precalculus

[quick bluff]

[quick bluff]

Feels like I have made some progress but I'm still getting y' in terms of both y and x, how can I find y'(1) ?

Oh wait

I think I'm close, substituting x=1 cancels out most terms

But I'm still left with dy/dx = (1 - cot y) / cosec^2 y

Nvm, I left a negative sign

OH WAIT

I GOT IT

THANK YOU

[quick bluff]

The 4th question, not sure how to start

[viscid thistle]

Questions, where both the base and the radix vary, are generally interesting in nature. Such questions are usually difficult to solve using the straightforward classic method. There are generally two methods to solve such problems. You can either express the core function in terms of logarithm or use the base e for their representation. If you can appreciate the relation between natural log and e, that would be the cherry on the top.

[viscid thistle]

Q#4 seems quite unsatisfactory and very misleading from the perspective of a high school student. I am not sure either how to approach it as of now. You should read this https://mathoverflow.net/questions/73492/how-misleading-is-it-to-regard-fracdydx-as-a-fraction/73496

[quick bluff]

Will look into it, thanks a lot

[viscid thistle]

The answer(to almost the same question) on this link seems better https://math.stackexchange.com/questions/21199/is-frac-textrmdy-textrmdx-not-a-ratio

[molten garnet]

how do u do 3log5 2 + 2log5 3
what do u do with the front 3 and 2

hey guys can someone help me

[blazing raven]

a Log_b(c) = log_b(c^a)

Still there? I can give you a short lesson “why” based on nice examples.

[molten garnet]

yes

sure

[blazing raven]

So what are the first 4 powers of 3?

3^1, 3^2, etc.

[molten garnet]

3,9,27,81

yes

[blazing raven]

Okay. Log_(3) 27 is … 3 to what power is 27? The base is in the subscript.

[molten garnet]

3^3 = 27

[blazing raven]

One more log_(3) 81 is 3 to what power is 81. The “usually big” number is to the right.

Log_(3) 81 = 4. That’s it.

As a rule adding logs multiplies the numbers inside log_3 (81) + log_3 (9) = log_3 (729)

This means coefficients become powers.
3 log_3 (9) = log_3 (9) + log_3 (9) + log_3 (9) = log_3 (9 x 9 x 9) = log_3 (9^3) = log_3 (729)

I used numbers here that you can verify by hand or even mentally.

[molten garnet]

ok thanks

[blazing raven]

Yep.

[molten garnet]

When u prooft that log 500 = 3 - log2

How is it written as log1000 - log2

[blazing raven]

Yeah.

Log(10) = 1.

Log(10) + log(10) + log(10) = 3

[neat osprey]

quick question, is there any formula or intuitive way to find the local min and local max of a function?

specifically a cubic function?

also is there any formula to find the point of symmetry of an odd function

specifically, again, a cubic function

?

[willow bear]

okay so you chose not to answer my question back in #help-9 so i'll ask it here again

how much calculus do you know?

@neat osprey

[dusk edge]

Set derivative equal to 0

[neat osprey]

oh ok thanks

[dusk edge]

Then solve for x

[neat osprey]

k

[dusk edge]

And then plug that x into the function

You get your coordinates

[neat osprey]

oh ok thanks

[neat osprey]

not too much, just like a handful

[willow bear]

...well i guess SRK managed to say it before me

to find the local extreme points of a function, what you're generally interested in is the zeros of its derivative

[dusk edge]

Because at the extreme points the slope of the tangent lines are equal to 0

[neat osprey]

oh yeah

thanks : )

I also needed some help with this. It was on one of my worksheets for pre calc (I had to cover up some things bc it disclosed some personal info such as my name and such).

[willow bear]

whats troubling you here?

[neat osprey]

the problem itself, idk how to solve it

:/

lol

[willow bear]

have you ever done problems involving objects in freefall before?

[neat osprey]

yeah

the thing is I keep getting a complex solution

when I try to solve this question

[willow bear]

oh, so you have work to show?

[neat osprey]

yeah

[willow bear]

well show it then

[neat osprey]

can I type out my work?

[willow bear]

if you can make it comprehensible, sure.

[neat osprey]

ok

so that's the formula I got

then I tried using quadratic formula

to solve it

but the discrimant was negative

:/

[willow bear]

to solve what?

[neat osprey]

oh solve for when the height is 1500

[willow bear]

this is the height of the rocket at time t.

okay

[neat osprey]

yeah sry for to put the =1500 part

[willow bear]

also, a note on handwriting: your t's look like plus signs

maybe make them not look like that

[neat osprey]

ok

[willow bear]

anyway

one moment

[neat osprey]

ok

[willow bear]

hm

i seem to be getting a negative discriminant too

hold on a minute

ah.

yeah, looks like it's not your fault actually.

[neat osprey]

oh ok

[willow bear]

the discriminant really will be negative here

the rocket simply never reaches a height of 1500 meters.

[neat osprey]

oh ok

thanks : )

[willow bear]

i just calculated its max height and got 1262.4

so yeah

[neat osprey]

ok thanks

[quick bluff]

The third question

I'm not sure how to take the inverse of that function or whether that is even necessary

[willow bear]

it is not necessary

but you do need to know the inverse rule for derivatives, or whatever you want to call it

[quick bluff]

I then tried using the fact that fog is the identity function

[willow bear]

$g'(x) = \frac{1}{f'(g(x))}$

[obsidian monolith]

Ann

[willow bear]

so really all you need is the value of g(1), which can be guessed as 0 here

[willow bear]

this can be derived from f(g(x)) = x

[quick bluff]

Oh I see

[quick bluff]

Sorry if this is a silly question, but how did you guess that g(1) is 0?

To find g(1) don't I need to find f^-1(1)?

[willow bear]

f(0) = 1 therefore g(1) = 0

g is the inverse of f

[quick bluff]

Oh

Oh yeah

Makes sense

Thanks Ann

[quick bluff]

The 4th question, the answers are given as B and C. I'm pretty sure I got B since I got LHD=RHD for all x, but how do I say that f'(x) is a constant?

