precalculus

[night herald]

see the first double angle formula

ig the answer is 0.25 or 1/4

[somber void]

you sum it for every x separately
concepts like amplitude, period, phase shift etc. don't apply in the way you think they do to the sums of trigonometric functions

[gritty ibex]

i see, but how do you find the extrema of sum trig function? Or do you need calculus for that i assume?

[willow skiff]

you need calculus

[gritty ibex]

ok, i guess ill stop worrying abt it for now. Just focus on working my way to there

thanks for answering my question!

[somber void]

I can think of some nightmare cases like sin(x) + sin((π/3)x+1)

no global maximum?

,w maximize sin(x) + sin((pi/3)x+1)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by cyclic_tree

[tame pine]

💀

Fr bro

[arctic meteor]

@night herald I got a different answer using a limit calculator.

$(cos \frac {x}{2} -1 ) (cos \frac {x}{2}+1)$

[obsidian monolith]

wolly5114

[arctic meteor]

How did this answer appeared?

[night herald]

Where is the answer?

[arctic meteor]

The one with cos(x/2)

$(cos \frac {x}{2} -1 ) (cos \frac {x}{2}+1)$

[obsidian monolith]

wolly5114

[somber void]

add fractions, difference of squares

[night herald]

this isn't the answer

but i get your point

if this is the final step, then after putting x = 0(i mean x -> 0) we get limit = 0

can you send me the link of the website you are using @arctic meteor

[arctic meteor]

https://mathdf.com/lim/

[night herald]

see, the answer is 1/4

[chrome barn]

Pretty sure this limit is beyond precalculus level

[winter comet]

precalc is anything under the sun 😵‍💫

[arctic meteor]

${\lim_{x \to 2} \frac{2^x-4}{x-2}}$

I know that ${\lim_{x \to 0}\frac{a^x-1}{x}= ln a$

[obsidian monolith]

wolly5114
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[arctic meteor]

Where can i find proof for the second limit and use this rule in the first limit?

[obsidian monolith]

wolly5114

[night herald]

i don't get the question, i think you mixed up first and second

[night herald]

you want proof of this

right?

[arctic meteor]

Ok proof on the first limit and based on that i need to calculate the limit of 2^x

[night herald]

use taylor series expansion for a^x

[arctic meteor]

Without calculus involved

I mean You can write $a^x-1=t$

[obsidian monolith]

wolly5114

[arctic meteor]

If I substitute I get $t \frac{ln(t+1)}{lna} $

[night herald]

what do you do after that

?

can i use a standard limit??

[arctic meteor]

This îs what i found but i don't know how to use this in my limit

Should I factor 4 as 2^2?

[night herald]

just use the standard limit directly

but first let x-2 = t

then take 4 common from numerator

and then apply the standard limit

[primal skiff]

I'm trying to get the formula of a cone via (sorry for the bad drawing):

[daring tapir]

consider a part of a very small height at a distance h from the vertex

[night herald]

yeah and then form a circle at the that height

[daring tapir]

yeaa

i cant really type it out nvm

[night herald]

then form a relation between the radius of circle formed and the height which you choose with help of angle b/w h and r

[daring tapir]

or similarity

whatever suits ya better

[night herald]

now to form the integral you can write

where h is the height vertical distance from vertex of cone

R is the radius of cone

H is the height of cone

and (hR)/H is the radius of circle formed

[primal skiff]

ok yeah its making sense

thank you

[night herald]

if you look closely, the $pi(\frac{hR}{H})^2\$ gives us the area of circle formed at height h, and multiplying by dh gives us its volume

[obsidian monolith]

Black_Gold

[night herald]

integrating this from 0 to H sums all the volumes, hence giving us the volume of cone

[stuck flame]

is this channel suitable to post about circuits and etcetera

[willow skiff]

no

you can ask in a help channel

#❓how-to-get-help

also are you aware of the physics server?

[night herald]

i am not

tell me

[stuck flame]

no I had no idea

[hardy latch]

does anyone know how to find the end behavior, degree, and leading coefficient of this graph? or if any of my answers are right

[meager hedge]

incomplete pic

[hardy latch]

it isn’t: i just took a picture of the part i was stuck on

in the first question it asks for the limit notation for when x approaches negative infinity and then the second asks for when it reaches positive infinity. the third question is just asking for the degree and leading coefficient

[winter comet]

the answers don't look right. Although there is an asymptote y = x, still as x increases y increases without bound, and as x decreases y decreases without bound, so lim x->infinity h(x) = infinity and lim x->-infinity h(x) = -infinity

[silent prism]

Nice precalc content

[hardy latch]

i’m not sure because my teacher said something about rational function and one of them approaching y=0 instead of an infinity

[abstract tiger]

do you guys know how to solve equations involving rational exponents. Can you also give me an example and worksheet to practice it?

[karmic oasis]

anyone having midterms this week

[abstract tiger]

not me

[winter comet]

If it approaches 0 then yes, but these don't approach 0

They don't approach any constant

[karmic oasis]

can somebody explain this

ik 3 is included but

[severe pond]

Can anyone help me understand why x > 1 and not x >= 1 for this composite function

[night herald]

because the first the function the input is going in is 'g' and g can't take 1

[river drift]

what would you get for x = 1?

[severe pond]

Ohhh. Jeez that seems so obvious idk why my smooth brain couldn’t figure that out lol

1 - 1 = 0 undefined denominator

[safe geode]

so like

next year right

I can test out of AP precalc AB or BC

and go straight to AP calc

should i do that?

cuz some of my friends say I need to take pre and others say its a filler

btw im aiming for an A in AP calc

and im doing this so I can go to linear algebra. If I don't ill be in calc 3

[winter comet]

precalc has AB and BC?

[winter comet]

is your question should you or can you?
i would assume you can since precalc isn't a prerequisite for calc (almost always)
if you should do that depends on how you are ready, if you know trig and algebra 2 you can probably do it. i didnt take precalc and just went straight to calc and did fine. I agree that its a filler in most cases, except some people need it, but honestly up to you if you think you can.

[silent prism]

Pre calc is a good class to fortify your algebra and gain exposure to more trig but yeah like he said if you think your prerequisites are strong enough then go for it

I personally needed pre calc

💔

[arctic meteor]

$\lim_{x \to a} \frac{log_a[1+(x-a)]}{x^2-a^2}$

[obsidian monolith]

wolly5114

[arctic meteor]

1/(2a*ln(a)) îs the answer

How can i reach that answer?

Also there îs another limit that appears from this limit

$\lim_{x \to a} \frac{log_a[1+(x-a)]}{x-a}$

[obsidian monolith]

wolly5114

[arctic meteor]

I don't know the answer to this limit

[chrome barn]

L’Hospital?

