precalculus

[smoky pagoda]

alright sorry yes

imagine drawing a vertical line at x =1

then your set is the right hand side of that

not including the line

[viscid thistle]

so I was right

ok

[smoky pagoda]

yeah sorry

[viscid thistle]

for
{(x, x+ y) : x ∈ R, y ∈ Z}
I drew this

[smoky pagoda]

okay so

x is free to do as it pleases

but y can only take discrete values

[viscid thistle]

what do you mean by discrete

[smoky pagoda]

not continuous

integers in this case

ive got to admit im a tad confused by this set

[viscid thistle]

no idea how to draw it then

[smoky pagoda]

i mean

since the second part of the pair is a real number + an integer

thas just another real number

[viscid thistle]

so i'm good?

[smoky pagoda]

okay no this is weird gimme a sec to think

let's take some cases

[hard hornet]

whats the question?

[smoky pagoda]

if x is 0

[viscid thistle]

for
{(x, x+ y) : x ∈ R, y ∈ Z}
I drew this
@viscid thistle

[smoky pagoda]

you have all the integers on the y axis

if x is 0.5

all the 0.5 offset integers on the line x = 0.5

okay i think i got it

i belive it's a striation of lines with slope of -1

each offset on their y intercept by integer values

[viscid thistle]

could you please draw it

[smoky pagoda]

yes lemme do it quickly

this is what drawing with a mouse gets you XD

this isnt a shaded area

the lines that arent the axes are part of the set

[viscid thistle]

how have you arrived at this conclusion

i belive it's a striation of lines with slope of -1
@smoky pagoda

[smoky pagoda]

checking certain cases of x

havent proved it or anything though

but intuition gets me this

because on any line x

like vertical line x = a

you can plot points on it that belong to the set but these points will be separated by integer distances

additionally the value of x applies an original offset

since we have x+y

and this offset will be the same for integer values of x

which is why the 0s are at x = ...-5,-4,-3,-2,-1,0,1,2,3,4,5,....

and otherwise between these intervals the change in offset is continuous

[viscid thistle]

makes sense
in the last two I have no idea what R^2 is supposed to mean.
51 {(x,y)∈R^2 : (y-x)(y+x) = 0}
here I would just do y^2-x^2 = 0
52. {(x,y)∈R^2 : (y-x^2)(y+x^2)=0}
not sure

[smoky pagoda]

R^2 is the real plane

think of it like a 2D R

similarly R^3 is the 3 dimensional real space

so it's the same as saying x belongs to R and y belongs to R

[viscid thistle]

haven't i been drawing R on a 2d sheet of paper

[smoky pagoda]

and for the last one you would do y^2 - x^4 = 0

yes youve been using R^2 up till now anyways

[viscid thistle]

but it told me to use R?

[smoky pagoda]

it's just your regular plane

no but your sets as subsets of R^2 by convention

the sets themselves arent always R^2 though

it's just R^2 is easily represented by a cartesian plane

so you're saying that (x,y) belong to R^2

which must mean the resulting set is at least a subset of R^2

If you know what a cartesian product is, R^2 is just shorthand for RXR

[viscid thistle]

could you please draw both - show me the difference between them?
{(x,y) : x, y ∈ R, x^2 +y^2 = 1}
{(x,y) : x, y ∈ R^2, x^2 +y^2 = 1}

cause from what you're saying i get that there's no difference in drawing

[smoky pagoda]

k=okay so your second set isnt exactly correct

elements of R^2 must be of a form (x,y) where x,y belong to R

x and y are real numbers

but the tuple (x,y) is an element of R^2

you cant say x belongs to R^2

or x would have to be its own tuple

R is the set of all real numbers

[viscid thistle]

what about
{(x,y) : (x, y) ∈ R, x^2 +y^2 = 1}
{(x,y) : (x, y) ∈ R^2, x^2 +y^2 = 1}
then

[smoky pagoda]

okay so now

the first set is wrong

[viscid thistle]

guessed it

[smoky pagoda]

because you say that a tuple belongs to R

[viscid thistle]

so when using tuples, I say that they ∈ to R^2, otherwise to R, yeah?

[smoky pagoda]

yes

[viscid thistle]

ok I get it

[smoky pagoda]

depending on the size of tuples of course

a tuple (x,y,z) belongs to R^3

[viscid thistle]

ok thanks

[smoky pagoda]

note this power notation can be applied to any set

in general A^2 = AXA

where X is the cartesian product

[viscid thistle]

have to go now, bye

[smoky pagoda]

alright bye

[viscid thistle]

The set T = {0,{1, 2, 3}, 4, 5} also has four elements, namely, the three integers 0, 4
and 5 and the set {1, 2, 3}. Even though 2 ∈ {1, 2, 3}, the number 2 is not an element of
T; that is, 2 ∈/ T.
why not? if A = {1,2,3}, A ∈ T, 2 ∈ A, then does it not logically follow that 2 ∈ T ?

[harsh smelt]

no

since A is considered as single element here

ur statement would be trou if u were talking about subsets

but not sets as elements

[viscid thistle]

thanks

how come

ac + adi + bci + bdi^2 = (ac − bd) + (ad + bc)i
isn't
(ad + bc)i = adi + bci
and that would mean that
(ac − bd) = ac +bdi^2
?

[sour eagle]

(ad+bc)i = adi + bc
(ad+bc)i = ad*i + bc*i = adi + bci

You forgot an i

[viscid thistle]

yeah, thanks

but I don't think that changes the fact that

(ac − bd) = ac +bdi^2

[sour eagle]

It is true

[viscid thistle]

how is it derived

[past meadow]

Are you asking how ac-bd=ac+bdi^2?

[viscid thistle]

yes

[past meadow]

ac-bd =ac+bd*(-1)=ac+bd*(i^2) since i^2 is defined to be -1

[viscid thistle]

oh makes sense, thanks

what about
Let S = {−2, −1, 0, 1, 2, 3}. Describe each of the following sets as {x ∈ S : p(x)}, where p(x) is some
condition on x.
D = {−2, 2, 3}.

i don't see a pattern here

[sour eagle]

p(x) : |x| > 1

[viscid thistle]

it wouldn't include -2 then, I think?

[sour eagle]

|x| is the absolute value of x

[viscid thistle]

oh right

[sour eagle]

😄

[viscid thistle]

also wouldn't 'your' set in full be written as
'{x ∈ S : p(x), |x| >1}'?