Oh wait

Can I say since the derivative is of the 0/0 indeterminate form that it gives a finite value?

[uneven bluff]

nah

[rustic linden]

I'm pinged here

[past meadow]

if it was a couple days ago check the pinned

[rustic linden]

Oh okay thanks

How can i understand you name

@past meadow

[blazing raven]

I know no one asked, but here's a hand-wavy why 0/0 is indeterminate and why 4/2 isn't.

Indeterminate forms are ... indeterminate. There's more than one way to approach them and you get different answers.

0/0 can be approached by 0.3/3, 0.03/0.3, 0.003/0.03 ... is 1/10.
0/0 can be approached by -0.3/3, -0.03/0.3, -0.003/0.03 ... is -1/10.
0/0 can be approached by 0.3/0.3, 0.003/0.03, 0.00003/0.003 ... is 0
0/0 can be approached by 0.3/0.3, 0.03/0.003, 0.003/0.00003, 0.0003/0.0000003 ... is infinity.

Also, 4/2 is NOT indeterminate. Why? Try this with 4/2 ... no matter what numbers you choose to get close to the numerator 4 and no matter what other numbers you choose to get close to the denominator 2, the ratio of the "approaching 4 number" and the "approaching 2 number" will always get close to 2.

[rustic linden]

@blazing raven highschool crisis hit?

[blazing raven]

I dunno. What are you babbling about?

[quick bluff]

For the function f(x) = floor( x^2 - 3 ) in the domain [-1/2 , 2], how is it discontinuous at 4 points and not 3?

[willow bear]

the last point of discontinuity is 2

$f(2) = 1$, but $\lim_{x \to 2^-} f(x) = 0$

[obsidian monolith]

Ann

[willow bear]

@quick bluff

[quick bluff]

Ohh I get you

Thanks Ann

Wait, can I write floor(x^2 - 3) as floor(x^2) - 3 ?

Oh yes I can, forgot about that property

[main hollow]

https://cdn.discordapp.com/attachments/644247088552738829/861010162340855808/image0.jpg

[vagrant meteor]

why do we need left and right hand limits can anyone explain pls?

[viscid thistle]

@vagrant meteor are you still wondering?

[vagrant meteor]

got an answer

[last parrot]

Anyone know a good resource for going over precalc

I have already taken the class and would like a brief recap of key concepts

[night moat]

Khan Academy

[haughty lance]

OpenStax offers a free book that I think is pretty comprehensive (decent practice problems, considering it's free): https://openstax.org/details/books/precalculus

OpenStax is a part of Rice U and is a 501c3

[last tinsel]

https://flexbooks.ck12.org/cbook/ck-12-precalculus-concepts-2.0/ CK-12 has pretty good stuff. The book has precal and uni concepts

[tardy ridge]

https://courseware.cemc.uwaterloo.ca/

don't use khan academy.

[dusk edge]

why not

?

khan academy is fine for a refresher

[last tinsel]

khan academy is good

at least as refresher

idt it's the best way to only use that as a reliable resource

[glossy cairn]

$n^{\ln n} = g(n) \iff e^{n^{\ln n}} = e^{g(n)} \iff e^{\ln(n) \cdot n} = n^n = e^{g(n)} \iff \ln(n) \cdot n = g(n)$

[obsidian monolith]

jabra

[glossy cairn]

where is the mistake?

[tardy ridge]

X^y is not xy

[obsidian monolith]

jabra
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[glossy cairn]

yeah hmm

[tardy ridge]

There are better resources

(x^y)^z =x^(yz)

But x^(y^z) is not x^(yz)

[glossy cairn]

yeah seems like my brain needs some sleep

[quick bluff]

The number of points where the tangent does not exist for y = sgn(x^2 - 1) is given as 0, how?

[void thorn]

yeah that seems wrong since x=1 and -1 there's no tangent

[quick bluff]

[nocturne urchin]

can anyone tell me whether i'm right or wrong, pls, I don't know the ans

[viscid thistle]

,w lim 2x(sqrt(x^2+1)-x) as x -> inf

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by kitchen1112#3264

[viscid thistle]

Yeah correct

[nocturne urchin]

thanks man

[nocturne urchin]

so that's hoe to type this question. Thanks again man

[woeful marsh]

lim 2x(sqrt(x^2+1)-x) as x -> inf

,help

[obsidian monolith]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[woeful marsh]

,help cmd

[obsidian monolith]

You really shouldn't take it literally :upside_down:. Please type ,help ping, for example!
The full command list may be found using ,list.

[woeful marsh]

,list

,text

,5x

,text 5x

, 5x

,text a

$n^{\ln n} = g(n) \iff e^{n^{\ln n}} = e^{g(n)} \iff e^{\ln(n) \cdot n} = n^n = e^{g(n)} \iff \ln(n) \cdot n = g(n)$

[sick steppe]

dont spam

[dire lintel]

can someone please solve this question for me

Differentiate:sqrt((1-tan x)/1+tan x))

[trim hemlock]

can you show your work?

[willow bear]

we don't do that here

[dire lintel]

Oh sorry..I forgot to send the working

my badd

Can smne help me frm here

My teacher differentiated 1-tan x/1+tan x

Every1 will have different answers when it comes to differentiating right?

[quick bluff]

No

(1-tanx)/(1+tanx) is nothing but tan(pi/4 - x) I'm pretty sure

This should make your problem much more easier

[wary mulch]

i need help in doing this sum
pls mention me if u r helping

[willow bear]

@wary mulch have you made any progress so far?

[wary mulch]

yea

[willow bear]

care to share?

[wary mulch]

yea sure

ignore the 1st 2 steps @willow bear

@willow bear is wat i did correct

[willow bear]

this whole cancel-fest is a little hard to follow

[wary mulch]

ohh ok

wait il do the sum without skipping any steps ok?

[wary mulch]

@willow bear is this sol easy to follow

[willow bear]

this looks a little sus now

easier to follow, and easier to see the reason why it's sus

you're trying to do selective substitution

which you should not

[wary mulch]

ohhh then wat shud i do

[willow bear]

one moment

i hope you don't mind if i write x instead of theta

[wary mulch]

ok i wont mind

[willow bear]

this is what i would have done

yes i know +∞ isn't one of the options listed

basically break your thing up into factors which approach finite, nonzero values

[wary mulch]

thanks

[dire lintel]

How

Can You please show me how you got this?

