you can reach a certain input for f without spending as much x as you would need to if you fed x directly into f.

[bronze rock]

Ah I see. That makes sense. Thanks

[muted shore]

oh ur talking about why 1/a is a stretch but multiplying by a is a compression

im guessing

ya does seem p reversed

[tired stone]

how do I get the instantaneous rate of change

?

[willow skiff]

you need calculus, more specifically take the derivative and sub in the x-value of your point

[umbral crow]

i need a bit of help with precal

the question is "graph f and g on the same set of coordinate axes (include two full periods)"

[willow skiff]

!da2a

[tender quest]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[willow skiff]

oh ok

[umbral crow]

f(x) = -2 sin x

g(x) = 4 sin x

i kind of forgot like everything so im struggling with how to graph individual ones

[willow skiff]

can you graph y = sin x?

like what's sin 0 for example

[umbral crow]

i have the graph memorized

but like im still kinda new on the entire concepts of graphs and stuff

sin 0 = 0?

[willow skiff]

oh right, that's good then

[willow skiff]

yes!

[willow skiff]

so you need to watch a video on graph transformations then

what the -2 and the 4 do when multiplied by sin x is that they stretch the y-direction

so 4 sin x stretches the y-direction by a factor of 4

in other words the max is now 4 and the min is now -4

[umbral crow]

ohhhh

[willow skiff]

-2 sin x is similar but after you stretch the graph by a factor of 2, you reflect it across the x-axis (flip it upside down)

you flip it upside down when there's a minus sign multiplied by sin x

[umbral crow]

wait so it's like normal rules of transformations

[willow skiff]

exactly

[umbral crow]

okay wait that made so much sense

[willow skiff]

cool sounds like something didn't click for you and that you actually know your stuff

[umbral crow]

thank you for the help 🙏

[hearty granite]

looking for some hlep in studying for cal final if anyone is interested

[shadow summit]

What sort of stuff is it?

[flat shuttle]

What does it want me to say when it asks for an explanation for my reasoning

[chrome ether]

why do you conclude the answer that you do?

(if it’s a polynomial, just state the parity of the degree (even/odd) and the sign of its leading coefficient (positive/negative))

[flat shuttle]

ok

[gritty ibex]

hi could anyone help me with this series problem in this thread?#1252945019342557204 message I cant ping helpers there due to this:

[autumn snow]

Hey could someone help me with this one question

[daring tapir]

!da2a

[tender quest]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[paper tartan]

#calculus is a better channel for that

right, you can't see that channel, can you?

take the Undergraduate Math Role in id:customize

that's good enough

the role is meant for people who study undergradate math

you don't necessarily have to be an UG for it

many are not

#calculus is the channel for calculus discussion

this one is suited for precalc

[ashen flame]

I need to solve for the critical values of cos(2x)+sinx+1=f(x). I set the derivative equal to zero and got 0.25 and 2.9. However, there are two other critical values that I don't understand how to get mathematically.

[summer ruin]

there are infinitely many critical points, not just two or four

you need to show your work

[low slate]

How far does pre-calculus go? When does it end? Like- i took the 5 rules of derivation and some integration techniques, am i at calculus 1 or pre- calculus

The last two lessons i was taught: how to differentiate trig functions and applications on derivatives

Next year first two lessons:how to differntiate functions in parametric form and implicit differentiation

[paper tartan]

you are most definitely in calculus, not precalculus

take the Undergraduate Math role in id:customize, and then go to #calculus if you have calculus-related problems

this channel is for the usual precalculus curriculum

[low slate]

I did i did (i usually ask questions in help channel) ok ty higher

Higher is like

The infinte version of high

[paper tartan]

higher!

[low slate]

You cant get higher than higher

[paper tartan]

I don't see your UG role, strange

[low slate]

So higher is infinity

[low slate]

Oh? Lemme check

[paper tartan]

now you have it

[low slate]

🤩

[tired stone]

did I do it correctly?

<@&286206848099549185>

[dire aspen]

ermmmmm technically akshually calculus is Pre-uni and not undergrad 🤓

[tired stone]

bro

pls help me

[paper tartan]

depends on the school

[paper tartan]

the graph looks mostly okay, but 3^4 = 81, not 51 like you've plotted

[proven night]

If two men are moving towards each other, the first one with speed 10 m/s and the second one with speed 7 m/s, how soon will they meet

And the distance between them is 100 m

I dont understand how to solve this kind of porblems

[stoic rivet]

100/(10+7)

If they move towards each other u add the speed
If they leave each other then subtract two speed

[proven night]

it makes sense

I did almost the same: 10t + 7t = 100

[proven night]

Hello

I think i should replace 4x (i dont have theta character in my keyboard) with t, so the equation is gonna look like this: cos t + cos 2t + cos 3t = - 1

Is there an easier way of solving this equation without cos 3t identity?

I mean i always confuse cos 3t identity with sin 3t identity and it's hard to remember it to me, so maybe one can solve the equation without it?

[sturdy token]

@proven night

i used 3 identities, cos x + cos y
cosx - cosy and sin2x

[proven night]

Thanks, but the equation has more solitions as far as i remember

I solved using cos 3t identity

However thanks, i should learn sum of cosines identity

[sturdy token]

yea it would have more, but keep in mind that theta is between 0 and pi/2

[tired stone]

do the numbers look correct?

[willow skiff]

,w log_3 (17)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by sooouth.

[willow skiff]

,w log_3 (36)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by sooouth.

[willow skiff]

I don't get how you came up with those values

you just draw horizontal lines y = 17 and y = 36 on your graph

cause $3^{\log_3 17} = 17$ and so on

[obsidian monolith]

southy

[tired stone]

oh

ok

that makes more sense

[willow skiff]

no worries

[tired stone]

log 3 (17)is 1.707 though

[willow skiff]

it's not

and you shouldn't get that value from your graph

should definitely be 2 point something

[willow skiff]

ah that's different

try typing _ after your log

[tired stone]

ohhhhhhhhhhhh

that makes more sense

thank you again

[willow skiff]

no worries again

[tired stone]

so it gives me 2.5 but how do I graph it?

[willow skiff]

oh, type in 3 ^ x

and then you can type in y = 17, y = 36 on different lines

[tired stone]

how did you get 17 and 36?

