precalculus

[golden cairn]

Okay wait

So first I got: xlog(3/2)=(2-x)log5

That’s the first step I did

[hushed sphinx]

That looks wring.

If I start by just taking the logarithms I'd get

x log3 = log2 + (2-x)log5

[golden cairn]

The teacher said if there is something like this we should divide by 2 on both sides: 2x•5^2-x

[golden cairn]

Yeah I know but the teacher says to divided by 2 on both sides first

In a similar type of question

That I did before

[hushed sphinx]

We could start with that, but then you have (3^x)/2 on the left-hand side, not (3/2)^x.

Which would give you x log3 - log2 instead of xlog(3/2).

[golden cairn]

Yeah but log(3/2)^x can become xlog(3/2)

[golden cairn]

Yeah

[hushed sphinx]

But (3/2)^x is what you don't get when you divide 3^x by 2.

[golden cairn]

Oh

Once she made 2•3^x=5 into xlog3=log(5/2)

So I thought this was like a similar type of thing

I get 1.49 as the answer of this question but the actual answer is 1.445

So I don’t know what I’m doing wrong

[winter comet]

if you divide 3^x by 2, what do you get?

[golden cairn]

(3/2)^x

?

Maybe

[winter comet]

nope

[golden cairn]

(3^x/2)

[winter comet]

close, you might want to be more specific with your parenthesis

i mean its correct but

you might wanna say

( (3^x)/2 )

[golden cairn]

Okay

[winter comet]

it may help if i write it 1 sec

[golden cairn]

Okay

[golden cairn]

Oh yeah because (3/2)^x would mean the 2 also has the exponent of x

[winter comet]

exactly

7 mutual servers is crazy 😂

[golden cairn]

Oh yeah

Haha

I didn’t even notice that

[winter comet]

XD

but now you can take the log

of both sides

yk

[golden cairn]

Okay

I don’t know what I’m doing wrong

But I am still getting 1.49 instead of 1.445

[winter comet]

what did you get

[golden cairn]

1.49

As the answer

[winter comet]

._.

i mean

how did you get it

[golden cairn]

Wait

xlog^3=log2(2log5-xlog5)

I don’t even know what I did

Is this step even right?

[winter comet]

uh

i.. don't think so

[golden cairn]

Yeah I also tried like doing log(3^x/2)=2log5-xlog5

[winter comet]

thats correct

now use log(m/n) rule

log(m/n) = log(m) - log(n)

[golden cairn]

I don’t think I can do that

[winter comet]

why not?

[golden cairn]

Because there is like a 2 infront of log5 so

I don’t know

[winter comet]

not that one

[golden cairn]

Can you?

[winter comet]

the other side

[golden cairn]

Oh okay

[winter comet]

log( (3^x)/2 ) i meant

[golden cairn]

So log3^x-log2?

On the other side

[winter comet]

yeh

and do you know what log(3^x) is in another form?

[golden cairn]

Xlog3?

[winter comet]

yeeeeh

so now what is your equation

[golden cairn]

Wait one second

3logx-log2=2log5-xlog5

[winter comet]

yep

[golden cairn]

Sorry it took me a while my work was like all over the page and messy

[winter comet]

XD

do you know how to solve for x here?

wait

[golden cairn]

I put the logs with x infront of them on the same side?

[winter comet]

the first term is xlog(3) you meant right

[golden cairn]

Yeah oops

[winter comet]

ye

and then factor out the x

[golden cairn]

Ok

Yup

It worked

Thanks

1.44459

Wow thank you

You helped me out a lot today

[winter comet]

👍

[stiff kelp]

can anyone help me understand what log is?

[cursive cairn]

maybe this will help: https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:logs/x2ec2f6f830c9fb89:log-intro/a/intro-to-logarithms

[mortal vine]

ok the question says Find the area of this triangle with the sides 25,30,42 m and largest anlgle 100 degrees but idk what angles to use to find the area

mvm got it

[winter comet]

do you know heron's formula

ohk

[mortal vine]

1/2 ab sin(C)?

[winter comet]

nah but that works better

[willow bear]

hang on a minute

is this data even consistent

rounding to the nearest degree, the sides imply an angle of 99°, not 100°.

