precalculus

[willow bear]

f(g(0))

are you able to calculate that

[viscid thistle]

no

[willow bear]

why

[viscid thistle]

we dont know what f is

[rocky bison]

rip quantum mathematics

[willow bear]

we have a graph

[languid crane]

Quantum math

[willow bear]

you're meant to do it based on the graph

[viscid thistle]

cotton gtfo out im trying to learn

jesus

alright how

thats what i dont get

is the confusing part

[willow bear]

i mean

do you not know how to read off the value of a function at a certain input from a graph?

[viscid thistle]

idk like because what i learn

[languid crane]

sorry Sorry

[viscid thistle]

i plug 0 into an f

[rocky bison]

Do them in order, what'sg(0) from the graph, when once you've got that work out what f(x) is based on the graph

[willow bear]

that said, you'd need to determine g(0) first

are you able to determine what g(0) is

[viscid thistle]

no

how do i using the graph

[willow bear]

do you know what the graph of a function even is?

[viscid thistle]

no

kind of

i know vertical line test

[willow bear]

go read up on function graphs

[viscid thistle]

i do

[willow bear]

since that's where your knowledge gap is

[viscid thistle]

know what functions are

lol

[willow bear]

you don't know what a graph is

[viscid thistle]

i do

i've passed algebra 2

so i def know function graphs

im just confused what your trying to get at

[willow bear]

if you did then you would know how to find g(0) given the graph of g

[viscid thistle]

idk

are you telling me

to convert the graph

into a algebraic expression

[willow bear]

no, i'm not

[viscid thistle]

or function epxression

[willow bear]

ok look\

[viscid thistle]

yes

hoooaaaaa billions on mehhh yeee

lol

just explain it @willow bear

my brain only understands from actually doing it

if you do it my brain will naturally pick up on it

its how i work

or i quickly put the puzzles together lol you just gotta do it

just do it

show me how to solve the problem

and then

after you show all the steps

i'll quickly pick up on it

[willow bear]

okay so let me give you an example not directly related to the problem at hand

here's the graph of a function, h

[viscid thistle]

ok

[willow bear]

from this graph, i want you to read off the value of h(1)

[viscid thistle]

2

[willow bear]

see, you can do it!

[viscid thistle]

oh

oh thanks buddy

i get it now

[willow bear]

i don't understand what the hell stopped you from doing it before i prompted you with that thing

[viscid thistle]

lol i work in a very strange way

@willow bear

is the answer B

[willow bear]

no

[viscid thistle]

one sec

let me reevulate

@willow bear

wait how

[willow bear]

how what

[viscid thistle]

how is it not 1

because its 0

You look at 0

You have 2 & -3

[willow bear]

option B is 2, not 1

did you mean for your answer to be f(g(0)) = 1

[viscid thistle]

ye

[willow bear]

bc in that case that is not option B

[viscid thistle]

wait

[willow bear]

option B would be 2, not 1

[viscid thistle]

yes but how

how is it 1

[willow bear]

you yourself said f(g(0)) = 1 though

why are you asking me about it now

[viscid thistle]

but why

[willow bear]

g(0) = -3

f(g(0)) = f(-3) = 1

[viscid thistle]

ok

ah ok

dis makes sense now

thank you!

[willow bear]

i am utterly puzzled, but you're welcome

[viscid thistle]

np

[viscid thistle]

lol

[viscid thistle]

Are there any good resources that could help me grasp functions and limits?

[royal gull]

yes

[viscid thistle]

Could you suggest any?

[royal gull]

have you tried khan academy?

[viscid thistle]

Yes

[willow bear]

"grasp functions" is kinda vague tbh

[viscid thistle]

@willow bear That covers it better

[rocky bison]

Just Google each one of those tbh

There will be s video that explains it

Then if you've got questions ask here

And try to find problems online

[viscid thistle]

Okay 🙂

[glacial island]

precalc broke into my house punched me and called me a lil bitch

[royal gull]

ur weak den friend

[glacial island]

[hybrid charm]

[stark inlet]

https://cdn.discordapp.com/attachments/326138772477575180/568266398917328906/image0.jpg

not getting the correct value

[hybrid charm]

y≠4

[last violet]

Is it just

where -4<a<4?

[willow bear]

no, there is no restriction on a. the condition -4 <= a <= 4 is the condition that would ensure that z is real

[last violet]

So how do I find the roots then?

[willow bear]

you have done that already

wait, i got that condition backwards anyway. that condition ensures z is not real.

[last violet]

Ye

But the question before

https://cdn.discordapp.com/attachments/281589009111580674/568299002886488074/unknown.png

It has the same roots for a and b?

Then what's the difference between the problems?

[willow bear]

in one you restrict z to be real, in the other you don't

[last violet]

So b is the same thing but no a restrictions?

in the root

[willow bear]

yes.

[last violet]

So the roots moving in the complex plane

Look something like this, right?

[willow bear]

that looks correct

[last violet]

Thanks for the help!

[raven basin]

hi i need help in this

[raven basin]

help

lol

[dense kettle]

@raven basin still need help?

[raven basin]

Nah lol I figured it out lol. Took me while lol

Thanks tho

[bold isle]

what's the proof of this ?

[frozen needle]

to prove this is to prove that ln is continuous at a particular point

[brisk zealot]

How does the general term 2n generate the series 1+2+3+..2n

[echo plaza]

who said 2n was the general term

[brisk zealot]

<@&286206848099549185>

[slender river]

where am i looking at tho

[eternal folio]

It simply doesn't. @brisk zealot

[brisk zealot]

Ok thank you

@echo plaza what else would it be

[echo plaza]

$\sum_{k=1}^{2n} k$

[obsidian monolith]

CaptainLightning:

[patent beacon]

@brisk zealot

[slender river]

sooqa pow

[brisk zealot]

oh

um

that actually makes sense I think?

yeah it wasn't a typo im just dumb

the first question asks if you can find what the sum of the sequence 1+3+5+...+(2n-1) =

@patent beacon

Is this a correct interpretation?

oh so 2n isn't a general term.