[quick bluff]

Do you know the sum and difference identities?

Guess I forgot the formula, sorry about that

Wait

tan(a-b) = (tan a - tan b)/(1+tan a. tan b) if I'm not mistaken

tan pi/4 is 1

[quick bluff]

By replacing the 1s in your question by tan(pi/4) you can see it's of the form tan(a-b)

Where a is pi/4 and b is x

So your question is reduced to sqrt(tan(pi/4 - x))

[ripe schooner]

Anyone got any ideas?

[molten island]

It’s C

Because it happens on a loop

Or periodically

[ripe schooner]

Ty

[wary mulch]

i need help in doin this sum

[molten island]

Alright, lemme send a pic

[wary mulch]

ok

[molten island]

I apologize for the messiness

So you assign an arbitrary variable to equal the limit so you can do stuff to it, then backtrack and figure out what that variable is

Does this make sense? @wary mulch

[wary mulch]

can u xplain wat did u do in this step

[molten island]

Yea basically I used L’Hôpital’s Rule

Do you know what that is?

[wary mulch]

yea

[molten island]

That’s what I did

Then went from there

[inland flower]

An antenna is situated on top of a 750 ft building. From a point in front of the building, you
measure the angle of elevation to the top of the building to be 53° and the angle of elevation to the
top of the antenna to be 56°. How tall is the antenna?

anyone know?

[molten island]

Can you send a pic of the problem?

[willow bear]

is the antenna attached to the edge of the building?

otherwise there... might not exactly be enough info to determine its height

like, the height of the antenna is going to be different depending on how far it is from the edge of the roof

[inland flower]

antenna is on top of the building

[willow bear]

"on top of the building" is vague, unless we are modeling the building as an upright, 750 ft tall, negligibly thin stick

i have a hunch that the intended diagram is this

E = the location from which you're observing the antenna (E for "eye")
B = base of the building
R = roof of the building
A = tip of the antenna

if i had access to some paper right now i could draw a more realistic diagram

[inland flower]

here we go

[willow bear]

though, again, the wording of the problem leaves room for doubt.

no, the building itself is 750 feet, not the building plus the antenna

[inland flower]

oh

wow

[willow bear]

and that 56 degree label is also misplaced

you do realize 53+56 would add up to more than a right angle, do you not?

[inland flower]

i had a brain fart for a sec

[wraith pike]

lol

[uneven bluff]

lol

[hoary jungle]

im dumb pls help

[sick steppe]

it's a common base problem

[hoary jungle]

ok

[uneven bluff]

,tex $x = \pm \sqrt{5}$

[obsidian monolith]

monkers

[sick steppe]

No, and dont just give answers

[uneven bluff]

Actually tho

[sick steppe]

If you dont plan to help dont comment

[uneven bluff]

Hey, leave me alone ok.

[tender pawn]

^

[hoary jungle]

^

[sick steppe]

I say that to everyone who just gives answers

[hoary jungle]

yeesh

anyways

thanks

[sick steppe]

as the answer itself means nothing, especially when it's wrong/incomplete

[uneven bluff]

@hoary jungle You can get that exponent in a better form by taking the natural log of each side. Properties of logs says that you can multiply by that exponent when its in a log.

[sick steppe]

You also dont need logs since it's common base

[obsidian monolith]

monkers

[hoary jungle]

he still helped more than you tho xd

but yea

you probably dont need logs

[sick steppe]

$3^{x^3}=243^x \ 3^{x^3}=(3^5)^x \ x^3=5x$

[obsidian monolith]

Mosh

[hoary jungle]

yes thank you

0, sqr rt of 5 and -5

[sick steppe]

no

x isnt sqrt(-5)

[hoary jungle]

-sqrt

sorry

[uneven bluff]

+/- sqrt(5)

[sick steppe]

yes, $x\in{-\sqrt{5},0,\sqrt{5}}$

[obsidian monolith]

Mosh

[uneven bluff]

Actually, @mosh, I'm not sure how you found 0 to be a solution.

[sick steppe]

x is a factor of x^3-5x

[hoary jungle]

what she said

[sick steppe]

he

[hoary jungle]

he

hmm

im just gonna use "they" from now on

[sick steppe]

yeah, use gender neutrals if you dont know someone's gender..

[hoary jungle]

sure

[silk escarp]

Guys can you help me find out where am I going wrong

[uneven bluff]

@silk escarp yeah, t = ln(2)/4ln(1+0.45/4)

so 5th and 6th lines are baaaad 🐑

[silk escarp]

Thanks monkers

[viscid thistle]

Sadly for some weird reason they want us to sketch functions from R² to R, and i suck at drawing, any advice?

[sick steppe]

well it's a surface.. so contour plots?

[modest basin]

3

wait no

[viscid thistle]

contour plots are tedious but not hard, however trying to draw an actual 3d thing always gets really ugly even tho i have picture of it in head

[hidden breach]

oh-

hello typing

[upper token]

We're on horizontal shifts, and this question came up and my teacher doesn't know how to explain it, can someone explain this question?

hi @hidden breach

original question

[hoary jungle]

nope, but nice try 🙂

[upper token]

First, set the bases equal to each other. Then, remove the bases by logging both sides of that base.

[hoary jungle]

ik i figured it out like a day ago but thx

sqrt and -sqrt of 5 and 0

[upper token]

np

[neat osprey]

can someone explain what logarithmic re-expression is?

[upper token]

b^a = c

logb(c) = a @neat osprey

it finds the exponent

[neat osprey]

wait

that's it?

[upper token]

yes

very simple

[neat osprey]

what are regression models related by logarithmic re-expression then?

[upper token]

can you send a screenshot

[neat osprey]

it's not a question

I just read it in my textbook

was kinda confused on it

it says linear regression natural logarithm regression exponential regression and power regression

underneath it

[upper token]

uhh

idk

im just taking precalc rn lol

[neat osprey]

oh ok

I am too

lol

[modest basin]

is there like a method on finding the log?

