precalculus

[hexed ermine]

In terms of sine

[ember crane]

I don't know?

[hexed ermine]

Do you know the pythagorean identity

[ember crane]

yes

[drifting bear]

So how can you rearrange the Pythagorean identity so you can express cosine in terms of sine?

[ember crane]

so cos = x and sin = y?

[hexed ermine]

No

[ember crane]

I don't get the question though

[drifting bear]

technically that's true

[hexed ermine]

well I guess you can look at it like that

[drifting bear]

But you have to understand why it's true in order to assume that

[hexed ermine]

tell me the identity

[ember crane]

sin = op/hyp and cos = adj/ hyp ?

[hexed ermine]

the identity

the pythagorean identity

[ember crane]

sin^2(x) + cos^2 (x) = 1

that?

[hexed ermine]

yep

so, if I wanted to find cosine in terms of sine, how would I go about it

[ember crane]

Oh.....move sine over

[hexed ermine]

yep

and do what

[ember crane]

square root the whole thing

[hexed ermine]

yep

[drifting bear]

Gj 👍

[hexed ermine]

so what would it look like

[ember crane]

leh me do it, wait

this?

[hexed ermine]

yep

[languid crane]

+-?

[hexed ermine]

well +-

[drifting bear]

no, its -

second quad

[languid crane]

oh didnt notice that

[hexed ermine]

oh did they mean to do that lol

[drifting bear]

Gj noticing that Nicaria 👍

i wouldve missed that personally

[ember crane]

thnks guys~

how do I find the end behaviors of this rational function?

[hexed ermine]

the degrees are the same

so what does that mean?

[ember crane]

hori asympt = 2

[hexed ermine]

yeah

[ember crane]

but the end behaviors is the troublesome part

[hexed ermine]

wyd end behaviors

that is the end behavior

at x tends to -inf or inf, the rational function tends to 2

[ember crane]

like how to find which directions are the end arrow of the lines go?

[hexed ermine]

they both go to 2 lol

horizontal asymptote means that as x tends to -inf or inf it converges to some value

[ember crane]

so how do u know if it goes to inf or -inf ?

[hexed ermine]

well im just saying, end behavior is the functions tendency as it goes in either direction

as it gets bigger it converges more to 2

and as it gets bigger in the other direction is also converges to 2

so that's what the end behavior means here

[torn oriole]

If you have the highest powers on the numerator and denominator being the same, then you can compare those @ember crane

[ember crane]

oh ok got it, thanks @hexed ermine @torn oriole

how do I do this problem?

[paper pecan]

Make a triangle having an angle x such that sin(x) = 3/5

Then tan (x) will give you what you need

[ember crane]

yeah, how can I fine x in Sin (x) = 3/5 without using a calculator?

is it possible?

[hexed ermine]

well you dont wanna calculate it

do what sid said

make a triangle with your known

[paper pecan]

You make a triangle, for which sin(x) = 3/5 (the opposite would be 3 and the hypotenuse would be 5)

[willow bear]

you do not need x itself

all you need is a triangle

perhaps with two of its sides labeled with lengths 3 and 5

in a way that makes one of the angles have a sine of 3/5

[ember crane]

OHh!!! okay

[still yew]

Hello!

I was wondering something.

Say I had the function, $f(x) = 2x^4 + 15x^3 - 58x^2 + 171x + 117$

[obsidian monolith]

Jon123276:

[still yew]

Is there an easier way to find all of the roots of it, as in real and imagionary?

[hexed ermine]

nope

you gotta use rational roots test

[still yew]

Without using synthetic Division.

Oh

[hexed ermine]

no nice way to do it unfortunately

[tender bolt]

@hexed ermine when we use rational roots test, does equation need to have form of $ax^n+bx^(n-1)+cx^(n-2) ...$

[obsidian monolith]

ZeusBotko:

[tender bolt]

tf is this bot

does it need to have like x^4 and then x^3 and then x^2

or it can be like x^5 + x^2 +4

[hexed ermine]

It can have any real coefficient

It can even be 0

So x^5+x^2+4 is the same as x^5+0x^4+0x^3+x^2+0x+4

[tender bolt]

so you can use this test on $x^5 + 2x^3 + 4$

[obsidian monolith]

ZeusBotko:

[past meadow]

also curly brackets for your tex

[hexed ermine]

Yes

[tender bolt]

thanks

i understand now

[daring yarrow]

2 curves touch one another in a certain point if this point ........ and the tangent line in this point to the 2 curves is ............

i think the last one is opposite

idk about the first tho

im tempted to say "if this point exists on both curves"

actually no shouldnt the last one be "equal"

[hexed ermine]

Yep

[daring yarrow]

slope of the tangent line btw its a rough translation

hell yeah :D

i have no clue for the first one tho, could you give me a hint?

[hexed ermine]

Wait wait wait for the second one

Can I have the original problem please

[daring yarrow]

its in dutch

Twee krommen raken elkaar in een bepaald punt als dit punt … ligt en de raaklijnen in dit punt aan de twee krommen ….

i have to fill in the dots

[hexed ermine]

So the tangent lines are necessarily equal at the point

Because the slope of one can be different from the other

What are the options for the first one

[daring yarrow]

holdon

here are the symbols

[hexed ermine]

Translate

[daring yarrow]

k1 and k2 touch in point P(x0;y0) if and only if

and the rest is in math notation

en = and

the y-values at the given x-value are the same...?

[hexed ermine]

If the two curves touch each other at that point the x and y values are the same yes

[daring yarrow]

that seems like a pretty obvious thing tho, is that really what i should write down?

[hexed ermine]

What does raken in mean

[daring yarrow]

touch

[hexed ermine]

So its asking if both points touch at the same value means the two functional values are the same AND the derivatives at the point are the same

Well the first is true

Derivatives arent necessarily true

[daring yarrow]

is it not?