[willow skiff]

the question says 17 and 36 here

[tired stone]

ohhhhhhhh

ok

thank you again

[tired stone]

How does that look?

[ashen flame]

how do i prove that the derivative of sin x is cos x

[chrome ether]

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

at least take a few seconds to google it first

[eager narwhal]

use the limit definition of the derivative

[hollow jungle]

and the sin addition formula, $sin(x + h) = sin(x)cos(h) + cos(x)sin(h)$

[obsidian monolith]

KJ

[proven night]

I see several identities at the same time, but it didnt help me at all

I just dont even know what should i start with

I tried expanding sin 4x and sin 2x but it was still useless

[daring tapir]

You know the sum to product identities?

[daring tapir]

Wait are you trying to prove this or find solutions?
Cuz if prove then its not possible as the statement is wrong (look at x=0)

[proven night]

It's an equation

[proven night]

I am trying to solve it

[proven night]

Yes and i already applied it

[daring tapir]

What did you get

[proven night]

i got 2sin(x)cos(3x) = 4cos^2(x) -3

[daring tapir]

Expand cos(3x)

[proven night]

As sum of 2x and x or as a cosine of triple angle?

[daring tapir]

Latter

[proven night]

Okay

i got 8sin(x)cos^3(x) - 6sin(x)cos(x) - 4cos^2(x) + 3 = 0

But i dont know what how exactly should i factor this

[daring tapir]

Ahem

Take 2sin(x) cos(x) common

[proven night]

Ok

[daring tapir]

(Please ping when responding)

[proven night]

Alright

@daring tapir

[daring tapir]

Hm

[proven night]

Thanks, @daring tapir

I factored it

[proven night]

By the way, are this identity and triple angle identities popular? Should i learn them by heart?

Because i hardly ever use them, however this equation used both

[daring tapir]

Eh i mean the derivation is easy enough jusf derive it

[proven night]

I dont know how to derive it

[daring tapir]

Look it up its easy enough

[proven night]

Okay

[near basin]

I have forgotten limits
Why is the limit as x approaches 0 36?

\frac{\left(\sin\left(6x\right)\right)^{2}}{x^{2}}

My first thought is infinity since {x^{2}} is in the denominator

[summer ruin]

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

[obsidian monolith]

Transparent Elemental

[near basin]

How is that proved?

[summer ruin]

you can google the proof

[near basin]

I mean the equation you gave me

[viscid thistle]

guy just tell me one thing

does product rule for integration exist or nah

cuz my physics teacher told me there aint nothing like tha

but a book i follow has a rule formula named "integration by parts"

and google says i exists too

[summer ruin]

product rule would say something about the integral of a product of two functions

which you can't say anything about

[viscid thistle]

so what do u mean??

[summer ruin]

that's what product rule for derivatives does

it relates derivative of a product to derivatives of something simpler

[viscid thistle]

oh

i somewhat get it somewhat dont ( im new to calculus thats why lol)

so like u get the answer but it isnt exactly what you would get

[abstract gazelle]

How to do this?

[round current]

could someone explain thsi for me

i dont get what i got wrong

its coming to zero from the right side

and i dont get how it works cuz there are holes

[winter comet]

The limit looks at behavior near the point rather than the point itself

The graph is approaching 0 from both sides as x -> 0

As far as the graph, there is a hole at (0,0) they didn't draw well or at least it's hard for me to see

But otherwise it wouldn't be a function

[round current]

ohhhhhh

thanks

[viscid thistle]

How can I apply taylor series to a limit?

For example, I wanna solve this limit when x tends to 0 of sin (x) divided by x using Taylor series.

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

[obsidian monolith]

Machholz

[fading monolith]

This will go to #calculus

But you just need to substitute the sin(x) with its power series at 0 and then divide it by x. Since the series converges uniformly then you can interchange limit with infinite sum and thats it

[viscid thistle]

Thanks a lot!

[agile sparrow]

f(x-1)= x+2
f(x-1+1)= (x+1)+2

f(x) =x+3

[narrow briar]

Hi!

Do you know any nice web where I can find integral excercices by substitution?

[random yoke]

https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/usubdirectory/USubstitution.html

[vapid frigate]

Could some explain the difference between chain rule and product rule?

I'm kinda stuck lol

[viscid thistle]

Chain rule is for differentiating functions of the form f(g(x)) while the product rule is for f(x)g(x)

Also this belongs in #calculus

[vapid frigate]

Oh thanks

Sorry btw

It wasn't in my channel list

[viscid thistle]

Hey, so I wanted to learn calculus and downloaded the book Calculus: Early Transcendentals by James Stewart, and realized that I have huge gaps in the knowledge required from the diagnostic test.

I am currently 12 months in my 14 month army service in Cyprus, and things are starting to get easier. Every 6 hours, I have a 3 hour number where I basically just sit on a desk and write the credentials of anyone that takes out our puts in a gun. That is a great time to study, and I am currently just wasting that time on my phone.

Can you recommend a good precalculus book that will cover everything extensively? If I can't finish it during the next two months, I am prepared to put in the work in University as well.

Thanks!

What about this one?
https://www.opentextbookstore.com/precalc/

[lime nymph]

Hello. This is 9th grade math (functions). I don't know how to do the exercise in the middle. It's in spanish so I'll try to translate it:
"According to the following conditions for the functions k(x) and t(x)
k(2)=4, k(4)=8, k(6)=2, k(8)=6
t(2)=4, t(4)=8, t(6)=8, t(8)=4
Determine.
a. (kot)(2)=
b. (tok)(2)=
c. (kot)(6)=
d. (kot)(4)=
e. (tok)(8)=
f. (tok)(6)="
Please help.

