precalculus

[prisma prairie]

hmmm

i dont see how

[uncut mulch]

doing a rough sketch on a number line may help you visualise it

[prisma prairie]

im thinking (-inf,inf)

wait no

[-7,2)U(2,inf)

-7,2 sorry

[uncut mulch]

why is it an open interval at 2?

[prisma prairie]

its an aymptote

not included

[uncut mulch]

i think you're getting your domain and range mixed up again

can y be 2?

[prisma prairie]

oh yeah i am

ARGHH

so

[-7,inf)

[uncut mulch]

yeh

[prisma prairie]

is the domain simplified ?

[uncut mulch]

yeh

[prisma prairie]

thx for the help!

im always getting my domain range confused lol

[uncut mulch]

practice a bit more

[prisma prairie]

yeah

im going to for sure

[long crown]

Euclid's Division Lemma https://youtu.be/TJ7E1QHkYOQ

[heady jewel]

Ramanujan

Class 10

a=bq+r

completely relates

[scenic musk]

am i allowed to use difference of cubes for irrational number. e.g. (x)^1/3 - 1

i mean irrational power

[pale kettle]

You can, it's not going to be super nice

[scenic musk]

ty

can someone help me understand the second solution on this page. i dont understand the algebra

https://www.slader.com/textbook/9781305480513-calculus-8th-edition/103/problem-solving/15/#

its the second subbmited answer on the page

[viscid thistle]

i'm getting solutions when i shouldn't |x-1|+|x+1| <1

split them up into two cases and two further sub cases where x-1 < 0 (then x+1 < 0 or x+1 > 0) and x+1 > 0

somehow i end up getting -1/2 < x < 1/2

graphing software shows no solutions

never mind didn't test x+1>0 and x-1<0

[reef rune]

if the red function is concaved up, and you reflect it across the y-axis, concavity doesn't change, right?

let's say x^a and a > 1

pretty sure its still concaved up just making sure

[patent beacon]

That blue line is concave down

[reef rune]

oh guess not lol

[patent beacon]

Y Axis? That won't change it

[reef rune]

x*

typo

[patent beacon]

Yus then concavity mirrors

[reef rune]

red function is x^5/2 and blue is -x^5/2

[patent beacon]

You can prove that with the derivative if needed

[reef rune]

my teacher told us it concaves up when a > 1

in a power function

in the format of x^a

[patent beacon]

Guess they didn't consider a negative on the coefficient

[reef rune]

mhm

[patent beacon]

-x² is concave down, for example

[reef rune]

yeah it makes sense

[near ledge]

Hey can anyone help ?

I’m struggling to derive the cosine rule

[viscid thistle]

I'm here again

[near ledge]

So I split the triangle into two right angle triangles

And label the line h

[viscid thistle]

Is there something wrong with my S_n?

Wow wow

I can't help you @near ledge

[near ledge]

Obviously not judging by that

[viscid thistle]

I'm sorry ':(

[near ledge]

I hate stats

Keep it away from me

[viscid thistle]

But there's no stats here 👀

<@&286206848099549185>

[near ledge]

Oh never mind

It’s series

[viscid thistle]

Yeah

[near ledge]

Haven’t learned that yet

[viscid thistle]

I haven't learned about the cousin rule either

[near ledge]

I’d recommend khan academy

He’ll teach ya

[viscid thistle]

But ai don't need it now

I need to not if me S_n is right

Because I:m getting 2 different answers

[odd helm]

Isn’t it supposed to be -63+360

Not 180

To go from negative to positive

[viscid thistle]

I can't help you razor, I'm sorry

[formal patrol]

@odd helm the 180 thing is a property of tangent specifically

[odd helm]

Heh?

Oh so

[formal patrol]

tan(x + 180deg) = tan(x)

[odd helm]

For tangent you use 180?

[formal patrol]

adding 360deg is always fine

[odd helm]

So only when I use tangent and get my angle I can add 180 to it?

Like when going from rectangular to polar you do arctan(y/x) so by adding 180 it’s the same thing

I’m confused sorry

Afk

Brb

[stark hawk]

Guys sorry to butt in on green_razors question but what do i do if i have a negative in my denominator in a logarithmic function? For example : (log(x^10))/(log(x^-18)). I know having a negative in my denominator makes it undefined but how should i go about removing it?

[viscid thistle]

Can I get help too!?

[stark hawk]

id offer help but i've literally never seen that before in my life xD

[viscid thistle]

Have you never seen sum of a geometric sequence?

I've never seen your question either

I feel bad about tagging the helpers but

<@&286206848099549185>

Everything seems fine but it isn't

[valid flint]

whats going on here

are there 3 questions at once?

[viscid thistle]

Kinda

[valid flint]

if so, pls go to one of the question channels and choose one unoccupied

[stark hawk]

tyvm will do

[viscid thistle]

Quick question suppose that ln5=r and ln6=s so I need to make an equation with r and s to make

ln1.2= something with those numbers where do I start

120 = 6 * 5 * 4

@viscid thistle

??

Can you see that 6/5=1.2?

Honestly did not notice that at all

I was going to suggest to multiply by 0.04

Wtf are you talking about Rudy.

@viscid thistle Can you see that 6/5=1.2?

ye

Rudy go to #bots

Now are you familiar with the log laws of division?

Yes

So how can you make $ln(1.2)=ln(x)$

[obsidian monolith]

leviosa:

[viscid thistle]

What would x be?

[stuck lark]

\ln tbh

[viscid thistle]

Fine.

:c

I got it

Just s-r right?

yea

👍

[stuck lark]

@steel sequoia yes, whatever you do to one term, you do to the other terms on the same side

[steel sequoia]

Was it necessary to multiply the right side by 2x+3 also?