[upper token]

wdym

the base of the log or the value in the parenthesis

[modest basin]

i mean, both

like if the answer is like some fraction, and if the base is 2, how ould work out without using a calculator?

[haughty lance]

you should review the properties of log

doing practice exercises with those should give you all the reasonable log finding tools you may need in the short term

[jolly raven]

Well the argument would have to be a root

Do you mean the argument would be in root form or decimal form

[jolly marten]

can someone help me

^^^^

[toxic flame]

do you know the unit circle?

[jolly marten]

yes

[fleet yew]

polar?

Or cartesian?

[jolly marten]

polar

[fleet yew]

What is meant by simplest standard

[toxic flame]

do you know what cos(7pi/6) is?

[fleet yew]

If its polar then you dont need to evaluate that

[jolly marten]

-rt3/2

[jolly marten]

wdym

[toxic flame]

ok. find sin(7pi/6) and simplify

then ur done

[fleet yew]

What do you mean by simplest standard form?

[jolly marten]

-1/2

[toxic flame]

substitute those values in and simplify

[fleet yew]

as in a+bi? Or re^(iθ)

[jolly marten]

i think a+bi

[toxic flame]

you should end up with ||-3/2 - i sqrt(3)/2||

[fleet yew]

Ok then keep listening to blue name

[jolly marten]

one sec

wait am i suppose to add -3rt/2 and -1/2?

or multiple

multiply

[toxic flame]

you have
cos(7pi/6) = -sqrt(3)/2
and
sin(7pi/6) = -1/2

[jolly marten]

yeah

[toxic flame]

so just plug those in to the original expression

[jolly marten]

also would the rt3 on the outside cancel out with the rt on the inside of the parenthesis

so it's -2+(-1/2)?

[toxic flame]

yea after u distribute the rt(3)

[jolly marten]

to both numbers?

[viscid thistle]

$\sqrt{3}(cos(\frac{7\pi}{6})+i*sin(\frac{7\pi}{6}))$

[obsidian monolith]

Cicopath

[toxic flame]

not sure what u mean

[viscid thistle]

$\sqrt{3}(-\frac{\sqrt{3}}{2}+i*-\frac{1}{2})$

[jolly marten]

answer is -2+i -rt3/2?

[toxic flame]

no

[viscid thistle]

You would distribute $\sqrt{3}$ to both fractions

[obsidian monolith]

Cicopath

[jolly marten]

yes

and it cos7pi/6 becomes -2

[toxic flame]

should be -root(3)/2

[viscid thistle]

Ah, thank you

[toxic flame]

-3/2. not 2

[obsidian monolith]

Cicopath

[jolly marten]

why

[toxic flame]

because root(3)•root(3)=3

[jolly marten]

oh i thought it cancelled out completely

ok

[viscid thistle]

$\sqrt{3}*\sqrt{3}=\sqrt{3}^2=3$

[obsidian monolith]

Cicopath

[toxic flame]

even if it did, that would become 1/2, not 2

[jolly marten]

yea my fault

[toxic flame]

no worries

[jolly marten]

so answer is -3/2 + i -rt3/2?

[toxic flame]

yup

[jolly marten]

wow thank u!

can i ask for more help lol

on smth else

[viscid thistle]

And for future notice, you are allowed to post these questions in one of the #question channels

[viscid thistle]

Of course, no need to ask!

[jolly marten]

oh ok

so go to a questions channels?

channel

[viscid thistle]

Usually that is where people will notice your questions, so yeah

[toxic flame]

just ask whatever whenever, respectfully tho.

[jolly marten]

ok

#help-6

[viscid thistle]

Before posting in a channel, make sure it's free ofc. If the last message was posted 30 mins ago, it's free

[vagrant meteor]

hello

can anyone help me with the domains and range finding of different functions i ask?

like after that i can do myself like i can do trig functions domain n range if u first explain me i would be grateful

i wanna learn it for all functions

if anyone willing to help please go on i have very short time

@toxic flame can u help me please

[quick bluff]

The 16th question

Is it correct if I say that the function will have the greatest slope when it's second derivative equals 0?

Oh no

Not second derivative

Third derivative?

[sick steppe]

you want to maximize the derivative, so 2nd

[vagrant meteor]

ash

[quick bluff]

Doesn't equating second derivative give points of maxima and minima of the FUNCTION and not the derivative?

[vagrant meteor]

r u pakistani ? studying for exams? hmm

[sick steppe]

derivative is a function

...

turning point may occur when f'(x)=0

[quick bluff]

Oh wait

[sick steppe]

im not interested in maximizing the function value, i want the maximum slope, so I want to maximize the derivative

[quick bluff]

Second derivative is for checking which point is the maxima or minima

[quick bluff]

Oh yeah makes sense, I confused the maxima and minima with the second derivative

Thanks

[vagrant meteor]

can anyone help me

[quick bluff]

But the problem is I'm getting x as +- sqrt(3) but the answer key gives it as (0,0)

[sick steppe]

,w 2nd derivative of x/(1+x^2)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by moshill1#9771

[quick bluff]

Wait did I make a mistake in taking the derivative?

[sick steppe]

looks like it

[quick bluff]

Oh wait

I cancelled out an x

[sick steppe]

,w solve -2x(1+x^2)^2-4x(1-x^2)(1+x^2)=0

[quick bluff]

x=0 is also a possible solution

[obsidian monolith]

Results provided by WolframAlpha

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Requested by moshill1#9771

[quick bluff]

Yeah yeah I got x = 0 too

[sick steppe]

yeah so only 0 and +-sqrt(3)

[quick bluff]

So I have to take the third derivative?

And use that?

[sick steppe]

now you just figure out which is the max

[quick bluff]

[sick steppe]

so plug it into the derivative

[quick bluff]

Wait wait

Which derivative?

The first derivative?

Don't I have to plug it into the third derivative?