[hexed ermine]

For what

[daring yarrow]

for the derivatives

[hexed ermine]

Nope

[daring yarrow]

its just touch btw not intersect

cuz we're not supposed to correct or anything, we're just supposed to put it into words

[hexed ermine]

Yeah the coordinates are the same

So the x and y values are the same

[daring yarrow]

there's a couple of examples and all the derivatives are the same

so i guess the derivatives should be equal (in that point)

[hexed ermine]

Oh I see, if they only touch

Not intersect

Like look at x^2 and x

They intersect at (1,1)

But the derivatives at the point are different

[daring yarrow]

yeah cuz they intersect

[hexed ermine]

So that implies that they cut across

[daring yarrow]

yea

[hexed ermine]

Hmm didnt know that

[daring yarrow]

so the derivatives should be the same

still dont fully get what they're trying to say with the first one

the function values are equal...?

at the given x-value

[hexed ermine]

Yep

[daring yarrow]

that seems like a pretty obvious thing but hey

[hexed ermine]

I guess that makes sense, if it touches it once that means they act like a tangent at that point

[daring yarrow]

yep

example in the book is f(x) = x²-6x+8 and g(x) = -(x-3)²-1

the derivative on the point of intersection (at x = 3) is equal (0)

hmm for the first one i dont really know how to word it

[hexed ermine]

The y values at x_0 are the same for both functions

[daring yarrow]

"2 curves touch eachother in a given point if this point is ....."

[hexed ermine]

Oh

[daring yarrow]

thats what i thought too but the "is" makes that a bad sentence

(even when translated into dutch)

[hexed ermine]

Idk what answer you're expecting

[daring yarrow]

the language barrier is rough bruh

it's talking about laying which is just like a point lays on a graph in dutch idk kinda weird to translate

so they're saying "2 curves touch in a given point if this point .... lays"

so i guess it should be "2 curves touch in a give point if this point lays on both graphs"?

[hexed ermine]

Hmm lol

[daring yarrow]

fuck it i'll go with that

[hexed ermine]

Yeah

[daring yarrow]

thanks alot man

[hexed ermine]

It's really awkward but I understand

Np

[leaden stratus]

I know that cos B is 1/3 and I need to find cos A. I thought that I can do cos2A = cos (180 - B). Then divide for 2?

[uncut mulch]

@leaden stratus wdym by divide for 2?

[leaden stratus]

@uncut mulch well, I think I had a stupid idea

[uncut mulch]

starting from cos(2A) = cos(180 - B) is fine
and then apply some double angle, shift identities

[leaden stratus]

@uncut mulch apply some double angle?

[ember crane]

first I substitute 2sin(x)cos(x) with sin2(x) and then I'm stuck with the next step

[uncut mulch]

double angle identity for cosine

to express cos(2A) in terms of cos(A)

[leaden stratus]

But cos(2a) is cos^2a - sin^2a

[uncut mulch]

there are 3 forms of that identity

and they should all be listed in your formula sheet

specifically:

$\cos(2\alpha) = 2\cos^2(\alpha) - 1$

[obsidian monolith]

ramonov:

[uncut mulch]

@ember crane factor out cos(theta)

[ember crane]

ohhhhh! got it.......

thnk u

[leaden stratus]

@uncut mulch I solved it, thanks. But in another way

alpha = (180° - beta)/2

So you have 90° - beta/2

Cos alpha is the same of sin beta/2

[loud vale]

can you distribute inside an absolute value?

like if (x+1)|x+2|

[craggy dune]

that depends

if x is positive you won't run into problems

but -1 |x + 2| is something different than |-x - 2|

[molten forge]

How do you do this?

Its an online worksheet but our teacher explained nothing.

[craggy dune]

well think about it for a second

what happens with f(x) if you let it run towards positive infinity

[molten forge]

it approaches positive infinity?

[craggy dune]

yes

and now g(x)

let that one run towards negative infinity where does it go?

[molten forge]

Negative infinity?

[craggy dune]

no

what happens if you enter 1 into g(x)

[molten forge]

its negative

-6

[craggy dune]

okay now put in -5

[molten forge]

60

oooh

it approaches positive infinity

?

[craggy dune]

yes

[molten forge]

@craggy dune

[craggy dune]

do you understand why?

[molten forge]

Kinda, yeah

as x->inf, f(x)->inf and g(x)->-inf

[craggy dune]

where does f(x) go if you let it run towards negative infinity?

[molten forge]

-inf

cuz its cubic

[craggy dune]

very good

[molten forge]

so how do I express that as an answer?

[craggy dune]

well you just did

you identify which factor grows the fastest

[molten forge]

How do I do that?

[craggy dune]

and then explain if you let that one grow towards infinity or negative infinity it behaves in a certain way

well how did you know x^3 is relevant in f(x)

?

[molten forge]

cuz thats the degree

[craggy dune]

yes the largest one

so you can find x for which x^3 is larger than x^2 if you go towards infinity

or x^3 smaller than x^2 if you let it run towards infinity

[molten forge]

ok wait

sorry thats a bit much

can you break it down?

like im still trying to understand the fundementals here

@craggy dune

[craggy dune]

if you let something run towards infinity

that means you keep putting in larger values in f(x)

[molten forge]

ok

[craggy dune]

so lim to 100 f(x) would be f(100)

assuming f is continuous

so if you let something go towards infinity you try to identify which factor grows the fastest

later it will get trickier but right now you just have to understand, that starting at a certain x

x^3 will be larger than x^2

so the higher the exponent the faster the function grows

[molten forge]

ok

[craggy dune]

h(x) = x^5 - 100x^3

if we let that go towards infinity

[molten forge]

wait, what is h(x)

or are you just making it as an example

[craggy dune]

I want to see if you understood what Im telling you

[molten forge]

ok so h(x) would approach negative infinity as x approached infinity @craggy dune

correct?

[craggy dune]

why do you think that?