[outer oxide]

Hello, i don't exactly know what kind of operator "o" is but I'll give you a general idea of what you have to do in these cases.
A:
(k o t)(2) = k(2) o t(2) = 4 o 4

B:
(t o k)(2) = t(2) o k(2) = 4 o 4

[lime nymph]

"o" is the function composition

[outer oxide]

then you can apply that (f o g)(x) = f(g(x))

[lime nymph]

I tried doing that, but every result ends up being 8 except 2.
I tried k(t(2)) = k(4)=8.
I also tried k(8)= 8(8)=64
or 2(2)(2)=4(2)=8

(f o g) is (k o t), I know that much

but the (2) instead of (x) messes me up

[outer oxide]

well it has to be that way, the 8s might be only coincidence

[lime nymph]

;( maybe

anyway thank you

[outer oxide]

no problem

[viscid thistle]

Pre calc humbles me

(Never done pre calc)

[winter comet]

whats special about precalc

[viscid thistle]

Idk js never done it

[winter comet]

what do u know about it

[viscid thistle]

Nun

Which is why it humbles me

[paper tartan]

everything is hard when you've never done it before

especially in mathematics

don't be too discouraged!

practice makes perfect, as they say

[elfin cargo]

What do you guys learn in pre-calc?

I don't think my country has pre-calc

[willow skiff]

trig, composite and inverse functions, rational functions, logarithms and exponentials

I'm not American and only American people call it precalc

[elfin cargo]

That just algebra in my country

[willow skiff]

factor and remainder theorems, polynomial long division

[tender mist]

Depending on your depiction of "america". It's not just the US, Canada has it too

[willow skiff]

indeed, there's no single definition of precalc

[willow skiff]

technically all countries have material that is covered before calculus

but Ontario calls it advanced functions

BC calls it precalculus I think

[tender mist]

MB calls it precalculus, yep

[willow skiff]

ahhhh right yeah just not Ontario fair enough

[willow skiff]

and whatever the frick Alberta calls it I'm not sure

[tender mist]

Who knows Even calculus courses go by different names in different places

[willow skiff]

exactly

[lapis sphinx]

We did so much in precalc

[elfin cargo]

Wait, calculus has different name?

[tender mist]

Anyway, yeah. Precalc by the cirriculum I had was teaching exponentials, logarithms, trig, rational functions, and inverse functions

[tender mist]

Okay to be fair, my analogy was actually incorrect, lol

It's because calc courses go by different names, but never mind, that has nothing to do with actual alternative math names

I often say calc 1/2/3, just because generally speaking, even if places do things a bit differently, most people are able to distinguish between what content goes inside of those

I guess saying differential calc, integral calc, and multivariable calc is even better though

[willow skiff]

yes, like different unis call it differently

there's not just single-variable calculus but also multivariable

[tender mist]

Not like I really know much about calc yet. I only recently finished calc 1, and have just brushed the surface of calc 2, lol

[elfin cargo]

Make sense

I also often call calc 1, 2, 3 but my university have different courses on them

But calc 1, 2, 3 look universal enough so everyone know what in it

[tender mist]

Calc 2 is the only one at my university that goes by that name

Calc 1 goes by "Introduction to Calculus", which is honestly a pretty fair name. And calc 3 goes by "Multivariable Calculus".

[proven night]

Hi

Can anyone help me with it

I tried applying difference of cubes formula and i also tried solving the system of equations but neither helped me

[raw hill]

Recall that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

[obsidian monolith]

Civil Service Pigeon

[proven night]

I did it

[winter comet]

thats the difference of cbes formula lol

[proven night]

And replace a-b with 9

And divided 27 by it

[winter comet]

you can replace a with b + 9 then

[raw hill]

^

you can do that or reduce it to a system in (a+b) and ab

Using the fact that:

• ||a^2 + b^2 = (a+b)^2 - 2ab||
• ||(a-b)^2 = (a+b)^2 - 4ab||

(Note that a and b are complex numbers that aren't particularly nice to work with)

[raw hill]

reduce it to a system in (a+b) and ab
so they probably intended you to do this

especially b/c they explicitly ask for a product

[proven night]

I dont understand it

What do you mean by reduce it to a system in a + b

[raw hill]

I should've said "rewrite a^3 - b^3 in terms of a+b and ab" tbh

[proven night]

But i dont have a^3 + b^3. Instead i have a^3 - b^3

[raw hill]

Typo, the idea still holds.

[winter comet]

don't divide 27

you should have 9(a^2+ab+b^2) = 27

[raw hill]

? they said "27 by it" not "it by 27"

[winter comet]

you can divide by 9

oh

um

don't do either?

oh

[raw hill]

And replace a-b with 9
And divided 27 by it

[winter comet]

wait do you mean divide 9 by 27>

[raw hill]

it's clear "it" refers to 9

[winter comet]

oh ok

yea then just replace a with b + 9 yeh

[raw hill]

yeah but that's a lot of extra work

[raw hill]

you're better off just doing this

[raw hill]

b/c this

[raw hill]

using this

[winter comet]

oh yeh tru

[warped shuttle]

i think -26

[warped shuttle]

Here is the formula i used

i think it's easier than doing a system

[viscid thistle]

is that pi in the second line?

[left crystal]

idts

it's a variable r. which is = ab that we have to find

a-b = 9

[warped shuttle]

no it's x which is ab

a simple variable

[livid hinge]

bird $$3+3*3$$

[obsidian monolith]

Mark

[livid hinge]

wow nice formatting

[jagged vine]

these r just homework questions I cannot figureo ut

would love some help

also how is this the wrong slope?

[winter comet]

what do you need help with?

[jagged vine]

i just dont understand them

the sqrt root really throw me off

[winter comet]

you can rewrite the square root(f(x)) as [f(x)]^(1/2)

and then you can use the chain rule

[winter comet]

what don't you understand?

how to differentiate it?

[jagged vine]

yes

[winter comet]

you know how sqrt(x) = x^(1/2) ?

[jagged vine]

yes

[winter comet]

so like sqrt(x+2y) = (x+2y)^(1/2)

right

and sqrt(3xy) = (3xy)^1/2

and then you can use chain rule right?

[jagged vine]

yeah i wasnt thinking ab it like that

ty ill try it

[cerulean canopy]

i need help am i cooked

-4 > |x| > 4

is that right??