[thick raptor]

ew

@brisk zealot (2i-1) = i^2 - (i-1)^2 tbh

[brisk zealot]

I don't know sum of RHS though @thick raptor

i don't know how to find sum of any part of RHS

[thick raptor]

?

Why would u do it like that tho

[brisk zealot]

i know sum for both of the ones i showed

[thick raptor]

Then what's your problem

[brisk zealot]

Got resolved but my problem was: Why was 2n the general term for the series 1+2+3... I was informed that I misread and it is the last term of the series.

But for your method that you pinged me with, how could you find sum of LHS with RHS?

assuming summation bounds are same on both sides

[thick raptor]

what

It just telescopes

[brisk zealot]

I know but finding a formula for the sum

oh wait

nvm lol you still find it

(i^2 - (i-1)^2) + ((i+1)^2 - (i)^2) + ... and the rest cancel out besides -(i-1)^2 roight?

same problem though lol doesnt tell the sum

does it?

no, def doesnt

[thick raptor]

?

Everything cancels

You just get n^2

[brisk zealot]

? what does "-(i-1)^2" cancel with?

the 2nd part of 1st parenthesis

@thick raptor

[thick raptor]

You summed it wrong?

It's from i=1 to i=n

no?

[brisk zealot]

1 to n

but doesnt make a difference if telescoping right?

[thick raptor]

?

All that's left is the n^2 and 0^2

[brisk zealot]

oh the last remaining terms are -(i-1)^2 and n^2

but i = 1

oh

ok

ty

I don't know what it is called but I did it when first attempting the problem

instead of plugging in values for every iteration , I substitute (i+1)

ill stop that

[warped steeple]

how do i evaluate sum from 1 to n of [(-1)^k]/2^k

[frozen needle]

It's geometric

if $r\neq 1$ then $\sum_{k=0}^nr^k=\frac {1-r^{n+1}}{1-r}$

[obsidian monolith]

Tuong:

[warped steeple]

How about $\sum _{k=1}^n:\frac{1}{4k^2-1}$

[obsidian monolith]

boilhats:

[eternal folio]

It's a telescoping sum if I'm right

yes it is

first you need to factorize, then split the fraction

and you can find the sum easily

[thick raptor]

[eternal folio]

what are those emojis smh

[thick raptor]

[eternal folio]

Pls

[warped steeple]

$\sum _{k=1}^n:k\cdot 2^{k-1}$

[obsidian monolith]

boilhats:

[warped steeple]

Doesnt anyone know how to solve this as it isn't a geometric or arithmetic series

[frozen needle]

try to see it as a derivative

with $f_n:x\longmapsto\sum_{k=1}^nx^k,\
f_n'(2)=\sum_{k=1}^nk2^{k-1}$

[obsidian monolith]

Tuong:

[frozen needle]

And you probably already know another way to calculate that derivative

@warped steeple

[willow bear]

clever

[thick raptor]

$S=1\cdot2^0+2\cdot2^1+3\cdot2^2+\dots+n\cdot2^{n-1}$\$G=2^0+2^1+2^2+\dots+2^{n-1}=2^n-1$\$2(S+G)=2\cdot2^1+3\cdot2^2+\dots+n\cdot2^{n-1}+(n+1)\cdot2^n$\$2(S+G)=(n+1)\cdot2^n-1+S$\$S=(n+1)\cdot2^n-1-2G=(n-1)\cdot2^n+1$

[obsidian monolith]

SAckalope:

[wild trout]

,rotate 180

[wild trout]

@warped steeple

[glacial island]

do i need to be able to solve natural logarithms without a calculator?
for example ln e^8
i was only able to solve it by plugging it in the calc to get 8, but i wouldnt know how to solve it on paper

[patent beacon]

ln(x) and eˣ are inverse functions. One being in the other undoes both

ln(eˣ) = x
e^(ln(x)) = x

Much like how
sin(arcsin(x)) = x

[glacial island]

ooh okay thank u

[willow bear]

if you didn't know $\ln(e^8) = 8$ then you don't know what $\ln$ is

[obsidian monolith]

Ann:

[glacial island]

its the natural base aka e so theyre inverses. im working through the lesson rn so its taking a lil bit of time to sink in

sorry for the questions 😅

[thick raptor]

e^ln(-1) = -1

[frozen needle]

[ruby otter]

$lim_{x \to 0} \floor{sinc(x)}$

It's zero right??

[obsidian monolith]

Radical Ninja:

[willow bear]

don't think so

floor(sinc(x)) dips below zero infinitely many times, so its floor is bumped down to -1

in fact $\floor{\mathrm{sinc}(x)} = -1$ on all intervals of the form $[(2k-1)\pi, 2k\pi]$ for all integer $k \geq 1$

[obsidian monolith]

Ann:

[willow bear]

the limit doesn't exist

@ruby otter

[ruby otter]

No @willow bear

[willow bear]

[ruby otter]

,w plot floor(sinx/x)

[obsidian monolith]

Display results online and refine query

Requested by Radical Ninja#7581

[willow bear]

,w plot floor(sin(x)/x) for 0 <= x <= 100

[obsidian monolith]

Display results online and refine query

Requested by Ann#8192

[willow bear]

see

[ruby otter]

But it tends to zero right??

[willow bear]

no, it does not.

[ruby otter]

I means the x tends to zero

[willow bear]

... oh god

i thought you were talking about x -> infinity for some reason

sorry

yes, then that's zero ofc

[ruby otter]

No prob it's zero right??