[winter comet]

how did you even calculate that ☠️

[willow bear]

cosine law lmao

[winter comet]

oh 😂

[willow bear]

if i had it my way i would have said the angle is 1.73 rad

[hollow rose]

a logarithm helps one find what power a number needs to be raised to in order to get a specific result

thinking about the formulas like this helped me:

logarithm form: log_a(b) = c
exponential form: a^c = b

(a, b, and c just make more intuitive sense to me than b, y, and x)

so
log_2(16) = 4
just means we need to raise 2 to the power of 4 to get 16, hence:
2^4 = 16

[novel creek]

looks good :3

[hollow rose]

thanks bro!

[gritty needle]

Yeah the logarithm is the inverse function for exponentials but like the square root function, it doesn't tell you really how to calculate such value for any number

[hollow rose]

hey beginner can you help me visualize this?

i understand that when we have an unknown argument you have to do the inverse of a log and on the other sign of the equal sign you have it in exponential form

but what exactly is the inverse of a log

like when we cancel it out what are we doing

[gritty needle]

An inverse function is just a function that "undoes" what the original function did, for example, if f(x)=y is your exponential function, then the inverse does this, f^-1(y)=x, say you got some function f(x)=2^x then f(4)=2^4=16 then say g is the inverse of f, then g(16)=g(2^4)=4

Now, we can leave g like that, but what g if we wanted to generalize for any "base" we could conveniently name this function and call it "logarithm" since it appears to be useful

And say g(x)=log2(x)

But of course, putting a name for an inverse function to be convenient doesnt mean you can magically plug any value in it and calculage the result

There are algorithms (different from logarithms) that help us calculate the values for anything we plug in those inverse functions and that's a whole different difficult subject but if you understand that what an inverse function does is to "undo" what the original function does, you are gold

Now, i believe you also want a visial explaination for what an exponent does to some number and what a logarithm does, well, different exponent do different things to number, the exponent 2, for example, makes a line into a square, or a cube if it were 3, root takes that square/cube and makes it into a line, the logarithm work with a fixed length line and takes some shape, be it a square, cube, etc and tells you its dimension if you want to think of it that way

I bet someone else has a better analogy, but this is what i came up on the go

[proven night]

Does anyone have any idea?

[queen shoal]

hmmm

are we finding x?

[queen shoal]

x=2

[proven night]

How

What did you begin with

[queen shoal]

i guessed 1 then 2 lol

💀

[proven night]

My first thought was dividing both sides by one of the number (14^x for example)

But it didnt give me anything

[proven night]

I had seen similar solution before

[queen shoal]

you could also maybe find a number that when rooted by idk like 5 or a different number will equal 10,11,12,13,14

for example 16 = 2^4 and 4^2

idk

probably wont wrok

[hushed sphinx]

Pretty sure there is not any nice analytic solution to equations of that shape in general.

[brittle lava]

impossible to solve without graphing or guessing or binomial theorem (it could help long and maybe it wont works too for the binomial)

[stiff kelp]

square both sides

oh wait nvm im stupid

[glad solar]

hi

im new

[glad solar]

maybe find log base x of both sides?

or am i stupid

idk

[hushed sphinx]

That won't help, because there are no nice rules for the logarithm of a sum.

[glad solar]

true

i think just graph it

or maybe differentiate both sides?

nvm that won't work either

[weak relic]

Somone give me a Calc bc differential equation difficulty

[stone blade]

make it into a n = 10 and the whole eqn as (n+1) and things and expand and see where that gets you?

idk its getting too complicated

[silent oasis]

why do horizontal asymptotes can be crossed?

[willow bear]

y = x/(x^2+1)

has a horizontal asymptote at y=0 and crosses it

@silent oasis does this answer your question?

[silent oasis]

Why can't vertical be crossed?

[silent oasis]

yeah.

[willow bear]

are you sure that you asked the right question just now

ah, knew it. typo on your part.