[viscid thistle]

And no it’s not correct

[viscid thistle]

Why would something be more than -4 and less than 4

[chrome ether]

absolute values cannot be negative

what you have there is satisfied by the entire real line

[viscid thistle]

The first message yes but not the second message

[cerulean canopy]

thanks

i got it

figured out \

👍 '

[white rapids]

Could somebody clear this up for me here? I dont get why my answer is wrong

I used system of equations to get my quadratic equation

[chrome ether]

check your work

you might have solved the system incorrectly

[white rapids]

I used points (1,6) and (2,1) because it seemed easier
So like:
a(1)^2 + b(1) + ~~c ~~= 6 --> a + b = 6
a(2)^2 + b(2) + c = 1 --> 4a + 2b = 1

c's cancel out

Multiply (a + b = 6) by 4 to cancel out one of the terms

4a + 2b = 1
-- (4a + 4b = 24)
-2b = -23
b = 11.5

Plug b into a + b = 6
a + 11.5 = 6
a = -5.5

Plug a & b into quadratic equation:

-5.5x^2 + 11.5x + c = y

Use one of the points to get c:
-5.5(1)^2 + 11.5(1) + c = 6
-5.5 + 11.5 + c = 6
c + 6 = 6
c = 0

Equation is y = -5.5x^2 + 11.5x

i really dont get it

[white rapids]

@winter comet I need you pls

[willow skiff]

you need three points to find the unique quadratic passing through these points

so a + b = 6 is already wrong, the c doesn't go anywhere

you have a + b + c = 6 and 4a + 2b + c = 1

if you subtract both equations, you get -3a - b = 5

you need another equation like with the point (6, -99), so 36a + 6b + c = -99

see how you go from there

[uneven dew]

Hi, If we know a function isnt polynomial, exponential or logistical. Can we safely say its linear? The derivative of the function is always an integer ≥ 0

And f is also an integer

[elfin cargo]

No

If the derivative of the function is always >= 0 then we know it is an increasing function

[uneven dew]

Not always > 0

Always ≥ 0

[elfin cargo]

still the same

[uneven dew]

But = 0 isnt increasing

[elfin cargo]

If the function is linear then its derivative is always equal to a constant

[elfin cargo]

increasing doesn't mean strictly increasing

increasing can be 0 but strictly increasing is always > 0

[uneven dew]

By linear, i mean when the scale at which this function is viewed at, is huge

So these scenarios where f'=0 seem to disappear

And f smooths out

[elfin cargo]

Do you mean when you zoom in on a function it will look like a straight line?

[uneven dew]

When you zoom out

[vapid plaza]

Linear is also polynomial

[uneven dew]

When you zoom in, it looks like increasing linear at some ranges, then constant 0 at some too

Ik, but by linear i mean its a line, not parabolas or similar

[elfin cargo]

when you zoom out large enough, a function might look like a straight line, but it doesn't mean it is linear

kind of like an illusion

[uneven dew]

Can we prove this illusion then

Thats what am looking for

[elfin cargo]

No

[uneven dew]

Like, if we are given properties of this function. Can we prove that it will assume linearity as its domain increases

[vapid plaza]

I have no idea what you’re expressing

What is “f is also an integer”?

[uneven dew]

...

[elfin cargo]

Does this count as spam?

[uneven dew]

Think so

[vapid plaza]

Anyway can you clarify what you mean

@uneven dew

[uneven dew]

You know like how pi(x) is approximated with logs

[elfin cargo]

I think he means for example, when you graph a function on a graphing calculator, and you zoom out large enough then the graph looks like a straight line

[vapid plaza]

Nah I don’t even understand the first thing he said

What “and f is always an integer”

[uneven dew]

f(x) is always an integer, f'(x) too

Both share the same domain, x is in [0, ...⟩

[elfin cargo]

if f(x) is always an integer, then f'(x) will be undefined

☠️

[uneven dew]

?

[elfin cargo]

Your domain is still in real number right?

[uneven dew]

No decimals

[elfin cargo]

like x = 0, 1, 2, 3,....?

[uneven dew]

f(x), f'(x) = 0,1,2,3,...

[elfin cargo]

I mean $x \in \mathbb{R}$ or $x \in \mathbb{N}$ ?

[obsidian monolith]

NKH

[uneven dew]

mainly N_0

But we can accept and approximate the output of decimals too

But as mentioned this is just an approximation

[elfin cargo]

I am confused about what you are trying to do. But if you want to prove that a function f(x) is linear like you said then you need to show that an increment of x at which the function is approximately linear?

[uneven dew]

I dont understand your statement, can you elaborate thx

[elfin cargo]

like: at every $x + \Delta x$ then $f'(x + \Delta x)$ is approximately equal?

[obsidian monolith]

NKH

[elfin cargo]

you need to find: $\Delta x$

[obsidian monolith]

NKH

[uneven dew]

Thats true for x ≤ 2000. But i dont know how to prove it for the other values

[elfin cargo]

Then just set that as a condition

[uneven dew]

Condition for what

[elfin cargo]

So if x <= 2000, f'(x) is approximately equal?

[uneven dew]

Yea

[elfin cargo]

Is this a problem in your school? I am very confused

[uneven dew]

Nah, its just something am working on independently

[viscid thistle]

I wanna know if there’s a calculator that can solve all types of calculus problems for high school
Btw its not typical highschool calculus, alittle more advanced

[uneven dew]

Wolfram?

[elfin cargo]

If you want a handheld calculator, you can try Texas calculator

It also has a computer algebra system like Wolfram

[viscid thistle]

Can it derive, integrate, find the bounded area and so?
Also these are examples of the problems I have

[viscid thistle]

Iam not se well-educated about the names and so sorry😭

[elfin cargo]

Yes

You can try wolfram first

[viscid thistle]

Any tutorial video?

[elfin cargo]

Tutorial for what?

[viscid thistle]

How it solves such equations as shown

i've got one

but it's in hindi

[viscid thistle]

Its alr ijust wanna see how it does things

its also 5 hours long , but what you are looking for is explained very clearly in it \

https://www.youtube.com/live/qgk0a-W0LQ4?si=r8BP-xLi-hxuzLl_

i can solve it but , im doing vectors and 3d rn

Imma watch what i need and get back to you, thanks man!

[normal shadow]

if im doing binomial expansion and the first variable has a coefficient, how do i do it

i tried to find some videos online explaining it

but nothing really shows up

any recommended vids??

