precalculus

[modern prairie]

yeah i know but the question is more formal than understanding
more of a convention i guess
the question is if the domain includes infinty or not

[viscid thistle]

Me neither, i only asked it in case the word infinitesimal changed the result

[willow bear]

no, infinity is normally not included in the domain of any real-valued function

and i still don't know what you mean by "infinitesimal function"

[modern prairie]

i see
thx

[willow bear]

what is "reaching infinity" supposed to mean?

[modern prairie]

that's what my teacher taught me to say so i kind of translated it from my language to english

[viscid thistle]

Oh lol

[willow bear]

what's your language

[modern prairie]

hebrew

[willow bear]

and do you have an actual definition anywhere

so that i could like

[modern prairie]

i actually have no idea how to say it in english

[willow bear]

see what it's meant to be

[upper kelp]

Do you mean diverges, Gilad?

[modern prairie]

according to google translate: "Strives for infinity" but im not sure about that

[willow bear]

sigh

[modern prairie]

@upper kelp yes

[willow bear]

this is gonna be hard

ungh

yeah no

i'm out

[modern prairie]

aight

@willow bear thx anyway you actually helped me

[upper kelp]

As far as I know, the "formal" definition of divergence to infinity means that whichever lower bound M that you choose, there exists a point where the function (or the series of numbers) grows beyond that "forever."

For example, let's say that you want to say that the function f(x) = x diverges to infinity as x → ∞.
In this case, let's pick an arbitrary M.
In this case, you get that starting x = M, the function always satisfies f(x) ≥ M for all x ≥ M.

However, if there is a value that the function "always comes back to," then the function doesn't diverge to infinity.

Are you familiar with the functions sine and cosine?

Regardless, you don't need to know the intimate details of these functions.
Let's take, for example, the function f(x) = x·sin (x).
This function goes up and down, and its values get as high as you can possibly want.
However, that function doesn't diverge to infinity.
Why? Because no matter how far you go, you eventually go below the blue line.

[modern prairie]

thx

[carmine elbow]

Can someone help me with this problem?

[serene heath]

,rotate

[serene heath]

what have u tried

[carmine elbow]

What I did on the right

I have tried looking up how to do it and couldn’t find anything

[proud sparrow]

where's your tan phi

@carmine elbow

[carmine elbow]

Tan?

I’m asked to rewrite the problem in sine and cosine form

[viscid thistle]

@carmine elbow which problem?

25?

[carmine elbow]

Yes

[waxen egret]

having a brain fart this morning. is 5.5 greater than 2e?

[strange estuary]

Yes

5.5 > 2e

2e = 5.43

[waxen egret]

how did you get that

just starting a pre calculus course just confused about that

[strange estuary]

Do you mean 2*(Napier's constant)

[waxen egret]

yes

[strange estuary]

Just multiply it

[waxen egret]

oh

simple enough thanks

[pale pond]

can someone help me with this

[upper kelp]

It seems that you're right about those being the general solutions.

Could it be that they asked for the solutions in a certain range, though?
Say, in [0, 2π)?

[willow bear]

@pale pond are you still here?

[pale pond]

yea, i had my question answered by someone else in the questions sub servers

[willow bear]

channels. they're called channels.

[pale pond]

thanks, term slipped out of my head

[carmine elbow]

I'm trying to figure out how to finish this

[tardy ridge]

a=arccos (that decimal number)

[carmine elbow]

I did that, but the answer was wrong

This is what the answer is supposed to be

Number 6

[uncut mulch]

where did the 4th line come from?

[carmine elbow]

What 4th line?

[uncut mulch]

49/9? = 9/9 cos(A)

[carmine elbow]

Oooh

I am trying to get cos(a) by itself

[uncut mulch]

that would be a very huge jump even for some quite good at algebra

[carmine elbow]

Hmmm 🤔

[uncut mulch]

however that implies that cos(A) > 1 and is incorrect

show all your steps

[carmine elbow]

Ok

[uncut mulch]

order of operations

[carmine elbow]

soo, I should I multiply 9 times Cos(A)?

[uncut mulch]

no

you shouldn't have that 9 there at all

because:

$a-bc \neq (a-b)c$

[obsidian monolith]

ramonov:

[carmine elbow]

ok

[uncut mulch]

that looks better

[carmine elbow]

should my calculator be in radians or degrees?

[uncut mulch]

the question seems to want degrees

[uncut mulch]

why are you entering 260/220 instead of 220/260

[carmine elbow]

I always forget which way to put the numbers when dividing with the calculator 🤦🏿‍♀️

[uncut mulch]

the same way as it's written on the page

also note the answer for A given by the book is inaccurate

[viscid thistle]

How would I do 23?

[tardy ridge]

do you know how to find averages

the trick is ignore t, substitute t as log_a (x) directly

(r+s)/2=log_a (x)

this is basically the answer.

@viscid thistle

[brave patrol]

hey guys i need some help. So i am given a function f(x)= 3/(2x-1) and i have to find the points where the slope of the tangent is -24

how do i do that

I know that i have to find the 2 x values from the function and then i can just get the y values for each respective x value by plugging them into the functions equation but i dont know how to get the possible x values

oh wait mb i forgot we had channels where u can ask for help sorry bout that

[tardy ridge]

wouldn't you find the derivative

and set that =+-24 to find x

[tardy ridge]

I'm getting 1/4 and 3/4

but 1/4 doesn't seem right.

[tiny locust]

hey

could anyone help me with vectors

im trying to figure out if two vectors are either parallel ortho or neither

vector u = <40,16> and vector u = <1/2, -5/2>

im doing the dot product rn

Its orthogonal right?

40(1/2) = 20

-5/4*

16*-5 = -80 / 4 = -20

20 + -20 = 0

its ortho?

[upper kelp]

They are indeed. 👍
Intuitively, you can note that their "slopes" have a product of -1.

[tiny locust]

Thank you !

[harsh cipher]

Hello

y= 3 root(-(x-5)) - 7

y= root (-9(x-5)) - 7

are they same? why?

how can we take out root(-9) ?

[true stone]

can black people swim

[harsh cipher]

what?

[true stone]

im asking a question

related to precalc

[harsh cipher]

how is that related to precalc?

[true stone]

that's personal information

[harsh cipher]

then what are you trying to say?