[chrome barn]

No L’Hospital?

[arctic meteor]

Yes no differentiation

[crude orchid]

Do the substitution $(x-a) = u$.

[obsidian monolith]

Taaha_Tariq

[cyan cliff]

Does anyone know how to solve this?

[glossy radish]

maybe its wrong because you gotta add x=0 which is not positive nor negative

[cyan cliff]

i dont think the format supports that, they ask for it to be in a comma separated list but tysm for trying to help 🙂

[glossy radish]

no, i meant that x=0 is also a real zero possible

sorry for the wording

[cyan cliff]

wouldn't that make it 0 possible real zeros? maybe im not understanding you correctly

[glossy radish]

i was thinking (number of real zeros possible) = (number of positive zeros possible) + (number of negative zeros possible) + 1 (when x=0)

[cyan cliff]

what youre saying could make sense, i just wish the professor provided a reference video or any resource

[arctic meteor]

Îs there any proof for my limit without using differentiation that the limit of log(a) equals 1/ln(a)?

How did that limit reached that answer?

[cyan cliff]

you are right

[crude orchid]

Use the substitution $x - a= u$

[obsidian monolith]

Taaha_Tariq

[crude orchid]

Then use the rules of logarithms.

[quasi elbow]

how so?

[willow skiff]

draw a right triangle with opposite side 4, adjacent side 3

hypotenuse must be 5

so adj/hyp = ........

[quasi elbow]

got it!

[supple girder]

I definetly deserve here. I'm 1 week into pre-calculus. Lim, function, derivative, etc

Hopefully I get the help I'm looking for, I'm a self study kinda guy, been using AI and youtube for most of it

but there is so much these things can do

[safe geode]

alr ty. Ill prolly take it since i struggle with trig

[native sparrow]

how do i use the continutiy test?

[winged dew]

Hi guys

Any help...

Idk if this is pre calc or not we're doing reviews before we get into the actual work THAT DONT MATTER IF U CAN HELP PLS HELP THANK U

[winged dew]

nvm

THANKS

[bright mauve]

i dont understand limits

or epsilon and the upside down A stuff

[tiny kestrel]

what about them dont you understand

[vivid nimbus]

Alright I am in calculus 1 but we had to do pre calculus for the first 2 weeks of class and I need help with remembering trig functions

[willow skiff]

what exactly do you need help remembering

[arctic meteor]

$\lim_{x \to 0} \frac{1+xsin(x)-cos(4x)}{sin^2(\sqrt{3})x}$

[obsidian monolith]

wolly5114

[willow skiff]

try dividing numerator and denominator each by x^2

then the limit of sin(x)/x tends to 1

[arctic meteor]

The answer îs 9

[willow skiff]

similarly, you can do $\frac{\sin^2 (\sqrt{3} x)}{(\sqrt{3} x)^2} \cdot 3x^2 \to 1^2 \cdot 3x^2 = 3x^2$

[arctic meteor]

How can I get that answer?Don't I use trigonometric formulas?

[obsidian monolith]

south

[willow skiff]

nope

you have to use famous limits

the limit of sin(x)/x

[arctic meteor]

@willow skiff So I don't calculate cos4x?

[willow skiff]

nah I checked on Desmos and the answer is 3 btw

[willow skiff]

no, if you sub in x = 0 directly you get 0/0

[arctic meteor]

@willow skiff Don't You write it like this?

$sin^{2}x+cos^{2}x+ xx \frac{sinx}{x}-cos^{4}x+6sin^{2}xcos^{2} x-sin^{4}x$

[willow skiff]

that doesn't help you solve the limit though

also what you wrote is pretty much unreadable, but I'm saying expanding doesn't help you

[willow skiff]

you should go through your textbook and YT videos

then practice some simpler limit problems

everything you have told me so far suggests you don't understand how to evaluate these limit problems at all

like not understanding why you can't sub in (cause 0/0)

that suggests you don't know the topic

[obsidian monolith]

wolly5114

[arctic meteor]

I just don't understand If this îs how You solve IT.My textbook does't have steps to solve only answers

[willow skiff]

get a better textbook with better worked examples then

https://www.youtube.com/watch?v=HbtuSC_WOW0

also check out these simpler examples

basically you need $\sin x \approx x$ and $\cos x \approx 1 - x^2/2$ then you can sub those approximations into the limit

[obsidian monolith]

south

[arctic meteor]

So what i did isn't how You solve it?I understand the fundamental limits and I understand how to use them.

I get 8/3 as the answer because I expanded cos4x

[willow skiff]

no, you don't expand

you get $\frac{1 + x(x) - (1 - (4x)^2 / 2)}{(\sqrt{3} x)^2}$

[obsidian monolith]

south

[willow skiff]

= $\frac{1 + x^2 - 1 + 8x^2}{3x^2} = \cdots$

[obsidian monolith]

south

[arctic meteor]

1-4x comes from cos4x?

You write cos4x as $\sqrt{1-sin^{2}4x}$

[willow skiff]

that's just not how you approach it

[obsidian monolith]

wolly5114

[arctic meteor]

Then how?I'm confused

[willow skiff]

.

[arctic meteor]

But cosx îs Taylor series in your answer.How did I solve it wrong?

I said without differentiation

[willow skiff]

you can derive the cos limit without Taylor series

[arctic meteor]

@willow skiff How?

[willow skiff]

https://math.stackexchange.com/questions/1851459/understanding-this-proof-that-lim-limits-h-to-0-frac-cosh-1h-0

[willow skiff]

well then the binomial expansion $(1 - u)^{1/2} \approx 1 - u/2$ is also technically based on Taylor series

[obsidian monolith]

south

[willow skiff]

there's no escaping differentiation you know

[willow skiff]

but this is the fundamental limit I was talking about

without calculus

[modern stump]

Someone, pls help me! I dont know how do this: lim(x->inf)(3n+2)/(5n-3)=3/5

[winter comet]

divide the numerator and denominator by n

[viscid thistle]

ellipse and hyperbola have the same steps and procedures right? hyperbola is more of an inverted form of an ellipse?

[winter comet]

ellipses and hyperbolas are shapes, not procedures, so im not sure what u mean

[viscid thistle]

like ellipse has a and b switched while hyperbola jus has x and y switched

[winter comet]

wdym "switched"

[viscid thistle]

im not making sense wait 😭

[winter comet]

💀

i mean im not understanding what you're trying to do

are you trying to graph them, or find information about them, or what

[winter comet]

or find information about them in order to draw an accurate graph lol

although arguably like foci and directrices don't really help for human accuracy imo lol

[elfin cargo]

I think he is talking about standard equations for ellipses and hyperbolas

☠️

[winter comet]

yeah i know but i'm trying to figure out what hes trying to do

lol

he asked if they "have the same steps and procedures"

i don't know what this means but if you're talking about finding information about them and then graphing them

then the procedures are honestly different ._.