[sour eagle]

Yes

D = {x ∈ S : |x| > 1}

You just subtitute p(x)

[viscid thistle]

yeah thanks

is there a method for questions like this?
The set E = {..., −4, −2, 0, 2, 4,...} of even integers can be described by means of a defining condition
by E = {y = 2x : x ∈ Z}={2x : x ∈ Z}. Describe the following sets in a similar manner.

they are really time-consuming for me

[willow bear]

no there is no method

[viscid thistle]

thanks

how do I solve this
x^2 − (2 + √2)x + 2√2 = 0?
I don't know how to combine -2x with -sqrt(2)x

[willow bear]

??

this is just a quadratic lol

don't overthink it

b here is -(2+sqrt(2))

[viscid thistle]

i guess

D = {x ∈ Q : x^2 − (2 + √2)x + 2√2 = 0}.
describe the set D in another manner

what is meant by this? like do
{x^2 − (2 + √2)x + 2√2 = 0 : x ∈ Q}
instead?

is it possible to describe it any other way besides these two?

List the elements of the set i.e. roots of the quadratic equation.

[willow bear]

the second is bad

[viscid thistle]

but in this form

[willow bear]

solve the equation

[viscid thistle]

2, sqrt(2)

{2, sqrt(2)}

so are there only 3 ways?

two ways

{x^2 − (2 + √2)x + 2√2 = 0 : x ∈ Q}
is incorrect, yes?

not exactly incorrect, but yeah

ok thanks

You're welcome ^^

For A = {2, 3, 5, 7, 8, 10, 13}, let
B = {x ∈ A : x = y + z, where y,z ∈ A} and C = {r ∈ B : r + s ∈ B for some s ∈ B}.
Determine C.

what is meant by determine C? like do
C = {r ∈ B : r + y+z ∈ B, where y,z ∈ A} ?

B = {5, 7, 8, 10, 13}
C = {10, 13}

why is C= {10}, what about 13? 5+8 = 13

Oops, my bad

thanks for all your help

[lilac pier]

what's |v| and |w|

should be 4sqrt(10)

[odd helm]

for this problem i get undefined because i can’t do arccos when the value i plug in is less than -1

so this problem doesn’t even have an answer am i making a mistake somewhere

[uncut mulch]

where are you getting sqrt(4),

[odd helm]

oh

lmao

it’s sqrt10 i just did 1+3

[junior sable]

Hello, I was wondering if someone would help me understand a classic interest rate problem. I couldn't follow their steps.

[remote veldt]

@junior sable increasing something by 5.75% is the same as taking your number, and adding 5.75% of that number

5.75% of something is that thing times 0.0575

so they just did that three times

[junior sable]

gotcha ty @remote veldt

[remote veldt]

and specifically with interest, after your money increases, you now have more money that will increase

[junior sable]

gotcha. it recurs

can I ask your help with this one? how do you quickly find the horizontal asymptote of this?

Do you just plug in values?

[remote veldt]

well a horizontal asymptote asks about behaviour as x gets really really big, either in the positive or negative direction

when x gets really really big in the positive direction, what happens to 2 * e^(x - 2) + 3?

[junior sable]

I suppose it gets bigger. e is getting raised by a bigger and bigger number

[remote veldt]

yes! it blows up to infinity

so there's no horizontal asymptote that way

because a horizontal asymptote, sort of by definition, requires flattening out

what about the other direction? when x gets more and more negative?

[junior sable]

It would get smaller and smaller. As it passes 2, it will start to look like a fraction, 1/e

[remote veldt]

yep! and then 1/(e^2) and then 1/(e^3). As the exponent gets bigger and bigger, the exponential term gets closer and closer to 0

2 * 0 is just 0

so as x gets closer and closer to -infinity, the function gets closer and closer to 0 + 3

which is just y = 3

[junior sable]

you said that there's no horizontal asymptote, but isn't it y=3?

[remote veldt]

there's no horizontal asymptote in the positive direction

sorry, the "that way" meant "in the positive direction"

you need to check both

and yes, the horizontal asymptote is y = 3

but it's important to check both the positive and negative direction

[junior sable]

sure. that's very helpful, thank you

[remote veldt]

because there might be more than one asymptote, or the asymptote may only exist on one "side"

for example, the inverse tangent function has two horizontal asymptotes, and the two are different in the positive and negative direction

[junior sable]

In this example, how do I know that there isn't a vertical asymptote as well at x = 4?

[remote veldt]

well, the sort of cheating way is to say that f(4) exists

[junior sable]

because asymptotes approach impossible numbers

?

[remote veldt]

that's a little imprecise, but it is definitely the right idea

a vertical asymptote is that, as x gets closer and closer to the point, f(x) gets closer and closer to either positive or negative infinity

if f(x) is continuous and is defined at a point, there isn't an asymptote there

[junior sable]

that's what you mean by f(4)

existing

[remote veldt]

yes!

because think about it, an asymptote kinda keeps going up

(or down towards negative infinity)

[junior sable]

There's a peak when it comes to vertical asymptotes, right? eventually you get over the hump and start coming down

[remote veldt]

no!! what makes it a vertical asymptote is that you keep going up

[junior sable]

and at the peak it's undefined?

[remote veldt]

there is no peak

[junior sable]

oh...oof.

[remote veldt]

consider f(x) = 1/x as x gets closer to 0

give me any real number k, and I can find a value of x so that 1/x > k

[junior sable]

98

oooh

[remote veldt]

98? ok 1/(0.01) > 98

if you had picked 183423874823748237482374238 I could say that 1/(0.00000000000000000000000000000001) > 183423874823748237482374238

I can always find a number 'close to zero' (I'm putting that in quotes because it's imprecise but I think that any reasonable person would know what I mean) so that it gets arbitrarily large

that's what makes it a vertical asymptote

so a little more precisely, you know that e^(x - 2) + 3 doesn't have a vertical asymptote, since near any point, it's bounded

[hard hornet]

wtf is going on here holy 👀

[remote veldt]

wha?

[hard hornet]

horizontal asymptote, is when you have a very large negative/positive x value, in which the function approaches some constant value

[junior sable]

I guess I still don't understand it. I mean in e^(x - 2) + 3 it does get progressively bigger and bigger, why isn't it an asymptote?