[true stone]

im asking if black people can swim

[harsh cipher]

Im asking how is that related to pre calc

[lost lintel]

answer is no

[upper kelp]

Mirrion, to your question: I believe so.
More generally speaking, for all two non-negative numbers a, b ≥ 0, we have their root as the product of the roots of each one.
√a·√b = √(ab)
In this case, assuming -(x - 5) ≥ 0, meaning x ≤ 5, then the two are defined and are equal to each other.

We can't take minuses out of roots. That will cause a problem of the root being possibly undefined.

[supple zephyr]

Hello, I need help with a MAC1140, Pre-Cal question, I don't know where to start

[harsh cipher]

Okay

discord is reading messages to me

hi

hello

[harsh cipher]

I have a question

[lilac pier]

@supple zephyr Let's say we have a difference of d in the arithmetic sequence. Let's start from a_21..

[supple zephyr]

Ok

[harsh cipher]

for the equation y= root((-x^2+2))

how is the range y greater than or equal to zero

[lilac pier]

a_24 = a_21 + 3(d)

Then to go to a _34, we know it's a geometric sequence, so a_34 = 2(a_24)

Or a_34 = (a_21 + 3(d)) (2)

[undone pawn]

@harsh cipher shift to free channel

[supple zephyr]

may i ask where did you get a_24?

[lilac pier]

well we had to go to a_34 somehow

but first I need to from a_21 to a_24

using arithmetic sequence

then using geometric sequence, we go from a_24 to a_34

[supple zephyr]

oh ok

[lilac pier]

@harsh cipher For the function to exist, assuming we're talking about real numbers here, sqrt only exists for numbers greater than or equal to 0

And sqrt of numbers which are greater than or equal to 0 is also greater than or equal to 0

@supple zephyr Did you figure it out? We have $a_{34}= a_{21} + 3(d)$

[obsidian monolith]

Sup?:

[lilac pier]

You have a_21

[supple zephyr]

Um not really

[lilac pier]

You have a_34

Find d

[supple zephyr]

oh ok

[lilac pier]

@harsh cipher For your function, the range is 0 <= y <= sqrt(2)

[harsh cipher]

where is free channel

oh right because graph doesn't exist below x-axis

so y is greater than or equal to zero

[daring yarrow]

is sign(x) = arcsin(x)?

we've never used sign before as notation so im kinda confused

or is it this? wtf

[uncut mulch]

they are different

as the name suggests it is the "sign" function and is not related to the trig function: "sine"

also known as the signum function

which is usually used to avoid that confusion

[daring yarrow]

thanks

we've never even mentoined this function in class before so was abit confused

and teacher said there's alot of notations for arcfunctions so thought it was another one of those

,factor x²+2x-15

(x+5)(x-3) it is

[viscid thistle]

,ask x²+2x-15

[obsidian monolith]

Display results online and refine query

Requested by Al3dium#4084

[viscid thistle]

@daring yarrow yep it is

[unique hill]

...

[viscid thistle]

Why did i read is it

[carmine elbow]

I need help solving this vector

[willow bear]

<3i+2j,4j+j> is fucked up notation

[carmine elbow]

I have no idea what I’m doing. I was trying to solve it

[willow bear]

I have no idea what I’m doing.
bad!

[carmine elbow]

Can you help me?

[willow bear]

v + w is just (3i + 4j) + (-2j)...

v + w is just (3i + 4j) + (-2j), or 3i + 2j after simplification

[carmine elbow]

That’s what I thought too, but when I checked the answer key, I realized that wasn’t it

[willow bear]

or if you want to use the tuple notation, v = <3,4>, w = <0,-2> and v+w = <3+0, 4-2>

can you show the answer key

[willow bear]

and v+w = <3+0, 4-2>

3+0=3, 4-2=2

[carmine elbow]

Where does the zero come from?

[willow bear]

w = <0,-2>

if you want, you can write w as 0i + (-2)j instead of simply -2j

[carmine elbow]

That would be easier

Soo, like this?

[willow bear]

jeez no

[carmine elbow]

I forgot to put an I on the zero

[willow bear]

no

there should not have been an i there at all

or a j

if you want to use <..., ...> notation
only NUMBERS can go inside

v isn't <3i, 4j> it's <3, 4>

[willow bear]

there we go.

[carmine elbow]

Thanks!

[carmine elbow]

I was able to do some of the other problems, but I got stuck on this one.

[willow bear]

$\nrm{\vec{v}}$ and $\nrm{\vec{w}}$ are the \textbf{lengths} of $\vec{v}$ and $\vec{w}$ respectively

[obsidian monolith]

Ann:

[willow bear]

$a^{bc} = (a^b)^c$

[obsidian monolith]

Ann:

[willow bear]

$a^{b \cdot b} = (a^b)^b$ where $b = \log_3(7)$

[obsidian monolith]

Ann:

[daring yarrow]

my friend doesnt understand this and i havent covered it in class yet so im hoping someone from here can help

they're asking to calculate the difference quotient on the interval [5,7] and [1,3]

[lilac pier]

@daring yarrow The difference quotient on the interval [5,7] is just $\frac{f(7) - f(5)}{7 - 5}$

[obsidian monolith]

Sup?:

[daring yarrow]

thats what i thought aswell but then i looked it up and came across this

which threw me off alot

[lilac pier]

Same thing, you can think of it this way: f(5+2) - f(5) / (2)

[daring yarrow]

ok alright, thanks!

[teal hound]

hey guys, i have to calculate the average slope and the average angle over an interval (i dont know if that makes sense, i translated it from dutch lol)

and im struggling with it sinc ei have no idea how to approach it

[teal hound]

<@&286206848099549185>

[tardy ridge]

wouldn't you use the slope formula.

and for angle you would use tan.