[arctic meteor]

$\lim_{x \to -1} \frac{\sqrt[3]{x}+1}{sin\pi(x+1)}$

[obsidian monolith]

wolly5114

[arctic meteor]

Do I use the limit of $\frac{sin(x)}{x}$?

[obsidian monolith]

wolly5114

[arctic meteor]

Cube Root of -1 îs -1?

I thought square roots and cube roots need to be positive

[rigid burrow]

Hi guys can i get an idea how to solve $\frac{x^1000 sin5x}{sin^999x}$?

[obsidian monolith]

Eccentricity

[elfin cargo]

Solve what? 🧐

Simplify or what?

[rigid burrow]

oh sorry, lim x -> 0

[elfin cargo]

$\lim_{x \to 0} \frac{x^{1000} \sin(5x)}{\sin^{999}(x)}$

[obsidian monolith]

҉C ҉l ҉ø ҉s ҉e ҉r

[rigid burrow]

yeah that's it, i'm getting stuck. Do i use derivation or what?

[winter comet]

pain

i think its 0

lim x->0 x^999 / sin^999 * xsin(5x)

[arctic meteor]

Do I substitute x with t-1?

[winter comet]

technically both limits exist so
[ lim x->0 x/sin(x) ]^999 * lim x->0 xsin(5x)

= 1^999 * 0 = 0

if thats legal 😄

☠️

[arctic meteor]

I mean that's what the limit calculator shows me

[winter comet]

what is?

[winter comet]

oh yea that might help

then maybe dividing by t

sorry, t

[arctic meteor]

Where does t-1 comes from?

[winter comet]

you had x + 1, it just gets rid of that to become t

t=x+1, x=t-1

[elfin cargo]

Looks valid enough

[rigid burrow]

thank you😄

[arctic meteor]

@winter comet can You also solve it using another subtitution?

$\sqrt[3]{x}=t$

[obsidian monolith]

wolly5114

[arctic meteor]

and

$\sqrt[3]{x}+1=t$

[obsidian monolith]

wolly5114

[arctic meteor]

Which one îs the right one?

[exotic barn]

You will end up with t³ in your sine term, idt you want that

[fluid field]

Whats the fastest way to find the avg rate of change of a function, im in ap pre cal and teach is telling me to do some big ass table to find the answer

[worldly schooner]

Hey i have a test coming up is about the first unit Function and transformation from pre calculus 12…can someone give me some exercises just to see if I’m ready

Please

[restive parcel]

I need help solving b and c

[willow skiff]

for b, do you know how to expand the quadratic R(c)? if so, complete the square so that you can put it into vertex form, a(x - c)^2 + h for some constants a, h

part c is to set R(c) = 0
the quadratic has already been factored for you, so use the zero product property

alternatively find the roots by doing part c first, then by symmetry, the value of c for the vertex is the average of the two roots

[arctic meteor]

$\lim_{x \to 0} \frac{arcsin(12x)}{sin(2x)arctan(6x)}$

[obsidian monolith]

wolly5114

[arctic meteor]

The answer in my textbook îs 2

Is it incorrect?

I used the steps from each limit of sine,arcsine and arctan and I got 1/x

[viscid thistle]

I get why this is "pure bullshit" now

[willow skiff]

yeah I get the same too

just don't trust your textbook

use Desmos to check

,w arcsin(12x) / (sin(2x ) * arctan(6x)) Maclaurin series

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by sooouth.

[arctic meteor]

$\lim_{x \to 0} \frac{tan(a+x)+tan(a-x)-2tan(a)}{x^2}$

[obsidian monolith]

wolly5114

[arctic meteor]

I know that

$\tan(a+x)=\frac{tan(a)+tan(x)}{1-tan(a)tan(x)}$

[obsidian monolith]

wolly5114

[arctic meteor]

and

$\tan(a-x)=\frac{tan(a)-tan(x)}{1+tan(a)tan(x)}$

[obsidian monolith]

wolly5114

[arctic meteor]

Do You expand both?

The answer îs $\frac{4sin(2a)}{cos^3(a)}$

[obsidian monolith]

wolly5114

[arctic meteor]

When I calculate I get$2tan(a)+tan^2(a)$

[obsidian monolith]

wolly5114

[night herald]

I would personally use series expansion for this question

[arctic meteor]

Well is this the answer to my limit?

$2tan(a)+tan^2(a)$

[obsidian monolith]

wolly5114

[viscid thistle]

L' Hopital

twice

[viscid thistle]

i think this is wrong

[barren holly]

hi, quick question, out of these 3 "advanced functions, calculus and vectors and mathematics of data management" which one is the easiest to learn for grade 12 or so?

[knotty mountain]

Can someone help me with this

and this

[exotic barn]

Do you know the epsilon delta definition of a limit

[knotty mountain]

I do not

[exotic barn]

https://youtu.be/DdtEQk_DHQs?si=U2AeP-oSFauaWQ39

[viscid thistle]

took me 4 months to understand it

[knotty mountain]

Oh my gosh. That's like sandpaper to my eyes.

[viscid thistle]

try watching khan academy's video

https://www.youtube.com/watch?v=-ejyeII0i5c&t=6s

[knotty mountain]

How can I tell

[exotic barn]

Do you know the definition of continuity?

[knotty mountain]

yes

and no

I just need someone to walk me through a problem to help me understand

can we start at 27

[exotic barn]

ok

[knotty mountain]

so we equals the denominator to zero

[exotic barn]

yes good start

[knotty mountain]

wait factor it into x+2 and x-2 and then = to zero

right?

[exotic barn]

yes very good

[knotty mountain]

idk now

[exotic barn]

Well is x = 2 or x = -2 allowed?

[knotty mountain]

in terms of what?

[exotic barn]

to plug them in?

no they are not

because then you would divide by 0

so it seems that x = 2 and x = -2 are critical

in fact

these are vertical asymptotes

well one of them

[knotty mountain]

wdym they divide by zero?

[exotic barn]

,, g(x) = \frac{x+2}{(x+2)(x-2)} = \frac{1}{x-2}

[obsidian monolith]

bacc

[knotty mountain]

so when you plug them back in

[exotic barn]

yes

you divide by 0 which is illegal

[knotty mountain]

oh

[exotic barn]

Now see here this is interesting

[knotty mountain]

go ahead

[exotic barn]

If the numerator and denominator have a common root, then it's a removable discontinuity but if the denominator as a root of its own, then that's a vertical asymptote

[knotty mountain]

lost me

[exotic barn]

So x = 2 is a vertical asymptote and x = -2 a removable discontinuity

[knotty mountain]

ok

[exotic barn]

,w plot (x+2)/[(x+2)(x-2)] between x = -3 and x = 3

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by adoniszgks

[exotic barn]

gosh

stupid program

[knotty mountain]

so is 2 non removable is -2 if removable

[exotic barn]

yes

you cannot fix the discontinuity at x = 2

but at x = -2 you can if you define f(-2) = -0.25

but at both these it's discontinuous

[knotty mountain]

My brain is fried from looking at math all day. What do you mean by fix?