Many are arguing that this is right

[willow bear]

what the heck is this meant to be

also, floor(lim sin(x)/x) is not the same as lim floor(sin(x)/x).

floor is discontinuous

[viscid thistle]

it's supposed to be $\lim_{x \to 0^+}$

[viscid thistle]

i suppose

cursed notation:

[warped steeple]

Show that $$1-\frac{1}{t}<\ln t < t-1 ,t>1$$

[obsidian monolith]

boilhats:

[thick raptor]

use definition of ln to show the upper bound

the lower bound follows from ln(1/t) = -ln(t)

[ruby otter]

How did we simplify that

n tends to infinity

[viscid thistle]

$$\sin x = 2^1 \sin \frac x 2 \cos \frac x 2$$
$$\sin x = 2^2 \sin \frac x 4 \cos \frac x 4 \cos \frac x 2$$
$$\sin x = 2^n \sin \frac{x}{2^n} \prod_{k=1}^{n} \cos \frac{x}{2^k}$$

[viscid thistle]

@ruby otter

[serene heath]

epic

[idle dust]

NANDESCAAAAA

[ruby otter]

how did we get cot in the RHS

thats my doubt

@viscid thistle

$$\frac{\sin x}{2^n \sin \frac{x}{2^n}} = \prod_{k=1}^{n} \cos \frac{x}{2^k}$$

[obsidian monolith]

Radical Ninja:

[ruby otter]

$$\frac{\cos x}{\frac{\sin x}{2^n \sin \frac{x}{2^n}}^2} $$

[obsidian monolith]

Radical Ninja:

[ruby otter]

$$\frac{(\cot x)(2^n \sin \frac{x}{2^n})^2 }{\sin x} $$

[obsidian monolith]

Radical Ninja:

[ruby otter]

$$\frac{(x^2 \cot x)( \sin \frac{x}{2^n})^2 }{(\sin x)\frac{x^2}{2^{2n}}} $$

[obsidian monolith]

Radical Ninja:

[ruby otter]

$$\frac{(x^2 \cot x) }{(\sin x)} $$

[obsidian monolith]

Radical Ninja:

[ruby otter]

this is what i get @viscid thistle

[unborn island]

Can someone help with complex algebra

Express z = (1-isqrt(3))^17 in polar form

[viscid thistle]

convert 1 - i sqrt 3 to polar first

[echo plaza]

$1-i\sqrt{3}\equiv 2 \cdot (\frac{1}{2}-i\frac{\sqrt{3}}{2})$

[unborn island]

I know that r=131072

[obsidian monolith]

CaptainLightning:

[unborn island]

If you express it r(cos x + i sin x)

[viscid thistle]

wait wat

how is r = 131072

oh i see

you raised it to 17 directly

[unborn island]

Yes

i thought of using demoivre’s theorem? I’m pretty sure thats the right way to go

[viscid thistle]

yeah

[unborn island]

It confuses me though, trig and complex numbers are both bad

[willow bear]

1/2 - i sqrt(3)/2 is a sixth root of unity

i mean you can use de moivre's if you want to, i'm just putting this out there

[unborn island]

What

what is a

sixth root of unity

[viscid thistle]

a number such that $z^6 - 1 = 0$

[unborn island]

Is unity 1?

[viscid thistle]

yeah

[unborn island]

Oh yeah i get it now

[echo plaza]

it's also in the form cos(x)+isin(x) (like all roots of unity)

[unborn island]

So just take it’s reciprocal? Because 18=6*3, and 17=18+(-1)

[echo plaza]

yeah that'd work pretty well

[unborn island]

K solved it, there’s one more

[ruby otter]

,w lim_x->infinity cos ( pi * sqrt (x^2 +x))

[obsidian monolith]

Display results online and refine query

Requested by Radical Ninja#7581

[ruby otter]

is this correct ?? i think answer is zero

[echo plaza]

cos is periodic so that limit doesn't exist

[ruby otter]

i think its zero

[echo plaza]

what makes you think it's 0?

[ruby otter]

well you guys start but never end

my previous question

noone answered it

[echo plaza]

oh for natural x?

[ruby otter]

ya its given x element of integer

[echo plaza]

0 sounds right in that case yeah

[ruby otter]

how?

[echo plaza]

$\cos(\pi \sqrt{x^2+x})=\cos(\pi x\sqrt{1+\frac{1}{x}})$

[obsidian monolith]

CaptainLightning:

[ruby otter]

well i did its gonna be pix alone

ohh got it thanks

i didn't notice that its element of I

[echo plaza]

but you're the one who told me that

[ruby otter]

no i actually noticed it after you asked it whether its natural

i re-read the question

and then my previous steps made sense

btw check my prev. question

https://cdn.discordapp.com/attachments/363224154469826562/569973748028014600/IMG-20190423-WA0001.jpg

i got $$\frac{(x^2 \cot x) }{(\sin x)} $$ is this correct??

[obsidian monolith]

Radical Ninja:

[echo plaza]

why is this here in the picture

[tawny nacelle]

[ruby otter]

why does that bother you ?? it's a part of q

[echo plaza]

looking at the line above it should be $\frac{\cos\theta}{\cos(\frac{\theta}{2^n})\prod_{k=1}^{n} \cos(\frac{\theta}{2^k})}$

[ruby otter]

well do the steps

i already did this

[obsidian monolith]

CaptainLightning:

[ruby otter]

$$\frac{(x^2 \cot x) }{(\sin x)} $$

[obsidian monolith]

Radical Ninja:

[ruby otter]

$$\frac{\cos x}{\frac{\sin x}{2^n \sin \frac{x}{2^n}}^2} $$
$$\frac{(\cot x)(2^n \sin \frac{x}{2^n})^2 }{\sin x} $$
$$\frac{(x^2 \cot x)( \sin \frac{x}{2^n})^2 }{(\sin x)\frac{x^2}{2^{2n}}} $$

[echo plaza]

weird it looks like it's just that line that says square

I'll ignore it

[obsidian monolith]

Radical Ninja:

[ruby otter]

see the steps

finally you'll get this
$$\frac{(x^2 \cot x) }{(\sin x)} $$

[obsidian monolith]

Radical Ninja:

[echo plaza]

[ruby otter]

what wrong in this

[echo plaza]

these don't cancel

[ruby otter]

sin squared

n tends to infinity

[echo plaza]

oh it's a limit?

[ruby otter]

sin squared will get cancelled

ya

@echo plaza is that correct ??

[echo plaza]

I definitely don't think the top is equal to the bottom here

most of the $\cos^2(\frac{\theta}{2^k})$ terms in the denominator cancel with $\cos(\frac{\theta}{2^k})$ terms in the numerator

[obsidian monolith]

CaptainLightning:

[ruby otter]

so you say my steps are wrong??