[prisma rampart]

I think what you're talking about is related somehow to this formula: F'(x)= (F(a)-F(b))/a-b

In ]a, b[

[teal hound]

hmm, mightt be

ill give it a go, thanks allie

[vernal spindle]

What am I doing wrong

I used desmos and it seems the graph goes from -oo to oo

[serene heath]

what happens if x=0

[viscid thistle]

Yeah

Or basically can go through $2x-3≥0$
so $x≥\frac{3}{2}$ so by this, you know more precisely where the domain doesn't reach to

[obsidian monolith]

Al3dium:

[viscid thistle]

@vernal spindle

So it'd be Domain f(x) = [3/2, infinite)

[obsidian monolith]

Al3dium:

[vernal spindle]

didnt work

[spring thunder]

they want the domain of f^-1 not f @viscid thistle @vernal spindle

[vernal spindle]

For f I got -oo to oo but its also not accepting that

[viscid thistle]

Oh my bad

[vernal spindle]

what am i doing wrong

[molten forge]

How do I do this?

[willow bear]

what is giving you trouble?

[molten forge]

Well, during elearning, our teacher just kinda threw this at us without even telling us what to do

So I have no clue at all

[willow bear]

you know that $f(0) = 2$, $f(1) = 4.7219$ and $f(+\infty) = 6$; this gives you three equations in three unknowns

[obsidian monolith]

Ann:

[willow bear]

$f(+\infty)$ of course is to be interpreted as $\lim_{x \to +\infty} f(x)$.

[obsidian monolith]

Ann:

[willow bear]

the system should not be all that hard to solve

[molten forge]

Ok. Literally all we did before quarantine was nothing related to this and she just gave us this without context

So I really kinda need the theory on how to do this

I'm sorry about that, but I really don't understand what to do here.

@willow bear

[willow bear]

there is very little theory here if any

[molten forge]

Ok

[willow bear]

do you understand the connection between "f(0) = 2" and "(0,2) is a point on the graph of f"

[molten forge]

Yes

[willow bear]

okay great

[molten forge]

It's mostly this part. What do all those variables represent?

[willow bear]

you don't need to know what they represent to be able to solve the problem

[molten forge]

Ok

[willow bear]

in fact the knowledge would be kind of excessive

$f(x) = \frac{L}{1 + Ce^{-kx}} \ f(0) = , ?$

[obsidian monolith]

Ann:

[molten forge]

Oooh

That makes more sense

So $f(0) is 2

(idk how to use the bot)

[willow bear]

no, i want you to forget about f(0)=2 for a moment

[molten forge]

Hm ok

[willow bear]

plug $x=0$ into $\frac{L}{1 + Ce^{-kx}}$; what do you get?

[obsidian monolith]

Ann:

[molten forge]

That would be $\frac{L}{2}}

[willow bear]

no

[molten forge]

Oh

[willow bear]

$Ce^{-kx} \neq (Ce)^{-kx}$

[obsidian monolith]

Ann:

[molten forge]

Oooh

[willow bear]

$f(0) = \frac{L}{1+C}$

[obsidian monolith]

Ann:

[molten forge]

Ok alsso how do I use the bot

[willow bear]

it uses a markup language called LaTeX

there's a cheatsheet in #resources

[molten forge]

Ok, Ill learn that later

:)

[willow bear]

anyway ok so like

$\lim_{x \to +\infty} f(x) = L$

[obsidian monolith]

Ann:

[willow bear]

and $f(1) = \frac{L}{1 + Ce^{-k}}$

[obsidian monolith]

Ann:

[willow bear]

this really is just a matter of plugging in points

like. almost without thinking

[molten forge]

So f(1) = L/1+Ce^-k right? And that's also equal to 4.7219?

[willow bear]

parentheses

f(1) = L/(1 + Ce^-k)

and yes that's equal to 4.7219

L = 6

L/(1+C) = 2

[molten forge]

Wait where did u get L is 6 from

[willow bear]

$6 = \lim_{x \to +\infty} f(x) = L$

[obsidian monolith]

Ann:

[molten forge]

?

[willow bear]

look at the graph

there's a very clearly indicated horizontal asymptote at y=6

[molten forge]

Oooh that's where it's approaching

[willow bear]

which your graph approaches as you go further and further right

[molten forge]

So now what do I have to do?

[willow bear]

you have three equations, you have three unknowns

L = 6
L/(1+C) = 2
L/(1+Ce^-k) = 4.7219

[molten forge]

Ok

And from this

How do I get the equation in the original form?

Do I have to solve for k and C?

@willow bear

[willow bear]

you are supposed to solve for L, C and k

and then plug them back into your equation for f

[molten forge]

Ok. And we have L now

[willow bear]

and that'll be your answer

[molten forge]

Ok cool!

[willow bear]

yes we already have L

[molten forge]

So C is 3

And k is about 1.999983

Yes? @willow bear

[willow bear]

it sounds like the intended value of k was 2

since the y coordinate of the original point was given to 4 decimal places it stands to reason that k should probably also be rounded to 4 decimal places

[molten forge]

I plugged it into mathway and it said it was -ln(0.13533747)

[willow bear]

which will give k=2

[molten forge]

Ok so it's 2?

So the final would be

$f(0) = \frac{6}{1+2e^{-2}}$

[obsidian monolith]

ThatOneCalculator:

[molten forge]

Oh why was mine dark

Also I didn't mean f(0)

$f(x) = \frac{6}{1+2e^{-2}}$

[obsidian monolith]

ThatOneCalculator:

[molten forge]

Also I meant ^-2x

Not ^-2

Anyways thank you for your help @willow bear :))

[willow bear]

there's an option in the bot's settings for background color

[molten forge]

Ooh ok

Tysm!!

[drowsy karma]

,w 8x^3 y^2+12x^2 yz-16x^4 y^6 z^2

[obsidian monolith]

Display results online and refine query

Requested by Stewie#8321

[ember crane]

I don't know how to find this

[fallen crane]

is this correct?

im not sure if my work is right

[viscid thistle]

You have an error on the 3rd line

If you cross multiply, you shouldn't do common denominator as from doing cross multiplication, the denominators dissapear and it doesnt have sense

@fallen crane

[drowsy karma]

w, 2x^2:+:x:-12

,w 2x^2 + x - 12

[obsidian monolith]

Display results online and refine query

Requested by Stewie#8321

[fallen crane]

my b

[drowsy karma]

(x+y)^2 - y^6 z^8

Can that be factored?

,w (x+y)^2-y^6 z^8

[obsidian monolith]

Display results online and refine query

Requested by Stewie#8321

[tardy ridge]

yes

it can be factored @drowsy karma

[drowsy karma]

Could you show me how please?