[stuck lark]

yep, whatever you do to one side, you always do to the other side

[steel sequoia]

I feel dumb

I knew that.

I got a presentation I gotta give tomorrow over this topic

and I'm just stressing

But ty mate

[stuck lark]

never a good idea to study while stressed

no prob

[steel sequoia]

I got a whole paper written up over this

that's one of the examples I used

I was reviewing it tonight

and startled myself thinking there was a simpler way to come to the answer

[stuck lark]

it happens to all of us time to time

[shrewd flame]

How to evaluate ( H - 1 )^3 I always get confused how to factor 3rd degree polynomials out

Yes ( H-1)(H+1)

I just know if its x- something it factors out to both + and -

[willow bear]

no @shrewd flame, (H-1)(H-1) is not the same thing as (H-1)(H+1).

[shrewd flame]

^2 is squaring it

oh oops

[willow bear]

||H^2 - 1^2||

[long crown]

🙏 Euclid's Division Algorithm 🙏 https://youtu.be/rYH2MmqArHU

[green zenith]

hello there! Can i have someone tutor me on rational functions? I can do it with mic.

[modern nymph]

???

Hmm english term

It is like switch numbers right?

[proud sparrow]

rational functions, i.e. polynomial divided by polynomial

[willow bear]

"switch numbers"?

[green zenith]

yes that is it. i need to graph them step by step.

[modern nymph]

+6-2=+4 -2+6=+4

Example

[proud sparrow]

you mean rearranging equations/expressions

@green zenith You might want to consider zeros of the numerator and denominator

also watch out for a common factor of the numerator and denominator

also watch for the sign

and possible horizontal/diagonal asymptotes

if all fails, calculate a few points and plot them

then connect them by smooth curves

[modern nymph]

Oh I got what you mean

I just got it yesterday

Wait

Post a photo about it?

[proud sparrow]

if you want to talk about it sure

[modern nymph]

You meant this?

Or something

[proud sparrow]

synthetic division seems to only work for linear factors (x-a)

[green zenith]

@proud sparrow the only problem that i have at the moment are x intercepts. i'm having wrong answers even though i solved it correctly. And locating the graph relative to the horizontal asymptote which requires dividing polynomials.

[proud sparrow]

how about show a graph you drawn of a rational function?

[green zenith]

okay

[proud sparrow]

that you may be unsure of

[green zenith]

im just doing it again so its much clearer.

here it is. i hope its understandable.

my problem is at step 1 which is the x intercept. and step 5.

i think im wrong at dividing and the x intercepts.

@proud sparrow i will take another picture if needed

[proud sparrow]

x intercept you have 1/sqrt(2) and -1/sqrt(2)

don't forget all the solutions

[green zenith]

oh yeah. it is +-. but it is still wrong according to the book

[proud sparrow]

also, it's -x+2

[green zenith]

you mean in the division?

[proud sparrow]

step 5

or in the long division

[green zenith]

the remainder?

is it because im substracting?

[proud sparrow]

yeah you subtracted x

[green zenith]

okay. that explain the intercept.

could you find any more mistake? on the x intercept?

[proud sparrow]

what happens when x=-2?

[green zenith]

im assuming thats the horizontal asymptote. thats the point of x where the graph passes the horizontal asymptote.

[limber bone]

y=-3

[green zenith]

is that correct?

[hoary valley]

@limber bone So if they said "perpendicular to the x-axis" then x=1 right?

[willow bear]

lines perpendicular to the x-axis have equations of the form x=c, so yes

[hoary valley]

Thanks Ann

[green zenith]

uhmm... i dont mean to be rude but im curious. is this off topic to what we were talking about?

[willow bear]

my last message was addressed to joshwa specifically

[green zenith]

oh okay

[hoary valley]

An equation for the line passes through (3,-1) and parallel to x-axis is ?

[green zenith]

yes

if anyone is still willing to help me, im still here.

im still looking for my lost x intercept

[modern nymph]

Hmm

Question?

[green zenith]

i have the image not far above and would gladly repost the image if needed

[modern nymph]

5?

[green zenith]

are you talking to me @modern nymph ?

[modern nymph]

Yes

[green zenith]

so my graph is f(x) = 2x^2 -1/2x^2+x-3

i have an image, do you want me to post it?

[modern nymph]

Yes

[green zenith]

voila

[modern nymph]

What do you need to do?

X,y graphic?

[green zenith]

as of now, i want to know the x intercept

that seems to be my problem on this graph

[modern nymph]

You have to find (x)?

Or something?

[green zenith]

yes, of the numerator in this case because im solving for the x intercept

so 2x^2-1=0

[modern nymph]

May you do like 2 numbers same you may line it

[green zenith]

what do you mean?

[modern nymph]

-2x^2-1/2x+x-3= -1/-3?

Wait

I meant

[uncut mulch]

are you still trying to find the x-intercepts?
what seems to be the problem when trying to solve 2x^2-1=0?

[green zenith]

@uncut mulch yes i still am

[uncut mulch]

isolate x like any other equation
pretty sure you should've encountered plenty of questions similar to that

[modern nymph]

-2x^2/2x^2= -x?

[green zenith]

so i have it solved which is x=+-1/sqrrt2 but the book have a different answer which is x=+-sqrrt2/2

which made me think i missed something

[uncut mulch]

rationalise the denominator

[green zenith]

no @modern nymph thats not it

[uncut mulch]

(we usually don't like radicals in the denominator)

[green zenith]

oooohhh

[uncut mulch]

all g?

[green zenith]

yes thank you very much

[modern nymph]

You solved x by step 2 right then you can find x^2?

@green zenith ???