[chrome ether]

you apply basic exponent laws

raise the coefficient to the appropriate powers as well

ex $(2x+1)^2=(2x)^2+2(2x)(1)+1^2=4x^2+4x+1$

[obsidian monolith]

elrichardo1337

[chrome ether]

more explicitly, $(ab)^x=a^x\cdot b^x$

[obsidian monolith]

elrichardo1337

[normal shadow]

is there a faster method to do this when you have a large exponent

I have to get the power to the ten

do i have to just do it by hand?

[random yoke]

Binomial theorem

[normal shadow]

yeah, like i mean

in the binomial theorem

how do i plug in the 2x for example

like where does the 2 go

in front of the "a" where a is to the power of n - k

[random yoke]

No

The 2 should also be raised to the power

You can imagine replacing 2x with another variable like u, and then substituting 2x back in once you've applied the binomial theorem

[normal shadow]

so i would go (u + b)^10, apply the theorem, get the final answer, and resubsitute "u" and 2x?

[random yoke]

Yes

[normal shadow]

and then just simplify from there

coolio

thanks

Is this right so far?

[elfin cargo]

<@&268886789983436800>

[high venture]

hello guys , i am stuck with this problem , so , someone can help me please it 's an EMERGENCY !! show that (n+1)*3-n is equal to the sum from 0 to n of 3-n pleaseeeeee 😦

[uncut mulch]

did you copy down the quesiton correctly?

that sounds wrong

[uneven dew]

Pretty sure you could change this into an arithmetic sequence

Where a_n = 2, 1, 0, -1 ..., -(n-3)

[uneven dew]

So d = -1

And the sum S_n = (2a_1 + dn - d)*n/2

Then you could maybe continue from here and try forming equations and manipulating stuff

[uneven dew]

(5n - n²)/2 = (n+1)(a_n)
(5n - n²)/2 = (n+1)(a_1 + dn - d)
(5n - n²)/2 = (n+1)(2 - n + 1)
(5n - n²)/2 = (n+1)(3 - n)
(5n - n²)/2 = 3n + 3 - n² - n
5n - n² = (3 - n² + 2n) * 2
5n - n² = 4n - 2n² + 6
Might have done something wrong, but there exists some values of n where LHS ≠ RHS. Therefore the two sum formulars arent the same

[prisma pewter]

wait, it's all conics?

[daring tapir]

@fickle cedar

[fickle cedar]

,rotate

no

,rotate

there

is this readable?

[daring tapir]

Deleted

[lusty harness]

confused if this is wrong because ive checked on my calculator and im pretty sure the answer is b

[chrome ether]

wolfram alpha agrees with you

[lusty harness]

= ?

[rigid cove]

I’m pretty sure the answer is b

[chrome ether]

hmm b and c are equal

this is just a poorly made question

they want the expression obtained via half angle (ignoring that it can be simplified to the form shown in answer choice b)

[rigid cove]

oh huh

I see..

[chrome ether]

i think their "intended" answer is c then

[viscid thistle]

pls help me with this question . ([] means greatest integer function)

[willow skiff]

this means that x + 1/3 has to round down to 2, but x + 1/2 has to round up to 3

So the endpoints are x = 2 + 1/2 and x = 2 + 2/3

but it doesn't include 2 + 2/3, cause you get 3 + 3 = 6 instead of 5

[viscid thistle]

how many values of x did you get?

[willow skiff]

8

[viscid thistle]

ohh thats correct

pls can you elaborate it again

[willow skiff]

so from 2 + 24/48 to 2 + 32/48, but not including 2 + 32/48

so 32 - 24 = 8

[viscid thistle]

how did you get those numbers?

[willow skiff]

this

x + 1/2 is always greater than x + 1/3
so [x + 1/2] has to be 3 and so [x + 1/2] has to be 2

that's the only possible way to make 5

[viscid thistle]

ohh ok

understood

but why not 4 and 1?

[willow skiff]

that's impossible

[x + 1/2] and [x + 1/3] can only be 1 apart as (1/2 - 1/3) < 1

[viscid thistle]

1 part meaning?

[willow skiff]

[x - 1/2] - [x + 1/3] can only be 0 or 1

[viscid thistle]

but why??

[willow skiff]

from the definition of the floor function

Like keep trying different values of x such as x = 1/6, 1/2, 1/3, 2/3 if you're not convinced

cause you're rounding down, so if you have x + 1/2 and you add 1/3 - 1/2 = 1/6 to it

then you can cross a number, so x + 1/2 can round down to 2 but x + 1/3 can round down to 3

you can't have that x + 1/2 rounds down to 2 but x + 1/3 rounds down to 4

[viscid thistle]

i think i need to learn some maths

[paper tartan]

[storm galleon]

Why do we take infinity as 1/0 in limits

[chilly plume]

Idk if I’m understanding what you’re asking but because 0 on the bottom would make it undefined, there is a asymptote at that spot

[white elbow]

I take 2/0

[rigid cove]

We don’t really take infinity as 1/0

[proven night]

Hello, guys

I need help

cos (10^x) = 0

My answer is x = log(pi/2 + pi*k), k is an integer

But

I doubt that k can be any integer. Because if it's a negative integer, then the argument of the logarithm is also negative

But it's beyond a log domain

So

Should i say that the answer is x = log(pi/2 + pi*k), but k is a positive integer including 0?

[rigid cove]

sure, so you could say $x = \log(\frac \pi 2 + k\pi)$ for $k \in \mathbb{Z}_{\geq 0}$

[obsidian monolith]

Pseudonium

[proven night]

But photomath says the opposite

[proven night]

It says this is right

Is it just x = log(pi/2 + k*pi)?

Where k is an integer

[proven night]

So is it the correct answer?

[willow skiff]

technically correct but it's redundant

[proven night]

But if k is negative than it's beyond the logarithm domain

[willow skiff]

they've split the list up as (pi/2, 5pi/2, 9pi/2....) and (3pi/2, 7pi/2, 11pi/2....)

[proven night]

I mean they say that k is any integrr including negative ones

[willow skiff]

yeah so you could have a complex logarithm

if you have a complex logarithm then yeah those solutions aren't correct, you would need to be adding on multiples of 2pi i / ln(10) as well

[elfin cargo]

yeah, complex variable will solve the problem

[willow skiff]

so yeah PhotoMaths doesn't understand maths logic

[elfin cargo]

Oftentimes, math software will give you solutions in complex domain

[proven night]

I dont know what are complex logarithms, sorry, so what's the final answer without those complex logarithms

[elfin cargo]

value of k that makes log non-negative

[proven night]

Okay, thanks. But are those complex logs learnt in high school?