[echo plaza]

just that one

[ruby otter]

then how it shud be done??

[echo plaza]

if you cancel the terms like this you should get $\frac{\cos\theta}{\cos(\frac{\theta}{2^n})\prod_{k=1}^{n} \cos(\frac{\theta}{2^k})}$

[obsidian monolith]

CaptainLightning:

[ruby otter]

ohh that whole sq. wont be there right??

[echo plaza]

yeah except for cos(theta/2^n)

which is squared

because there's no term that it cancels with

[ruby otter]

ya got it'

stil

lol i got that photo and left some extra space now it helped me 😂

[ruby geode]

If I have a polynomial function, how do I algebraically find the turning points, instead of relying on the trace feature of my TI-84?

[serene heath]

set derivative equal to 0 n solve

[torn swift]

But this is precalc channel, does this person know about derivatives yet?

[thick raptor]

[viscid thistle]

🤔

You did derivatives in Precalc?

[torn swift]

no

[past shell]

A rectangular dog pen is to be fened with 16m of fencing. Determine the maximum area and the width of this rectangle?

Not sure what to do

[torn swift]

This is a typical optimization problem

But for real you can do this without derivatives and stuff?

[past shell]

I haven't learned derivatives so I assume so

[torn swift]

Darn it

[fickle quiver]

it can be done

ezily

[torn swift]

I never learned it the precalc way, so please show

[fickle quiver]

:O

a,b be sides of the rectangle

a+b = 8

we need to maximize ab

ab = a(8-a)

8a - a^2 -16 + 16 = 16 -(a-4)^2

[torn swift]

Darn, should have thought of something like that

[fickle quiver]

dwai

you hadn't learnt it the precalc way, so it is alright XD

[torn swift]

Let a,b be the sides of the rectangle:\
\$ a+b=8$\
\We need to maximize $ab$:\
\$ab=a(8-a)$\
\$8a-a^{2}-16+16=16-(a-4)^{2}$

[fickle quiver]

:o

[obsidian monolith]

⚡Amphy⚡:

[torn swift]

I know

[past shell]

idk how you got to the third equation

[willow bear]

it's just completing the square

[past shell]

i see

[glacial island]

log (x + 4) (x + 1) = 1
can be rewritten as
(x + 4 ) (x + 1) = 10

the 10 is because you raise the 1 to the tenth power right?
because natural log base is 10

[willow bear]

no, natural log is base e, not 10

also, no, 1^10 would be 1, not 10

also, no, log_10( (x+4)(x+1) ) = 1 would be (x+4)(x+1) = 10^1

[fickle quiver]

welp, diff conventions

in some places log(x) is read as logorithm x base 10 and ln(x) is read as logarithm of x base e

[glacial island]

how did they get rid of log and make it into 10?

[willow bear]

do you know the definition of log

[glacial island]

the answer is raising the base to what power

other than that definition i dont

[willow bear]

$\log_{10}( (x+4)(x+1)) = 1 \ 10^1 = (x+4)(x+1)$

[obsidian monolith]

Ann:

[glacial island]

OOOOH

oh my god

that makes sense

thank you lmao

[viscid thistle]

I have no clue where to start with this one. The only thing I know is I didn't manage to find the exact formula of f-1 yet and I'm not sure if it's needed here

[willow bear]

...what's the num of the third fraction supposed to be

[viscid thistle]

Oh..

Sorry

Sin(πa)

[willow bear]

$\frac{f(a)}{x-1} + \frac{f^{-1}(a)}{x-2} + \frac{\sin(\pi a)}{x} = 0$

[obsidian monolith]

Ann:

[willow bear]

is this the eq?

[viscid thistle]

Yup

[willow bear]

oof

[torn swift]

We can get rid of sin(pi a) though

Wait no

[viscid thistle]

My words exactly. If you can just tell me if I'll need to find the formula of f-1 or not i can try to start from there as it will take a while

[eternal folio]

I sense the Lambert W function...

[viscid thistle]

I just need to prove they exist and it's exactly two. So some Bolzano shit mixed with monotony(if thats the word)

Idk who that is

Time to find out I guess

[viscid thistle]

That Lambert thing is some witchery i don't understand anything

[hybrid charm]

Yes that is difficult

[signal ermine]

@viscid thistle dont think you need W

Differentiate the LHS on the two intervals and show that the function restricted to either of the intervals is monotonic and onto

[viscid thistle]

LHS?

No idea what that is but you gave me some sweet sweet ideas I think I'll manage to deal with it now.. Thank you I've been stuck for days

[hybrid charm]

Left hand side

[viscid thistle]

....😅

Thanks

[hybrid pewter]

Hi

i can’t figure it out

[torn swift]

,w derivative (2/3)(ln(1+3(cos(x)^2))

[obsidian monolith]

Display results online and refine query

Requested by ⚡Amphy⚡#0697

[hybrid pewter]

i got that at the very top

but trying to change it to tan x is where i had issues

[torn swift]

thonking about this right now

I don't know

[final rover]

I have an idea

Here is something you could use

[hybrid pewter]

Shit

I dont know too

[final rover]

$\sin(2a)=2\sin(a)\cos(a)$

[obsidian monolith]

TSM64CM:

[hybrid pewter]

yeah

[final rover]

The other thing is $\cos(2a)=2\cos^2(a) - 1 = 1 - 2\sin^2(a) = \cos^2(a)-\sin^2(a)$

[obsidian monolith]

TSM64CM:

[final rover]

Maybe try to change your 1+3 cos^2(x) ?

[hybrid pewter]

^ i did this

but i'll try the top one

sin 2x

i'll try that

yeah idk

again

[limber bone]

,rotate 90

[limber bone]

,rotate 180

[limber bone]

whats the question please?

[native sequoia]

Divide top and bottom by $\cos^2(x)$ and use $\sec^2(x)=1+\tan^2(x)$ (or use $1=\sin^2(x)+\cos^2(x)$ first).