[green zenith]

no, in step 1.

step 2 is the vertical asymptote

[modern nymph]

But you solved x = -3/2 so you can find x^2 and do -1/ under numbers?

[modern nymph]

Eliminate

[odd badge]

Yup

If they want a vertical line it’ll always be the x-coordinate

[hoary valley]

Yes, I'm just not confident in my answers.. Trying to build confident slowly but surely c.c

[odd badge]

Just think about it geometrically

[hoary valley]

Yeah I imagined it

[odd badge]

Envision a point on the graph and see its x-coordinate and the line going through it as x stays constant

[hoary valley]

Sometimes I do basic mistakes, kinda ruins my mood and kills my confident 😦

[odd badge]

Yeah happens to everyone

[hoary valley]

Right?

[odd badge]

Yup

Both vertical

If you had x= something and y= something they’d be perpendicular

[modern nymph]

Hmm

Think like x= n = y line y=n = x line

[hoary valley]

Correct ?

[modern nymph]

Wait?

Is that parabol?

That meant a=a right?

Like 2a+bx=2a+cx=ax something

[hoary valley]

This is a y-intercept equation

[modern nymph]

If that is parabol

2 lijn

Lines

Then it must have 2 same a?

A in ax+by=c

a= y/x

[hoary valley]

These lines must be parallel, right?

[modern nymph]

Lol

You can draw it

Holy

Do like (x,y) (x=-1 y=n)

Then (-1,0) (-1,1) (-1,2) (-1,3)... so on

Same like x=1

So

X slide -1 form line y x slide 1 form line y

If that is parabol then direction coefficient is same or a=a

[hoary valley]

Thanks man, I'm taking notes..

[modern nymph]

Hello are you there?

Did you draw it?

[viral imp]

,, log_2 16-3^\frac{1}{log_2 3}

[obsidian monolith]

Umma.Gumma:

[viral imp]

hi, I've got some trouble solving this

would appreciate a few hints

[willow bear]

did you mean $\log_2(16) - 3^{\frac{1}{\log_2(3)}}$?

[obsidian monolith]

Ann:

[willow bear]

in any case

which part of this is giving you trouble

[viral imp]

yes, well no parts in particular actually, I thought I knew log properties reasonably well

still, can't get to the solution

[willow bear]

where are you stuck

[viral imp]

give me a min to cope with tex

,, (4) log_2 (2) - 3^{log_3 (2)}

[obsidian monolith]

Umma.Gumma:

[viral imp]

I got here a few times

possibly that log in the exponent annoys me

[willow bear]

$4 \log_2(2) - 3^{\log_3(2)}$, ok great

[obsidian monolith]

Ann:

[willow bear]

both $\log_2(2)$ and $3^{\log_3(2)}$ can be simplified

[obsidian monolith]

Ann:

[viral imp]

yes the first is one

[willow bear]

indeed

[viral imp]

the second is less obvious to me

[willow bear]

it is just as obvious if you know what log is

[obsidian monolith]

euler/fermat:

[viral imp]

ok let me think a while about it

[willow bear]

log_3(2) is, BY DEFINITION no less, the exponent to which 3 must be raised to give 2.

log_3(2) is, BY DEFINITION no less, the exponent to which 3 must be raised to give 2.

there is nothing else to think about here.

[viral imp]

fock me, then il would be

,, 4-3^{log_3 (2)} = 4 -2 = 2

[obsidian monolith]

Umma.Gumma:

[willow bear]

\log!!!

[viral imp]

got it sorry, thanks!

[scenic musk]

can some explain why the function f(x) =|x| when turned into a piecewise function is f(x) = -x when x < 0

doesnt the function state that it has to be positive Y

so how can it become negative

[spring thunder]

if you take the opposite of a number that is already negative

you get a positive one

[scenic musk]

oh

its -(x)

i forgot about paranthesis so if x is -5, then its -(-5)

thank you

[random tulip]

im trying to check if these are inverse equations,
f(x)= (Sqrt x-3) g(x)= (x^2+3), ~~~~ x >= 0
I'm unsure how to set up the equation of f(g(x))

are any of these three the proper way to set it up or am i more confused than i think i am

not sure why its sideways sorry

[sleek pawn]

,rotate 90

[random tulip]

<@&286206848099549185>

[smoky dirge]

@random tulip what are you trying to ask?

[random tulip]

im not sure how to get the equation for f(g(x))

[serene heath]

The third one

Wait is $f(x)=\sqrt{x-3}$ or $\sqrt{x}-3$

[obsidian monolith]

the one n only:

[random tulip]

the first one

[serene heath]

Then the second expression u wrote is f(g(x))

[random tulip]

ohhh okokok thank you very much

[viscid thistle]

Hello guys, I have a situation

[merry sphinx]

Post it

[viscid thistle]

Like, why does the last term minus the first term time the inversed ratio gives the sum of the Geo seq

I think it all has to do with the fact that the ratio is being inversed

But I don't know how can this affects the result

Like, why does "putting the ratio upside down" affects the equation

[merry sphinx]

So you are asking why the geometric sequence sum formula works?

[viscid thistle]

Basically

I "discovered" that the fumula consist of this when adding the 3 first term

You have a number that correspond to the 4th term in the sequence - the first term of the sequence × inversed ratio

Guys?

[merry island]

I am trying to find the range of the function f(x)=(x^2 + 3x - 10)/(x^2 - 3x + 2) and I get the answer
(-infinity, 1)U(1,7)U(7,infinity) because there is an asymptote at y = 1 which is not crossed and there is a hole at (2,7) (and there is no other value of x which could produce an output of 7). Wolfram Alpha agrees with me, however Mathway and desmos say that 7 can be in the range. So am I right, or is 7 not in the range? Any help is greatly appreciated!