[elfin cargo]

I don't think so, high school just gives you an introduction to complex number

[proven night]

You meant log's argument?

[elfin cargo]

Yeah, if you want you can solve $\frac{\pi}{2} + k\pi \geq 0$

[obsidian monolith]

Closer

[elfin cargo]

to find value of k

[rigid cove]

this is just $k \geq 0$ anyway

[obsidian monolith]

Pseudonium

[rigid cove]

cause k is an integer

[proven night]

Okay, thanks, i just want to clarify everything so that i have a solid understanding of it

[rigid cove]

mhm

[proven night]

And i am struggling with this equation

I tried using logs on both sides

But i think that after that and applying square of sum formula i only complicated this

[elfin cargo]

I think you can try applying ln on both sides of an equation

so you have this: $ln(3^{\sqrt{10-\sqrt{x}}}) = \frac{3}{2} ln(10-\sqrt{x})$

[viscid thistle]

help

[obsidian monolith]

Closer

[proven night]

I still cannot figure out what to do next

[elfin cargo]

Actually, nvm

😂

[proven night]

Maybe i should let 10-√x be t

[proven night]

Why

[elfin cargo]

From that I can make it to this: $\frac{2}{3}ln(3) = \frac{ln(10-\sqrt{x})}{ln(e^{{\sqrt{10-\sqrt{x}}}})}$

[obsidian monolith]

Closer

[elfin cargo]

I don't know how to simplify it further

[daring tapir]

uh i dont have a perfect proof but here we go
$\frac{2}{3}ln(3) = \frac{2*ln(\sqrt{10-\sqrt{x}}}{\sqrt{10-\sqrt{x}}}$

[obsidian monolith]

Ender Doesn't Mind

[daring tapir]

now we can divide by 2

and we see that both are of form ln(t)/t

hence by simple comparison

sqrt(10-sqrt(x)) = 3

ie x = 1

but there is probably a much more formal proof

[proven night]

Thanks

By the way why ln x^c = c * ln |x| when c is even?

Is it because √a^2 = |a|?

[proven night]

But any even root

[daring tapir]

wdym?
when c is even it allows x to be negative however if we take it out it will be ln(negative number) which is not allowed hence we take it as ln|x|

[daring tapir]

yea pretty much

[proven night]

But when c is odd, we left x as x, but not as |x|, because x^c (c is odd) cannot be negative in logarithm by default?

I mean is it a reason why this rule doesnt apply to c when it's odd

[daring tapir]

if c is odd and x is negative then it isnt defined anyways

[proven night]

Okay, i understand it

[elfin cargo]

Where does the number 2 come from

Ah, nvm

[proven night]

Can anyone help with it?

I need to factor it, but ln |x| and ln x are different, right?

[willow skiff]

Note that you can't have x < 0 however, cause ln x would be undefined

[dire locust]

hoi

[willow skiff]

so all your solutions must have x >= 0 cause of the RHS only being defined for x >=0

[proven night]

So what should be done next?

[willow skiff]

so you can remove the absolute value

4 ln x - ln^3 x = 0 and now set ln x = u

[proven night]

But i dont understand why we can remove absolute value

[willow skiff]

the right hand side is (ln x)^3

[proven night]

i still dont understand why rhs matters

[willow skiff]

cause ln x is not defined for when x is negative

so (ln x)^3 wouldn't be defined either

[proven night]

So you're saying that x cannot be negative by default since we have this RHS?

[willow skiff]

yes

[proven night]

So there is no need to use abs value

[willow skiff]

yeah so if there are any solutions, they must all satisfy x >= 0

and so you can drop the | |

[proven night]

But if it was 4ln |x| = 0, we wouldn't be allowed to drop the abs value?

Cause x can be any number

[willow skiff]

yeah that's correct

you'd be better off doing ln(x^4) = 0 implies ln(x^4) = ln(1) and then ln is 1-to-1

[proven night]

Okay, thanks, i got it

[willow skiff]

np

[elfin cargo]

Or you can set: $4ln(x) - ln(x)^3 = 0 (x \geq 0)$ so you don't have to deal with absolute shi*t

[obsidian monolith]

Closer

[elfin cargo]

Absolute is really annoying

[willow skiff]

you have to provide a reason why you can drop the absolute value sign though

you can't just drop it randomly

[elfin cargo]

I don't drop it randomly, I have provided a condition so it is still valid

I saw this a lot in textbook, though

[willow skiff]

oh shit didn't read x >= 0, sorry

[elfin cargo]

No problem.

But setting conditions like this will not give you a general solution

[willow skiff]

I mean it gives you all solutions in the real numbers so

[elfin cargo]

In this problem, yes. But I am pretty sure there are a lot of problems where setting conditions for absolute value like this will not give you general solution even when it is in real number but it makes your life easier 😂

[willow skiff]

oh true yeah splitting into cases then

[elfin cargo]

Yea, that is why absolute value is really annoying

[sour phoenix]

please why is it that when you use the u-substitution in an integral like this:

Int(1/x^2(root of x^2-(9))

It doesn't work??

[raw hill]

because it doesn't? (at least w/ the substitutions you've prob tried)

if you substitute u = x^2 or x^2 - 9

du = 2x dx

you don't have a factor of x

hence it doesn't work

[sour phoenix]

ohh yeahhhh

It's either using à trig fxn or factorisation

Like setting x=asectheta

Or just factor out .

[raw hill]

Btw, questions like this belong in #calculus, not #precalculus

[raw hill]

Yeah my first instinct is trig sub

[proven night]

Hi

I am totally clueless what i must do to solve it

[raw hill]

$\sin x=a, \sin y=2a, \cos x=3b, \cos y=2b$

[obsidian monolith]

Civil Service Pigeon

[raw hill]

Now, consider what identity relate sin and cos

You should think of ||the Pythagorean theorem||

[proven night]

How did you figure this out?