[tardy ridge]

differentce of squares

[drowsy karma]

but the -Y^6 Z^8 part?

[tardy ridge]

what is $y^3*z^4$

[obsidian monolith]

Fishraider:

[tardy ridge]

squared.

[drowsy karma]

ohhh im dumb

I forgot when they multiply they add

[harsh cipher]

question

how do I know the horizontal compression or expansion here

[lost cove]

@harsh cipher https://www.desmos.com/calculator/into8z3lnz There you can play around with the graph

[tender sail]

got it nvm

[vernal spindle]

I can't find out my mistake in this

[lost cove]

The domain of f^-1(x) is the range of f(x)

[uncut mulch]

is oo how you're supposed to be entering inf?

[vernal spindle]

Yeah

[fleet yew]

Oo

Bro just use $\infty$

[viscid thistle]

How would I do this

[obsidian monolith]

AMD:

[uncut mulch]

what have you tried?

[tardy ridge]

Try multiplying by the conjugate.

[ember crane]

Question 27, I don't know how to find t

[uncut mulch]

you don't actually need to find t

[ember crane]

how do I do it?

oh! i got it

[supple zephyr]

Hello, i got this homework question for precal (mac1140), can someone help me figure out how to do this?

[narrow peak]

just sub in bro

[willow bear]

@supple zephyr you might consider writing down the general equation of a line passing through (2, -1) and look at which slopes make it only intersect once with the circle

[acoustic laurel]

Rip

[radiant spruce]

substitute into equation of circle: (x-1)² + (m(x-2)-3)² = 2
expand and form quadratic in x: (m²+1)x² - 2(2m+1)(m+1)x + 4(m+1)(m+2) = 0
set Δ=0: m² - 6m + 7 = 0 -> m = -1 or 7
--------------------------------
find length of tangent from touching point to P: use the m=-1 tangent
substitute into line: y+1 = -x+2 -> y-2 = -x-1
substitute into equation of circle: (x-1)² + (-x-1)² = 0 -> 2x² = 0 -> x = 0
find A: (0-1)² + (y-2)² = 2 -> y = 1 or 3(rejected, consider circle location, P and slope of tangent) -> A(0, 1)
|PA| = √( (2-0)² + (-1-1)² ) = 2√(2)```
||

how do i do the spoiler thing

[acoustic laurel]

Two ||

|| in beginning and end

Thanks for the answer! I couldn’t figure it out too lol

[uncut mulch]

(m²-1)x² - 2(2m+1)(m+1)x + 4(m+1)(m+2) = 0
should be (m^2 + 1)

for the distance, you could also find the distance from P to the center and apply pythagoras.

[radiant spruce]

oh sorry typo

for the distance, you could also find the distance from P to the center and apply pythagoras.

distance from P to center of circle: √( (2-1)² + (-1-2)² ) = √10
distance from P to A: √( 10 - r² ) = √(10-2) = 2√2

[stable pasture]

source for hard problems in limits and continuity?

[fleet yew]

@stable pasture khan

[stable pasture]

thats ez

[fleet yew]

@stable pasture

[stable pasture]

uh didnt you mean khan acad @fleet yew ?

[fleet yew]

khan academy yeah

[wind tiger]

can someone help me understand this question?

i am lost on what im suppose to do

[radiant spruce]

oh these questions are fun

take a) for example

wait let me draw

so if you see an inverse trig function you set up a triangle like this

for these types of questions you need to remember what sides correspond to what

like for example sine is opposite/hypotenuse

then you set up the top(x) on a leg and the bottom(√(x² + 3²)) on the hypotenuse

oh by the way we assume an angle to be θ, in this case the angle on the right

so sin θ would be x/√(x² + 3²), which matches your question

then you take sin⁻¹(x/√(x² + 3²)), then you get θ back
θ = sin⁻¹(x/√(x² + 3²))

because tangent is opposite/adjacent, you can look back at your triangle

the side opposite to θ is x

the side adjacent to θ is 3

therefore tanθ is x/3

[wind tiger]

so it's sin = x/sqrt(x^2 + 3^2)

that's for sin

[radiant spruce]

yeah

[wind tiger]

so it same thing without the inverse?

[radiant spruce]

no uhhh

we let this angle be θ

so sinθ is x/√(x² + 3²) because sine is opposite over hypotenuse

we take sin⁻¹ on both sides and we get
sin⁻¹(sinθ) = sin⁻¹(x/√(x² + 3²))

but sin⁻¹(sinθ) cancels out to just θ because we assumed the angle is < pi/2

so we have θ = sin⁻¹(x/√(x² + 3²)) which matches the whole thing in the question

then you take the tangent on both sides

so tanθ = tan(sin⁻¹(x/√(x² + 3²)))

but from the triangle you constructed tanθ = opposite/adjacent which is x/3

therefore tan(sin⁻¹(x/√(x² + 3²))) = x/3

[wind tiger]

i think i understand...

[wind tiger]

@radiant spruce

[radiant spruce]

i think you need to use pythagoras to figure out the other leg

also uh question says (-1-x)

this is what i got

[willow bear]

-1-x?

[radiant spruce]

https://cdn.discordapp.com/attachments/363224154469826562/702861862030671872/unknown.png

yes, -1-x

wait no

i forgot the square root

my bad

[willow bear]

you'll wanna be very careful indeed with negatives here

[radiant spruce]

true

[willow bear]

-1-x and x cannot both be positive

since you are making a triangle

you want your lengths to be positive

ok wait hang on

[radiant spruce]

there's no value for both to be true

[willow bear]

ALRIGHT SO

we're told to

assume that x is positive

[radiant spruce]

think unit circle

lengths can be negative i guess

[willow bear]

but then (-1-x)/x will be less than -1

so the answer to b is undefined

plain and simple

the assumption that x > 0 leaves us taking the inverse cos of a number not in [-1, 1]

[wind tiger]

so would it be 1 + x/-1

[radiant spruce]

oh domain

[tardy ridge]

It would be -1 -1/x. - which doesn't work.

[willow bear]

okay fine

y'all continue not listening to me

100% fine

keep taking arccos of numbers outside its domain

[radiant spruce]

i'm reading

[willow bear]

perfectly fine, no issue

/s

[viscid thistle]

How does this work?