[pseudo sonnet]

@viscid thistle thats hard to read

@merry island I think you meant to say vertical asymptote at x = 1

the domain is

(-infinity, 1) U (1, 2) U (2, infinity)

your mistake was you used Y values in your domain, domain consists of X values

range is Y values

[merry island]

@pseudo sonnet I am talking about range here, not domain

[uncut mulch]

you would be correct

[merry island]

Thanks

[uncut mulch]

removeable discontinuities aren't displayed well on desmos

[merry island]

Not only was it not displayed but when I typed in f(x)=7 it provided an answer which I thought was weird

[uncut mulch]

if you click on the exact point, it will say undefined
f(x)=7 will draw a horizontal line
entering f(2) in desmos will return undefined

[pseudo sonnet]

ah gotcha my bad. Like Ramanov said before your work is correct

[merry island]

fascinating, thanks for your help!

[whole socket]

how would i approach a question like this?

[whole socket]

aight

[trim acorn]

uh

wouldnt it be 2 m?

because the hypotenuse of the triangle is 4m

which is basically the diameter

divide by 2 to get the radius

2 m

im not sure if im correct though

@whole socket

[whole socket]

?

it would only be the diameter if it goes through the center point

which it doesnt

@trim acorn

[trim acorn]

u right

like i said

im not the brightest

[heady jewel]

bruh

you dont need to be bright to figure this oit

[proud sparrow]

@whole socket You have all 3 sides

so you can use cosine rule, then you can find the diameter using sine rule

[prisma prairie]

why am i getting this wrong?

[merry sphinx]

Which is lower, negative infinity or -6

@prisma prairie

[prisma prairie]

-infinite

[merry sphinx]

Yes, so which should come first in interval notation?

[prisma prairie]

-inf

[merry sphinx]

Alright, you should be able to know the answer now. What is it?

[prisma prairie]

(-inf,6]

[merry sphinx]

Be careful with inclusive/exclusive

Yep, the edited answer should be correct

[prisma prairie]

ok

[prisma prairie]

how do i know if a quadratic equation is already a perfect square?

[merry sphinx]

Try to factor it as a perfect square? I don't understand

[prisma prairie]

okay so i have this question

x^2+2x-8

how do i tex it

=tex x^2+2x-8

$x^2+2x-8$

[obsidian monolith]

The-Elite:

[prisma prairie]

ok

so i have this

[merry sphinx]

And what do you want to know? if it is a perfect square trinomial?

[prisma prairie]

yeah

so that i can make it in (x+number)(x+number)

or whatever

i want to make it into that form

[merry sphinx]

Do the 2 "numbers" need to be the same

Cause then you are looking to just factor it, not make a perfect square

[prisma prairie]

no

i want to find the zeros

of the equation

[merry sphinx]

Okay

Factor it by finding 2 numbers such that they sum to +2 and their product is -8

[prisma prairie]

is that the only option i have to solve this?

[merry sphinx]

You can use the quadratic formula

[prisma prairie]

anything else?

[merry sphinx]

Besides those, not really off the top of my head

There are a couple of special cases

[prisma prairie]

i watched a video on this guy completing the square but when i do that way i get (x+1)(x+1)-9

which are not the zeros of the equation

[merry sphinx]

Ohh, so you are trying to complete the square

[prisma prairie]

does completing the square find me the zeros of a quadratic equation?

[serene heath]

You can complete the square too yea

[merry sphinx]

Yes it can

[prisma prairie]

oh but then why am i getting (x+1)(x+1)-9

maybe i am doing it wrong?

[merry sphinx]

No that is correct

First I would suggest writing it as (x+1)^2 - 9

Second, you know this is equal to 0 so solve it

[prisma prairie]

but when i go on demos, the zeros are at x = 2 and x = -4

[merry sphinx]

wait a second

Yes? the roots are 2 and -4

$(x+1)^{2}-9=0$

[obsidian monolith]

pocofrosty12:

[prisma prairie]

oh i just have to solve for 0 now

when you square root both sides, you have to do +- right

• or -

[merry sphinx]

Yes, so solve it once as -3 and again as 3

[prisma prairie]

ok yeah i got the answer

so what is the reason for making it + or -

only for quadratic equations right?

[merry sphinx]

For any even powers or roots really

[prisma prairie]

oh yeah

true

that makes sense

okay i get this now

thanks for the help

[merry sphinx]

Cause (-3)^2 = 9 and so does 3^2

[prisma prairie]

yeah

[merry sphinx]

no problem

[prisma prairie]

how would you complete the square if the coefficient in front of the x is an odd number

$x^2+7x+12$

[obsidian monolith]

The-Elite:

[stuck lark]

complete as you normally would

[prisma prairie]

even if i get a fraction?

[stuck lark]

what do you have against fractions?

[merry sphinx]

Yeah, normally you will get fractions, especially if you can't factor normally

[prisma prairie]

hehe

okay

fractions are spooky

[kind moth]

Anyone here familiar with Interest formulas and stuff?

https://gyazo.com/c8ea7909ebf54df405eeb53c54809064

[proud sparrow]

just write out any formulas you may need

[kind moth]

This is the problem I have, I plugged in the correct information to the equation and it seems I cant get the right answers

[proud sparrow]

let's see your equations?

[kind moth]

sure one second

my phones almost dead I have to grab a charger

So I made Y (treasury note 2) = 2000 + x

since that note had a profit of whatever X was + 2000

I plugged in the interest rates

and used the total amount I would get from both

which is 7470

but after leaving X alone, I get a very high number that isnt right

7470/.17 for instance is 43,941.175

[proud sparrow]

hmm

[kind moth]

Looking it now I think I plugged in the wrong equation

its asking for the amount for each, which I am unsure how to find

[prisma prairie]

in a rational function, the common factor would be the location of the hole?