[winter comet]

That's just the ratios

[shadow ether]

u dont need abs

oh wait

old prob

nvm

[white elbow]

Right, you don't need abs, strength is more important these fitness freaks making abs overrated

[flat shuttle]

Which average rate of change is greater -0.04 or -0.4?

[ivory vale]

-0.04

[flat shuttle]

yeah I'm sure that generally -0.04 is greater but when a question says which has greater rate of change wouldn't you say -0.4 because it changes the most?

[ivory vale]

Oh,yes yes my bad.

[flat shuttle]

Ok thank you

[normal yoke]

Can someone help me with these

[ivory vale]

Have you tried all these questions at least once?

[normal yoke]

No i dont understand the topic that well

So i kind of skipped them

[ivory vale]

You should then first get a grasp on the graphs of quadratic functions

Are you an 11th grader?

[normal yoke]

10th

[shadow ether]

for the first one

remember k is the y intercept

so find the y intercept of -x^2+3x

and use that for the others

[daring tapir]

<@&268886789983436800>

[viscid thistle]

I'm just joking..

[slender river]

Hurb

[viscid thistle]

sorry alr

[slender river]

lets not please, thanks

[viscid thistle]

oh ok

[north nebula]

It will be C or D?

[subtle wren]

try them out

like say try x=-2 and x=2

[viscid thistle]

not exactly a good advice but

[viscid thistle]

It's valid only if 1/sqrt(6-3x) is defined

[subtle wren]

sorry

[viscid thistle]

a fraction is defined when the denominator is not zero so sqrt(6-3x) cannot be zero thus 6-3x cannot be zero

and the square root inside is only defined when 6-3x is non-negative

this leaves you to solve for values where 6-3x is positive

[subtle wren]

basically you can see that $\sqrt{6-3x}>0$, you can say that $6-3x>0$, so $6>3x$, then $2>x$, if you srebt sure its absolure or not, just try a number where one is correct and the other isnt, so for example -3, you can see that the value is valid

[obsidian monolith]

Skill_Issue

[north nebula]

But someone wrote thus

[subtle wren]

i think they are wrong

[viscid thistle]

Yes

[willow skiff]

No that's correct

,w 1/sqrt(6 - 3x) Maclaurin series

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by sooouth.

[willow skiff]

Not true, a function's power series can diverge even if the function itself exists

This is calculus @north nebula

[north nebula]

And this question is?

[willow skiff]

Your question is calculus

[viscid thistle]

Oh wait I just saw the expansion oops

[willow skiff]

WA won't tell me the radius of convergence goddamjt

[pastel swallow]

With this question about the conversion, the general rule is that x< absolute (1) right?
But is there an expansion where x doesn't have to be in that range?
It feels a bit useless having an expansion that is only valid for such a small amount of x values right?

[willow skiff]

the expansion doesn't converge to the function elsewhere

that's the largest possible domain that it converges

[pastel swallow]

But if we use for example (1+2x)^-1
Then we would use this formula if im not mistaken
We would have to say The absolute value of 2X is < 1, but couldn't there be greater values for x than 1/2 that would still yield answers?

would the answer we get from above formula be plain wrong if x were to be greater than 1/2

[summer ruin]

you can just google whether the series for (1+x)^{-1} converges for x >= 1

[summer ruin]

different series converge in different ranges

you can't just always say |x| < 1

[pastel swallow]

I have this alpha mathematics book, I hope im understanding this right

so i think if there is (1 + (X) )^-1 then whatever is in place of X, that sould be < 1

[viscid thistle]

How do I differentiate

X^2ln(lnX)

Can anyone just me me what to do with the ln(lnX) part?

I am very confused

[winter comet]

the chain rule

[viscid thistle]

Ahhhhh

That makes sense

Hold up

Tell me if it's correct

[viscid thistle]

Wait what

Chain rule???

[winter comet]

yes...

[viscid thistle]

But how can I use that here?

Wouldn't product rule be better?

[winter comet]

for ln(ln(x))?

[viscid thistle]

No

[winter comet]

or for x^2 ln(ln(x))

[viscid thistle]

Yes

[winter comet]

yes you have to use the product rule for the entire thing

but you asked "can someone just tell me what to do with the ln(ln(x)) part" and you have to use the chain rule to differentiate that

[viscid thistle]

Ahh so you were talking about that part

[winter comet]

yeah

[viscid thistle]

I thought I had to differentiate them separately

Anyways lemme see

[winter comet]

i mean you will eventually differentiate ln(ln(x)) and use it in the formula for the product rule

[viscid thistle]

Uhhh

I am faced with a little problem

[winter comet]

what is the problem

[viscid thistle]

I used product rule

I got

X^2 * 1/x +2x*lnx

[winter comet]

you did sum wrong lol

[viscid thistle]

But how can I multiply 2x and lnX?

[winter comet]

why X?

[winter comet]

you don't, and also its not correct

[viscid thistle]

So I keep it as it is?

Btw it's lnx

Not lnX

That was my keyboard auto correcting me

[winter comet]

d/dx x^2 ln(ln(x)) = x^2 * d/dx [ln(ln(x)))] + 2x * ln(ln(x)) by the product rule

right

[viscid thistle]

Yes

[winter comet]

d/dx [ln(ln(x))] = 1/ln(x) * d/dx ln(x) = 1/ln(x) * 1/x = 1/[xln(x)]

right

[viscid thistle]

Ahhhh right

[winter comet]

and then you can plug that into here

[viscid thistle]

That makes sense but didn't make sense at first

[winter comet]

yea

the nested ln are a lil trippy lol

[viscid thistle]

So that's where I made the mistake ig

I did

A d/dx B + B d/dx A

But I took A as x^2ln

[winter comet]

😭

ln is a function

[viscid thistle]

Yea

[winter comet]

ln(ln(x)) is like f(g(x))

and xf(g(x)) is not xf * g(x) like its not multiplied

its a function of

doesnt really maek sense to do :l

[viscid thistle]

Yeppie

Imma start functions after I am done with differentiation 2

[winter comet]

😭 😭 😭

bro is doing it in the wrong order 😭

[viscid thistle]

Am just doing the hard stuff first then imma start with the easy stuff

Oh right

Another problem....