[obsidian monolith]

EpicGuy4227:

[native sequoia]

@hybrid pewter

[hybrid pewter]

@old sentinel thank you!

[safe tinsel]

Help with 34

No one responded in the help section

<@&286206848099549185>

[dense zealot]

Hi

So

The zeroes

And a constant multiple

Makes it

f(x) = a(x^2 - 2)(x^2 - 1)

Plugging in x=2,

-20 = 6a , a= -10/3

therefore the polynomial is f(x) = (-10/3)(x^2 - 2)(x^2 - 1) = (-10/3)(x^4 - 3x^2 + 2) =
(-10/3)x^4 + 10x^2 - (20/3)

Or something idk

[safe tinsel]

Aight thanks

[swift pier]

i need to calculate f(x) + f(1/x),if f(x) = log6(x) + 3log3(9x)

base of first log is 6 and the second is 3, i just dont know how to subscript

[willow bear]

log_6, log_3

so what exactly is giving you trouble here?

[swift pier]

well i dont know how to do these

[willow bear]

"these"?

[swift pier]

types of problems

havent encountered them before

[willow bear]

not to sound snarky but have you tried thinking rather than going "oh i haven't encountered this kind of problem before let me just not read it at all and go beg for help"

you're given a function (here called f) and are asked to simplify an expression involving it

[swift pier]

yes i have, my result isn't the same as the one in my book therefore i turned here for help

[hoary yoke]

😂

[willow bear]

okay

now we're getting somewhere

what was your result?

[hoary yoke]

What does ur book say Virtual

[swift pier]

12

is mine

[willow bear]

and the book?

[hoary yoke]

What is the book answer uwu

[swift pier]

log_3 sqrt(2)

which doesnt make any sense whatsoever

[willow bear]

...are you sure you're looking at the answer to the right problem

bc what you got is correct

[hoary yoke]

I thought the answer was 0

[swift pier]

yes

[hoary yoke]

Wait are those powers

[willow bear]

but what you claim is the book's answer is nowhere near correct

[swift pier]

yes im sure

doesnt matter

[hoary yoke]

Oh its bases

[swift pier]

my result is correct

thanks you very much

❤

btw is there a name for problems of this type, id like to practice them a bit more

this is the only one in my book

[violet latch]

I don't understand how to do this limit problem

lim (t^3 + t^2 - 5t + 3)/(t^3 - 3t + 2) as t approaches one

[viscid thistle]

First, see if the function is defined at 0 by plugging in 0 everytime you see a t.

[violet latch]

my mistake, it is as t approaches 1

where it is not defined

[viscid thistle]

Oh

[thick raptor]

Just factor out t-1

[violet latch]

why do you suggest t-1

[thick raptor]

So you can do what that psycho told you?

||always listen to strangers without question man||

[viscid thistle]

DOE changed the question haha

[thick raptor]

Well obviously don't use zero

Just use 1 instead

[hybrid charm]

(t-1) is the villain here that's why 😖

[violet latch]

when I factor out (t-1) for the top, I got (t-1)(t^2 + 2t - 3)

I think im doing something wrong

okay nvm, thank you

so If the limit to a number becomes 0/0, you take out x - (number)??

[viscid thistle]

Not necessarily

The first thing you want to do is see if the numerator and denominator have any factors in common. (After checking to see if the actual numbet itself works).

[hybrid pewter]

hi

where did i go wrong

[rocky bison]

,rotate 90

[rocky bison]

,rotate 180

smh

[willow bear]

@hybrid pewter the derivative of x^2 is not x

[hybrid pewter]

oh

my god

right

thank you

[valid vector]

I'm confused as to how to solve these two. Can anyone help me?

[serene heath]

you could refer to the unit circle

or u can expand using compound angle formulas

[valid vector]

Still confused, lol

[torn swift]

you can just look up identities if you are that stuck
no shame in that, usually most of us are given identities to memorize the first time around

but think about the two acute angles in a right triangle

$\alpha + \beta=90$\
\
What are a pair of angles whose sum is 90 called?

[obsidian monolith]

⚡Amphy⚡:

[valid vector]

Well complementary angles

[torn swift]

ah man this is easier to show visually

let me see if I can get an online figure that shows what I want to tell you

https://www.khanacademy.org/math/geometry/hs-geo-trig/hs-geo-complementary-angles/v/showing-relationship-between-cosine-and-sine-of-complements

this should really help you see it

[valid vector]

I kinda get it, but still quite confused, lol.

Nvm, got it, oof! Thanks!

[torn swift]

definitely use all your resources

you can learn a whole lot from different people

[valid vector]

Yeah, thanks a bunch!

[torn swift]

you know the technique of drawing a triangle?

[valid vector]

OK, so this one has me stuck as well. I get cos's value, but they are only giving me one and I don't understand how to get the other ones.

I don't think so

[torn swift]

search up "find the ix trig functions by drawing a triangle" or something like that

there are plenty of resources there

[valid vector]

OK, will do!

Can you help me understand this problem, BTW?

[torn swift]

first read up what you can because I think you'll understand it, actually I think Khan Academy has videos on it too

[valid vector]

OK, will do now.

[torn swift]

https://www.youtube.com/watch?v=GWdQ9nfyN3Y

this is a good video

[valid vector]

Lol, that's the exact video I watched! Thanks, I got it now!

[torn swift]

great!

[valid vector]

You were of huge help! Thank you a lot! I'll probs come back in a little bit with something else I might not understand 😅

[torn swift]

always use all available resources, and yup we're one of them

sometimes it's a matter of asking the right question

[valid vector]

If I have for example 3/(2*sqrt{2})
I know I have to rationalize, but do I multiply only by the sqrt{2} or by the whole denominator including the 2 outside the sqrt?

Why is this wrong?

[valid vector]

IDK, tbh, I started from the basic formula and somehow ended up there. But yeah it was the 5th option, thank you so much!

[viscid thistle]

How should I approach improving my trig identity skills? I know basic ones like pythagorean and double angle but haven't used them too much. What would be a good way to practice and improve them for Calc 2 coming up?