[rose locust]

can someone help me with graphing calculators? I dont know what my window should be set to based on equations such as -490t^2 +75t + 12

[scenic meteor]

can someone help me with business precalculus stuff?

dm me if you can

its 2-3 questions

[patent beacon]

I won't dm, but feel free to post it here

[scenic meteor]

An investment grows by 45% over a 5 year period. What is its effective annual percent growth rate?

An investment grows by 4% per year for 40 years. By what percent does it increase over the 40 year period?

thank you !

[proud sparrow]

@scenic meteor first one, simple or compond?

Second one, I suppose compound

can you set up your equations?

[scenic meteor]

yeah

for the 2nd one

i tried

.04^40

[grave gate]

1.04^40?

[scenic meteor]

the 1st one i tried .45^1/5

yeah

i meant 1.04^40

[grave gate]

Ok

[scenic meteor]

its online homework and i have to round 3 decimal places

for the 1st one

[grave gate]

You can probably use a calculator

[scenic meteor]

i did

[grave gate]

So you're good?

[scenic meteor]

its asking for % tho

no it still says its wrong

want me to share screens?

[grave gate]

Nah its fine

When you get 1.0blahblah do you subtract the one and multiply by 100?

[scenic meteor]

when do i get 1.0blahblah

[grave gate]

When you take the fifth root of 1.45?

[scenic meteor]

ohhhh

i've been doing .45

instead of 1.45

[grave gate]

Oh ok

[scenic meteor]

OMG

TY

1.0blah blah -1

i get my percentage

7.714%

thanks

ok

now one more problemo

[grave gate]

Ok

[prisma prairie]

in a rational function, the common factor would be the location of the hole?

reposting

[grave gate]

What do you mean by common factor?

[proud sparrow]

if the numerator and denominator have a common factor...

[prisma prairie]

yeah

[proud sparrow]

I suppose @prisma prairie

depends

[prisma prairie]

what element said

[grave gate]

How can a common factor be a location?

[proud sparrow]

$\frac{x^3+1}{x^4+x^2+1}$

[prisma prairie]

when you make it = 0

[obsidian monolith]

Element118:

[proud sparrow]

find the common factor

it doesn't create a hole in the function

[prisma prairie]

true

so when would it?

[grave gate]

When the denominator is 0, due to one of its factor being 0

[prisma prairie]

hmmmmmm

[scenic meteor]

An investment grows by 4% per year for 40 years. By what percent does it increase over the 40 year period?

need help with thjs last one

[grave gate]

By 1.04^40 -1

And then to percent

[prisma prairie]

@grave gate

$\frac{\left(x^{2}+7x+12\right)}{x^{2}+2x-8}$

[obsidian monolith]

The-Elite:

[grave gate]

What about it

[prisma prairie]

$\frac{\left(x+3\right)\left(x+4\right)}{\left(x+4\right)\left(x-2\right)}$

[grave gate]

Yes

[prisma prairie]

but why did this have a hole at x = -4

[uncut mulch]

bottom doenst look right

[grave gate]

Ye

It should be different

[flint stirrup]

plug in negative 4 to original equation

also hi elite

[obsidian monolith]

The-Elite:

[prisma prairie]

ok so why is there a hole in x = -4

[grave gate]

Oh I get what you were referring to

[prisma prairie]

yeah

[flint stirrup]

are you doing calc rn?

[grave gate]

The discontinuity because if 0/0

[prisma prairie]

hi galilei

[flint stirrup]

spoiler alert

[prisma prairie]

yeah calc

[flint stirrup]

i thot you was linear alg

[prisma prairie]

yeah but this is another course

im taking 4 courses

[flint stirrup]

woah

dayum

[prisma prairie]

statistics, computer science, linear algebra and calculus

[flint stirrup]

whats your major?

[prisma prairie]

CS

[flint stirrup]

ooh me too

i feel like we had this talk

[prisma prairie]

yeah

we did

lol

[flint stirrup]

rip

[prisma prairie]

wait

wth

[flint stirrup]

:/

[prisma prairie]

im confused

@grave gate The discontinuity because if 0/0 what?

[flint stirrup]

If you need help with any of those ask me in the Tagu server, it needs attention

ill take over

[prisma prairie]

oh ok

[flint stirrup]

basically whenever a function has a 0/0 at a point

that point is a hole

try plugging in -4 into the original equation

[hidden nimbus]

im being big dumb

so this is a velocity question

they only give me the equation

how would you find the time when the object thrown reaches the height

like not with a graph

[grave gate]

With subtracting the height from both sides and using the quadratic formula?

[delicate thistle]

I need help with finding tan(θ) my bros

[uncut mulch]

in what context?

[delicate thistle]

i just need help finding tan θ if sec θ =√(42)/6 and sin θ < 0

[smoky dirge]

Recall that $1 + \tan^{2}(x) = \sec^{2}(x)$

[obsidian monolith]

hadeedji:

[uncut mulch]

and use the given properties to determine whether the solution should be positive or negative

[delicate thistle]

not to sound dummy but the x is theta or the x is the results after the equals sign?

[radiant spade]

x is the given angle

[delicate thistle]

oh!!

lol!!

[delicate thistle]

how do you find the given angle?

whatever I'm doing for x isn't working I guess

[prisma prairie]

how can i find the domain of this?

[merry sphinx]

Assume that the domain is all reals

Then look for numbers that would give an undefined/impossible value

Example is division by 0, square root of a negative, etc.