[uncut mulch]

multiplying by "1" to get a common denominator

and then combine both terms into a single fraction

[radiant spruce]

i think you can just add it up normally

the bottom becomes $(1 - sin^2 x)$

[obsidian monolith]

Conan:

[radiant spruce]

which is just cos^2 x

and the top cancels out to 2

so you get 2 sec^2 x

[fleet yew]

@radiant spruce O Brien or the Barbarian?

[vague minnow]

hey, I need an idea on how to start solving this equation

actually... I think I have an idea

[rugged zenith]

Im completely stuck on solving this equation, does anyone have any ideas?

[willow bear]

i feel like it might not have any nice solutions lol

[rugged zenith]

Yeah, same. I used my calculator to solve it and none of them were clean answers so I was trying to find a way to do it algebraically. I’m doing this for an application project and I accidentally did way too much

[copper vigil]

can't you just treat that as a quadratic in x

and then get two linear equations

[rugged zenith]

Can you explain further? I’m not sure I follow

[copper vigil]

x^2 - pi*x + tan(x) = 0

a=1

b= -pi

c=tan(x)

[rugged zenith]

Ah okay, thank you! that makes sense!!

[vague minnow]

hey, kinda stuck on this problem here. I need to find all a where the distance between the equation's roots is less than 6/25

[fleet yew]

you got anymore pixels?

[tardy ridge]

is that ax^7 as a base?

[vague minnow]

ohhh hmm I didn't even notice that

[fleet yew]

looks like ax^2?

idk what it says

[vague minnow]

I just wrote ax

but it might be ax^2

[fleet yew]

there's definitely something there

but can't read it lol

[vague minnow]

yeah with ax wolframalpha says no solutions

with with ax^2 there are some

[fleet yew]

ok

i think it's a 2

[vague minnow]

yeah probably

[fleet yew]

ok

log_a(b)=log(b)/log(a)

[spare badge]

I'm so lost.

(I'm not in precalc, idk anything abt vectors) but I need help with this

oh wait theres a help channels sorry

[fleet yew]

why are you doing this if you don't know what vectors are

[spare badge]

need to finish pre-calc units

for credit

[vague minnow]

log_a(b)=log(b)/log(a)
@fleet yew thanks a lot, I think I got it now! 😃

[verbal stump]

@vague minnow
I solved it but idk if answer is correct

[vague minnow]

it's almost correct

if you also look at the fact that the right side of the initial equation must be >= 0

then -arccos(-1/sqrt(5)) won't be a solution

[verbal stump]

Oh yeah, forget to check that

Anyway idea of solution is right xD

[vague minnow]

yeah I have pretty much the exact same solution 😄 thanks

[unborn willow]

Hey guys I had a quick question so I got the answer to the problem but I am having a hard time putting the answers in order. If anyone can help me I would appreciate it.

@verbal stump would u mind helping me?

[verbal stump]

I don't understand why these are solutions

[unborn willow]

I just need help re ordering them on a number line from lowest to greatest

[verbal stump]

Oops

I understoos

I'm dumb

Ye u right

[unborn willow]

is it in the right order

or am i doing it wrong

[verbal stump]

Anyway √2 is not answer

Omg

Why am I so dumb

Yeah you're right

It's right order

[unborn willow]

thanks brother I appreciate your help

[muted steeple]

Hi, when deriving the quadratic formula, how did they got the +- sign?

I understand the reason for it's existences etc, i'm just curious about deriving it.

When deriving it, i got to the part

$x+\frac{b}{2a} = \frac{\sqrt[]{b^2-4ac}}{2a}$

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

Now, $x = \frac{\sqrt[]{b^2-4ac}-b}{2a}$ should be the next logical step.

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

If i move the b term to the left, shouldn't the radical always give me a positive root, hence $-b+\sqrt[]{b^2-4ac}$?

[obsidian monolith]

꧁༺Vocal༻꧂:

[tardy ridge]

when you square root both sides, you have to do |absolute value|

or +-

[muted steeple]

hmm, i didn't get that.

So you are referring to this point

$\left(x+\frac{b}{2a}\right)^2 = \frac{-4a+b^2}{4a^2}$?

[obsidian monolith]

꧁༺Vocal༻꧂:

[fleet yew]

Yes

if x^2=25, what is x

[muted steeple]

$5$

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

😄

[fleet yew]

Theres another solution

[muted steeple]

oh -5

[fleet yew]

Yeah

[muted steeple]

but the radical supposed to be concerned with a positive root?

[fleet yew]

If you only use the positive root you will only have 1 solution

Parabolas are second degree polynomials

So they require 2 solutions

Hence the $\pm$

[obsidian monolith]

AMD:

[muted steeple]

a quadratic is a parabola?

[fleet yew]

Yes

[muted steeple]

Hmm, very insightful. Thanks.

[fleet yew]

You should look into FTA

[muted steeple]

I would prefer to have it in the fractional exponent notation tbf

FTA?

[fleet yew]

Fundamental Theorem of Algebra

[muted steeple]

I've been hopping from book to book, and I'm not sure i've seen them talk about it.

Quite frustrating.

[fleet yew]

Its important

At this stage you wont be able to prove it

But you should know that it is true

It states that any $n$th degree polynomial must have $n$ complex roots (that are not necessarily distinct)

[obsidian monolith]

AMD:

[muted steeple]

oh

so 2 degree's 2 roots

[fleet yew]

Yes

[muted steeple]

why "complex roots"?

[fleet yew]

solve x^2=-1

[muted steeple]

$x = \sqrt[]{-1}$

[obsidian monolith]

꧁༺Vocal༻꧂:

[fleet yew]

$\pm i$ which is a complex number

[obsidian monolith]

AMD:

[muted steeple]

I know about complex numbers, but i don't know how they tie into this.

[fleet yew]

Because not every polynomial has real solutions

[muted steeple]

Ah, so it generalises the statement to non-real solutions?

[fleet yew]

It generalizes the statement to all polynomials

[muted steeple]

I guess i haven't encountered 2nd degree polynomials with non-real solutions.

I'll CTRL+F and see if there's a mention of fundamental theorem of algebra 😄

"Complex Zeros and the Fundamental Theorem of Algebra"

3 chapters away.