[lost cipher]

deriving some of them might be a good idea, although that's only if you have the time

[viscid thistle]

Basically I'm looking for a pragmatic plan to approach the subject

I can make the time

It feels like memorization at the moment working with them

like, they don't all feel like they mesh together like the rules of algebra do

[torn swift]

Khan Academy will be a big help

at least it was for me

very visual and explained nice and slow

[tawny nacelle]

this post was made by khan academy gang

[fickle quiver]

it was too slow

using 2x made it fine tho

[viscid thistle]

idk guys

maybe I'll just slam through a ton of practice problems

time to dig out the old trig textbook

[torn swift]

confused about that?

[viscid thistle]

well, how does my boi here decide to make sin(x) = u

[willow bear]

substitution is always a little bit of guess and check

but with experience you start seeing certain patterns

[patent beacon]

You have to choose such that dx can get transformed

A sensible way that could happen is if
cos(x) dx = du

But for that to happen,
sin(x) = u

So ask yourself if that's a reasonable u-sub. You'd have ∫ 1/u² du, so that works well

I like to choose what "goes with dx" first, it can make it easy

[viscid thistle]

So we were just introduced to u-sub a few days ago and I have the mechanics down but none of the nuance.

[languid crane]

isnt this precalc channel?

[viscid thistle]

when you say so dx can get transformed what do you mean

ahh

haha yeah

I had a precalc quesiton and then a calc one popped into my head and i forgot to switch over

[languid crane]

ok :D

[patent beacon]

You have two goals with a u-sub

1. Get rid of every x
2. Turn dx into du

You should have one of those goals in mind when proposing a sub, then check if the other goal is met

Usually Turning dx into du is the easy way to think up a u-sub

[viscid thistle]

But we have to turn dx into the right du or its useless right?

like when choosing my u i need to think how it will end up cancelling when i sub it in

[patent beacon]

Think backwards. Rather than trying to predict how dx will transform, simply write dx's transformation, then get the u-sub from that

In the example above, it's somewhat clear that cos(x) dx = du will happen, because how else could you get dx out?

You could write 1/sin(x) dx = du, but that's far more complicated. Easier to check the other option first

[viscid thistle]

So, let me just write it out to make sure I understand you correctly. We know we would like to cancel out our cos(x) in the numerator to give us a nice easy 1/u^2 in the denominator. This leads us to think it would be nice if du= dx * cos(x) since that would cancel nicely. Then we go look for a u that can help us in accomplishing that goal, namely sin(x)

[patent beacon]

No, instead we have to turn dx into du somehow. The easiest way to do that is to write cos(x) dx = du

Then sin(x) = u follows, and this is a working u-sub

[viscid thistle]

I think I understand what you are saying and just worded my post poorly

[patent beacon]

Fair lol. Have any other examples? Maybe we can reason through one

∫ ln(x) / x dx
Is a good one

[viscid thistle]

hmm, ok one moment. I'll try to work it out

is it x^ln(x)?

[patent beacon]

Is what?
And no, that's not right

[viscid thistle]

I tried to do it just shooting from the hip as I couldn't decide what to make u equal

1/x ?

[patent beacon]

There is a u-sub necessary here, and you only have a few options. Just like the other one, it's easiest to find what pairs well with dx to get rid of it

Your first reaction might be to try
ln(x) dx = du

But that doesn't work as ln(x)(x - 1) = u is ugly

So your only other option is 1/x dx = du

It follows that ln(x) = u
And the integral becomes ∫ u du

[viscid thistle]

I see it now, that's really helpful!

I wasn't even working in the right direction beofore! I'm sure it will take practice but at least now I have a toehold. I appreciate you taking the time, I'll be around tommorow working on more integrals, hopefully I see you around 😃

For now I have to sleep though, take care

[swift glacier]

How would I use given points to find a rational function?

They aren't intercepts, just points on the function

[torn swift]

need example

[swift glacier]

Because of the hole it becomes a linear equation, so I guess the two points give the slope

But I'm not sure how it'll help

[torn swift]

can you use the window+shift+S shortcut to crop the screen then go back to discord and press control+v into the chat bar to post a high quality screenshot?

[serene heath]

snipping tool

[swift glacier]

OOOH

MAGIC

[torn swift]

thanks lol

[swift glacier]

I NEVER KNEW THAT

[torn swift]

looks much better right?

[swift glacier]

Thank you for teaching me

yes

It wouldn't have a horizontal asymptote, right?

Is the range is 4 exclusive because of the hole

?

[torn swift]

not very sure how the range restriction will come in for now, but since you know d already you can plug in the (x,y) points to get expressions for a,b, and c

[swift glacier]

So would it be a system of equations?

[torn swift]

but you cannot have asymptotes

[swift glacier]

yeah

[torn swift]

so that means d can't be -1

hmm

let me think about that one

a hole means there is a removeable discontinuity

so this has to be factorable so that a factor of (x-1) appears both on tht top and bottom

[swift glacier]

yeah, that was what I was thinking with the -1

[swift glacier]

Is there a specific process to solve it?

[torn swift]

no idea, I don't think I ever did a problem like that

but so far what your attempted answers looked like?

[swift glacier]

Uh, it's around the same place as before

I did try plugging the points in, but I couldn't go anywhere with it

wouldn't d be negative one is (x-1), or is that what you meant

[torn swift]

you worked on it for three hours...?

[swift glacier]

No, I came back after dinner + worked on other problems in the problem set, lol

I've also been trying to search up methods online but they all require finding asymptotes, which this doesn't seem to have

Could it be possible that it's oblique?

[torn swift]

you cannot have an oblique asymptote in this set up

[swift glacier]

because it cancels?

[torn swift]

slant only happens if you have a numerator whos leading term is one less than the denominator

this set up is x^2 / x so that will never happen

[swift glacier]

I thought it was the other way around? Like if the numerator has a higher degree than the denominator

[torn swift]

perhaps that is correct, let me refresh on that

excuse me, you were correct

got that mixed up

[hybrid charm]

a=log(80)=log(8*10)=log8+log10
=log8+1=log(2^3)+1
Now you can do the rest

[vapid torrent]

is this not 144?