[molten forge]

oh wait

would it kinda be like this shape?

@craggy dune

[craggy dune]

something like that

but where does it go?

[molten forge]

as x -> inf, h(x) -> and as x -> -inf, h(x) -> -inf

[craggy dune]

good

do you understand why?

[molten forge]

cuz the left side is going down and the right side is going upo

[craggy dune]

yes but you can see it just by looking at the term

you're learning limits so you can graph functions without a calculator

[molten forge]

ok

looking at the term

the degree is 5

meaning it will be in that kind of shape

[craggy dune]

yes

l(x) = -x^2 + 1000000

what happens if we let that go towards negative infinity?

[molten forge]

it would approach -inf

even if the y intercept is huge, itll be a parabola facing down cuz of the -x^2

[craggy dune]

good

[molten forge]

So the answer to the problem would be +inf, +inf?

[craggy dune]

which one?

oh ya

[molten forge]

How do you do this?
https://cdn.discordapp.com/attachments/363224154469826562/710554489815302194/unknown.png
@molten forge

[craggy dune]

the starting one

[molten forge]

the original

[craggy dune]

yes thats +inf +inf

[molten forge]

ok cool

@craggy dune so this would be right?

[craggy dune]

Try to write a little more

the point of those questions is to show that you understood what is happening

[molten forge]

ok

[flat hemlock]

Daniel knows 40% of the material for his final. When he has taken tests before, if he knows the material for the question, he has a 95% chance of getting the correct answer. If he does not know the material for the question, he has a 25% of getting the question correct. What is the probability of Daniel not knowing the material for a question and getting the answer correct? Leave your answer in decimal form to 4 places.

need a quick hand

[rain mulch]

Can anyone help with #3?

[lilac pier]

@rain mulch it's correct

[rain mulch]

Yes

But how to get it

[lilac pier]

ah

[rain mulch]

That’s the study guide

Answer key

[lilac pier]

Well you use one of the properties of mean

E(cX) = cE(X)

[rain mulch]

Wtf are those lol

[lilac pier]

okay nvm that

have you studied some kind of formula for mean

@flat hemlock you can try and make a tree diagram, it's not neccessary but will help you

[rain mulch]

No

[lilac pier]

@rain mulch Okay donald, well, what you can do is write down 10 values: x1, x2, .... x10

[flat hemlock]

i was wrong stll on that

[lilac pier]

One thing that is given to you is that x1 + x2 + x3 + .... + x10 = 60

To find the mean, what you do is divide this sum by the total no. of values(which is 10)

So divide both sides by 10, you get (x1 + x2 + ... + x10) / 10 = 6

Your mean is 6, you already know that I hope. Let's find the new mean.

Every value in the data was multiplied by 3

That means x1 is now 3x1, x2 is now 3x2

So you have (3x1 + 3x2 + .... + 3x10) / 10

You can factor 3 from numerator, you get 3(x1 + x2 + ... + x10 ) / 10

And you already know what (x1 + x2 + .... + x10) / 10 is, that's 6

So now you have 6 * 3 = 18 your new mean

@rain mulch

[flat hemlock]

anyone know this

[tardy ridge]

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

Find values for each using Pythagoras

[flat hemlock]

yeaa

i dont have time

5 minutes left

but ty

[viscid thistle]

i dont have time
@flat hemlock EXCUSE ME?

<@&268886789983436800> thats the most clearest evidence of an exam i have seen in my life

[fringe stream]

banned

thanks

[viscid thistle]

Np :)

[lilac pier]

lmao

and i was trying to help him on a question

😡

[viscid thistle]

Lol

anyone able to lend a hand here? I attempted to isolate u, but that answer was incorrect

[rain mulch]

@lilac pier

I was late, but I got it thanks

[tardy ridge]

I'm not sure how to express u as an exact answer without tan^-1 in there.

[kind dust]

well

I need a full explanation on this

[willow bear]

this is the triangle they want you to sketch

θ = arctan(x)

[viscid thistle]

hey guys!

how would I evaluate arcsin(sqrt(3)*cos(pi/6))

would it be arcsin(sqrt(3)*(sqrt(3)/2))

[willow bear]

indeed, and as stated it looks like this reduces to arcsin(3/2), which is undefined.

are you sure you copied the problem down correctly?

[viscid thistle]

Yea im sure! my professor tends to like questions like this because they truly test understanding

thank you for the confirmation that i was on the right track though! i really appreciate it

[sullen canyon]

if i want to simplify 4sin(5x)cos(5x)

is that 2(2 sin 10x) or is that simply 2 sin 10x

[past meadow]

uh

oh sorry

the latter

wait

no

[sullen canyon]

its double angle identity

[willow bear]

$4\sin(5x)\cos(5x) = 2\sin(10x)$

[obsidian monolith]

Ann:

[sullen canyon]

i don't understand why its 2 though

cuz if there is a 4

[willow bear]

$4 \sin(5x) \cos(5x) = 2 \times (2 \sin(5x)\cos(5x))$

[obsidian monolith]

Ann:

[sullen canyon]

isn't it sin2x = 2cosxsinx

[willow bear]

[sullen canyon]

wait what

[willow bear]

parentheses

you've missed three pairs of 'em

sin(2x) = 2sin(x)cos(x) yes that is correct

[sullen canyon]

ok i see it

so if it was sin 10x, then it would be (2sin(5x)cos(5x))

but since its 2sin10x, then its multiplied by 2

making it (4sin(5x)cos(5x))

[median quail]

this might sound a bit too simple, but I don't understand the definition of limits 😅 . Does a limit mean approaching or equal to? Like lim x -> infinity 1/x , does the limit every equal zero?