[fleet yew]

What textbook?

[muted steeple]

James Stewart Precalculus Mathematics for calculus.

7th edition.

[fleet yew]

Never used that

But yeah sure read up on it

The concept itself is pretty simple

[harsh cipher]

Hi I have a question.

my teacher is offering me to not take mid term..and just do the remaining unit test

should I do what he says or take the mid term?

my current grade is 84%

[regal nest]

@harsh cipher I have no idea what a 'unit test' is. I myself always prefer homework assignment(s) over midterm, because I can work at my own pace and use Google 🙂

[harsh cipher]

unit test is end of chapter test

[regal nest]

Now with remote teaching, we wrote a midterm test this monday and we were officially allowed to use lecture notes, which is great

[fleet yew]

Bro is it open note?

[harsh cipher]

no its not

[fleet yew]

Take the midterm for an easy 100

[harsh cipher]

he said no penalty

if I dont write it

[fleet yew]

Bro just look up the answers if you have to lol

No way he'll be able to know

[harsh cipher]

lol?

[fleet yew]

Corona virus has made school 10x easier

[regal nest]

@fleet yew I disagree, I am in 1st year masters program Cybernetics and Robotics, we have tons of homework assignments, lectures, mettups etc.

I love it that I don't have to go out of the apartment 😄 But the school is not easier

[harsh cipher]

lets assume AMD meant younger students.

lol

[fleet yew]

I mean grades wise at least

Because everything has shifted to open note

@regal nest why masters?

[regal nest]

because I have graduated my bachelor's degree last year and continued with masters 😄

[fleet yew]

Nice

[regal nest]

We didn't do much robotics in the bachelors, it was mostly math and control theory

[torn iron]

does anybody have good online resources to use for practicing verifying identities and solving for trig functions?

[fleet yew]

Kahn

[torn iron]

does kahn do stuff with verifying identities? I went through the whole precalc course and didn't find much of that

[fleet yew]

I think they have a separate thing for trig

Lemme check

Yeah they have an entire trig section separate from the precalc section

[torn iron]

they do but it doesn't really do much because the most advanced stuff in there is the stuff from the precalc section, the rest of it is just knowing what a unit circle is and how to use em with triangles

you dont really get any challenging solving or verifying problems from there in my experience at least

[fleet yew]

Go to he trig section

At the bottom they literally have "challenging problems"

[torn iron]

Sorry for not explaining myself better earlier but I've already finished everything in precalc, that entire module with the challenging problems stuff I've already looked at because it is also in the precalc section.

[fleet yew]

Ok

If you really find that stuff easy then its time to move on

Trig identities are not even that fun by themselves

They can be used for a lot of cool stuff though

[torn iron]

I just want to make sure its all solid in my head before I do anything else

Im going through some review at the moment

[wide rover]

How do I write cot(t)in terms of cos(t) in quadrant III?

[tardy ridge]

would it be cos(t)/(sqrt(1-cos^2(t))?

and add a negative sign

[full garden]

hey guys I have a question on getting the equation for this sequence

I got tn = n^2+6n-2

but I dont know what the next step is can someone please help me

[rigid sun]

uhhhh

[dark stump]

is anyone good at polar equations and graphing them

so what does the "zero of r" mean? there's a little after it saying, "r=0 when theta = pi/3" but like what does it actually mean?

is it like when the angle is 60 degrees, then the length is 0

[radiant spruce]

yeah i think that's what it means

it's like a polynomial, "the zeroes of the function"

but instead you have the function loop around

[dark stump]

function loop around?

sorry can you explain a little if you have time

[radiant spruce]

polar functions are like

they're functions with r as the dependent variable and θ as the independent variable

for normal functions with x and y, you're going left and right along the graph and getting different shapes

but for polar you're pushing the pole around the origin

[dark stump]

ohh hm thank you

and also one more thing, when making a chart for the points, how do I know which theta values to give?

[radiant spruce]

you usually pick nice numbers

[dark stump]

on my notes, the only examples given are when theta goes from 0 to pi and then you fill in some r values in between

hm

[radiant spruce]

like π/3 or π/2 or π/6

[dark stump]

how do I know the greatest theta value?

or is that a thing

idk

something about the symmetry

how you don't need every single value because you can just reflect it across its line of symmetry?

thank you tho

[radiant spruce]

greatest theta value?

in what context though

[dark stump]

like this it goes from 0 to pi

is there a reason it goes up to pi

[radiant spruce]

what function is this

sinθ?

[dark stump]

r = 1-2costheta

its symmetric about the polar axis

[radiant spruce]

yes that's why

you see it through the cos part

because cosine is symmetric about x=π

look at the graph

[dark stump]

oh

[radiant spruce]

cos(θ) = cos(2π - θ)

[dark stump]

thank you im not certain about this subject yet but i think i understand better

thanks

[muted steeple]

Hi, can something like a cubic formula be derived the same way as a quadratic one?

[willow bear]

a cubic formula does exist but it's fuckoff-complicated

why do you ask

[muted steeple]

Yesterday I was proud to be able to derive the quadratic formula. Thought the same principle could be applied to the cubic one, but it's not immediately clear to me that it can.

[willow bear]

yeah no like. the derivation for the cubic formula is significantly more complicated

sufficiently so that i probably would need to look it up

[muted steeple]

I see. Maybe some day in the future.

[radiant spruce]

bombelli had some fun with the formula

but if you wanna do cubic you're better off using complex numebrs

[viscid thistle]

I can't seem to see why this is true

oh whoops that sentence isn't finished

trying to show F=G^-1

all I seem to get is $(G\circ F)(x)=\frac{x}{1-|x|+x}$

[obsidian monolith]

exele:

[frozen needle]

maybe try again but slower

[viscid thistle]

$(G\circ F)(x) = \frac{x}{1-|x|} \bigg/\left(1+ \frac{x}{1-|x|}\right)$

[obsidian monolith]

exele:

[viscid thistle]

$\frac{x}{1-|x|} \bigg/\left(1+ \frac{x}{1-|x|}\right) = \frac{x}{1-|x|} \bigg/\left(\frac{1-|x| + x}{1-|x|}\right)$

[obsidian monolith]

exele:

[viscid thistle]

$\frac{x}{1-|x|} \bigg/\left(\frac{1-|x| + x}{1-|x|}\right) = \frac{x}{1-|x|+x}$

[obsidian monolith]

exele:

[viscid thistle]

@frozen needle?