[subtle turtle]

@vapid torrent i havent learnt 3d vectors but i think it would be 25+16+9=50

[spring thunder]

@vapid torrent how'd you get 144?

[viscid thistle]

I never learned vectors in precalc

[half axle]

https://gyazo.com/db5076bd31ec16f6e6c845e6aafc7ea7

anyone know how to do these?

[viscid thistle]

I do

[half axle]

Can you explain ❤

[viscid thistle]

So first, it states that π/2 is the only solution from 0<=theta<2π.
Then it states that the period is 2π.

The first one is immediately out because it gives π as a solution if you plug in 2 for k, which doesn't work.
The second assumes a time period of just π, which goes against what the problem told you.
Think you can see which one is right between C and D?

[half axle]

D?

[viscid thistle]

For both of them, you tried 0 and 1, right?

[half axle]

I thought the third one would be considered π

[viscid thistle]

Since the time period is 2π, adding 2π just loops back around.
To make it slightly more intuitive, in this case, 2π is just going around 360 degrees. AKA back to where you started

[half axle]

so wouldnt that be incorrect

[viscid thistle]

Wouldn't what be incorrect?

[half axle]

C

sorry im terrible at this stuff, I don't need this class so im just trying to get through this last week

[viscid thistle]

Your fine

With C, plugging in 0 gives π/2.
Plugging in 1 gives π/2 + 2π. Since 2π is the time period, you're essentially back at π/2, though you went around the circle an extra time (please ask if this makes no sense.)

With D, plugging in 0 gives π/2.
Plugging in 1 gives (π+π)/2, which then is 2π/2, which then is just π.

There is a way to do this without looking at the amswers though.

You first have to look at what solutions will work. Then, the minimum time period in which none would be skipped.
So, for this one, only π/2 works and you want it to count every 2π. Counting every π would count solutions that do not work.

[half axle]

thank you

appreciate it ❤

[viscid thistle]

NP :)

[viscid thistle]

If I'm given this equation, how can I modifiy it for k = j-1? Would the sum just become equal to (n(n+1))/2 - n

I think it would be as I would just distribute the summation across the minus sign, which I think I can do.

[torn swift]

x^3 > x^2 as x goes to infinity

[echo plaza]

that isn't enough tho

[torn swift]

no?

figured it had to be more rigorous

[echo plaza]

if you replace 1/2 with 1/4 it doesn't work for example

just expand it

$\left(x+\frac{1}{2}\right)^3 \equiv x^3+\frac{3x^2}{2}+\frac{3x}{4}+\frac{1}{8}$

[obsidian monolith]

CaptainLightning:

[echo plaza]

the 1/8s are going to cancel allowing you to reduce it to a quadratic

[warm nebula]

Yes and once you have that use power rule and then prove this is true with variation table!🤯

👌

What would the conditions be for solving this problem?

🤔

[solid reef]

Pls help

[torn swift]

Looks almost like these two are inverses, can’t verify that just using mental math though lol

Of course you have to identify where the asymptotes are though lol

[solid reef]

Yea but how

[limber bone]

asymtote of y = b^(x-1) + 2

is lim x-->-inf

y=2

asymtote of log(x-1) +2

is x=1

ig

[limber bone]

fuck that prob killed me https://media.discordapp.net/attachments/363224154469826562/572304176009445376/image0.png

[compact tendon]

This is ez

He says the intersection of the asymptotea

It is at (1,2)=(x,y)

Find the asymptotes of both functions

[clear coral]

Eeeeeee can someone halp meeee

How tf do I graph a rectangle w a polar equation

I need to do it 4 a art project loll

[serene heath]

,w plot abs(y/3+x/2)+abs(y/3-x/2)=3

lmao

[obsidian monolith]

Display results online and refine query

Requested by the one n only#7133

[serene heath]

convert it into polar ig

[clear coral]

Howw

[torn swift]

x=rcos(theta), y=rsin(theta)

[clear coral]

Ayeeee tysm

[viscid thistle]

\

So the answer is how my teacher worded it, is this the same as saying
1 cycle / 3 seconds ?

I think I got what he meant by 1/3 cycle per second

Just not sure

[brave frigate]

how do i do this

[viscid thistle]

<@&286206848099549185>

[dense zealot]

omg

ok

1 cycle per 3 seconds = 1/3 cycles per second

@brave frigate think about it

[viscid thistle]

I think I see how

[dense zealot]

ur swapping 2nd and 3rd row

[viscid thistle]

1/3 is .333/1

[dense zealot]

since rows/column order doesnt matter in a matrice

the rest is too small to read but its probably self explanatory

[brave frigate]

2. multiply original 2nd row by -1/2
3. add -2 of original 2nd row to 3rd row

i dont know what it means by original row

like before swap is original?

[dense zealot]

original meaning the same line

so the original 2nd is the3rd after swap

etc

[brave frigate]

ok ty

[void quiver]

Help with 5

<@&286206848099549185>

[compact tendon]

The product of these is the sum of the argument

The answer is...

4sqrt(2)(1+i)

Simple

[void quiver]

Uh

wtf

[compact tendon]

Lol

What is it?

[void quiver]

NOpe

[compact tendon]

Oh

[void quiver]

Thats wrong

[compact tendon]

Trigonometric form

So

Answer is :

[void quiver]

I have no clue

[compact tendon]

Just take 2 and 4 out

Multiply

8

Times this thing

Which is the sum of the arguments

[void quiver]

so its 69

[compact tendon]

You know the cis function right?