[sullen canyon]

wait so, based on that

how would i do 2cos^2(3x)-1

[willow bear]

apply the double angle identity for cos

and get cos(6x)

[sullen canyon]

ohhhh

cos2(3x)

so if 2cos^2(x)-1 = cos2x, then 2cos^2(3x)-1=cos6x

cheers

[median quail]

As in, is it alright to say that a limit tends to infinity for a function means that it is equal to zero. Hopefully I will try to make more sense. For f(x) = 1/x, If f(x) → 0 as x → ∞ , then why does limx→∞
f(x) = 0. (Because from how I understand f(x) only tends to zero and never reaches zero.)

[echo knot]

U are not making sense

[median quail]

Sorry I don't really know how to explain my question, but what I mean is the limit of a function as x approaches c means c is infinitesimally small. What does this tell about the function f(x)? Like for example, does it affect the graph in any way, for example?

[echo knot]

When we say f(x) → l as x →c ...by this we mean...as x is made closer and closer to the number c f(x) goes closer and closer to the number l

@median quail

[median quail]

ok thank you for your help

Sorry if it was confusing

[vague minnow]

hey I'm having trouble figuring out how to solve this equation

[echo knot]

Shift x in the right then multiply by 6 and thereafter apply cos

[vague minnow]

hmm ok I got a simpler equation, but I think the cos(15x) is causing problems 🤔

[ember crane]

this is the question

an this is how I do the question, am I wrong?

[glacial coyote]

don't cancel out cosine because there's a chance that you lose a factor

in this case you do

first factor out cosine to get (cos(theta))(sqrt(2) - 2*sin(theta))

then you can solve for the zeroes of both of those factors

[tardy ridge]

also cos(insert theta) could be 0 and by canceling you are doing illegal

[ember crane]

Ohhhh ok..... got it

[latent turtle]

i beleive this gives one of the equations. shud be able to find the others

i count a total of 3 answers for that question

[elder plume]

can someone help

[viscid thistle]

Formula of a circle is: $\$
$(x-h)^2+(y-k)^2=r^2\$
Where...$\$
$(h,k)$ is the center. $\$
And when $r$ is the radius.

[obsidian monolith]

leviosa:

[viscid thistle]

@elder plume

[elder plume]

woah

[odd helm]

How do I convert my answer into one of the answer options

[past meadow]

multiply the numerator and the denominator denominator by the 1-sqrt(3)'s conjugate, 1+sqrt(3)

[odd helm]

Alright thank you I got it now

[remote bison]

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO I LOST MASTERY POINTS ON KHANACADEMY

IT HAPPENED AGAIN ON THE FIRST QUESTION NOO

WHYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY DOES THIS HAPPEN

KHANACADEMY SHOULD HAVE A 'FORFEIT' BUTTON ON QUIZZES

[serene heath]

ok

[remote bison]

shut up you don't understand the pain at all

[paper pecan]

wht

[willow bear]

bruh

[remote bison]

have you been on khanacademy? ever?

[willow bear]

if you still mess up often enough for this to happen you can't rly claim you've mastered the concept

[remote bison]

you get 1 question wrong even by mistake and sal is like: "YOU KNOW NOTHING"

[willow bear]

what no

bruh just retake the quiz

also like

be more attentive ig

you're kinda overreacting here

[remote bison]

you can't "leave" in between

you have to suffer pointlessly through the rest and still have 1 wrong

[paper pecan]

It's merely 4 questions bruh

[remote bison]

ye and I've got lazy during quarantine

do you people have no sense of memery out here

you're kinda overreacting here

wait...... ohhhhh

I ranted on the wrong server lol

[viscid thistle]

also failing questions are good for you to learn not that it is horrible

[remote bison]

ik lol

[viscid thistle]

----------(free channel)---------

[paper pecan]

----------(occupied channel)---------

bye
----------(free channel)---------

[remote bison]

--------(paid channel. Monthly subscription: $π)---------

[paper pecan]

Bruh, that's irrational

[remote bison]

Like every tv and radio channel rn

--------(paid channel. Monthly subscription: $π(cos90 + i sin90)---------

ye now you can't even pay

[willow bear]

ugh

y'all

quit it

[remote bison]

quits it

resumes

[paper pecan]

​​​​

[remote bison]

--------(paid channel. Daily refreshed subscription: $π(ncos45 + i nsin45), where n = number of days subscribed) ---------

quits it for real now

[willow bear]

ugh.

[remote bison]

okay, "aNn"

but idc

I shall continue

with an inverter

so that mean I won't continue lol

but if I don't continue, the inverter will do the job

hence I will have to

quit it

k imma leave

[upbeat moss]

Do anyone know what this means?

[paper pecan]

It means you're asked to find the difference quotient given by $\frac{f(x+h) - f(h)}h$

[obsidian monolith]

Sid:

[paper pecan]

@upbeat moss

[remote bison]

thanks captain obvious

[paper pecan]

There's nothing more to tell, unless they have more questions

[upbeat moss]

I never was asked a question like this so, I wasn't sure how should I answer

[craggy dune]

you take the difference quotient of your function and then look for the lim

[pale bison]

No

[paper pecan]

They haven't asked for a limit here

It's just a difference quotient

That's it

[craggy dune]

oh you don't even need that

then just put in your function

[paper pecan]

You're just asked to find the value of $f(x+h)$ subtract $f(h)$ from it and then divide what you get by $h$.

[obsidian monolith]

Sid:

[upbeat moss]

And I can use any values I want since none are given correct?