[hard hornet]

for your function g(x) = 2e^{x-2} + 3, it's obvious that for a very large positive number, it doesn't approach any constant value, it just keeps going up, but for a very large negative number, the e portion approaches a very VERY small value, almost negligible, thus resulting in a constant value of 3

there is a horizontal asymptote

[junior sable]

okay that's helpful

[remote veldt]

@hard hornet yes, we established there is a horizontal asymptote

[hard hornet]

choose a very large negative x value

[remote veldt]

that's done

there is a horizontal asypmtote at y = 3

[junior sable]

but I guess I still don't see why f(4) isn't an asymptote, is it just that the function doesn't approach a specific point there?

[hard hornet]

if you're still confused, take a look at this function instead, not sure if this'll help

,w graph y = e^x

[obsidian monolith]

Display results online and refine query

Requested by Cytis#6174

[hard hornet]

asymptotes are values the function never reaches when you approach negative infinity or positive infinity

we talk about asymptotes in the context of VERY LARGE numbers, i.e infinity

[remote veldt]

buncho, what I'm replying to is

In this example, how do I know that there isn't a vertical asymptote as well at x = 4?

the answer is that f(x) is bounded near 4, or alternatively, that f(4) exists and f(x) is continuous

since for there to be a vertical asymptote, f(x) must be unbounded near a point

[hard hornet]

similar idea, except instead of x approaching infinity, it's more like y not being defined, and y blowing up to infinity as you get closer to x = 4

[remote veldt]

...and that is exactly what I said

[hard hornet]

if you're trying to figure out easy, simple tests for asymptotes, for horizontal asymptotes, choose a very large positive/negative value, and for vertical asymptotes, usually it's with functions with fractions, because that's how you get undefined values. Check the point in which it's undefined

[junior sable]

@remote veldt are you a teacher? You have a real gift for encouragment.

[remote veldt]

I'm unsure if you're being sarcastic or not ;-;

[junior sable]

no!

don't let the internet poison you!

I mean it!

[remote veldt]

I'm an undergrad student, decidedly not a teacher

I want to do it some day

[junior sable]

Well it suits you!

[remote veldt]

I'm just rather confused about the "wtf is going on here holy"

thank you!

[junior sable]

I got another one for you if you're ready

[remote veldt]

of course!

actually there was one quick thing I did want to answer

[junior sable]

?

[remote veldt]

it does get progressively bigger and bigger, why isn't it an asymptote?

an asymptote needs to be getting bigger without bound at a specific point

for e^x or any transformation thereof, for any point you pick, e^x is a finite number, and e^x is finite near that point

what makes it an asymptote is if it goes to +/- infinity at a point

not if it goes to +/-infinity as x goes to +/- infinity

[junior sable]

and it's because there will be an x value which is undefined as it approaches it?

[remote veldt]

there is a vertical asymptote if there is an x value where the function goes to infinity as you get close to that x value

e^x has no such x value

so e^x has no vertical asymptotes

e^x has a horizontal asympote, since as x goes to negative infinity, y gets closer and closer to 3

[junior sable]

okay, i think i get it now

[remote veldt]

awesome!

[junior sable]

yeah. cool!

I got it.

Here's the one I'm stuck on now. I just don't understand the "Grammar" of logarithms.

does this mean that 4^(gx) = x+1?

I tried to plug in -1 for x and then I got confused

[remote veldt]

rearranging it into that form isn't going to be particularly helpful/meaningful; your simplifcation is a little bit wrong though.

I would hope that you got confused if you plugged in -1 for x, since the function is only defined for x > -1

[junior sable]

how so?

[viscid thistle]

Try graphing the parent function and then transforming it.

[hard hornet]

^^^

,w graph y = ln(x)

[obsidian monolith]

Display results online and refine query

Requested by Cytis#6174

[viscid thistle]

the (x+1) moves the graph left 1 I believe.

[remote veldt]

yeah. This is a series of transformations applied to log_4(x)

[which is really just ln(x) multiplied by a constant but whatever]

[viscid thistle]

true.

@junior sable do you think you understand?

[junior sable]

I think I need to understand graph transformations....i hate graph transformations....

And on the website I'm working with, they don't let you choose a log shape and then play with it

you have to plug in x and y coordinates

So you do need to make your way through the function algebraically

that's what I'm stuck on right now

[remote veldt]

use desmos

there is something so helpful desmos has

[junior sable]

I mean I have a final on wednesday that I can't use Desmos for

[remote veldt]

but to gain understanding

[junior sable]

sure

[remote veldt]

set something like this up

and play with the sliders

[viscid thistle]

When doing so don’t just make it give you the answer, try to understand the nature of the function you’re transforming. (e.g how it transforms based on the transformation.)

[remote veldt]

yeah^^ of course. But if you play with the sliders one at a time, you can get a lot of intuition

[viscid thistle]

I agree.

[junior sable]

just gotta play with it.

[viscid thistle]

[junior sable]

Does it ever seem impossible to try to map out the geometric and visual knowledge onto the mathematical, algebraic knowledge?

I can play with a map and go "Oooh, pretty shapes" but that's a whole different skill set than "How do these numbers work"?

And I'm not entirely sure I can keep both in my head at the same time.

[viscid thistle]

I don’t know if I understand what you mean... like trying to retain geometric and algebraic intuition? or smthn

[junior sable]

Well you can understand math visually, with graphs, or you can do it with pen and paper and logic.

[viscid thistle]

I mean they go hand in hand, you just have to study their relationship.

it took me a month to understand why y = Mx + b somehow turned into a / or \ . I just had to ask why the formula did what it did...

I guess I’m not too good at e x p l a i n i n g stuff huh?

[remote veldt]

@junior sable make sure there's a method to how you play

you should be able to answer

• what does increasing this do geometrically
• what does it matter if this is positive or negative
• what does decreasing this do geometrically
  for all the parameters

[viscid thistle]

that’s much better ^^^ yes

[remote veldt]

ie start by playing

then start asking specific quetsions about what happens when you play

[junior sable]

https://tenor.com/view/youre-good-robert-de-niro-gif-11622680

[viscid thistle]

that’s a spectacular gif

[junior sable]

@remote veldt I'm telling you, you're gonna be the next 3blue1brown

[remote veldt]

LMAO no way

[viscid thistle]

Who is 3b1b

[junior sable]

you gotta check him out. he's the bob ross of math

[viscid thistle]

ahh yes

Well I agree

@remote veldt you might just be.