[exotic barn]

We can make it continuous

as the name suggestes removable discontinuity

we can literally patch it

and thus make it continuous

[knotty mountain]

but how can I know it's removable or non removable without looking at the graph and by just doing the equation

[exotic barn]

.

[knotty mountain]

so in this case

[exotic barn]

yes!

[knotty mountain]

because you physically remove them

[exotic barn]

that's not why we say removable

we say removable in a sense of removing the discontinuity

here it's still not continuous because f(-2) is undefined

but that's the thing

we can remove this discontinuity by defining here f(-2) = -0.25 reasonably and thus make it continuous

,, g_{new}(x) = \begin{cases} \frac{x+2}{(x+2)(x-2)} & x \neq \pm 2 \ -0.25 & x = -2 \end{cases}

[obsidian monolith]

bacc

[exotic barn]

This is what I mean

we can patch this removable discontinuity by defining the value there reasonably

and thus make it continuous

[knotty mountain]

So you're just changing it essentially

[exotic barn]

adjusting is the better term

fixing or patching it

[knotty mountain]

ok another question is this

[exotic barn]

obviously the limit exists at x=-2 so what's missing to make it continuous is to define a reasonable function value at x=-2 and that reasonable value is the limit itself which is -0.25

[knotty mountain]

for this part, is -2 the removable because when you equal it to zero it ends up as -2

[exotic barn]

I didn't understand that part

[knotty mountain]

my question or

[exotic barn]

yes

[knotty mountain]

I'm trying to articulate it better

[exotic barn]

take your time

[knotty mountain]

x+2=0 ---> -2

therefore, wouldn't I say the removable is -2 or x+2

[exotic barn]

So what you are saying is, if x+2=0 then x=-2 is removable?

[knotty mountain]

so yeah, we crossed out x+2, so because -2 is the answer to x+2=0, wouldn't the removable be -2 and not x+2?

even though we crossed out x+2

it's not a big deal

is this 0

because cos(x)/x+2 ---> cos(0)/0+2 ---> 1/2

[exotic barn]

woah

ok wait so you jump to a new question

[knotty mountain]

haha

[exotic barn]

then you ask if this is 0 but say it's 1/2

[knotty mountain]

do the next one ig

my professor has 0 down

idk how she got it

[exotic barn]

it's 1/2 yes

[knotty mountain]

well she does say to find it algebraically. Idk if that changes the fact that it should be 1/2

[exotic barn]

no it doesn't

[knotty mountain]

Well, anyway. Have a good one. Thanks for the help

[arctic meteor]

$\lim_{x \to 0}\frac{cos(x)-cos(2x)}{(e^x-1)sin(2x)cos(4x)}$

[obsidian monolith]

wolly5114

[arctic meteor]

You write this as

$\lim_{x \to 0}\frac{cos(x)-cos(2x)}{(e^x-1)sin(2x)}$=$\lim_{x \to 0}\frac{cos(x)-cos(2x)}{(e^x-1)2x}$

[obsidian monolith]

wolly5114

[arctic meteor]

?

[viscid thistle]

this is right

not sure if the first to second step conversion is right

[arctic meteor]

Because $\lim_{x \to 0}\frac{sin(2x)}{2x}=1$ and $\lim_{x \to 0}\frac{1}{cos(4x)}=1$

[obsidian monolith]

wolly5114

[viscid thistle]

again

the second part not sure

you can only split the limits only if both exist

so idk

[viscid thistle]

jee?

[viscid thistle]

oh wait

@arctic meteor

you can split it

mb

like
im sure

[arctic meteor]

$\lim_{x \to a}\frac{a^{sin(x)}-a^{sin(a)}}{x^2-a^2}$

[obsidian monolith]

wolly5114

[obsidian monolith]

wolly5114

[arctic meteor]

$\lim_{x \to a}\frac{sin{(e^x)}-sin{(e^a)}}{sin(x^3-a^3)}$

[obsidian monolith]

wolly5114

[viscid thistle]

What's your question?

[arctic meteor]

What are the steps to solve the limit?

Do You use sina-sinb?

[elder cosmos]

simba

[arctic meteor]

@elder cosmos What's Simba?

[elder cosmos]

lion kig

[lusty notch]

cos(0)/cos(0)=1/1?

how good is it?

[exotic barn]

as good as sin+cos=tan in your bio

[lusty notch]

how

error math

my bio is troll

[arctic meteor]

$\lim_{x \to a}\frac{sin{(e^x)}-sin{(e^a)}}{sin(x^3-a^3)}$

[obsidian monolith]

wolly5114

[arctic meteor]

How do I solve it?

[viscid thistle]

@arctic meteor

l'hopital

Since it's a 0/0 form

[viscid thistle]

Are you familiar with this method ?

[elder cosmos]

🤔🔍

[obsidian monolith]

bacc

[exotic barn]

@arctic meteor

[agile sparrow]

Isn't this just e^a.cos(e^a)

[worldly crown]

what is integration

[willow skiff]

have you Googled it

[worldly crown]

yeah

[willow skiff]

so?

[arctic meteor]

$\lim_{x \to 0}\frac{e^{4x}-2e^{2x}+1}{4e^{4x}sin^{2}\frac{(x)}{(4)}}$

How do I factor the numerator?

Îs there a formula?

[obsidian monolith]

wolly5114

[willow skiff]

it suffices to use the first terms of the Taylor expansions for $\ln(1+x)$, so $\lim_{x \to 0} \frac{1}{x} - \frac{1}{x + \sqrt{1 + x^2} - 1} = \lim_{x \to 0} \frac{\sqrt{1+x^2}-1}{x(x + \sqrt{1 + x^2} - 1)}$

[obsidian monolith]

south

[willow skiff]

now use the binomial approximation $\sqrt{1 + x^2} = 1 + x^2/2 + o(x^4)$

[obsidian monolith]

south

[willow skiff]

I'm not sure how to think of this rigorously, but note that 1/x - 1/(x + o(x^2)) cause of the binomial approximation will have error with order o(x^2)

[arctic meteor]

I just applied something randomly $(1-e^{2x})(1-e^{2x})$

[obsidian monolith]

wolly5114

[arctic meteor]

Oh wait I figured it

[proven night]

Hi, how do i prove that red angles are 45 degrees

[night herald]

Do you want the blue area or your your question?