[paper pecan]

I am not sure I understand

If $f(x) = x^2$ (say), then $f(h) = h^2$

[obsidian monolith]

Sid:

[upbeat moss]

O

Nvm

[viscid thistle]

----------(occupied channel)---------
@paper pecan bruh

[paper pecan]

Bruh that

that's dead

You're late

[viscid thistle]

lol

-------(free channel)-----------

[willow bear]

y'all this stopped being funny 40 mins ago

[viscid thistle]

@willow bear i dont do it bc of fun, i do it to be clear its a free channel

[paper pecan]

She didn't refer to the ----(free channel)---- thingy

[viscid thistle]

Oh lol mb

[viscid thistle]

any idea on what f(x) and g(x) could be

[craggy dune]

ya

f(x) = e^(1/x)

that should get you g(x) if you think about it

[patent beacon]

Think e definition
(1 + 1/x)^x approaches e
@viscid thistle

[tardy ridge]

hm

[thorn mountain]

Yes, (at least) two different ways to accomplish that

[desert socket]

i am utterly confused at whats happening here

when i calculated the inverse tangent of <-2, 2sqrt(3)> i get -60 degrees or 300

but this site and my math book says its supposed to be 120

is my formula wrong?

isnt theta = inversetan of y/x?

in regards to vectors

ping me pls

okay I think i have it figured out

so depending on the points you get a general direction of the vector (quad 1-4) and then you do quadrant angle - the inverse tan. in this case 180 - atan(2sqrt(3)/-2).

[hexed ermine]

Right obviously here the angle is negative because it's in the 2nd quadrant, the reference angle is given by 60 so you know that it is in the 2nd quadrant with a reference angle of 60

Which is 180-60 degrees

[vernal wasp]

could someone please help with this one cause i have the biggest brain fog rn and ive forgotten about injective functions completely

[craggy dune]

(e,5)

[willow bear]

f is not a function from X to Y at all

[vernal wasp]

wait really

why not

i thought it was one to one

[willow bear]

e ∉ X

f is not a function from X to Y, period

[vernal wasp]

omg

i did not read properly

thank you so much

i could not for the life of me understand why i was getting it wrong

[sweet niche]

Why can any complex number or function be split up in two parts: $x(z) = x_R (z) + i x_I (z)$

[obsidian monolith]

meerio:

[serene heath]

thats how they're defined?

[acoustic laurel]

Complex numbers are defined like that, but I’m not aware if there’s any underlying reason to that. There probably is. I’ll have a look.

I can’t really get beyond this, but someone more qualified might be able to give some context. Maybe read up on the history of complex numbers, but I’m assuming that you need the underlying logic for it: which is, I think just because of complex numbers being defined that way.

[harsh smelt]

you can think of complex number as a vector on the plane

with real component and imaginary

[smoky fractal]

Can someone tell my why im not given 44.1° even when i plugged in its sine on the inverse?

[viscid thistle]

radians or degrees?

[smoky fractal]

Sorry its sideqays

I think my calc is set on radians

[viscid thistle]

oh I didn't even look at the pic

I just really thought it would be a not on right setting issue

[smoky fractal]

Lol its ok

[past meadow]

Wait whats the issue? The .79 is 44.1 deg in radians

[smoky fractal]

OH IM STUPID

thank t3ou so much

You*

[past meadow]

Lmao np

[viscid thistle]

so it was almost a not on right setting issue

[smoky fractal]

lol ig

[steel brook]

<@&286206848099549185> I am having the hardest time with a rational functions question.

Conditions:
-Exactly two x-intercepts: (5, 0) and (-2, 0).
-Exactly one vertical asymptote: x = 3

1. Write a formula for a rational function in factored form that meets the conditions above and has no horizontal asymptote. Call this function f.

2. Write a formula for a rational function in factored from that meets the two conditions above and has a horizontal asymptote of y = 1. Call this function g.

For problem #1, I did the following, which seems to work out. But I can't figure out how to do #2 at all.

f(x)=

(x-5)(x+2)
/
(x-3)

There doesn't seem to be anything I can fiddle with to get it to hit both x-intercepts while only having a single vertical asymptote AND a horizontal asymptote. I can get it with multiple vertical asymptotes but that doesn't fit the constraint.

[willow bear]

swap out the (x-3) for a (x-3)^2

[steel brook]

That doesn't seem to work out when I throw it into a graph.

Crosses the horizontal asymptote

[willow bear]

yeah so what

nothing wrong with a function crossing its own horizontal asymptote

[steel brook]

That just... doesn't feel right.

Isn't that what a asymptote is all about? Can't cross it?

[uncut mulch]

that's a misconception

[willow bear]

x/(x^2+1): am i a joke to you

[steel brook]

Yeah, googled around and that seems to be right. Egg on my face for being so untrusting.

[maiden pebble]

https://i.imgur.com/auei3BA.png

These are some exmaples that my teacher gave to me

In the first one, I was wondering where the 5 comes from (the 5 from 5cos)

[willow bear]

the derivative of 5x

is 5

[maiden pebble]

how?

I thought it was 1

[willow bear]

you thought the derivative of 5x was 1?

[maiden pebble]

Yes

I did

[willow bear]

what did you think the derivative of 14x was

or of 7x

or of 1000x

[maiden pebble]

Yea I was doing my math wrong

[willow bear]

[maiden pebble]

I took the derivative of 5 and x

[willow bear]

bruh

$(f \cdot g)' \neq f' \cdot g'$

[obsidian monolith]

Ann:

[willow bear]

also the derivative of 5 isn't 5

[maiden pebble]

yea its 0

[willow bear]

0*1 isn't 1

[maiden pebble]

yes thats why i corrected myself

then deleted that message for some reason

...

https://i.imgur.com/VkksVms.png

5cos - 3cos

?

$f5cos - 3cos

[past meadow]

$5 \cos(t)-3 \cos(t)$?

[obsidian monolith]

Sneaky:

[past meadow]

for the derivative?

[maiden pebble]

Yes

is that the derivative of my question

[viscid thistle]

Check again

[maiden pebble]

Wait

without the t's

[past meadow]

hm?

[maiden pebble]

oh i wrote that wrong

[viscid thistle]

Still check again

[maiden pebble]

my bda

-3sin

[past meadow]

write the variable with it

-3sint

[maiden pebble]

but why would the t still be there

[viscid thistle]

Hmmm?

[past meadow]

sin is a function

not a constant

[maiden pebble]

ok....