[sick steppe]

yes, the function you maximized

[quick bluff]

The function which I maximised was the first derivative, so to figure out which is the max or min don't I have to substitute the value I got from the second derivative into the third derivative and see which one gives a positive/negative value?

Or am I confusing something again?

[sick steppe]

I mean you can, but have fun differentiating again

[quick bluff]

True, but I don't understand how plugging it into the first derivative gives us the answer

[sick steppe]

however if you got values say x=1,2,3 and got f(1)=0, f(2)=-3 and f(3)=3, clearly x=3 is the max

[quick bluff]

Oh wait

So I got the possible x coordinates for max or min slope

Now I just verify which one gives the highest one with the first derivative right?

Isn't that what you just said?

[sick steppe]

yes, you just need the function value which is the biggest

[quick bluff]

Thanks again dude

Yup I got it

thanks for the third time

[dry pier]

Can someone help with 7b I’m not sure how to do it and where I went wrong

[haughty lance]

even in the start it's a bit iffy

i recommend doing the multiplication in parenthesis first and then multiplying by -1 to be clean

also the subtraction operation disappeared...

[winged kettle]

i dont understand how to do irrational roots from polynomials could someone help

[dry pier]

I tried doing it again but I’m not sure what to do and I’m pretty sure that I messed up somewhere again

[viscid thistle]

Paying $20 for someone to help me with an Online Pre calc 11 test. 20 multi choice, 5 written. 10 upfront 10 after. Please msg me!!!

It's all radical equations and absolute value stuff

[sick steppe]

Again, no one is going to help you cheat.

[uneven bluff]

smh

Its over for math. I think we may be witnessing the last generations of solid mathematicians.

[drowsy karma]

what is this even saying

[analog bridge]

When f(x) = y, then inverse of that function means that f(y) = x

So in this case you have x = a*y as your inverse function

And you need to find values of a that your normal function and inverse of it are the same for every value

[cunning lantern]

Can someone please help me on this question

[sick steppe]

what have you tried?

[cunning lantern]

Idek how to start

[sick steppe]

do you know how to do transformations?

[cunning lantern]

yes

[sick steppe]

ok, so given y=f(x), what do you do to reflect it in the y axis?

[cunning lantern]

y=f(-x)

[sick steppe]

right, and x axis?

[cunning lantern]

y= -f(x)

[sick steppe]

so together?

[cunning lantern]

y=-f(-x)

[sick steppe]

$y=-f(-x)$ yes

[obsidian monolith]

Mosh

[sick steppe]

so do that for this f(x)

[cunning lantern]

ok

is that all?

[sick steppe]

yes

[cunning lantern]

oh okay thanks

[quick bluff]

How does the function f(x) = x^(1/3) and it's inverse meet at 3 points?

Doesn't it meet only at 2 points?

[willow bear]

$x^{1/3} = x$ has solutions $x = -1, 0, 1$

[obsidian monolith]

Ann

[quick bluff]

Oh yes

Wait why does the graph of x^(1/3) only have positive numbers with 0 as it's domain?

[willow bear]

are you graphing it with WA?

[quick bluff]

Ye

[willow bear]

WA is interpreting $x^{1/3}$ as the principal cube root rather than the real cube root, and as such it gives complex values for $x < 0$

[obsidian monolith]

Ann

[quick bluff]

oh ok

[willow bear]

specifically, $x^{1/3} = |x| e^{2\pi i/3}$ for $x < 0$ in WA's notation

[obsidian monolith]

Ann

[quick bluff]

Woah

Okok thanks a lot Ann

[nocturne urchin]

can anyone give me a full solve of this math

[sick steppe]

what have you tried?

also why are there 2 little circles?

[nocturne urchin]

that's the problem those are degrees

[sick steppe]

ok well you dont do calc in degrees

nor do you write the degree sign for a variable

but anyway, what have you tried was the more pressing question...

[nocturne urchin]

o so i'll convert the degrees into radian

[sick steppe]

$f(x)=2x\cos(3x) \ f'(x)=?$

[obsidian monolith]

Mosh

[nocturne urchin]

i tried to do the differentiate with degrees

[sick steppe]

what did you get for the derivative?

[nocturne urchin]

nothing cuz i was unable to differentiate the degrees

[sick steppe]

so what rule did you try and use?

[nocturne urchin]

tried to create a impllict function

[sick steppe]

????

[nocturne urchin]

and then differentiate it

[sick steppe]

why would you make the question harder

[nocturne urchin]

i mean a teacher told me i can solve any differentiation with implict function

[sick steppe]

you can differentiate an implicit function, yes

but this is clearly explicit

[nocturne urchin]

so i tried to do this in this way

[sick steppe]

what rule do you use when it's a product of 2 functions?

[nocturne urchin]

i can say Y = the qus can't I

[nocturne urchin]

sorry but can't understand ur qus

[sick steppe]

what differentiation rule do you use when you differentiate a function which is a product of 2 other functions?

[nocturne urchin]

i do now understand ur qus but actually i don't know the name of the rule

[sick steppe]

product rule......

[nocturne urchin]

is the ans "2(cos3x-3xsin3x)"

@sick steppe

[sick steppe]

$f'(x)=2\cos(3x)+(2x)(3)(-\sin(3x))$

[obsidian monolith]

Mosh

[nocturne urchin]

can't i simplify it by taking a common "2"

[sick steppe]

you can

[nocturne urchin]

@sick steppe i really don't get it what did we do to the degrees?

[sick steppe]

you dont do calculus in degrees

and you dont write a degree symbol on a variable

[uncut mulch]

convert it to radians

[sick steppe]

both of which I said already

[uncut mulch]

if the values are in degrees, you'd need the conversion, you can't just drop it

[nocturne urchin]

so why the qus even exist with degrees

[nocturne urchin]

how can i convert

[uncut mulch]

do you know your trig?

[nocturne urchin]

very bad at trig

[uncut mulch]

to convert from degrees to radians, multiply by pi/180°

[nocturne urchin]

o thanks

so the actual actual qus will be "pi/90 cospi/60"

[uncut mulch]

qus?