[junior sable]

@viscid thistle https://www.youtube.com/channel/UCYO_jab_esuFRV4b17AJtAw pick a topic go nuts

[viscid thistle]

Yea just looked him up

Might just become one of my idols

alongside Khan and Gauss

[junior sable]

I could have used someone like him in my life 10 years ago

Teaching math with patience and wonder

[viscid thistle]

well maybe I can use him now

[junior sable]

definitely. it's never too late. he covers topics way more advanced than what I'm ready for but maybe you'll find something your liking

[viscid thistle]

he seems like a genius,

I will have to check him out

mathematics is the language that which we speak to the universe in.

[junior sable]

thanks everybody for the help! sorry I got lost in desmos

[junior sable]

@remote veldt I'm going to try to tackle it now algebraically now, any pointers? it looks like a real tangled mess

If i put 4 in for x I get 3 log 4 (4+1) -1...what order do I begin to evaluate such an expression?

[viscid thistle]

Lol what.

Bad notation.

$3 \log_{4}(4+1)-1$

?

[obsidian monolith]

leviosa:

[junior sable]

yes

[viscid thistle]

Just follow PEMDAS.

Parantheses:

$3 \log_{4}(5)-1$

[obsidian monolith]

leviosa:

[viscid thistle]

,w log_4(5)

[obsidian monolith]

Display results online and refine query

Requested by leviosa#9955

[viscid thistle]

Then you have.

$3(\text{whatever the fuck this equals to})-1$

[obsidian monolith]

leviosa:

[viscid thistle]

And multiply.

Then minus 1.

And bam.

Ez claps.

@junior sable

[junior sable]

how would you solve it if you didn't have access to a calculator?

[hard hornet]

solve what?

[junior sable]

$3 \log_{4}(4+1)-1$

[obsidian monolith]

Jandro:

[past meadow]

Generally you would just leave it as 3log_4(5)-1

[hard hornet]

i woiuldnt be able to give you the exact value, but i could give you some approximation

[junior sable]

I don't know what my class is expecting of me

[hard hornet]

this is precalculus, they wont ask you to give the exact value

[junior sable]

They give me that as g(x) and then say "graph it" no graphing calculator

[hard hornet]

when they say graph it

it's more like

understanding the general structure of it, and some important points

[junior sable]

no they literally want me to graph it

[hard hornet]

if you gave me $g(x) = 3\log_{4}{x+1} - 1$, i would base it off parent function

[viscid thistle]

You want to plug something in then.

[junior sable]

yes

[viscid thistle]

You can do a few things.

[obsidian monolith]

Cytis:

[viscid thistle]

You can do the inverse of the function.

You'd then have to deal with a function of y rather than a function with x.

(Literally just turn your paper 90 degrees LOL.)

[junior sable]

computer class. sure I'll turn the monitor 90 degrees

[viscid thistle]

Or, you can plug in certain x's that will make it so that log_4(??) would turn out to be a constant number.

[hard hornet]

well, you'll still need to understand some form of the parent graph

if you understand your parent graph

then this question is trivial, it ends up being an easy transformation

[viscid thistle]

Bruh really.

Wait yeah.

[hard hornet]

.-.

[viscid thistle]

Parent graphs probs easiest.

LOL.

[junior sable]

I understand the parent graph, i just need some stable numbers to graph onto

[hard hornet]

you don't

here

[junior sable]

and they won't let me get those without a calculator

[hard hornet]

let me show you some magic

the parent function is f(x) = log_4(x) right

[viscid thistle]

@hard hornet Nah. Shitty classes like the ones I take asks you to make sure you have 5 dots plotted 🤡

[hard hornet]

still easy

[viscid thistle]

Yeah.

Might as well just plot the 5 points to get the graph.

[hard hornet]

Jandro, you following?

[viscid thistle]

Anyways.

[junior sable]

yeah, the parent function is right

[hard hornet]

your parent function would be f(x) = log_4(x) agree?

graph it

[junior sable]

okay, how do I do that without desmos

[viscid thistle]

[hard hornet]

its obvious you have some points already from the log function

(1, 0)
(4, 1)
(16, 2)
(64, 3) etc etc

[viscid thistle]

You plug in numbers.

^

[hard hornet]

and there's a vertical asymptote

so there, boom, you got the structure of the graph already

[junior sable]

There is nothing boom about that.

I do not understand how you got those numbers

[hard hornet]

do you know what the logarithm function is?

[junior sable]

It's a measure of magnitude

[hard hornet]

give me the definition of the logarithm function in terms of exponents

[junior sable]

a^b = c, log a c= b

[hard hornet]

yup

log_4(x)

means if you plug in powers of 4

you get the exponent

(4^0, 0)
(4^1, 1)
(4^2, 2)
(4^3, 3)
(4^4, 4)
etc

that's how you get the numbers

[junior sable]

and to be clear, we're talking 4 as the a value for log, right?

[hard hornet]

yes, log_4

it's logarithm base

4

[junior sable]

okay I'm not 100% sure I get you but I'm starting to get there

[hard hornet]

ok

using the definition of logarithm u stated

come up with your own points

ping me when you feel like you've got it

[junior sable]

well okay @hard hornet 16, 2 and 64 and 3

16 and 64 are the c values, 2 and 3 are the b values

are these coordinate points?

[hard hornet]

for the function f(x) = log_4(x)

yes

,w graph log_4(x)

[obsidian monolith]

Display results online and refine query

Requested by Cytis#6174

[hard hornet]

so ur parent function looks like this, right?

[junior sable]

yes but hold on

you're saying that the points 2, 16 and 3, 64 are on that line?

[hard hornet]

(16, 2)
(64, 3)

(x, y)

are on that line

its the graph of the function y = log_4(x)

[junior sable]

please continue

[hard hornet]

now

let's start adding some components to it, and slowly build out way to the original function

we know what y = log_4(x) is, now try to figure out what happens with y = log_4(x-1)

hint: it's a shift of the graph

think it through, no answers for you

[junior sable]

It's a move to the right, but only because I was playing with it on Desmos earlier in the night

[hard hornet]

👍

makes sense right?

since it's x-1 instead of x, you need to increase the value of x by 1 to maintain it

now, let's move on, this one is going to be a bit harder

now, try y = 3log_4(x-1)

hint: it's a shrinking of the graph

[viscid thistle]

$y-h=f(x-|k|) \ \text{is a translation k units right and h units up}$

Wait.