[past meadow]

are you talking about how the derivative of 5x is 5?

[maiden pebble]

you agree that x^2 would become 2x

[past meadow]

yes, the derivative of x^2 is 2x

[maiden pebble]

So if we have t, then wouldnt that become 1t^0

So it would be 5 cos times 1

[past meadow]

=1 yes. but what does that have to do with sin(t)?

oh

[willow bear]

COS IS NOT A NUMBER

cos(t) is not the product of t with a number called "cos"

[maiden pebble]

https://tenor.com/view/calm-down-gif-5362498

[thorn mountain]

"5 cos" is a meaningless, nonsense collection of symbols

cosine is a function

[willow bear]

cos is a function, not a number.

cos is a function, not a number.

[thorn mountain]

it REQUIRES an input variable

[maiden pebble]

input variable?

[craggy dune]

something you put into it

[thorn mountain]

cos(t) cos(x) cos(something)

[maiden pebble]

ok yes

So why do derivatives act different with functions?

[willow bear]

[maiden pebble]

5t^2 = 10t

[willow bear]

no

you are not misusing the equals sign like this

5t^2 and 10t are not the same thing

[maiden pebble]

yea, but I assumed yall were smart enoguh to know what im talking about

[willow bear]

this is no excuse

THIS IS NO EXCUSE FOR IMPROPER NOTATION.

[viscid thistle]

[willow bear]

the equals sign does NOT mean "next step".

[maiden pebble]

ppft, proper notation lol

[willow bear]

if you want to write that the derivative of 5t^2 is 10t

[willow bear]

then you write that the derivative of 5t^2 is 10t.

d/dt (5t^2) = 10t, perhaps.

[maiden pebble]

ok fine Ann, 5t^2 --> 10t

[thorn mountain]

still no

[maiden pebble]

better Ma'am?

[willow bear]

ok, that's better.

not ideal, but better.

at least now you're no longer claiming 5t^2 and 10t are the same thing.

[maiden pebble]

better = good enough

[willow bear]

[thorn mountain]

it seems math isn't your strong suit

lets make an analogy

[willow bear]

why do ppl find it hard to understand that notation matters

[maiden pebble]

well if t = 2 then id be correct

[thorn mountain]

if I want you to know that the sky is blue, does this quote let you know that? "apoidghb-iapoingo[ia"

[maiden pebble]

lol math isnt my strong suit

really roasting me

Since math is my only suit

[willow bear]

[maiden pebble]

ok so annyway

Why do I have to treat functions differently from constants?

https://i.imgur.com/5ecD3Th.png

[willow bear]

'cause they're entirely differnt objects maybe?????

[maiden pebble]

Cuz in that example sinx^(n-1) occurs

[willow bear]

ok so like

malix

you're not helping rn

[maiden pebble]

Sorry malix im kinda ignoring you

Atleast your being nicer than Ann tho

[willow bear]

$\sin^4(x)$ is the composition of two functions

[obsidian monolith]

Ann:

[maiden pebble]

I appreciate it

[thorn mountain]

@maiden pebble you're ignoring everyone

[willow bear]

it's $[\sin(x)]^4$

[obsidian monolith]

Ann:

[maiden pebble]

@thorn mountain I take back what I said

[willow bear]

i.e. it's the composition of (u ↦ u^4) with (x ↦ sin(x)), in that order

and what's being used there is the chain rule

but it sounds like you're not nearly familiar with that

given that mere minutes ago you thought functions were the same thing as numbers and not a completely different object

[maiden pebble]

So would it be incorrect notation to write sinx^4 instead of sin^4 x

[willow bear]

sinx^4 is just bad.

sin(x)^4 is ok

as is sin(x^4), but this now refers to a different function obviously

but sinx^4 is bad

in general you should always write parentheses around arguments of trig functions

sin(x), not sin x.

cos(t), not cos t.

[maiden pebble]

Also I do know the chain rule

[willow bear]

tan(θ), not tan θ.

[maiden pebble]

ok, so sin^4(x) is different from sin(x^4)

which would mean that its never the x being squred, cubed, etc, but the sin, cos, etc

[thorn mountain]

sin^4(x)=(sin(x))^4, which is different than sin(x^4) yes

[maiden pebble]

Well now my homework makes sense

Thanks for the help

https://i.imgur.com/VFObxDp.png

how would you write this with parenthesis?

(the first one)

[past meadow]

sin(5x) is fine

[maiden pebble]

why does the 5 go inside the parenthesis with the x, but when its an exponent it doesnt

[willow bear]

because stupid notational conventions that's why

[maiden pebble]

this is what originally confused me because what I did with 5x followed the rules I knew but when there was an exponent they didnt work

[thorn mountain]

it's an unfortunate habit that math software and textbooks have with notation for these types of functions

it should have been written y=sin(5x) from the beginning

[maiden pebble]

what is the derivative of tan?

[thorn mountain]

that is not a thing

[past meadow]

remember just "tan" doesnt mean anything

[maiden pebble]

no wonder my teach told us to do #4 different;y

[thorn mountain]

the derivative of tan(x) with respect to x is sec^2(x)

[maiden pebble]

https://i.imgur.com/OxZ6n4S.png

[thorn mountain]

ok. so do that

[maiden pebble]

easier said than done

[thorn mountain]

tan(x) = some fraction you should know

what is that fraction

[maiden pebble]

sin/cos

[thorn mountain]

again and for the last time

that is not a thing

[maiden pebble]

yea yea

sin(x)/cos(x)

[thorn mountain]

thank you

now use the quotient rule

[maiden pebble]

your welcome

[thorn mountain]

it's not my welcome

i can go picky on things in math or in english

[maiden pebble]

hope you dopnt correct peoples grammar online

[thorn mountain]

had it not been an issue with the functions, it wouldn't have come up here

you good with the quotient rule?