[granite quiver]

How would you write this as a piecewise function?

[summer ruin]

cos(x), when cos(x) >= 0, and -cos(x) when cos(x) < 0

[nimble bear]

achieved it? ill try to do it now

[nimble bear]

@proven night

probably theres an easier way

but I tried to make 0 assumptions

[proven night]

@nimble bear but how did you get it?

[nimble bear]

look at T1

[proven night]

Ok

[nimble bear]

this quadrilateral

so

the sum of all angles is 360

I know alpha+theta = 90

so I summed 90 + (90 + alpha) + theta + beta = 360

90 + 90 = 180, alpha + theta = 90

180 + 90 = 270

360 - 270 = 90

therefore beta = 90

so if beta = 90 everything matches up

as I said, probably theres a way better way to determine this

but this is what ive done

[quiet pine]

this calc?

[nimble bear]

nop, just geometry

[proven night]

But i mean, why alpha = theta = 45. We dont know the values of alpha and theta separately, we only know if we add them then we will have 90

[nimble bear]

if beta is 90, then this line is a bissectrice

and these angles are equal

@proven night got it?

[proven night]

Yeas

Thanks

[nimble bear]

np

[proven night]

I didn't pay attention to the fact that you drew the diagonal from the upper angle, and therefore if b=90°, then, the second line is also a diagonal and then is a bissectrice

[low yoke]

can someone explain why the above is true?

[river drift]

context?

[tender mist]

What's the context here?

[quiet pine]

this is the general form (the name doesnt matter and changes depending on where u are) of a linear equation. Its not a question of true or false , but a parametric equation for a linear relation between x and y

in this instance, the metrics are A,B,C. by changing them you can modify the your linear graph

[vestal jetty]

Can someone help me find out whats the actual answer

idk if 27 is right

[nimble bear]

3^3x = 4, (3^3x)^3, therefore 4^3

[nimble bear]

hey

[nimble bear]

@vestal jetty want further explanations?

[vestal jetty]

Yep

[nimble bear]

k

27 is 3^3 right?

so lets swap

$(3^3)^x=4$

[obsidian monolith]

daniel_zein

[nimble bear]

theres this property

in which you multiply the exponents

when theres a power of a power you multiply them

so

$3^{3x}=4$

[obsidian monolith]

daniel_zein

[nimble bear]

ok

so you keep the info that $3^{3x}=4$

[obsidian monolith]

daniel_zein

[nimble bear]

ok so lets get to the $3^{9x}=?$

[obsidian monolith]

daniel_zein

[nimble bear]

$3^{9x} = (3^{3x})^3$

[obsidian monolith]

daniel_zein

[nimble bear]

its just the same property

buuuut

you know the value of $3^{3x}=4$

[obsidian monolith]

daniel_zein

[nimble bear]

just substitute

$(4)^3=64$

[obsidian monolith]

daniel_zein

[nimble bear]

@vestal jetty

[echo verge]

I'm trying to understand factoring by grouping, and here in the second step, he inerted a negative one, i'm confused how this is allowed? I know when you do equation = equation, what ever you do on one side you need to do to the other, but im confused how when your adding things together, you are allowed to insert a -1? Does inserting the negative one not modify the equations value?

It's just a way to rewrite the equation right?

I think I understand now, that i typed out my question but just want to double check

[tacit grotto]

compare the (3-y) to when you distribute -1(-3+y)

what do you notice?

[echo verge]

Oh, -1 * -3 = 3

Adding a -1 just undos it?

[tacit grotto]

yeah. In essence, you can rewrite expressions like these in different ways as long as it's always equal

[echo verge]

O

Lowkey starting to get interested in math

Planning on majoring in computer science, so gonna have to learn a lotta math

[tacit grotto]

another example for that is when you have an equation like 7/12 + x/4 = 1, you can use 12/12 for 1 so that you can bring 7/12 to the other side and start solving for x like that

[tacit grotto]

gl

[echo verge]

Thanks

[warm cove]

I failed the ap exam for computer science (2) I wish you the best of luck

[echo verge]

I got a 3

Took it sophmore year

Such a good decission to take it sophm,ore year, becaue as an ap elective it boosted my csu / uc gpa calculations

[warm cove]

I'm taking ap precalc now and so far it doesn't look good for me atm so I might have to go do tutoring

[echo verge]

I'm taking ap pre calc too

Senior year rn for me wby

[warm cove]

Junior

[echo verge]

It doesn't seem to diffiult personally, but I dont pay attention in class so I don't know most of the things, but i'm studying it right now becaue I got a test, and i'm like learning it

[echo verge]

That's a hard year

Atleast for me

My most difficult classes were during junior year

[echo verge]

Try watching tutorials on youtube it helps a lot

Since my teacher lowkey sucks all she does is talk and talk, and doesn't explain anything

And chat gpt to help u figure out how to solve something

I learned how to factor by grouping just now by using it

[warm cove]

Will give it a shot, got any channel recommendations or just watch anything in general?

[stuck oak]

its just a way of rewriting things, for an example you can always rewrite -2 as -1(2), in that case he did it just to be able to factor the y-3

[echo verge]

Not really, just search up the topic / problem and bunch of result should pop up, I choose videos that have high view counts and are relatively short

[warm cove]

Alright

[echo verge]

But trust me Chat GPt helps a lot, it's litterly a tool

[warm cove]

Yeah, the update it got last year was a hit

[echo verge]

The 4.0 or what eveR?

[warm cove]

Idk the version

[echo verge]

Apparently the paid version is really good

It's nice that you can have it look through links

[warm cove]

Oh wow

[warm cove]

I had a similar thing in my alg 1 class a long time ago, then I got a engineering teacher shortly after. He carried the class

[echo verge]

Nice

It does such when most math teachers follow the routine of just talking, and copying down. The good thing about engineering teachers or like programming teachers is that they understand that it helps to learn from doing it themselves and understanding why things work

Well i need to get back to studying nice chat wit u though and good luck

[warm cove]

Alright, have fun

[static adder]

hi

[stuck oak]

how do i prove that 5^n - 1 is always a multiple of 24 for every even n?

i tried doing n = 2k and got to (5^k + 1)(5^k - 1) but i don't know what to do after that

[viscid thistle]

cuh, 5^2-1 = 25^1-1, 5^4-1 = 2= 25^2-1

https://tenor.com/view/drake-notebook-gif-20708336

[low yoke]

Yeah, but, why it is equal to 0?

[dark garden]

It is basically a convention we follow in maths, even if you have 2x+3=4y, many would convert it into 2x+3-4y=0 when solving, suppose for simultaneous equations, because it's just easier

In conclusion.. it's a convention

[low yoke]

oh

okay

thank you

[tame pine]

Yo

(1-(n^2)/(n^2+x))

[willow skiff]

there are multiple forms of a line!