[junior sable]

I thought it was an inflation of the graph. I thought the graph has higher values with a higher a value

[viscid thistle]

I mixed up the letters.

Oh well.

[obsidian monolith]

leviosa:

[viscid thistle]

Okay.

[hard hornet]

ah yes, you're right, it should be an inflation of the graph

,w graph 3log_4(x-1)

[obsidian monolith]

Display results online and refine query

Requested by Cytis#6174

[viscid thistle]

@hard hornet Your pfp cute.

[hard hornet]

yes you're right

its stretched upwards/downwards

ty ❤️

my mistake, good catch on that

now

the final part

y = 3log_4(x-1) + 1

i think that was the original querstion

[junior sable]

-1

[hard hornet]

how does it shift?

[junior sable]

down 1.

[hard hornet]

,w graph y = 3log_4(x-1) + 1

[viscid thistle]

That silence basically means no.

[obsidian monolith]

Display results online and refine query

Requested by Cytis#6174

[hard hornet]

well it was a 50% chance

it's a rise of 1

because you're adding 1 to the function

[junior sable]

no i mean you had the wrong function

[hard hornet]

oh

well

it's a shift down of 1

[junior sable]

it's -1, not plus 1

[hard hornet]

my mistake

but you got the idea now?

[junior sable]

the coin flips in my favor!

[hard hornet]

indeed it does

if you need more practice

[junior sable]

well what do we do with the data points now

that's what I need to graph this thing

[hard hornet]

well, you could reverse engineer

we know that the data points originalyl is

(4^0, 0)
(4^1, 1)
(4^2, 2)
(4^3, 3)

right?

[junior sable]

right

[hard hornet]

in the first part, this gives us these new points

(4^0 + 1, 0)
(4^1 + 1, 1)
(4^2 + 1, 2)
(4^3 + 1, 3)
for the function f(x) = log_4(x-1)

agree?

[junior sable]

hmm

hold on

[hard hornet]

you can test the values

reverse engineering

[junior sable]

here's where I get lost

the new y values, how did you get them?

wait

I mean the new x values I misread it

[hard hornet]

we went from a change of x to x-1 right?

[junior sable]

right

[hard hornet]

to maintain the same y value

we have to negate it

by adding 1 to each x value

do it on apper

paper

and work the math out

[junior sable]

yeah that type of motion I never understood

[hard hornet]

it's kind of realizing, oh shit the function needs an increase of 1 for the x value to maintain the same y value

so, just add 1 to each input, and keep the y value the same

[junior sable]

ohhhh

yeah

f(6) will be the new f(5)

[hard hornet]

basically

[junior sable]

so it's gonna shift to the right

[hard hornet]

well more like f(5) is the new f(4)

but you got the iea

[viscid thistle]

Oh wait wtf.

I thought your pfp was a bunny.

But it's a fucking bird bunny.

I take what I said previously back.

[hard hornet]

lmao

[viscid thistle]

Ugly asf.

[hard hornet]

😦

[viscid thistle]

[junior sable]

ban this fool.

[viscid thistle]

[hard hornet]

[viscid thistle]

[hard hornet]

so far so good @junior sable

[junior sable]

wait, i'm almost there

[hard hornet]

ping me when ur ready to move on

[junior sable]

so you're saying that instead of a point being at (16,2) it's gonna be at (17,2)

@hard hornet

[hard hornet]

yup

its a shift to the right

by 1 space

[junior sable]

Well alright then

I have some data points to work with

[viscid thistle]

Are you graphing this on paper

This man aboutta graph (65, 4)

What a legend.

Oh wait.

(65, 5)

Mb.

[hard hornet]

65, 3

u okay there

[junior sable]

yeah

I only need 3 points

it's a -10 to 10 field

[viscid thistle]

@hard hornet Bruh you okay.

(64+1,4+1)

Totally 65, 3.

[hard hornet]

y = log_4(x-1) is what we're doing rn

[viscid thistle]

Where tf you get that 3.

[hard hornet]

log_4(65 - 1) = log_4(64) = 3

(65, 3)

[viscid thistle]

Oh wait.

[hard hornet]

are you okay?

[viscid thistle]

For a good moment.

I thought 4^4 was 64.

[hard hornet]

i know

[viscid thistle]

My mind has been changed.

I'm a new man.

Thank you Buncho Goons.

[hard hornet]

np

[junior sable]

i have a -10 by 10 field on a computer screen. what points can I use?

[hard hornet]

so u srsly need 10 points?

i mean 3 points

[junior sable]

yeah

[hard hornet]

like does it have to be THAT ACCURATE???

[junior sable]

yeah!!!

[hard hornet]

ok that's stupid imo

on a 10*10 screen

thats really stupid

[junior sable]

i think they expect me to plug in answers that have e's in them

[hard hornet]

that's if you're dealing with natural logs

hoenstly at this point, just use a freakin calculator,

on a test, you can approximate as long as you label ur points

[junior sable]

I can't! That would be cheating.

I can't! it's on a computer.

[hard hornet]

if it has to be super accurate, everyone is gonna fail anyways, especially with a log function, on a 10*10 screen

you can prolly graph like 2-3 points RELIABLY

more like 2

[junior sable]

4^(3/2)?

[hard hornet]

(4^0, 0)
(4^1, 1)

are like the only two points you can reliably graph

cuz im assuming only integer points are allowed, anything fraction will give u trouble graphing

[junior sable]

no fractions are okay

there's a field for that

you plug in the fractions and it'll plot it for you

[hard hornet]

this homework assignment is giving me a headache

[junior sable]

[hard hornet]

ok i legit don't know how to do it anymore. graphing homeworks shouldnt be done this way, it's silly

i gave you a lot of integer points to work with, and showed you the math behind it

try your best, if there's still anything you need, i'll try

[junior sable]

I appreciate your help.

Thank you. Logs are making sense to me now

do you own a parakeet?

[hard hornet]

no, want to tho

[junior sable]

I hear they're hilarious birds

[hard hornet]

they r

[junior sable]

i hope you get one

I gotta go to sleep. thanks for the help, internet denizen.

Good night!

[blazing raven]

heyo ... Prove, for any distinct, positive a and b, (ln(a) + ln(b))/2 cannot equal ln((a + b)/2)

[willow bear]

strict concavity of ln

[blazing raven]

yep yep. Man I don't think I've ever outsmarted you in math.

[junior sable]

can someone walk me through how to do this problem?