[prisma prairie]

true

okay

working on it

thx

[viscid thistle]

Hi, noob here, can someone give me the most catchiest, simplest explanation for differentiating functions and relations?

[flint stirrup]

look for vertical asymptotes and holes

@viscid thistle is this for some class cool thing

or for understanding

[viscid thistle]

Nah, just for mnemonics

[flint stirrup]

oh cool

[viscid thistle]

Yeah understanding

[flint stirrup]

oh

[viscid thistle]

Yep

[flint stirrup]

is my goal here to sound cool or be thorough

[viscid thistle]

Just catchy, and specific, if that makes sense

[flint stirrup]

hmm

"differentiating functions and relations"

like finding instant rate of change at a point?

[viscid thistle]

No, just pointing out the difference between them

Not much to it haha

[flint stirrup]

i dont think im hip enough for this

good luck lol

[viscid thistle]

Ah fine

Or it doesnt have to be catchy

Just straightforward then

[flint stirrup]

i dont even know exactly what differentiating functions and relations means exactly

OHHH

FINDING THE DIFFERENCE BETWEEN THEM

[viscid thistle]

Yes

[flint stirrup]

i was thinking of finding the derivative (calculus)

[viscid thistle]

Sorry if im somehow not clear

[flint stirrup]

no thats totally mt wrong perception

[willow bear]

you could've said "telling apart"

for clarity

[flint stirrup]

nah, he isnt in calc

[viscid thistle]

Arent they the same thing?

[flint stirrup]

i dont think he is very familiar with differentiating in math terms

anyway

[viscid thistle]

Or is differentiating, a jargon here

[flint stirrup]

you will learn the mathematical definition to "differentiating" in your next math class

anyway

[viscid thistle]

I see

[flint stirrup]

lets first define "relations" and "functions" seperately

[viscid thistle]

Sure

[flint stirrup]

a relation has an input and output. they are related as the output depends on the input

but in a relation, an input could have multiple outputs

for example, lets say someone walks into a store with 20$. with that money, they couldve bought 1 burger, 2 burgers, 3 burgers, etc

there are multiple different possible outputs for this one input of 20$

now a function is a relation, however there is only one output for each input

[viscid thistle]

Mhm

[flint stirrup]

for example, if I work 40 hours this week, i will get exactly 200$ in my paycheck, no more no less

basically

relation: input leads to 1+ outputs
function: input leads to exactly 1 output

[viscid thistle]

Ah

[flint stirrup]

now, a catchy phrase

how long

like a sentence or two?

[viscid thistle]

Doesnt have to be too catchy

Perhaps something that gets the point across

Less sentences as possible, so it’ll stick

[flint stirrup]

a relation can have multiple outputs per input, related and fun. a function has one output for each input, like a relation but one.

heck i should not be a rapper good luck dont use this its bad cya good luck ahhh

i tried to rhyme

[viscid thistle]

Doesnt have to rhyme haha

But thanks

[flint stirrup]

:c

[viscid thistle]

I’ll try to chop some words out to fit my needs

Thanks again

[flint stirrup]

:/

[frozen merlin]

So i just finished my first out of two minor tests leading up to finals at the end of this year. One of the A-questions i just couldn't wrap my head around, and i can't get it out of my head. The question reads as follows:

Given the function y=1/x. Prove that no matter which point on the curve, the tangent will together with the positive x- and y-axis form a triangle with an area of 2 a.u. Please denote (x , y) as (a , 1/a).

I couldn't find anyone in my class that was confident that they answered this one correctly. Would appreciate some light on this one. Thanks!

[merry sphinx]

I like proofs

What is a.u.? @frozen merlin

[frozen merlin]

area units

[merry sphinx]

What is area units? Like a 1 x 1 square of the cartesian plane?

Can I get a drawing as an example triangle?

[frozen merlin]

Let's say you take two steps to the right and 3 steps up to form a rectangle, assuming all steps are equally large, that rectangle would have an area of 6 area units.

just think of it as base x height / 2 = 2

in this case

[merry sphinx]

So we don't care about the negative curve in this case

[frozen merlin]

correct

[uncut mulch]

are you able to find the equation of the tangent at (a, 1/a)?

@frozen merlin

[merry sphinx]

This is not rigorous, but think about the relationship between the coordinates and triangle formed

Think about splitting up the big triangle into 4 similar ones with equal area

Find the area of the first triangle (Hint: (a, 1/a)) and multiply that by 4

I think that works

[frozen merlin]

Oh yeah, i can visualise it now. Still, what if i wanted to do it strictly algebraically?

[uncut mulch]

equation of tangent line
find where it intercepts the axes
area of a triangle

[frozen merlin]

y' = -1 / x^2 , y = kx + m , y = m - 1/x , m = 2/a , y = 2/a - 1/x , does that seem correct?

and the intersections are the m-value and x-value when y=o ?

[uncut mulch]

no

firstly you should follow convention and use m for the gradient/slope

you found the derivative
y' = -1/x^2
what would the slope be at the point (a, 1/a)?

[frozen merlin]

-1 ?

[uncut mulch]

what is the value of x at (a, 1/a)?

[frozen merlin]

a

[uncut mulch]

and what would the slope be when x=a?

[frozen merlin]

• 1/a^2

[uncut mulch]

yes

now you have the slope at a given point
are you able to find the equation of the tangent line using that information?