[maiden pebble]

do I leave out the 3

[thorn mountain]

also...why are we in the precalculus channel talking about derivatives so much?

[maiden pebble]

Or do I include that with the sinx

Because im in PreCalc

and this is my homeowkr

[thorn mountain]

the 3 is a constant multiple on the tangent function

[maiden pebble]

and last time I used #calculus people told me to move to here

[thorn mountain]

I'm not a mod since I just joined, but that seems weird unless the calculus channel was occupied at the time

but then, it seems weird that you are doing so many derivatives in precalculus too so /shrug

[maiden pebble]

They didnt want me using it because then people in that chat would assume that I had already learned everything from preCalc

such as the tangent function

[thorn mountain]

to be fair, you should have learned what the tangent function is before trying to find its derivative...

but the tangent referred to in calculus ("the slope of the tangent line") and the tangent function from trig are not exactly the same things

[maiden pebble]

ok i got 3(-sin^2(x) - cos^2 (x))/cos^2 (x)

[thorn mountain]

where'd the minus signs come from?

[maiden pebble]

the subtraction or negative?

[thorn mountain]

yes, the - symbols

what caused them to appear?

[maiden pebble]

well I did sinx times the derivative of cosx

[thorn mountain]

what is the quotient rule?

[maiden pebble]

which i believe is -sin^2x

low d high - high d low / low^2

[thorn mountain]

nope

[maiden pebble]

bruh

[thorn mountain]

wait, maybe

i hate that notation

what part are you supposed to take the derivative of in "low d high"

[maiden pebble]

low, deriv of high - high, deriv of low

[thorn mountain]

so why did you do "high d low" as your first part?

[maiden pebble]

ahhhh

3(-cos^2(x) - sin^2 (x))/cos^2 (x)

[thorn mountain]

again, why are there minus signs?

[maiden pebble]

-2 x 2 = -4

which is equal to -(2^2)

[thorn mountain]

i'm questioning your derivative taking, not your multiplication (yet)

[maiden pebble]

lol

[thorn mountain]

d/dx (sin(x))*cos(x) = cos(x) cos(x) = cos^2(x)

no minus sign

[maiden pebble]

isnt the derivative of cos, -sin

or is it the other way around

[thorn mountain]

there is no derivative of cos

[maiden pebble]

WHAT

[past meadow]

cos(x)

it needs

an input

[maiden pebble]

umm, x?

[thorn mountain]

the derivative of cos(x) is -sin(x), that would be correct

but that is "d low"

[maiden pebble]

low d high, cos(x) times cos(x)

[thorn mountain]

uh huh

[maiden pebble]

3(cos^2(x) + sin^2 (x))/cos^2 (x)

[thorn mountain]

good. now does the stuff inside the parenthesis in the numerator look familiar?

[maiden pebble]

yea

= 1

[thorn mountain]

yep

[maiden pebble]

so have I learned the tangent function now?

[thorn mountain]

not really

you've learned how to find the derivative of the tangent function

assuming you could repeat this process without someone guiding you step by step

[maiden pebble]

yall ruthless

[thorn mountain]

Telling someone lies that make them comfortable is a quick way to lead them to destruction

[maiden pebble]

the man who has no imagination has no wings

[thorn mountain]

sure

[maiden pebble]

https://i.imgur.com/ODpQQ5c.png

[thorn mountain]

also fair

i wasn't criticizing you when i said those things

[maiden pebble]

sure sounded liek it

[thorn mountain]

¯_(ツ)_/¯

sorry it felt that way

[maiden pebble]

it just so happens that I have another question

[thorn mountain]

ask away

[maiden pebble]

https://i.imgur.com/mzNCkm5.png

Would this become e^20x^3

i assume that wouldnt be the final step though

[thorn mountain]

use the chain rule

there will be a 20x^3 somewhere

there will be an e somewhere

but the derivative will not be e^(20x^3)

[maiden pebble]

the chain rule?

but there is only one variable

(i believe that is a varaible)

[thorn mountain]

that's not what determines whether you should use the chain rule or not

the chain rule is used to take the derivative of a function which is a composition of functions

$ \frac {dy}{dx} = \frac {dy}{du} \frac{du}{dx} $

[obsidian monolith]

Malix:

[thorn mountain]

is probably the prettiest way to write the chain rule, but it isn't the only way

[maiden pebble]

how does that help me wiht my problem?

[thorn mountain]

another way to write the chain rule would be: If f(x) = O(I(x)), then f'(x)=O'(I(x))(I'(x))

find the functions which are composed to make the function you want to take a derivative of

[maiden pebble]

I understand what the chain rule is, i just dont know how to use it with this problem

[thorn mountain]

find the functions which are composed to make the function you want to take a derivative of

as some more hint...that's an exponential function. what's the most basic exponential function?

@maiden pebble did you give up?

[maiden pebble]

no

[serene slate]

I need to turn this into

this

hwo do i do that?

[maiden pebble]

im just very lost at what ur saying Malix

[thorn mountain]

channel occupied, try one of the questions channels

[maiden pebble]

me or thomas?

[thorn mountain]

thomas

you were here first

what is a basic exponential function?

not descriptively, just give a formula for one

[maiden pebble]

x^2

[thorn mountain]

that is not an exponential function

exponential functions have the variable in the exponent

[maiden pebble]

so x^x?

[thorn mountain]

that's overly complicated

[maiden pebble]

1^x

[thorn mountain]

and that's too simple since no matter what the exponent on 1 is you always get 1

[maiden pebble]

i mean, you asked for basic

[thorn mountain]

but 3^x (or e^x) would be what I'm talking about

since 3 and e are both positive constants that are not 1

[maiden pebble]

ok

[thorn mountain]

so e^x is an exponential function, can you see another function you could compose with that to get your f(x)?

[maiden pebble]

Another function I can compose with that?

x^e

[thorn mountain]

well e^x isn't the f(x) in your problem

do you know what function composition is?