[viscid thistle]

So

First consider taking ln on both sides @junior sable

[junior sable]

is ln and log one of those operations you have to do on both sides?

i would assume so

[viscid thistle]

... of course

But also take into account to choose the base of ln in our benefit

I need you active for this ass-long problem

@junior sable

[junior sable]

okay

I'm ready

[viscid thistle]

First consider taking ln on both sides
But also take into account to choose the base of ln in our benefit

[junior sable]

why ln instead of log 10?

[viscid thistle]

Yeah that's what i meant

Take into consideration the base

To our benefit

What would be the base for our benefit

Of log

[junior sable]

i think log 10 would be easiest

log10 is the default, it goes with everything

[viscid thistle]

Look at the bases of the equation

11 and 9

What would be the base for our benefit
Try again

[junior sable]

I don't know. 11 and 9 don't share factors

[viscid thistle]

Do you know why im saying this?

[junior sable]

no i don't

[viscid thistle]

$\log_a(a)=1$ this will happen here on this problem

[obsidian monolith]

Al𝟛dium:

[junior sable]

oh

log 11 and log 9

[viscid thistle]

So try again

One of them

Lets keep with 11

So log_11 on both sides

So that

$$11^{x-9}=9^{6x}$$ taking $\log_{11}$ on both sides $$\log_{11}(11^{x-9})=\log_{11}(9^{6x})$$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Everything fine until here?

[junior sable]

yes

[viscid thistle]

Good.

Now consider this

$9^{ax}=(3²)^{ax}=3^{2(ax)}$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Everything good?

[junior sable]

well hypothetically sure, but we don't have a's in our equation

[viscid thistle]

It was an example dw

Apply it to the RHS

[junior sable]

the what?

[viscid thistle]

RightHandSide

Lol

The right side of the eqn

[junior sable]

3^12x

[viscid thistle]

Yep

With

But ok

[junior sable]

so 2(6x)

[viscid thistle]

$$11^{x-9}=9^{6x}$$ taking $\log_{11}$ on both sides $$\log_{11}(11^{x-9})=\log_{11}(9^{6x})$$ $$11^{x-9}=9^{6x}$$ taking $\log_{11}$ on both sides $$x-9=\log_{11}(3^{12x})$$

so 2(6x)
@junior sable no like 3^(12x) not 3^12x

Bc its ambiguous

[junior sable]

?

[obsidian monolith]

Al𝟛dium:

[junior sable]

ok

[viscid thistle]

Hold on

Uh

Forget it lol

So let me see

[junior sable]

haha ok

[viscid thistle]

Yeah ok

So take the 12x out of the log

[junior sable]

12x log 11 3

[viscid thistle]

Yeah

$$11^{x-9}=9^{6x}$$ taking $\log{11}$ on both sides $$\log{11}(11^{x-9})=\log{11}(9^{6x})$$ $$11^{x-9}=9^{6x}$$ taking $\log{11}$ on both sides $$x-9=\log_{11}(3^{12x})$$ taking 12x out $$x-9=12x\log_{11}(3)$$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Im having this so that you can write it on your paper

Later

[junior sable]

i'm writing it down as we speak

[viscid thistle]

So you can focus

Uh

Well whatevr

[junior sable]

the steps are important ty

[viscid thistle]

So move the x term/s at one side and non-x term/s at the other

Yep

We are really close

[viscid thistle]

Uh not like that

From the prev step

Isolate 9

[viscid thistle]

$$11^{x-9}=9^{6x}$$ taking $\log_{11}$ on both sides $$\log_{11}(11^{x-9})=\log_{11}(9^{6x})$$ $$11^{x-9}=9^{6x}$$ taking $\log_{11}$ on both sides $$x-9=\log_{11}(3^{12x})$$ taking 12x out $$x-9=12x\log_{11}(3)$$ moving terms $$x-12x\log_{11}(3)=9$$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Well yeah fair

Now factor out x and isolate x and that's it :)

As it is the final step let me write it down

$$11^{x-9}=9^{6x}$$ taking $\log_{11}$ on both sides $$\log_{11}(11^{x-9})=\log_{11}(9^{6x})$$ $$11^{x-9}=9^{6x}$$ taking $\log_{11}$ on both sides $$x-9=\log_{11}(3^{12x})$$ taking 12x out $$x-9=12x\log_{11}(3)$$ moving terms $$x-12x\log_{11}(3)=9$$ factor out x $$x(1-12\log_{11}(3))=9$$ finally $$x=\frac{9}{1-12\log_{11}(3)}≈-2$$

@junior sable

[obsidian monolith]

Al𝟛dium:

[junior sable]

my answer does not look like yours

[viscid thistle]

Welp you did different stuff on the very few last steps

Post a better pic

Also on the last step you didn't isolated x correctly

[viscid thistle]

Also on the last step you didn't isolated x correctly
.

Pretty basic algebra

$a=x(b)$ isolate x is $x=\frac{a}{b}$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Not whatever you did on the last step

@junior sable ok?

[junior sable]

no, how should i have done it differently?

[viscid thistle]

From there

Factor x

$-9=x(12\log_{11}(3)-1)$ once again

[obsidian monolith]

Al𝟛dium:

[junior sable]

yes that's what I have

[viscid thistle]

Isolate x appropiately

$a=x(b)$ isolate x is $x=\frac{a}{b}$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Pretty darn basic algebra

[junior sable]

no thanks

[viscid thistle]

AGAIN?

Are you listening to me?

[junior sable]

Apparently I'm not getting you.

[viscid thistle]

The last step is not what we want

Do you know what isolate is?

[junior sable]

oh i see

[viscid thistle]

Post it

[viscid thistle]

Yes.

,w (-9)/(12log_(11)(3)-1)

[obsidian monolith]

Display results online and refine query

Requested by Al𝟛dium#4084

[viscid thistle]

Yep correct.

$$11^{x-9}=9^{6x}$$ taking $\log_{11}$ on both sides $$\log_{11}(11^{x-9})=\log_{11}(9^{6x})$$ $$11^{x-9}=9^{6x}$$ taking $\log_{11}$ on both sides $$x-9=\log_{11}(3^{12x})$$ taking 12x out $$x-9=12x\log_{11}(3)$$ moving terms $$x-12x\log_{11}(3)=9$$ factor out x $$x(1-12\log_{11}(3))=9$$ finally $$x=\frac{9}{1-12\log_{11}(3)}≈-2$$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

We got the same

What the hell is that?