1. Ax + By + C = 0
   this form is useful for finding the shortest perpendicular distance from a point to a line, among many other things

2. y = mx + c

3. y - y0 = m (x - x0), slope intercept form

4. x/a + y/b = 1, intercept form

there are also forms in polar coordinates and other coordinate systems more generally

each of these is useful for different situations

[tame pine]

So we get x-n^2ln(n^2+x)

[willow skiff]

oh yeah in 3D there's parametric form as well as Cartesian ((x - 2)/2 = y = (z + 3)/(-1)) for example

[tame pine]

And we get

n-n^2ln(n^2+n)-0+n^2ln(n^2)

So we get

n+n^2[ln(n^2/(n^2+n))]

n+n^2ln(1/(1+1/n))

n^2(1+1/n)ln(1/(1+1/n))

1+1/n=t

n tend to inf

t tend to 1

1/(t-1)^2=n^2

So

t/(t-1)^2ln(1/t)

ln(t)/(t-1)^2

1//2t(t-1)

[willow skiff]

from $n(n \ln(n/(n+1)) + 1)$ might be easier to sub $n \to \frac{1}{u}$

[obsidian monolith]

south

[tame pine]

Yes bro but me checkin

Once

[willow skiff]

Wolfram says your work is correct up to that step

[tame pine]

Wait

Bro

[willow skiff]

don't know about the rest

[tame pine]

Yo southy

[willow skiff]

yeah

[tame pine]

Fr

Shit

I did a mistake

1/2

?

southy bro is it 1/2

Fr

@willow skiff Brooo

[willow skiff]

it is!

you can check yourself using Desmos

[tame pine]

Wohoo

Woohoo*

[tame pine]

This also correct then

[willow skiff]

fair enough

[tame pine]

Cr

Fr

[warm cove]

Apparently my brother failed college algebra so he won't be joining me on precalculus

I wanted to work together to hopefully make it easier

[willow skiff]

twin brother?

[tame pine]

Aah

💀☠️

[warm cove]

Ye

[warm cove]

I'm starting my first precalc tutor today

Got a test on Thursday to, hope I can get prepared enough in time

[neon quartz]

i started pre calc today

i have a test too on Thursday

[open glacier]

Do you need help?

@neon quartz

[neon quartz]

i do yes

i need to simplify this one problem

[open glacier]

Do you have any exercise that you need explanations for?

Okay, what is the problem ?

[neon quartz]

2/3 = 100/1+7e^0.069 (x)

i need to simplify it

sorry not simply

find x

i got stuck after bringing the (1+7e^0.069(x)) to the left side

i then also tried to bring 2/3 to multiply it by the 100

ignore C, i need help w D

[open glacier]

is -0.069(x)?

or +0.069(x)?

2/3 = (100/(1+7e^0.069x)) => 2(1+7e^0.069x) = 3 x 100 => 2+14e^0.069x = 300 =>14e^0.069x = 300 - 2 => 14e^0.069x = 298 => e^0.069x = 298/14 => e^0.069x = 149/7 => ln(e^0.069x) = ln(149/7) => 0.069x = ln149 - ln7 => x= (ln149-ln7)/0.069=>x=~44.32

ln(a/b) = lna - lnb

Do you understand better if I write with a digital pen on a digital sheet?

[next shard]

hi guys need one quick answer for a general graphing question. can someone hop on call wiht me

[heady brook]

so this shows the original function and the final inverse of the function but then it describes the domain as [0, infinity). where does 0 come from? i know you need to cut off some of the function but how do you actually get 0

[tender mist]

I could be wrong, but I think the implication that the original function has a range restriction (only being at and above the x axis) is the reason why the domain of the inverse has to be restricted as well. While the inverse itself isn't necessarily restricting negative values of x, they only belong to the - root, whereas what you have is strictly the + root.

I'm probably horribly wrong about that though

[exotic barn]

The range of a square root function is [0,oo)

2x-3 = 2(x-3/2) just implies a horizontal compression and a shift by 3/2 units to the right

but that doesn't impact the range

[arctic meteor]

$\lim_{x \to 0}\frac{\sqrt{cos(x)}-\sqrt[3]{cos(x)}}{sin(x)ln(1+sin(x))}$

[obsidian monolith]

wolly5114

[arctic meteor]

How do I conjugate the cube Root?

I mean If there were square roots I could conjugate them but in this case I don't know

[exotic barn]

I think you might either split the fraction or consider multiply by x/x

[arctic meteor]

@exotic barn So You can't conjugate it?

[exotic barn]

Found something

[obsidian monolith]

bacc

[exotic barn]

Try this maybe it helps

[tranquil raven]

Why didn't they find the height by finding the magnitude of projection from the tip of a onto vector b? Why doesn't that work?

[elfin cargo]

Because the projection of a on b doesn't equal to height h

If you imagine a plane that goes through vector a and b then you will know why

[tranquil raven]

No I mean like the line from tip of a to be that's perpendicular to b

Because that's how they proved the geometrical meaning of cross product so I thought why that logic wouldn't apply here

[elfin cargo]

Well, yeah. I mean this

How do we call this, again? orthogonal component of projection?

[elfin cargo]

In short. If you imagine a plane that goes through vector a and b then the orthogonal component of the projection of vector a on b doesn't not equal to h

Because the plane is tilted

[tranquil raven]

Oh I get it

I thought the sides weren't slant

Thanks

[warm cove]

Bad news: I have a precalculus test on Thursday

Good news: there is a hurricane coming to my area on Thursday

[elfin cargo]

Math teacher will still come to school that day

[warm cove]

[daring tapir]

Hope the test papers arent waterproof and prof prints them at home for some reason

[unborn brook]

Is an odd-function's X and Y reflections always the same?

[winter comet]

yup

[echo axle]

Isn't that basically by definition

Cause if an x reflection is the same as it's y reflection, that would imply -f(x)=f(-x), which is the definition of an odd function

[frozen adder]

can somebody help

[echo axle]

Do you know how to determine if a function by looking at it?

Perhaps "asymptote" rings a bell?

[frozen adder]

ik what those are

but i dont get this

[echo axle]

What don't you get about it specifically?

[frozen adder]

the second one mostly

[echo axle]

Well, there are just two values it can't be instead of one

[cunning cedar]

Help guys

Idk

What are the points of inflection?

[marble night]

The point where value of derivative is 0

Or by graph, the point of local maxima or minima

Wait no

[cunning cedar]

So would it be the ones I circled

If you can see it?

welp

[marble night]

Ye and also -0.5

[cunning cedar]

huh?

so like I would write it as

(-3,-8), (2,2), (4,0)

is that what I would write?