[frozen needle]

$$G(F(x))=\frac{\frac x{1-|x|}}{1+\abs{\frac x{1-|x|}}}$$

[obsidian monolith]

Tuong:

[viscid thistle]

oh whoops

[haughty cave]

So I have the point (2,4) on the graph of quadratic function f(x) = x^2

The question says "Find the slope of m1 of the line joining (2,4) and (3,9). Is the slope of the tangent line at (2,4) greater than or less than the slope of the line through (2,4) and (3,9)? Could someone please assist me with this?

[tardy ridge]

You can find the slope of m1 using the slope formula and you can use the derivative thing to find the tangent slope.

but if you don't need to explain that then it should be intuitive

[haughty cave]

The derivative of x^2 would just be 2x right?

[viscid thistle]

Yep

[haughty cave]

Okay and then just one last question, another portion of it is asking me to find the slope mh of the line joining (2,4) and (2+h, f(2+h)). Am I correct in finding the slope to be f(2+h)-4/h or am I missing a step? I'm not quite sure on how to work the function into this

Would I just use that specific function once given a value for H? So then the slopes equation would just be 2+(h value)-4/(h value) once I have an actual value?

[merry sphinx]

So, the slope would be f(2+h)-4/h

I don't quite get your next question though?

f is a function, so you would need to know the function to calculate slope

In which case, I don't see how the slope equation becomes 2 + h - 4 / h

because f(2 + h) does not have to equal 2 + h

[haughty cave]

Ohhhh so the function in the problem is f(x) = x^2, so f(2+h) would become (2+h)^2?

and then I would plug the h values into that within the slope?

Yeah I think I understand now thank you all, sorry for so many questions

[merry sphinx]

yes, that is how it would work

[viscid thistle]

Which

4&5

How am i supposed to know what 4) is asking, may i ask

4 is asking “write each trigonometric expression as an algebraic equation”

[willow bear]

consider making a triangle where one of the angles is arctan(x).

[viscid thistle]

So would the triangle be like this

[tardy ridge]

yes.

[viscid thistle]

Ok thank you

Any idea for 5)?

[uncut mulch]

express cos(2theta) in terms of sin(theta) and solve the resultant quadratic

[viscid thistle]

Would the final answer be
theta=pi/2+2kpi?

[tardy ridge]

the k is not needed.

[viscid thistle]

May I ask why?

[tardy ridge]

the question restrictions.

[willow bear]

0 ≤ θ < 2π

[viscid thistle]

Oh I see

Thank you for explaining

[vague minnow]

hey um
I need a little help on starting this equation system

is there a simple/elegant way or do I just ram through?

[stark vale]

hey guys how do i have to finish this problem. i got to half of it and dont know what to do with the y'(2)

do i just replace the x with 2 ?

[tardy ridge]

@vague minnow I have not yet tried it, but I see something factorable in the second equation, maybe try factoring them?

[vague minnow]

do i just replace the x with 2 ?
@stark vale yes

@vague minnow I have not yet tried it, but I see something factorable in the second equation, maybe try factoring them?
@tardy ridge hmm I might try finding smth thanks

[stark vale]

k thx

[flint viper]

Can someone help me with a quick question please?

[rocky prism]

Could somebody explain, how they get the arctan(-3/5) to be -31 degrees? isnt arctan the 2nd key of tan on the calculator? I get a decimal

[lilac pier]

@rocky prism make sure you have degree mode on

[rocky prism]

ahh I forgot that was a thing, thanks

[warped dagger]

hi folks

x^2 + 6x + 9
(x+3)^2

I literally have no clue how does this become (x+3)^2, any point in the right direction is appreciated

[willow bear]

are you familiar with the identity $(a+b)^2 = a^2 + 2ab + b^2$

[obsidian monolith]

Ann:

[warped dagger]

thats the binomial one right

or difference of squares?

[willow bear]

do you see a square subtracted from another square?

no this is not a difference of squares

this is the square of a sum

[warped dagger]

thanks

[midnight lake]

"Sketch the reciprocal of each function shown."

pls help

<@&286206848099549185>

[unique hill]

how quick was that copswing reaction lol

[midnight lake]

yea ikr

[unique hill]

15mins rule

[midnight lake]

as soon as I sent haha

[unique hill]

yea cuz u pinged helpers before 15mins when posting a question

[midnight lake]

ahahaha

what

[unique hill]

fishraider that's enough bread for today

[stuck lark]

i was here before the ping, that's how

[midnight lake]

wait what 15min rule

[unique hill]

read #❓how-to-get-help

[midnight lake]

oh no

sorry

[stuck lark]

#❓how-to-get-help needs to be renamed to #rules-click-this-before-anything-else

[midnight lake]

i read the "readme" and came here....

soz

[unique hill]

woog should make that thing where u have to scroll all the way down that #❓how-to-get-help and pop a reaction to gain access to rest of server whenever a new person joins

no worries

[midnight lake]

;-;

[stuck lark]

rename #rules to #rules-1 and #❓how-to-get-help to #rules-2

[midnight lake]

do all math discords have this feature?

cos this is my first time using a discord like this...

[unique hill]

also for graphing reciprocals, all x-intercepts become vertical asymptotes and the y-values are reciprocated @midnight lake

[midnight lake]

Im not sure what to do with the y=2 reciprocal

[unique hill]

reciprocal of 2 is...?

[midnight lake]

1/2

[unique hill]

there's an asymptote at y=1/2 for the horizontal asymptote for the reciprocal function

[midnight lake]

ah ok

[unique hill]

and x intercepts become vertical asymptotes due to reciprocal transformations

[midnight lake]

understood

[unique hill]

remember every y value is reciprocated

[midnight lake]

so the smaller values become huge and the bigger values become smol

so would there be a negative parabolic curve beneath the asymptote?

[unique hill]

yes indeed

[midnight lake]

oh.

[unique hill]

wait

between the x=1 and x=3

the neg parabolic curve

has to be aligned for the x values

[midnight lake]

ah ok

got it

then is the part above the asymptote a exponential curve?