[uncut mulch]

wdym by

so the actual actual qus will be "pi/90 cospi/60"

[nocturne urchin]

this qus will convert into "pi/90 cospi/60"

[uncut mulch]

no

whered the x go

[nocturne urchin]

i don't know

[uncut mulch]

the x remains

after the conversion

[nocturne urchin]

"xpi/90 cosxpi/60"

[uncut mulch]

from degrees to radians

$f(x) = \frac{\pi x}{90} \cdot \cos{\frac{\pi x}{60}}$

[obsidian monolith]

ℝamonov

[nocturne urchin]

i do understand now thanks for ur time

@uncut mulch

[quick bluff]

Is it correct if I say that if a function f(x) is increasing in the interval (a,b) then the greatest value of f(x) tends to b and the least value tends to a?

[viscid thistle]

No

[quick bluff]

why not?

OH SORRY I typed it incorrectly

I meant to say does the greatest value tend to f(b) and not b

And least value tend to f(a) and not a

[quick bluff]

The second question

One sec I'm sending what I did so far

How can I show that cos(sin x) . cos x is >= 0 in that interval?

The answers are given are options A and B

[shut timber]

what is h(1) refering to?

the first index? the y value for the x value of 1?

[echo wagon]

The function value of 1 w/r/t h

Yep

[shut timber]

in h, the output would be 2

[echo wagon]

Yep

[shut timber]

ohh, okay thnx

[echo wagon]

Oops, edited original message

Np

[drowsy karma]

It's II and III no?

[ivory loom]

is I and III not the exact same thing? xD
and yes, i'm pretty sure your answer is correct

[drowsy karma]

well 4 is on top

[ivory loom]

yeah but the terms 'compression' and 'expansion', unless its just worded weirdly

like if u were to compress something by a factor of 1/2, its equivalent to expanding it by a factor of 2 (kinda like if u divide a term by 1/2, its the same as multiplying that term by 2)

[drowsy karma]

Ye but it's vertical

So if it's under 1 its compression

Wb this?

Just an inversion question

[ivory loom]

yeah that's correct

[drowsy karma]

dope thanks

ohhhh i see what you mean for the compressing and expansion

yeah it is worded weirdly

[viscid thistle]

help

[uncut mulch]

what have you tried?

[gritty crest]

I cannot believe that I just learned that a log can be used to find the dimensions of a fractal. My I’m dumb.

[rain current]

What app is tthis

[drowsy karma]

Not sure it was on the assignment

[undone dust]

Do I need to know all of these factors formulas for precal? I only learned like 4 of them last year.

[willow bear]

you should not be looking at these like pieces of info to memorize

they all come in very handy in many circumstances, but you must know where they all come from

[viscid thistle]

If I've got 3 equations: x,y,z ; What do I do in order to get y(x,z) or f(x,y,z)?

[mild cloud]

Please be more precise

[viscid thistle]

I want y ( one of the equations ) to be a function of the two other functions

if I have y = at
and x = bt
then t= x/b
so y(x) = a(x/b)

So the same but for 3 functions instead of 2

[willow bear]

this is kind of vague??

but also if i understand you correctly it doesn't matter what order you do it in

[viscid thistle]

Well y(x) = (a/b)x , so you can view a/b as a constant

and if we had another function z = ct, then t=z/c and we get y(z) = (a/c)z
So the question is, how do I then combine the two to get a function of both x and z?

it's supposed to be a 3 dimensional eq.

[willow bear]

are you trying to describe a line in three-dimensional space?

[viscid thistle]

yes

[willow bear]

you will not do that with one equation

you will need two

[viscid thistle]

I actually want to describe motion in 3D, so I've got 3 equations for the x,y,z directions

[willow bear]

the plot thickens

okay

[viscid thistle]

its quite easy to do when you only consider y and x

[willow bear]

so you have your equations of motion

how is that not good enough?

what do you want out of your description?

an implicit equation describing the curve that your particle moves along?

[viscid thistle]

yes

[willow bear]

yeah as i said you can't do that with just one equation

one equation can only give you a surface (in most cases)

so you will need two

also for future reference you really really really should NOT overabstract your questions like this. if you don't know what details matter, you WILL omit the crucial ones and you WILL be misunderstood

http://xyproblem.info/ and all that

[viscid thistle]

I've got the two equations, but I'm still not quite sure how to relate them to get y(x,z)

and Vcos\theta, Vsin\theta, Vsinp terms all remain constant

[willow bear]

what do you want your hypothetical y(x,z) to output when (x,z) is a point not on the shadow of the ball's trajectory?

[viscid thistle]

it should model the path taken by the projectile

in theory it should work the same as when I consider z = 0, so a normal 2D parabola which maps the yx positions over the projectiles path

so each point in y gives a corresponding point in x ( y,x being the ball's position coordinates in space )

so y(x,z) should then give me the coordinates for both x and z , which would also be parabolic, just into the z plane as well

This for example is also parabolic because all terms remain constant, and this models both the x and y motion of the ball.

[mild cloud]

Transform your coordinate system so that the parabol is only a 2D movement, then transform it back

https://pages.mtu.edu/~shene/COURSES/cs3621/NOTES/geometry/geo-tran.html

[viscid thistle]

Thank you

It makes the problem much easier!

[mild cloud]

np, did you know how to get the inverse of a matrix?

[viscid thistle]

no not quite, but I'm guessing I have to set up a matrix for the variables in my problem?

[mild cloud]

Let me explain it so:
You have a 3D space and in this you will make a parabolic movement

[viscid thistle]

yes

[mild cloud]

And I think we are d'accord that in 3D this is shit to make

[viscid thistle]

definitely

[mild cloud]

But, when the start would be on (0, 0, 0) it will become a little bit easier

and it would be very nice, when the one of the movement directions would be along one of the axis

[viscid thistle]

the initial conditions start at (0, 0, 0)

so now all I need is to transform the function so that Z = 0?

[mild cloud]

when you mean it so that you rotate and/or transpose it so, that every z value goes to zero, then yeah, then it should be 2D

[drowsy karma]

In graphs, do you do the stretch first or the horizontal,vertical shift?

pretty sure its the shift no?