[marble night]

Yes

But also include (-0.5,-3)

[cunning cedar]

why?

[marble night]

Points where graph changes from concave up to concave down or vise versa

[marble night]

Because before that point its concave up, afterwards its concave down

[cunning cedar]

so my answer would be (-0.5,-3), (3,1)?

?

[marble night]

(-3,-8), (2,2), (4,0) (-0.5,-3),(3,1)

I didnt do that good of a job explaining so https://www.youtube.com/watch?v=UK2shgCXALo

[cunning cedar]

okay thx

[undone sage]

is the lim h-> 0 ah-1/h ln?

bing said it but idk

[orchid garden]

I can't handle my pre cal anymore huhu

[arctic meteor]

$\lim_{x \to \infty}\left(\frac{x+1}{x-1}\right)^x$

[obsidian monolith]

wolly5114

[viscid thistle]

Any tries ?

[viscid thistle]

It is of a particular form

[obsidian monolith]

bacc

[onyx flicker]

So sorry for bother, but could someone briefly explain one-sided limits? I was gone on the day of notes.

[exotic barn]

Sure what is your doubt

[onyx flicker]

like wahts the formula for them?

[exotic barn]

There is not really a formula where you copy and paste things and get a result

The idea with one sided limits is as the name suggests, you approach a limit from one side instead of both sides

[obsidian monolith]

bacc

[exotic barn]

This is very important if your function is restricted, for example 1/x

If you approach 1/x from the right as x goes to 0 then you end up with +inf

however if you approach 1/x from the left as x goes to 0 then you end up with -inf

and this is very important, to investigate the limits individually from both sides, because this tells us if the "both-sided" or usual limit exists

it may be, especially with piecewise functions that you get different results from both sides, and then by that you can tell that the limit from both sides does not exist and it is very important when it comes analyzing continuity or differentiability

[arctic meteor]

$$\lim_{x \to 3}(13-4x)^\frac{1}{x-3}$$

[obsidian monolith]

wolly5114

[arctic meteor]

It needs to look like this but I don't know the denominator

[exotic barn]

Maybe factoring x works

[arctic meteor]

Îs it based on the exponent?

[obsidian monolith]

bacc

[exotic barn]

yea this works out elegantly well

@arctic meteor

[serene geode]

ce faci

ce clasa esti

[charred oasis]

can someone explain to me the math on this

like how do I figure it out

[winter comet]

after n half lives? the heck

isn't the half life only supposed to happen once ._.

[river drift]

after 1 half-life, there is 1/2 left. after two half lives there is half of that, 1/4 left. after 3 half lives, half of that, 1/8 left, and so on

[winter comet]

oh

i thought there was one singular half life of a substance 💀

i would call that a fourth life

nah but idk chemistry or whatever LOL

[river drift]

let's say you have 1 kg of radioactive substance with a half life of 1 year. you put it in a room with a timer and write down "this substance started out as 1 kg". after 1 year you come back and there's 0.5 kg left

but the substance is continuously decaying, it was before you got it, and it is after you sell it

so you sell it to someone else, who puts it in his own room, with a timer, and writes down "this substance started out as 0.5 kg". after 1 year he comes back and there's 0.25 kg left

you could have bought it from a guy who had 2 kg of it the year before that

[primal summit]

hey

[daring tapir]

Heyyy i learnt that like last month so yes those terms exist too but theyre not as common cuz well standards lol

[river drift]

well there's no point in having a 'fourth life' when it's just 2 half lives

[daring tapir]

We had notations like t_0.75 and stuff to denote when 75% of the reaction is complete

[winter comet]

but its a fourth of a life :<

it isn't but :<

[daring tapir]

😭 you just made the molecules alive

What if they revolt against us

[undone sage]

how do you prove a^h-1/h is natural log without using circular reasoning? i.e proof it gives the loge of the number h, ive already proved that it equals 1 when a is e

[undone sage]

jo worries i figured it out

[onyx flicker]

ohhh, thank you! I got 100% on my assignment :]

[tame pine]

Yo

[languid pollen]

hello

[uncut mantle]

If a function stops at a point such that the point's y value is the highest/lowest, is it a local extrema?

[river drift]

an endpoint of a function's domain is usually a local extremum, yes

[drowsy sable]

where are points of inflection? there should be 3.

im putting an answer of b,d,e but getting marked wrong.

[severe pond]

I’m about a month in to my precalc class and all we have done is linear functions… identifying slope, y-int, and intersection points, and now demand functions which are also linear

Is this common for a precalc curriculum?

[narrow moss]

can someone please help me with my work

((bx^(a)+7)(4x-c))/((3x^(2)+5)(2x-12)(x-d))

i know the answers are
a=3
b=12
c=24
d=5

[spiral sonnet]

Can someone tell me if this is a typo or not? I am wildly confused.

[winter comet]

no, (f o f o f)(x) = f(f(f(x)))

[spiral sonnet]

Oh, I felt like I was having a stroke reading it.

Thank yo u

[stone geyser]

What chess move is this???

[winter comet]

interesting move

but you fell right into my trap

now i can sacrifice THE ROOOK

[stone geyser]

NOOOOOOOOOOOOOOOOOOOOOOO

[lucid scaffold]

if you had -5x^3/4 using chain rule would it be -3/4(5x)^-1/4 * 5 or 3/5(-5x)^-1/4 * -5?

wait sorry

wait nvm i got it

[delicate goblet]

l

[exotic barn]

well done :)

[quasi elbow]

how do we solve this question?

[dusky saffron]

ans is option a??

[arctic meteor]

$$\lim_{x \to a}[\log_b(\log_a(x^b)]^\frac{1}{x-a}$$

[obsidian monolith]

wolly5114

[arctic meteor]

Do I change the base of log with ln?

[quasi elbow]

it is. how did you get it, though?

[willow skiff]

https://byjus.com/question-answer/if-sin-4-a-a-plus-cos-4-a-b-1-a-plus-b-then-the-value-of/

there's a derivation online

[quasi elbow]

thank you!

[willow skiff]

no worries!

[quasi elbow]

probably a stupid question but how did we get this step? i just do not see it

oh nevermind

how does this imply that sin(x) is an odd function?

[willow skiff]

the definition of an odd function is that $f(-x) = -f(x)$ for all real $x$

[obsidian monolith]

higher's secret brother

[quasi elbow]

right

[quasi elbow]

how do we find the values of sec(x) at certain points from this?

[exotic barn]

you mean the vertical asymptotes?

[quasi elbow]

yeah

[exotic barn]

recall that sec(x) = 1/cos(x)

the vertical asymptotes happen if cosine becomes 0

because you would divide by 0

so all vertical asymptotes are the solutions of cos(x) = 0

[quasi elbow]

but shouldn't that be undefined, then?