[void quiver]

I have no clue my dude

[compact tendon]

Lol come on

[void quiver]

I jut want the asnwer get me out of this

[compact tendon]

You must know this

How would you do well in examinations then? 😂 😂

[void quiver]

Don't worry about that

ill fuck my teacher ez pz

LMFAOO

[compact tendon]

8 * ( cos(pi/4)+i *sin(pi/4)))

Dammit

Well it is up to you here

But I recommend you to be serious

And leave Naruto behind when shit gets real

jk

[void quiver]

Thanks

Thanks

[serene heath]

[void quiver]

Cna someone help with both problems hehe

@compact tendon

[old marten]

Hey, could someone please peer review my answer for this precalc question?
Prompt:
"Two forces with magnitudes of 25 and 30 pounds act on an object at angles of 10° and 100°, respectively. Find the direction and magnitude of the resultant force. Round to two decimal places in all intermediate steps and in your final answer. "
My answer:
Given: 25lbs at 10degs, 30lbs at 100degs, The angle that is btwn the forces is 100deg-10deg=90deg
So, based on the derived angle we find the magnitude of the resultant force, and the angles between the resultant force and the 2 given forces. Leaving out the x&y axis', the 2 forces and their resultant force can be drawn as a rectangle with magnitude (in lb) of the resultant, R, is the lengths of the hypotenuse of a right triangle with legs measuring 25 and 30: sqrt(30^3+25^2) = sqrt(900+325) = sqrt(1525) = 61sqrt5 = 39.05 (ish)
The angle R makes with the 25lb force is theta as tan(theta)=30/25=1.2 --> theta=50.19deg (ish)
So, the angle between the +ve x-axis and the resultant measures about 10deg+50.19deg=60.19deg (ish) and the magnitude of the resultant is roughly 39.05lbs.

[hexed ermine]

Just turn them both into vector form (i and j components)

Add them

And then convert back

[old marten]

ok thank you very much

[rigid beacon]

What's the difference between a dot product being something like <x1, y,> • <x2, y2> = x1x2 + y1y2 vs something like Work = |force|cos(x)|∆x|

The first way I learned in my Precalc class last year

But other way I learned in physics this year

And both are called a dot product

[serene heath]

they're equal

same thing

[vast karma]

One is between vectors and the other one is between numbers? I dunno what you mean

[fading token]

Same 👏 thing 👏

[fallen cloud]

work = vector, force = vector, cos(x)deltax is scalar

right?

[tawny nacelle]

rite

[ruby otter]

$\lim_{x \to 0} \frac{sinx}{x}$

[obsidian monolith]

Radical Ninja:

[ruby otter]

is 1

[willow bear]

yes, but write $\sin(x)$, not $sinx$

[obsidian monolith]

Ann:

[viscid thistle]

[willow bear]

$lim_x->0 sinx/x$

[obsidian monolith]

Ann:

[willow bear]

bad

[rigid beacon]

Oof

[ruby otter]

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

[obsidian monolith]

Radical Ninja:

[willow bear]

\sin

[ruby otter]

this must also be 1 right??

[viscid thistle]

ye

[ruby otter]

but why is it 0??

ya

[willow bear]

no

NO!

LH does not apply here

the limit is not 1

[ruby otter]

ya its zero but why??

[willow bear]

this is not in the form f(x)/g(x)

[echo plaza]

sin(x)/x never reaches 1

and remember what floor does

[ruby otter]

$= \floor{\lim_{x \to 0} \frac{sinx}{x}}$

[willow bear]

no

[obsidian monolith]

Radical Ninja:

[willow bear]

floor isn't continuous

also, \sin

not sin

[ruby otter]

why it is zero not 1

[willow bear]

because when x is between 0 and pi, 0 < sin(x)/x < 1

[viscid thistle]

[willow bear]

@viscid thistle but FLOOR(sin(x)/x) is not the same as sin(x)/x itself!

[viscid thistle]

maybe u can explain me dis after cuz i dont know what u mean

lemme get outta here for now fren

[willow bear]

$\lim_{x \to 0} \floor{\frac{\sin(x)}{x}} \neq \floor{\lim_{x \to 0} \frac{\sin(x)}{x}}$

[obsidian monolith]

Ann:

[willow bear]

floor is not continuous at 1

[ruby otter]

$but \lim_{x \to 0} f(g(x)) = f( \lim_{x \to 0} g(x)) $

[obsidian monolith]

Radical Ninja:

[willow bear]

no

that's only true if f is continuous at lim[x->0] g(x)

which in this case IS NOT TRUE

[ruby otter]

ohh so we can use only if f is continuous okay

$but RHL = LHL \neq f(0)$

[obsidian monolith]

Radical Ninja:

[willow bear]

what

[ruby otter]

for sinc(x)

[willow bear]

sinc(0) is defined to be 1

that fills in the hole at 0

if however by f(x) you mean floor(sinc(x)) then yes what you said is true

[ruby otter]

but you say sinc(x) is not sinx / x

right??

[willow bear]

when did i say that

[ruby otter]

i said that

[willow bear]

but you say sinc(x) is not sinx / x

you are claiming that i said "sinc(x) is not sin(x)/x"

[ruby otter]

sinc(x) is defined at x=0

but sinx / x is not

[willow bear]

yes, in that sense they're different

that's the only difference, in fact. they are definitionally equal everywhere else.

[ruby otter]

so how do i solve my prob

floor ( sinx / x) is not continuous ??

so its zero ??

[viscid thistle]

@willow bear

@ruby otter I'm learning why im wrong fren but it's supposedly different, something about floor functions

<@&286206848099549185> help dis boi finish

What

Pung

[knotty spear]

whomst pungg

[viscid thistle]

Where's the fire

[knotty spear]

[tawny nacelle]

pung me

i kill u

[viscid thistle]

[tawny nacelle]

i sue u

bish

[knotty spear]

also whats wrong with sin(x)/x

[viscid thistle]

@ruby otter floor/ceil functions are always discreet. What are you asking papi?

[ruby otter]

floor of that is zero hence lim is zero right ??

[knotty spear]

floor(sinc(0)) = 1 but lim x to 0 floor(sinc(x)) = 0

[ruby otter]

but i don't understand why are you ragin at him

[viscid thistle]

f(g(x)) can't be defined when g(x) isn't defined. If sinx/x isn't defined at x=0, then floor(sinx/x) isn't either.

[willow bear]

floor(sinc(x)) = 0 for all x such that x != 0 and |x| < 1

[ruby otter]

where is sinc(x) here??

[knotty spear]

well floor(sinc(x)) =0 for all x!=0 right?

wait no

negative ofc

[ruby otter]

no it can also be -1