[viscid thistle]

Graphs will look as if they’ve been stretched, then shifted

Because the constants always stay the same

They get added last

How do you go about solving this?

Oh wait it’s literally just chain rule

yeah, precisely.

Wait or is it product rule

G is a differentiable function

But we don’t know it

So product rule doesn’t work

why would it be the product rule? there is no product here, you essentially have a case of g(f(x)), whose derivative by the chain rule is g'(f(x))f'(x). try seeing which option would correspond to it.

[viscid thistle]

youre right

but how do you know dy/du

if y = g(u)

[willow bear]

g'(u)

[viscid thistle]

yes

yes it is

[opaque locust]

can someone help me with verifying an identity?

[viscid thistle]

would that be correct?

(-infinity,infinity)

or would it be (-infinity, -3)U(-3,infinity)

considering its right on the vertex origin

[narrow lichen]

Does anyone know how to solve this?

I got a in terms of b,n, and m, but I can't get rid of b

[echo wagon]

Similarly to how you wrote a = b^(m/n), you could say b = a^(n/m)

[narrow lichen]

hmm, yes. But I still don't know how this helps me in any way, maybe I'm missing something

[echo wagon]

You can replace b with a^(n/m)

Then you have an equation with only a, m, and n

So you can isolate a

In that equation

[narrow lichen]

I see, I see

thank you

[echo wagon]

Np

[narrow lichen]

Did I do something wrong?

[echo wagon]

I can't tell, you have a lot of unnecessary stuff. I didn't mean plug it into that equation though. Plug it into the a = one

I'm not sure where the first equation that you plugged it into comes from

Just use the a = 2bm/(3n) one

[obsidian monolith]

Xi64

[viscid thistle]

@opaque locust if you have further questions, ping me

(pardon me for the ugly latex)

[narrow lichen]

yep, it worked

[narrow lichen]

I didn't know what to do, so there is a lot of stuff

thank you

how did you know where to plug it in tho?

[echo wagon]

Because I could see that I could use exponent rules to simplify in that one

And its the simplest equation

[brazen birch]

hello

[viscid thistle]

What is synthetic division

[sick steppe]

polynomial division but slightly quicker, but only works for linear divisors

[surreal river]

Can anyone help me figure this question out?

[sand dune]

draw the ellipse and you'll see that it's true

[junior thunder]

can somebody help me with this question

[mild cloud]

Which part of it

[junior thunder]

a,b and c

[mild cloud]

a= how many time was passed since initial state?

[junior thunder]

what?

[fleet yew]

For a, find W(0)

For b, set W(t) = 0.3 * W(0)

For c, find 1 - (W(1500))/(W(0))

[junior thunder]

oh ok thank u

[shut timber]

how would u do this

[willow bear]

one step at a time

remember how to do shifts and stretches along each axis

write down the equation you get at every step

(though it isn't entirely clear what's meant by "invert")

[shut timber]

like just f(x)^-1

first i start off with f(x) = 1/x, shift right, so f(x-2) , so 1/x-2

horizontal stretch, so g(1/2x) , -> 1/1/2x - 2

then i invert it to get

2/x + 4

then i multiply everything by 2 for vertical stretch, 4/2x + 8, then add 2 to shift left

4/2x + 10

inverted its x/2 + 5, and then i have to multiply everything by by 1/2

but the answer is not on the included options

[willow bear]

parentheses! you meant 1/(x-2), not (1/x) - 2.

[willow bear]

are you sure you don't mean f^-1(x)

[shut timber]

oh yea

[viscid thistle]

someone help me pls

[blazing raven]

Sec(x) … ?

[blazing raven]

@viscid thistle still need this?

[vestal vessel]

yall

i have multiple choice test about integrals

any way to eliminate choices?

[junior ridge]

is it about Riemann sums?

[viscid thistle]

@blazing raven yes please

[junior ridge]

@viscid thistle Try to apply the AM-GM inequality to the formula for the surface area

As a further hint

you might need to rewrite S in a different way

Motivation for this will come from the cube root in the formula from the problem

[drowsy karma]

HOW do you do these again

[blissful ridge]

The way you did it the first time

Joking aside
Let the number of shots taken in the second be x and you know that all the shots are hit

Form the equation from the given facts and solve for x

[drowsy karma]

KEKW

Yeah I figured it out thankyou

@blissful ridgeCould you help me tackle an equation on Asymptotes?

how do you find the equation for that

I have y = 1/x + 2 so far

so I got the asymptotes down

[uncut mulch]

how are you getting y = 1/x + 2?

[drowsy karma]

Because +2 for the horizontal since the variable values are the same. and x has to equal 0 so it's just x?

[uncut mulch]

whut?

[drowsy karma]

for the asymptotes

[uncut mulch]

wdym by

and x has to equal 0 so it's just x?
and why are you adding 2?

[drowsy karma]

wait

could I just do

y = 2x/x

[uncut mulch]

that would be y=2 (with a hole at x=0)

[drowsy karma]

wouldn't that be right?

[uncut mulch]

no

[drowsy karma]

heckle

could you nudge me a little bit?

[uncut mulch]

consider starting with:
$$y = \frac kx + c$$

[obsidian monolith]

ℝamonov

[drowsy karma]

right

[uncut mulch]

(ignore the hole for now, account for that after determing the values of k and c)

[drowsy karma]

well the vertical asymptote is 0

so wouldn't x just be by itself?

[uncut mulch]

wdym be by itself?

[drowsy karma]

it would just be x

[uncut mulch]

not sure what you mean

x is x

[drowsy karma]

like its not x-1 or x+1 for example

to determine the vertical asymptot

[uncut mulch]

x is alone in the denominator, like I implied above

[drowsy karma]

yeah

could we just break down the final equation?

[uncut mulch]

wdym by break down?

[drowsy karma]

I think I'd understand better

Like give the equation of the asymptote and break down the different parts

[uncut mulch]

be more specific

[drowsy karma]

Like give the final answer

and break it down

break down each part

[uncut mulch]

do you see why i started with that equation?