[exotic barn]

they are

it's just these values of x are then vertical asymptotes

[quasi elbow]

makes sense, thank you!

[hollow plover]

log to the base of e is ln

[arctic meteor]

@hollow plover that's what I asked

Do I change the base with e?

[hollow plover]

Yeah

For example log 8 to the base of e is the same as ln(8)

[quasi elbow]

what is happening here

[viscid thistle]

dang it can't find another g

[quasi elbow]

what do you mean?

[viscid thistle]

What is your question?

Oh I see

[viscid thistle]

What they are doing here is converting degrees to radians

[quasi elbow]

I get that but

[viscid thistle]

And here what they are trying to say is

left side function(2n+1 pi/2+theta) = right side function(theta)

[viscid thistle]

Mhm?

[quasi elbow]

yes, but how so? how are they using the stuff written in the left to conclude that? if that makes sense

[viscid thistle]

Oh

They are just subbing n=0,1,2,3,4, so on

Subbing=substituting

[quasi elbow]

oh

wow, that was such a stupid question

but thank you!

[viscid thistle]

No worries
Have a good day

[quasi elbow]

likewise

[dense dagger]

guys

if statistics is stats

then isn't mathematics maths

[topaz swallow]

Yes…

[somber void]

In the US, you only math once

[topaz swallow]

?

[winter comet]

and yet you don't physic

go figure

[echo axle]

Here's a bit of linguistics for you (if anyone cares)
In a British accent (which is usually where math is pronounced with the s), during the th, the tongue tends to be further in your mouth. That and the taller vowels, it's much more natural to end on an s.
On the other hand, Americans tend to put their tongue further out, as well as having wider vowels. This means it takes extra extra effort to pull your tongue back for the s. It's just more natural to want to and on more of an uh vowel
(I'm not an expert, so I may not be super accurate)

[devout plover]

Hello! Could someone explain to me where the 1/4 came from and why imaginary parts changed to 2x+1? (Teachers work btw) thank you!

[river drift]

well we know that if two roots are [ x = \frac 12(-1\pm \sqrt{19}i) ] then two factors will be [ \ab(x - \frac 12(-1 + \sqrt{19}i)) \ab(x - \frac 12(-1 - \sqrt{19}i)) ]
this is a little cumbersome, so we multiply by 2 inside the () to clear the fraction, and divide by 2 on the outside to cancel that out

[obsidian monolith]

cloud

[devout plover]

Ohhh okay that makes sense tysm 😭

[quasi elbow]

in this question,

what mistake am I making here?

option b is the correct answer

[shadow summit]

It shouldn’t be -sec(θ)

Or -x

[willow skiff]

yes, $\frac{1}{\tan \theta + \sec \theta}$ is not $\tan \theta - \sec \theta}$

[obsidian monolith]

higher's secret brother
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[willow skiff]

you should have $\sec \theta - \tan \theta = e^{-x}$

so $2 \sec \theta = e^x + e^{-x}$ and the result follows

[obsidian monolith]

higher's secret brother

[quasi elbow]

oh

I get it now

thank you

[willow skiff]

np

[quasi elbow]

how do you solve this?

[willow skiff]

x + 1/x >= 2 for positive x

and x + 1/x <= -2 for negative x

now what is the range of cos theta?

[quasi elbow]

-1 to 1?

[willow skiff]

yes

and so.............

[quasi elbow]

oh

no value of theta is possible

got you, thank you so much

[willow skiff]

exactly, cheers!

[quasi elbow]

for this question,

i have got this

and this

neither of which I think is correct

how do you approach this question

[quasi elbow]

how do we proceed further?

[flat minnow]

Can someone please give me a crash course in precalculus?

[willow skiff]

https://tutorial.math.lamar.edu/Classes/Alg/Alg.aspx

[simple urchin]

How do I find the directrix and focus of a parabola but not written in this strange standard form, but written in the form of f(x)=ax^2+bx+c

Like they are finding the focus, but they are doing it in this strange form that is not like a typical quadratic. Written in descending order.

[willow skiff]

you need to complete the square first

[simple urchin]

So I take my function in descending order, then do complete the square to put it in standard form?

[willow skiff]

no, you complete the square first

let's say you get y = m (x - n)^2 + r for example

then you can rearrange this to get 1/m * (y - r) = (x - n)^2

so 1/m = 4p or that p = 1/(4m)

k = r

and h = n

[dry nest]

Can anyone help me with these differential equations?

[winter comet]

#odes-and-pdes

[night herald]

for the first one,

just differentiate the solution

the easier method

but i'll try to solve it normal way

[quasi elbow]

.

[night herald]

what's the question

[quasi elbow]

second part of the second question

[dry nest]

Here U go buddy 🙂

[quasi elbow]

got it, thank you

[fathom sluice]

"Use F40 = 63, 245, 986 and F38 = 39, 088, 169 to find the value of F39 using Fibonacci sequence formula" isn't the question wrong because after finding the answer, F38 > F39 which isn't supposed to be because it's an increasing sequence?

[unborn brook]

Are the order of operations and order of transformations-of-functions arbitrarily set (considering the convenience issues as well?)

[summer ruin]

no

you don't even get the same results if you change the order

[unborn brook]

That's not what I meant

Are they the way because we put them that way or is there an underlying reason for those particular orders

[willow skiff]

what do you mean by this

[unborn brook]

Why are those steps ordered the way they are?

[willow skiff]

cause you get different functions if you change the order of transformations

okay let me give you an example

if we start with f(x) = x^2

and we want to translate this function 3 units to the right and then compress it by a factor of 2

we get $x^2 \mapsto (x - 3)^2 \mapsto (2x - 3)^2$

[obsidian monolith]

higher's secret twin brother

[willow skiff]

https://www.desmos.com/calculator/6fdlcinhzu

have a look for yourself

[unborn brook]

I am completely aware of that. What I am looking for is an explanation for this specific order:

Is there a specific reason for why we defined it this way? Or is it just arbitrarily set?

[willow skiff]

it's more arbitrary than you think

the one rule that I'd say matters the most is that you do the stretching or shrinking before the translation

[willow skiff]

cause this isn't shifted 3 units to the right: it's been shifted 3/2 units

[willow skiff]

and then the vertical and horizontal transformations are independent
you can do them in any order if you follow this rule

[unborn brook]

So it's basically a set of rules we legislated for conventional purposes?

[willow skiff]

I would not approach maths in general as memorising a set of rules

like PEDMAS or whatever you call it is a dumb idea

we never use $\div$ in real life

[obsidian monolith]

higher's secret twin brother

[willow skiff]

we just use fractions instead cause it's so much clearer