Just type it onto the calculator and you'll get ≈-2

End of the problem

@junior sable ?????

Can you write instead of posting pics with no context at all

[junior sable]

I'm asking if the answer if we got is the same as 9log11/(log11-6log9) which is the answer i'm supposed to get

[viscid thistle]

,w (9log(11))/(log(11)-6log(9))

[obsidian monolith]

Display results online and refine query

Requested by Al𝟛dium#4084

[viscid thistle]

Yep its the same

[junior sable]

good

thanks for your help

[viscid thistle]

Np!

[junior sable]

I'm trying my best. try not to be angry at people who are trying their best.

[viscid thistle]

You could have said "hold on" or whatever but if i see you inactive i lose my patience

[junior sable]

oh it said that?

I'm new to discord

I was here the whole time

maybe I didn't move the mouse enough

[viscid thistle]

And i said before that i was writing the latex for you to be focused, not writing it on your paper

[junior sable]

what do you mean, be focused?

And i'm taking notes, what's wrong with that?

[viscid thistle]

Like being participative

And i'm taking notes, what's wrong with that?
Again. I was writing latex for you to be focused thats it

So the "notes" could be taken at the end

[junior sable]

it's not easy doing this over discord. I had no idea you were angry. if this were irl it'd be totally different.

[viscid thistle]

So that the attention goes to the conversation and not half-brain writing half-listening which is less efficient.

[junior sable]

even voice chat would have been easier

[viscid thistle]

Nah i wouldn't be able to do so

If i could i would have done it

Just don't worry, lets keep this channel open

See ya and good luck in the future @junior sable

[junior sable]

ty @viscid thistle

[viscid thistle]

[reef jasper]

how do u change this polar equation to rectangular coordinates

what do i change the x^2 to?

nvm i got it

[alpine basin]

Im trying to find the formula for this problem and I got 32,500 * 0.05^x

but when I try to plugin in 12 to get the value of it, it comes out wrong, help would be appreciated

[viscid thistle]

The general formula for this is $$\text{depreciation}=G(1-\frac{r}{100})^n$$ where G is the initial value, r the rate and n the years

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Try plugging the values of G, r and n as you have them again with the formula above and you should have it. @alpine basin

[alpine basin]

thanks

@viscid thistle Do you know how I'd match these functions? the differences seems so subtle and my teacher's video doesnt explain it well

[viscid thistle]

So

Lol.

@alpine basin Still need help?

Hold ln

On

Im writing it

[alpine basin]

ok

[viscid thistle]

@viscid thistle let me with him

Pog.

Okay.

I was just gonna say, that you could definitely start by substituting x with 0.

Then everything else is easy claps.

[alpine basin]

lmao

[viscid thistle]

You should know that $$f(x)=a(b)^x$$ is an exponential function. Consider the fact that a will decide the y intercept of the function as when a function meets the y axis, x=0. So basically $$f(0)=a(b)^0$$ and you should know that $b^0=1$ so we have $$f(0)=a$$ so from there we get that the a value on the function will act always as the y-intercept. (Notice that some functions have a=2 and a=4, look at the y-intercept to differentiate some). Now for the b meaning, the b will decide if the function decreases or increases. If: $$f(x)=a(b)^x, b>1$$ we know that the function will increase bc of the $b>1$ and viceversa we have $$f(x)=a(b)^x, b<1$$ we know that the function is decreasing. The last tip to differ these functions will depend on the b value, where i will explain through an example: Take 2 functions, $$f(x)=3(1.4)^x\ g(x)=3(3.2)^x$$ now notice that their a values are equivalent so, we can differentiate them by looking at the b value. On f(x): b=1.4 and on g(x):b=3.2 So what we will expect to watch on their representation is that f(x) will be way less curvy than g(x). As when the b value comes closer to smaller values, the curve will be smaller than when b is greater.

@alpine basin

[obsidian monolith]

Al𝟛dium:

[alpine basin]

so

the larger the b value is the more curve there is?

And a is y intercept

[viscid thistle]

Read the whole thing there is one more tip you are missing

,w graph 3*1.01^x

Look

[obsidian monolith]

Display results online and refine query

Requested by Al𝟛dium#4084

[viscid thistle]

The larger the b is, the faster the slope "goes up" as you go along the x axis.

[alpine basin]

yeah the way the line is flipped or not right?

[viscid thistle]

,w graph y=3*4^x

[obsidian monolith]

Display results online and refine query

Requested by Al𝟛dium#4084

[viscid thistle]

yeah the way the line is flipped or not right?
@alpine basin decreasing or increasing as well matters

Notice how both functions have different b values and the way the curve differs

[alpine basin]

ye

[viscid thistle]

Right

So any doubt?

The larger the b is, the faster the slope "goes up" as you go along the x axis.
@viscid thistle yeah that may be a better way of expressing it

@alpine basin all good?

[alpine basin]

Yeah

Thanks for the help

[viscid thistle]

Let's compare two cases.
$$y=(0.5)^x$$
and
$$y=(2)^x$$
$$ $$
In the first case, as x goes bigger, you eventually get: $\$
$\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ for when x=2 $\$
$\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$ for when x=3. $\$
$$ $$
In the 2nd case, as x goes bigger you eventually get:$\$
$2 \cdot 2 = 4$ for when $x=2 \$
$2 \cdot 2 \cdot 2$ for when $x=3 \$
$$ $$
With this, you can conclude that in the function of: $\$
$$y=(b)^x \$$
When $0<b<1$, the y value decreases for when the x value increases. $\$
When $b>1$, the y value increases when the x value increases $\$

[obsidian monolith]

leviosa:

[viscid thistle]

Damn this is shit notation.

Whatever.

It got solved already lol

Yeah whatever.

:)

Np! @alpine basin

[alpine basin]

this man said easy clap

lol

[viscid thistle]

Big easy claps.

[pseudo hull]

what is the name of the time speed and distance chat?

[viscid thistle]

#prealg-and-algebra

[alpine basin]

How would I show the horizontal asymptotes for these?

[viscid thistle]

Just draw it

Use your big brain knowledge on translations.

And then use common sense.

[alpine basin]

yeah for 26 for example I graphed the line

[viscid thistle]

You don't have to graph it tho...

But okay.

[alpine basin]

yteah its -3 right

[viscid thistle]

Yeah.