[drowsy karma]

yes

[uncut mulch]

ok.

from that you should be able to identify the value of c pretty much directly

and then you can sub in a point on the curve to determine the value of k

[drowsy karma]

which would be

[uncut mulch]

for an equation in the form:
$$y = \frac kx + c$$
$c$ will be your horizontal asymptote

[obsidian monolith]

ℝamonov

[drowsy karma]

right

[uncut mulch]

can you identify the value of c from the graph?

[drowsy karma]

2?

[uncut mulch]

why 2?

[drowsy karma]

idk thats why i wanted to break it down

[uncut mulch]

the horizontal asymptote is represented by that purple dashed line

(and unless specified otherwise, each grid interval is implied to have a length of 1 unit)

what's the equation of that purple dashed line?

[drowsy karma]

x-2

[uncut mulch]

no

[drowsy karma]

idk man

[uncut mulch]

x-2 is not an equation

don't overthink this

[drowsy karma]

i honestly dont know

im pretty tired and i dont know

[uncut mulch]

what does the equation of a horizontal line look like?

(give me an example of one)

[drowsy karma]

telling me to give you the equation of the line is the same as me giving the example of one

i wouldn't know

[uncut mulch]

this is a major issue then

[drowsy karma]

i know horizontal is determined by the numerator and denominator

[uncut mulch]

ignore that

what i'm asking for atm is a lot simple than that

eg if I asked you to draw the line y=4 would you be able to do it?

[drowsy karma]

Yeah

[uncut mulch]

ok

what would that look like?

[drowsy karma]

c would just be 4 no?

[uncut mulch]

c would just be 4 no?
what does c represent there?

[drowsy karma]

the stretch

[uncut mulch]

no

[drowsy karma]

mf uh

[uncut mulch]

don't overthink this

draw y=4 for me

[drowsy karma]

this method isn't really workin for me

telling me to do something I don't know how to do won't make me know

how to do it

[uncut mulch]

do you know what a horizontal line is?

[drowsy karma]

y = 4x/x

[uncut mulch]

no

ignore that

what i'm asking for atm is a lot simple than that

[drowsy karma]

yes i know what a horizontal line is

well make it more complicated

[uncut mulch]

are you able to draw a horzintal line?

[drowsy karma]

i dont know what answer you're looking for tbh

[uncut mulch]

these are currently yes/no questions

[drowsy karma]

yes

[uncut mulch]

to identify where the issue is

[drowsy karma]

i can physically draw one yes

[uncut mulch]

if i were to give you the equation of a horizontal] line e.g would you be able to draw it?

[drowsy karma]

yes

[uncut mulch]

applying the same principles, if you were given the graph of a horizntal line,
would you be able to identify its equation?

[drowsy karma]

yes

[uncut mulch]

ok good

now to get to the main point

what is the equation of the purple dashed line?

[drowsy karma]

y = 1

[uncut mulch]

yes

finally

that was all i needed

[drowsy karma]

bruh thats like

ok

[uncut mulch]

and that's your value of c (the location of your horizontal asymptote)

don't overthink it

[drowsy karma]

ok

[uncut mulch]

so now you'll have
$$y = \frac kx + 1$$

[obsidian monolith]

ℝamonov

[uncut mulch]

and then sub in a point into that equation to determine the value of k

the black dot indicating the x-intercept would probably be the most convenient

[drowsy karma]

okay -2 and 0

k = 2

[uncut mulch]

yes

now that you have the main shape of your graph figured out,
its time to account for the hole

[drowsy karma]

point of discontinuity

[uncut mulch]

combine the right side into a single fraction, and the introduce a linear factor into the numerator and denominator

[drowsy karma]

it would just be x-3 on the numerator and denomintor

[uncut mulch]

no

[drowsy karma]

x-3

mb

[uncut mulch]

yes, youd multiply the numerator and denominator by (x-3)

[drowsy karma]

yeah

[quick bluff]

Is this conversion from the integration to derivative correct?

[agile gyro]

indeed

oh wait

no

If you change the 1 at the bottom, do you agree that F(x) changes by a constant amount?

[quick bluff]

Yes

[agile gyro]

And what's the derivative of a constant?

[quick bluff]

And isn't that constant amount g(y) where y is the new lower bound?

[agile gyro]

(not a trick question)

[quick bluff]

0

[agile gyro]

yeah, so basically F'(x) doesn't depend on the lower value of the integral : it'll only shift it by a constant, which disappears in the derivative

so then why does your expression depend on it

[quick bluff]

so the lower bound only shifts the graph?

*translate the graph I mean

[agile gyro]

yup

[quick bluff]

Oh ok

[agile gyro]

lemme show why that is

[quick bluff]

Okok

[agile gyro]

$\int_0^x f(t)dt = \underbrace{\int_0^1 f(t)dt}_\text{constant} + \int_1^x f(t)dt$

[obsidian monolith]

Syst3ms

[agile gyro]

there you have it, changing the lower bound only shifts it by a constant

[quick bluff]

Oh so we can keep repeating that "splitting" till x?

[agile gyro]

that's not my point

[quick bluff]

Oh

Oh yeah I get you

[agile gyro]

My point is that if you change the lower bound, it only shifts the graph

So the derivative has to be the same

[quick bluff]

Yes

Got it

[agile gyro]

So then why does the expression you gave me depend on the lower bound

Here's another way to see it

[quick bluff]

What would be the general process for converting these types of integrations into derivative?

[agile gyro]

Set F such that F'=f

[quick bluff]

I replace the variable within the integration with the variable in the upper bound then multiply by the derivative of the upper bound?

[agile gyro]

(an antiderivative)

[quick bluff]

Ok

[agile gyro]

then $\int_a^b f(t)dt = F(b)-F(a)$

[obsidian monolith]

Syst3ms

[quick bluff]

Yes

[agile gyro]

Using some notation you may be more familiar with, $[F(t)]^b_a$

[obsidian monolith]

Syst3ms

[quick bluff]

NP the previous notation is what I'm used to

[agile gyro]

ok then

So then, i'll rename your function F to G in order to avoid naming issues

[quick bluff]

Ok