[frozen merlin]

2/a ?

is m

so, 2/a - (1/a^2)x = y

[uncut mulch]

firstly you should follow convention and use m for the gradient/slope

but your equation is correct

the variable m is pretty much reserved for slopes

[frozen merlin]

oh

[uncut mulch]

the y-intercept form is usually written as y = mx + b

[frozen merlin]

in sweden we write y = kx + m

but i figured i could just write x and y in terms of a, so i got m - (1/a^2)a = 1/a

and derived m to be 2/a

and form there it's pretty straight forward

thank you btw! and poco too

[merry sphinx]

That was pretty cool question

[uncut mulch]

oooh didn't know why they do that in sweden

i wouldn't mind as much if they didn't use m for the y-intercept

[frozen merlin]

well, thanks anyways!

[proud raven]

in sweden we write kx + m

savages

[uncut mulch]

worse than non metric units

[proud raven]

i got triggered for a second

[flint river]

Uh, Im clueless on what to do here, translation is something like:

Given f has a (?) continuous derivative (?) such that lim x-> ... blah (expression) = 8...

Find f(0), f'(0)

any hints???

[proud raven]

[flint river]

[obsidian monolith]

jan Niku:

[proud raven]

does that mean anything

that e^f(x) and f(x) being present makes me think theres some sort of differentiation trickery afoot

[serene heath]

Hmm

Try splitting up that limit maybe

[proud raven]

have you guys covered limits at infinity

i wonder if they want you to get it into some weird form

0/0 or something

[flint river]

yep we did

[serene heath]

I think splitting it helps

Do you know what the answers are?

[flint river]

nope 😦

[serene heath]

Did u try what I suggested

[flint river]

still got nowhere

let me write it

you mean ((1st term from numerator)/ x) + ((2nd term idem)/ x) ???

sorry i'd take ages to latex it

[serene heath]

Yea

So u can cancel that first x on the top

[flint river]

hmm you mean to solve for f(x)??? I thought operating arithmetically on such a limit equation wasn't allowed...
like...

[obsidian monolith]

ike98:

[serene heath]

Why infinity

[flint river]

yeah, sorry, 0

[serene heath]

Oh is that an example

[flint river]

yep

[serene heath]

It's fine to split if u know both limits exist

But like we might as well try this approach and look for a way to justify the manipulation

[flint river]

well I "solved for f(x)" and got the same result above which Jan Niku posted... but that doesn't make sense to me

😦

[green zenith]

Hey guys! Quick quesh. What is it called when you solve (x-3)^3?

[flint river]

cube of a binomial?

[green zenith]

that is it!

thanks!

[serene heath]

So u have $\lim_{x \to 0} f(x)-7+3\left[\frac{e^{f(x)}-2}{x}\right]=8$

[obsidian monolith]

the one n only:

[flint river]

right

[serene heath]

We know f is differentiable

So that means e^f is also differentiable and we can split it like that

Ie that square bracket limit alone exists

U with me ?

[flint river]

Im here but lost honestly

[serene heath]

Which part

I just split it up into 2 bits

[flint river]

yeah but I don't know how to continue

[serene heath]

I'm asking if ur good so far

[flint river]

I'm all ears 🙂

[serene heath]

Ok so we know that square limit exists

Which means the numerator has to also approach 0

Ie e^f(0)=2

[flint river]

woulnd't that be indeterminate 0/0 ?

[serene heath]

Well yea

But we know the limit exists

[flint river]

ok

[serene heath]

If the top didnt approach 0 but rather a finite value while the bottom went to 0 then it would be dne

So now we know f(0)

[flint river]

If the top didnt approach 0 but rather a finite value while the bottom went to 0 then it would be dne
oh, makes sense!

[serene heath]

Ok

So we know f(0) now yes?

[flint river]

indeed

[serene heath]

Can u work out that square limit now

[flint river]

let me see

square limit, wdym?

[serene heath]

The one inside the square bracket

[flint river]

ok

hmmm

so, some L'hopital there and we get e^(f(x)) ???

[serene heath]

Do u?

What's the derivative of e^f(x)

[flint river]

oh right, e^f(x) * (f(x))' ?

[serene heath]

Yea

So what's ur limit

[flint river]

but f'(x) is unknow

im stuck

[serene heath]

In terms of f'(0) what is it

[flint river]

idk

anyways I guess it's too tough for me this one, I'm giving up... thanks for the help!

[serene heath]

Its f'(0)e^f(0)

But np

[green zenith]

hey guys! i need some help with some basic quadratic equations

[merry sphinx]

Post it then

[green zenith]

so if i have x^2-4^2 in an equation, would that turn into x^2-16 or x^2+16?

[merry sphinx]

Depends on the specific notation of the equation

This is very important

[green zenith]

yeah im starting to realise it now

[merry sphinx]

(-x)^2 = x^2, or -(x^2) = -x^2

Same thing applies to numbers

In this case I would assume it is the first equation

[green zenith]

so it is x^2-16?

i dont really have any parenthesis in this equation

what i have is f(x)=sqrt(x^2-4^2)

[merry sphinx]

It is generally assumed that it is -(x^2) if there are no parenthesis

Cause -x^2 is -1 times (x^2)

-1 is just the coefficient

[green zenith]

yeah

okay i will just assume that it is then

and one other question

and im sorry to be bothering you with simple quadratic equations

[merry sphinx]

no problem, quadratics are important baselines for the future

[green zenith]

okay so im looking for the values above 0 of 4-sqrt(x^2-4^2)

so 4-sqrt(x^2-4^2) >/= 0

and then i have -sqrt(x^2-16)=-4

now when i square both sides, do i remove the minus sign in front of the square root?