[alpine basin]

and 25 is 1

[viscid thistle]

$y=a(b)^x+h \ \text{h is your horizontal asymptote}$

[obsidian monolith]

leviosa:

[viscid thistle]

"sketch the graph"

[alpine basin]

yeah

what if theres no h

like in 27

[viscid thistle]

Then it's +0.

[alpine basin]

hmm mk

[viscid thistle]

a(b)^x+0

[viscid thistle]

Would anyone be able to help explain a function decomposition problem for me

Post it

oops thats wrong

Given the composition h(x)=f(g(x))

If h(x)=(x+5)^8 and f(x)=x^8 find the function g(x).

Well so

A little bit of logic can be used here

We have $h(x)=(x+5)⁸$ and $f(x)=x⁸$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

And we know that h(x) is f(g(x))

yeah

So

$f(g(x))=(g(x))⁸$

Hold on a sec

thank you

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Now

And we also know that h(x) is basically f(g(x))

From logic reasoning, we have from the question that $f(x)=x⁸$ so let's plug g(x) on the function f(x) so that $$f(x)=x⁸$$ to $$f(g(x))=(g(x))⁸$$ And we also know that $h(x)=f(g(x))$ THEN as we are given that $$f(g(x))=(x+5)⁸$$ we can form this $$\begin{cases}f(g(x))=(x+5)⁸\ f(g(x))=(g(x))⁸ \end{cases}$$ to solve the system.

@viscid thistle

I'm trying to understand

Okok

the answer ends up being 5+x btw

Yep

Read it carefully

The very first line has little context hold on

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

i think i understand

Awesome

thanks so much

So we are basically done

yeah

so when you put the (g(x))^8

Yes

why isn't it (g(x))x^8

We put g(x) not (g(x))⁸

Wait where

Which line?

nvm i understand

thanks a lot

Good

Np!

[lofty prism]

x = 11
f(x) ~ 3x
g(x) ~ 9x
point f(g(x)) is a point of line (D)
determine the equation of line (D

please can.you help me

[viscid thistle]

...

Post a pic of the q

[lofty prism]

what q?

theres no line q

also there is no sketch

if thats what you asked for

i corrected value of x in the exercise

[viscid thistle]

what q?
@lofty prism question

Post a pic of it

Im pretty sure you are missing stuff

[lofty prism]

what

[remote veldt]

post a picture of the question

[lofty prism]

im sorry i cant but i can write iy

for you

[viscid thistle]

Do it

[lofty prism]

do what

oh take apicture

im sorry i said i cant

[remote veldt]

why can't you take a picture

but if you really can't, write out the exact wording of the question, and don't leave anything out

[viscid thistle]

^^

do what
@lofty prism write it for me literally without missing ANYTHING

[lofty prism]

a = 5
f(x) = 3x
g(x) = 8x
(D) is perpendicular to (Cg)
f(g(a)) is a point of (D)
determine equation of (D)

i really hope i didnt miss something

i know how to do this f(g(a)) = f(a) * g(a) = f(5) * g(5) = 5/3 * 5/8 = 25/24
but idk what to do next lol

i think i need to use (D) perpendicular to (Cg)

but idk what to do with it

[viscid thistle]

What is Cg???

Uhhh

[lofty prism]

line of g

i think

[viscid thistle]

If you can't post a picture of it, i won't help. It looks really messy as you stated and very lacky of information.

[remote veldt]

I would also note that f(g(a)) is not the same thing as f(a) * g(a)

[lofty prism]

i am really sorry i will try to rewrite it and see if i missed something like really im sorry for wasting your time

I would also note that f(g(a)) is not the same thing as f(a) * g(a)
@remote veldt oh okay

Let a = 5
f(x) = 3x
g(x) = 8x + 4 and (C_g) its graphic representation.
(D) is a line of equation y = px + q with p and q real numbers.

Statements:
(D) is perpendicular to (C_g)
M(a;f(g(a))) is a point of line (D)

Task:
Find the values of both p and q.

is this okay now.i verified everything i think its complete?

so

is f(g(a)) equal to
f(g)*g(a) or something f(a)/g(a)

[remote veldt]

no, f(g(a)) is just f(g(a)). there's no multiplication or division

if g(a) = 3, then f(g(a)) = f(3). If g(a) = 12, then f(g(a)) = 12

[lofty prism]

how do you calxulate it

[remote veldt]

f(g(a)) means "take the number g(a) and use it as an input for the function f"

[lofty prism]

but g(a) is a function

i mean an output

[remote veldt]

g is a function, g(a) is evaluating that function at a specific point a

[lofty prism]

its an output

but what you enter in a function is an input

[remote veldt]

... either way it's a real number

f(g(a)) takes the output of g(a) and uses it as an input for f

[lofty prism]

i am not sure you are correct we will wait for someone else to tell

[past meadow]

He is correct. I tell

[remote veldt]

f(x) = 3x
g(x) = 8x + 4

f(g(2)) = f(8(2) + 4)
= f(20)

I am absolutely correct lol you have serious misconceptions about what functions are

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

[lofty prism]

but g(x) is output you cant say its an input for f

you must understand

[remote veldt]

again, you have serious misconceptions about what functions are

[lofty prism]

oh

[finite wraith]

a function g(x) takes x as it's input and whatever the function does to it, it spits one output g(x)

[remote veldt]

that output can then be used as an input to f(x)

[lofty prism]

you give functions a number they give you another. and this another number cant be inputed in another function thats what i know it belongs to the "output class" so it cant be input

[finite wraith]

g(x) is an output

think of the function sin

[remote veldt]

there is no such thing as an "output class"

[lofty prism]

just to illustrate

[remote veldt]

I will ask you a question

[lofty prism]

a function g(x) takes x as it's input and whatever the function does to it, it spits one output g(x)
@finite wraith huh

[remote veldt]

what is g(10)

[finite wraith]

sin(x) if i give it 0 as an input, its output will be sin(0) which is just 0. aka a number.

[remote veldt]

the numerical value

what????

[finite wraith]

what the fuck lol

$g(x)=8x+4$

[obsidian monolith]

trece:

[finite wraith]

just input 10 instead of x to the function g.

[lofty prism]

i know

[remote veldt]

no

g(10) = 8 * 10 + 4
= 84

[lofty prism]

it's 84

yes

[remote veldt]

great. What is f(84)

[lofty prism]

g(10)=84