[unique hill]

just made a quick one

wdym exponential

the reciprocal of the horizontal asymptote is at y=1/2

[midnight lake]

oh yea

[unique hill]

remember that it's still a HA

right i gtg sleep

[midnight lake]

i meant as in the shape like the one shown on the very right

ah ok

gn

thanks for the help man

[unique hill]

np

[midnight lake]

and sorry for breaking the rules;-p;

;-;*

[unique hill]

somebody will def help u later on

no worries

[midnight lake]

thanks

i got it now

[brittle fable]

What is F composed of here?

[willow bear]

wdym

what do you mean, "composed of"

[brittle fable]

i want to map x and y to and x and y by the identity and z to f(x,y,z)

to get F(x,y,z) = (x,y,f(x,y,z))

[willow bear]

i mean what's wrong with just saying F(x,y,z) = (x,y,f(x,y,z))

[brittle fable]

hmm maybe im overthinking

yes im overthinking sorry

[viscid thistle]

Could someone help me expand this using Pascal’s triangle? I already know the coefficients, it’s actually the exponents that are confusing me.

[blazing parrot]

do you mean the binomal theorem?

[viscid thistle]

No

Well idk is it called that

[blazing parrot]

i think it is

well the exponent determines the line of Pascal's Triangle you have to look at

[viscid thistle]

I’m talking about Pascal’s triangle before you integrate it with binomial theorem I think

[blazing parrot]

so how are the exponents confusing you

[viscid thistle]

I am trying to make sure I can do it right

So 1/x^5 is first term

What is the easiest way to go about the next term who’s coefficient I know is 5

[acoustic hound]

Next coefficient is 10

Then 10 again

Then 5

Then 1

[viscid thistle]

I know that

I am having trouble getting what is after coefficient actually

[willow bear]

(1/x)^4 * (sqrt(x))^1
then (1/x)^3 * (sqrt(x))^2
then (1/x)^2 * (sqrt(x))^3
then (1/x)^1 * (sqrt(x))^4
then (sqrt(x))^5

[acoustic hound]

Yup

[blazing parrot]

for every term the power of of the first variable goes down by one while the power of the second increases by 1

[viscid thistle]

Okay, I get that. Why can the second term be written as - 5/x^(7/2)

Oh I see it now, thank you

[fleet yew]

@viscid thistle if you have an expression in the form (a+b)^n, then the powers of each term add up to n

as you can see, in this case n is 5

so you have

5, 4+1, 3+2, 2+3, 1+4, 5

[viscid thistle]

how would I find the limit of this sequence?

[fleet yew]

lhopitals?

good luck diffrentiating that sum

[viscid thistle]

i’m supposed to use a theorem

idk what its name was

oh

squeeze theorem

and i’ve looked for the answer and i don’t get anything

[rapid furnace]

To use the Squeeze Theorem, you have to find sequences that are upper and lower bounds of this sequence.

[viscid thistle]

it tells me this

and then it says that the limit is ln a -_-

i understand the first part

but how do I find the limit for (ln a^n)/n

[frozen needle]

you rewrite ln(aⁿ) as nln(a)

[viscid thistle]

oh wtf yeah

what about the second one?

ln(n*a^n)/n

it’s supposed to be ln a too

[tardy ridge]

I don't understand the question but I know that is ln(n)+nln(a)

if that helps..

[viscid thistle]

yeah

i found that too

then i guess it would be ln(n)/n + ln a?

[tardy ridge]

is n approaching infinity?

then you will be left with ln a

don't take my word for it though, ask the YEE profile picture guy.

[viscid thistle]

yeah

it is

so the answer should be ln a

[tardy ridge]

sounds logical to me.

[viscid thistle]

I did this question and the answer online says the answer is r=4 and theta=pi/4

[unique hill]

is this precalc?

[viscid thistle]

My question is why can’t theta be 5pi/4

Yes

[past meadow]

5pi/4 is in the wrong quadrant

oh wait

nbm

[upper kelp]

As Sneaky implies, you have to consider the signs of cosine and sine.
In the third quadrant, where 1.25π lies, both cosine and sine are negative.

[viscid thistle]

Oh I see

[past meadow]

actually yes im right

[viscid thistle]

So then it doesn’t fit restriction

[past meadow]

as roi siad

xd

[viscid thistle]

Got it 🙏

[upper kelp]

Just in case you find it helpful:

[fleet yew]

u rly only need to know it for sin and cos

or even just sin

because you can just take sine and shift it by a quadrant

[karmic wave]

depends how much conscious thought vs mindless memorization you want to put in

[upper kelp]

As someone who has studied it for quite some time, I think I have a good intuition for it, but I sometimes get mixed up, so things like that could be helpful at least for starters.
I can also admit that it took me a really long time to remember that csc and sec are the inverses of sin and cos, respectively.

[smoky needle]

Hi can someone help me with a maximum value problem?

This is what I have so far

[fleet yew]

Just take the derivative?

wait no

you can just find the max of the quadratic

@smoky needle

you know how to find max of quadratic?

[smoky needle]

Hi, and no. Not for this problem. I’m actually stuck because of the 5th root.

I tried getting rid of it, but then I get stuck with what to do after.

@fleet yew

[fleet yew]

let's just look at the quadratic

you have

$-x^2+4x+28$

[obsidian monolith]

AMD:

[smoky needle]

Yes

[fleet yew]

do you know calculus

pls say yes it's gonna be really hard to explain otherwise

[smoky needle]

It’s been a while... that’s why I decided to take precalc😅

But I’ve taken calc before

[fleet yew]

ok

[smoky needle]

So if it’s derivatives then I know how to do that

[fleet yew]

the extrema of the function are at the roots of the derivative

[smoky needle]

Yes, so will I have to find the derivative?

[fleet yew]

yes

[smoky needle]

Of just that right? I leave y^5 alone

?

[fleet yew]

just take the derivative of that expression

the quadratic

[smoky needle]

-2x+4

[fleet yew]

find the roots

or the root in this case

[smoky needle]

Sorry I’m kinda slow, what do you mean by roots?

In this case

[fleet yew]

set that expression equal to 0

[smoky needle]

Okay, and I’ll just solve for x correct?

[unique hill]

by roots, he meant x intercepts

[smoky needle]

Thank you!

X=2