[merry sphinx]

I would transfer the -1 by dividing or multiplying both sides by -1

Remember that you need to switch the greater than or less than signs if you divide/multiply by a negative

Then you can square both sides

[green zenith]

oh

so that sneaky little -1 there is going to be divided

[merry sphinx]

Yep

[green zenith]

before i square

huh

okay

thank you very much

[merry sphinx]

In this case, you can technically square, but build some good habits by always removing the coefficient before squaring

In the long run, this is useful

[green zenith]

okay

i will take a note of that

thanks

[formal iris]

Guys where can I learn all of preclac

[patent beacon]

Lang's Basic Mathematics is a good book, and will teach you everything you need to enter university mathematics

[formal iris]

facts?

[patent beacon]

And it's free online ofc. Just search the pdf

[formal iris]

any youtube lists?

[patent beacon]

For YouTube, you can't go wrong with Khan

I don't know if prof Leonard does precalc, but it's worth looking

[formal iris]

he does but he is recording this year

I have never proved anything so time to cry

any thing else I got that book

[patent beacon]

Hmm. Not too sure? For a particular thing, patrickjmt is very good, but you have to know what you're searching for

[formal iris]

like entire preclac

My teacher only got to Chapter 4 out of 12

Im in calc1 now

yes I know this thats why I asked for youtube first I have a book already

[formal iris]

If I send a link here can someome see the book for me and see if it is good for me to study frm

from

[proud sparrow]

we probably need the context of what you have learnt already

[green zenith]

Hey guys. I have a question.

If i want to inverse the function f(x)=12x^3-12, do i include the cube with the x when inverting?

Or do i leave it there for y?

[limber bone]

12x^3-12=y

12x^3=y+12

x^3=(y+12)/12

x=cuberoot((y+12)/12)

[green zenith]

Or x=cuberoot(y/12) + 1 to simplify?

Nvm guys i got it

Thanks @limber bone

[limber bone]

ofc

anything else?

[formal iris]

Lets Hypothetically say near nothing most of alge2

I know up to Trig Identities

[green zenith]

That is all for now

[green zenith]

hello guys i need another help

[limber bone]

jusut asl

just ask*

[green zenith]

so how do i simplify 33/4/4

sorry im really just not familiar with this type of form of fractions

[patent beacon]

/ isn't associative, you have to show the order
Is it (33/4)/4 or 33/(4/4)?

I'm guessing it's not the second one lol

[green zenith]

i think its (33/4)/4 4x>/_33/4

ah sorry, its from 4x>/=33/4

and again, sorry for a very noob question

[patent beacon]

4x = 33/4
Divide both sides by 4
x = 33/16

[green zenith]

so is it because you reciprocated the from under?

ie (33/16)4 => (33/16)(1/4)?

i should go to other channels for this one

ooopsie

[hoary valley]

Why it doesn't have 2cos x ?

...

This is the original formula..

[kindred rock]

We'll learn this together, and try to go through it.

[frozen needle]

it has 2cos x tho

[kindred rock]

$sin 2x = sin (x + x)$

[obsidian monolith]

vypr:

[kindred rock]

$sin(x + x) = sin(x) cos(x) + cos (x) sin (x)$

[obsidian monolith]

vypr:

[kindred rock]

which in turn equals $2 sin(x) cos(x)$

[obsidian monolith]

vypr:

[kindred rock]

so $sin(2x) = 2sin(x)cos(x)$

[obsidian monolith]

vypr:

[kindred rock]

make sense?

@hoary valley

forgive my bad latex, getting used to the thing

[uncut mulch]

\sin, \cos in the future

[hoary valley]

@kindred rock Give me a min I'm thinking about it, but thanks a lot

[kindred rock]

@uncut mulch MUCH appreciated.

@hoary valley Think of x = y

and as we know, 1+1 = 2(1), 2+2 = 2(2), etc.

[hoary valley]

Yeah we did that to the sin... but why we didn't do the same to cos

[kindred rock]

because it's $2(\sin(x) \cos(x))$

[obsidian monolith]

vypr:

[kindred rock]

it multiplies the whole thing

in latex it just removes that parenthesis to be more readable

because the whole thing is being multiplied

$2 * \sin(x) * \cos(x) = 2(\sin(x) \cos(x)) = 2 \sin(x) \cos(x)$

[obsidian monolith]

vypr:

[kindred rock]

no matter how you multiply it, it will end up being equal

@hoary valley

[hoary valley]

I get it now

@kindred rock Thank you so much ❤

[kindred rock]

exactly

not a problem ❤

[heady jewel]

lol

weird

[viscid thistle]

bruh

wholesome discord

[hoary valley]

Guys I verified this formula and I think it's true... but somehow it says that my answer is wrong...

[serene heath]

Its false

[frozen needle]

indeed

[hoary valley]

I will try verifying it again ..

[pale kettle]

how did you verfiy it?

[hoary valley]

I replaced cos^2x by 1-sin^2x

[frozen needle]

you can unverify it by plugging in 0 for example

[pale kettle]

What cos^2x did you replace

There's no cos^2(x) in this equation

[hoary valley]

Cos(2x)=sinx^2 - cosx^2

[uncut mulch]

missing the (x) also that isn't the correct identity

[hoary valley]

I replaced that cosx^2 with [1-sin^2x]

[uncut mulch]

$\cos(2x) \neq \sin^2(x) - \cos^2(x)$

[obsidian monolith]

ramonov:

[hoary valley]

Oh

Cos(2x)= cosx^2 - sinx^2 ?

I switched sin with cos ?

[uncut mulch]

write it as cos^2(x) - sin^2(x)

[hoary valley]

^ that's not equal to sin^2(x) - cos^2 (x) right?

[uncut mulch]

in plain text

its off by a factor of -1

[scenic musk]

how does one rationalize cube roots in numerator. For example (x^1/3)/5.

Do i multiply that rational by x^1/3 twice?

[patent beacon]

"That irrational"?

[scenic musk]

the one in my example

the line above