[winter comet]

the chain rule is literally differentiating a function

a composite function

[viscid thistle]

I got x^2 * 1/lnx + ln(lnx) *2x

From which

I did

x^2/lnx + 2xln(lnx)

[winter comet]

x^2 * 1/ln(x) should be x^2 * 1/(xln(x)) -> x/ln(x)

otherwise its right

[viscid thistle]

Yeah it did come as x/lnx

[winter comet]

ok

[viscid thistle]

But can you explain the middle part?

Like x^2 *1/ xln(x)

[winter comet]

wdym

oh i meant

x^2 * 1/ln(x) is wrong, its actually x^2 * 1/ xln(x) which simplifies to x/ln(x)

[viscid thistle]

But why though?

[winter comet]

the derivative of ln(ln(x)) is 1/ln(x) * d/dx [ln(x)] = 1/ln(x) * 1/x = 1/xln(x)

chain rule

[viscid thistle]

Ye

Ohhh

Alright

So it has to be differentiated till there is no ln left

[winter comet]

i mean there is still an ln left

in 1/xln(x)

[viscid thistle]

Ehhhhhh

[winter comet]

you just have to use the chain rule :l

[white rapids]

Can somebody help me explain the symmetry here

x^3 - 4 also looks symmetric but also x^4 - 4 when i plug it into desmos

Whats the difference here

[whole void]

x^4 - 4 is not symmetric

[burnt yarrow]

could someone explain to me how this equation matches up with the graph e and not graph d?

[uncut mulch]

why do you think it should be d

[burnt yarrow]

because the asymptotes lined up

but i know it’s e i’m just confused

[uncut mulch]

asymptotes are the same for botg

[burnt yarrow]

yeah so like how can i tell the answer 😭

[uncut mulch]

sub in a few x coods for points

to see which section the curves will be in

e.g for something between the asymptotes
what happens when x=3

[burnt yarrow]

ohhh

okay that’s smart

thank you

[white rapids]

How come

Its symmetric on both sides

[visual pier]

Hi guys, I'm having trouble with arithmetic, does anyone have a book to recommend?

[whole void]

both sides of what?

[white rapids]

Positive and negative on the x axis

[whole void]

wdym?

[white rapids]

Ok,

So like on a quartic or a quadratic it goes to infinity for either -infinity and infinity on the x axis

[whole void]

what is x^4 - 4 symmetric about

[white rapids]

I really dont know

Explain to me

[whole void]

what is y = x^2 symmetric about?

[white rapids]

????

[whole void]

why the question marks?

[white rapids]

Because I dont know man

If you could explain that'll be great

If not i can just ask someone else

[whole void]

without looking it up (cause if u dont know i can explain it better) what does it mean for something to be symmetric

[white rapids]

Equal on both sides evenly

[whole void]

in terms of shapes and geometry, more specifically

[white rapids]

Equal in size??

[whole void]

no

do u want me to explain?

[white rapids]

Yes

[whole void]

two shapes could look very differant and equal in area btw

[whole void]

ok so

sorry my parents kept bothering me

a shape/curve/graph is symmetric about a particular point/line (etc.) if for every point on that graph, the reflection of that point about the point/line of symmetry yields an image that also lies on the graph

try to understand this

@white rapids

[whole void]

this u should rlly know

its called the axis of symmetry of the parabola

heres a quick sketch of this

[white rapids]

Right

So why cant a quartic function be symmetric even though at the same x value, its 2 of the same y values

[whole void]

what is y = x^4 - 4 symmetric about, if anything?

[white rapids]

Lets say

[whole void]

and why

[white rapids]

For x^4 - 4, for 1.25 and it's reflected x value, and we plug it into the function

Both the x value and its reflected x value, are the same y value

Both points are equidistant from the axis of symmetry

[whole void]

.

[white rapids]

It doesnt have to be x^4 - 4 but literally any transfomation or shifts it has

[whole void]

wdym?

what is x^4 - 4 symmetric about?

[white rapids]

Because this still applies

[white rapids]

Reflected x values, same y value

[whole void]

that doesnt answer my question what is it symmetric about

[white rapids]

For any reflected x value, it'll be the same y value

It has the same reflected image whether its going x negative or positive infinity

[whole void]

that still doesnt answer the question

what is x^4 - 4 symmetric about

@white rapids

[white rapids]

Its equidistant from axis of symmetry from both points

I can keep going

[whole void]

u r never answering the question i dont know why

what line/point/shape is the graph of x^4 - 4 symmetric about

[white rapids]

@winter comet

[whole void]

can u answer pls?

[winter comet]

yo

[white rapids]

Im not sure if im misinterpreting this

[winter comet]

its symmetric about x = 0

[whole void]

yes i wanted to get them to figure that out

they wouldnt answer

[white rapids]

Axis of symmetry i thought i said that

[whole void]

ok so now we have figured one thing the graph is symmetric about
but why is x^4 - 4 not symmetric about (0, -4)

[whole void]

i meant what in particular

[winter comet]

symmetric about a point i'm pretty sure means the same distance, 180 degrees away

[whole void]

a reflection

[whole void]

this should fit

[winter comet]

except about a point instead of a line but yeh...

[whole void]

nope

it can be either

[whole void]

@white rapids

[winter comet]

thats why the odd function reflects about the point and the even function doesn't

[whole void]

an odd function is specifically defined to mean symmetric about the origin i.e. a point

but im pretty sure u have heard of lines of symmetry in a square for example

[white rapids]

How do you get 180 degrees

Do I gotta look at the z axis or something

[whole void]

z axis?

its just a reflection over the point

[winter comet]

just by definition of "reflecting over a point

[winter comet]

nah wut

this is all just 2 dimensional

[whole void]

@white rapids

[white rapids]

I still dont understand

[whole void]

why

[winter comet]

alr uh

[whole void]

what dont u get

[white rapids]

Are we looking at x = 0 to know whether it determines

[whole void]

no

[winter comet]

no

otherwise x^4 - 4 would be correct

[whole void]

it says a point not a line

why is x^4 - 4 not symmetric about (0, -4) @white rapids and not vicious viper pls answer this

[white rapids]

Because

Its not at y = 0

[winter comet]

"Because"
-greatest answer