[maiden pebble]

Not by name

[thorn mountain]

f(g(x))

[maiden pebble]

yes

[thorn mountain]

so, can you see a function that could be plugged into e^x to make the function in your problem?

[maiden pebble]

x = 0

plug in 0

[thorn mountain]

e^0 = 1

[maiden pebble]

yes

[thorn mountain]

that is not f(x) in the problem you posted

[maiden pebble]

???

f(x) = e^5x^3

[thorn mountain]

https://i.imgur.com/mzNCkm5.png

[maiden pebble]

Yea

thats the quetion

[thorn mountain]

i know.

what function could you replace the x in e^x with to get that function?

[maiden pebble]

So you want me to fine something equal to e^x and plug it into f(x)?

[thorn mountain]

no

what function could you replace the x in e^x with to get that function?

[maiden pebble]

OOOOHHHH

[thorn mountain]

that is what i want you to do

[maiden pebble]

5x^4

[thorn mountain]

ok. that is the "inside function" for the chain rule

the function we plugged it into (e^x) is the "outside function"

[maiden pebble]

e^20x(5)

[thorn mountain]

NO

USE THE CHAIN RULE

[maiden pebble]

I DID

[thorn mountain]

you did no such thing

[maiden pebble]

well i forgot the ^3

4(5x)^3 times 5

[thorn mountain]

that is a thing you forgot, but that wouldn't fix the problem of not using the chain rule

also, what you just put is still not correct

[maiden pebble]

what do you think the chain rule is?

[thorn mountain]

another way to write the chain rule would be: If f(x) = O(I(x)), then f'(x)=O'(I(x))(I'(x))

or

$ \frac {dy}{dx} = \frac {dy}{du} \frac{du}{dx} $

[obsidian monolith]

Malix:

[maiden pebble]

wait whats the inside outside thing

like when you have (2x+1)^2

Whats that claled

[thorn mountain]

in that example, the inside function would be 2x+1 and the outside function would be x^2

that is function composition

which is what you use the chain rule to find the derivative of

[maiden pebble]

yea, so arnt we trying to use that to solve this problem?

[thorn mountain]

i am trying to get you to do so, yes

you have instead decided to use: If f(x)=O(I(x)), then f'(x)=O(I'(x))

which is false

[maiden pebble]

I dont see how what I did was wrong

I did the derivative of the outside, time the inside

Times the derivative of the inside

[thorn mountain]

you claimed the derivative would be e^(20x^3)

[maiden pebble]

https://i.imgur.com/iNeK4F8.png

[thorn mountain]

(5x)^3 = 125x^3 all on its own

[maiden pebble]

so 3000x^3

[thorn mountain]

NO

you don't need the chain rule to find the derivative of 5x^4

that derivative IS 20x^3 as you said from the beginning

[maiden pebble]

Yes

agreed

[thorn mountain]

what is the derivative of e^x?

[maiden pebble]

xe

Wait

let me think of that one more

[thorn mountain]

It will not follow the rule for polynomial derivatives

[maiden pebble]

xe^x-1

[thorn mountain]

no

that is the rule for the derivative of x^n

which doesn't work for things that aren't x^n

[maiden pebble]

I dont know how to do e^x then

as ^n is the only notatino i know

[thorn mountain]

if your class has gotten to this point, either the teacher has covered the derivative of e^x, or they are not actually teaching the course

are you familiar with the definition of derivative?

and limits?

[maiden pebble]

d/dx

[thorn mountain]

that's a symbol

[maiden pebble]

or dy/dx i believe

[thorn mountain]

not the definition

[maiden pebble]

a tiny change in y over a timy change in x

does this picture help https://i.imgur.com/oVMxMMl.png

[thorn mountain]

this is one form of the definition

the picture you posted is the "shortcut" for the chain rule with exponential functions

you could use it to answer your original question

but you won't have any idea why it works

[maiden pebble]

i mean

its US Highschool

were taught that stuff works, not why it works

[thorn mountain]

i teach in the US

i know that

but I am here to help you understand

not simply say "here's a formula, it works cause I say so"

that last quote is doing a disservice to students

[maiden pebble]

thank you for your service

on and offline

[thorn mountain]

haha

[maiden pebble]

according to my "shortcut" would the next step be e^5x^3 time 20x

[thorn mountain]

no

the exponent part will not change at all

[maiden pebble]

did I change it?

[thorn mountain]

yes

[maiden pebble]

ohm y ba

my bad

e^5x^4 time 20x

[thorn mountain]

what's the derivative of 5x^4?

[maiden pebble]

20x

[thorn mountain]

no

[maiden pebble]

yea chill

20x^3

[thorn mountain]

there ya go

so (20x^3)e^(5x^4)

[maiden pebble]

wait what

oh yes, i see

Would that be my final answer?

its like 10x messier than the original

[thorn mountain]

that would be the final answer

no one said taking derivatives made things "cleaner"

[maiden pebble]

well i wish they did

[thorn mountain]

fair enough

[maiden pebble]

thanks for the help

turns out that when you know how to do something it is...easier

[pseudo lance]

hi this isnt really calculus but it is the closest to calculus so: is air resistance maximal at terminal velocity or at minimal (downwards) acceleration

[rare pagoda]

Both?

Terminal velocity occurs when air resistance = mg so there will be no downwards acceleration because a = 0

@pseudo lance

[pseudo lance]

Yeah but in my graph the terminal velocity is different from minimal downwards acceleration

Cos its about a skydiver who suddenly opens his parachute

[rare pagoda]

You reach terminal velocity when a = 0

[desert socket]

I have a question regarding the magnitude of a vector. Are there specific situation when |v| = sqrt(a^2 + b^2 +c^2) and others when it’s |v| = ai + bj + ck

I had a question in my textbook that asked to find the volume of a parallelepiped

[rare pagoda]

That's fun