[silent oasis]

my bad

[willow bear]

for a vertical asymptote of x=a to be crossed, your function would have to be defined at a (which in the overwhelming majority of cases it isn't) and also be continuous at a -- but for a vertical asymptote, by definition, lim[x -> a±] f(x) = ±∞

so having that limit be equal to f(a) is out of the question

[silent oasis]

uhh. ok. thanks

[queen shoal]

I learnt integration today so I tried it with a triangle but it does not work pls help

[willow bear]

f(x) = 5 - x, not -x.

you forgot the constant of integration.

also you want to integrate f and not f'.

also you might wanna get some better paper than napkins.

[queen shoal]

Ok

[silent oasis]

How do I do this?

[queen shoal]

How do I figure out the constant of integration

[willow bear]

ok one person at a time

@queen shoal move to #calculus

[queen shoal]

Lol ok

[willow bear]

@silent oasis do you know in general how to calculate limits at infinity? y/n

[silent oasis]

y

[willow bear]

ok

ok, in that case: factor out 4x^4 inside the root.

4x^4 - x = 4x^4 (1 - 1/4 x^-3)

and bring that 4x^4 outside the root, but mind your signs.

[silent oasis]

and then?

[willow bear]

show me what you get

write it out in full and show me.

[silent oasis]

ok. one moment

[willow bear]

also in general

idk about other helpers

but when i give out an instruction

the expectation is that you, as the recipient, will do one of three things:

1. follow it, then send me the result, and then optionally ask for the next step
2. attempt you best to follow it, then send me the result along with clarification like "i did not understand fully what you wanted me to do, but i tried anyway"
3. if you cannot understand my instruction enough to follow it, then say so, such as "i don't understand <part of instruction>. what do you mean by this?"

understood? @silent oasis

[silent oasis]

yes.

[silent oasis]

actually i haven't been able to do anything after what you said here.

i think i am lacking basics of finding limits at infinity

[willow bear]

and bring that 4x^4 outside the root, but mind your signs.

you know $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$ yes?

[obsidian monolith]

|Ann⟩

[silent oasis]

yes

[willow bear]

so $\sqrt{4x^4 - x} = \sqrt{4x^4} \cdot \sqrt{1 - \frac14x^{-3}}$

[obsidian monolith]

|Ann⟩

[silent oasis]

is that the principal root?

[willow bear]

we're in the real numbers

so yes obviously the principal root

$\sqrt{4x^4}$ can be simplified --- mind that $x \to -\infty$, so $x$ is negative. you will need this info to simplify $\sqrt{4x^4}$.

[obsidian monolith]

|Ann⟩

[silent oasis]

2x^2

[willow bear]

correct

so now your limit equals $\lim_{x \to -\infty} \frac{2x^2 \sqrt{1 - \frac14 x^{-3}}}{2x^2 + 3}$

[obsidian monolith]

|Ann⟩

[willow bear]

do you see how to proceed, or should i give you another hint

[silent oasis]

after this should i divide both the numerator and denominator by 2x^2

[willow bear]

depends on what you mean when you say "should"

if you mean "is there an obligation to do this, to the exclusion of all else": no.
if you mean "is it legal to do this": yes.
if you mean "will doing this help me": maybe. try it without asking for my blessing!

[silent oasis]

i meant all three.

[willow bear]

then i have given you answers to all 3.

[silent oasis]

indeed

[willow bear]

ok so to be explicit

do what you suggested (divide num and denom by 2x^2)
but not because of an obligation
instead, do it because you want to see where it takes you.

[obsidian monolith]

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

[silent oasis]

$\lim_{x \to -\infty} \frac{ \sqrt{1 - \frac14 x^{-3}}}{1 + \frac{3}{2x^2}}$

[obsidian monolith]

Nischal

[silent oasis]

i don't know what to do with roots

[willow bear]

alright, good.

this is exactly where i wanted you to get to.

now remember: what is the first thing that one ought to try when looking at a limit problem like this?

just any limit problem in general

the very first and simplest thing to try

[silent oasis]

substitution?

[willow bear]

right.

can you substitute x = -∞ here?

[silent oasis]

no

[willow bear]

why not?

[silent oasis]

what's -infinity to the power -3?

[willow bear]

0, of course

$\frac{1}{\infty} = 0$. all that's left of your limits is \textit{negative powers} of $x$, which approach zero!

[obsidian monolith]

|Ann⟩

[queen shoal]

you can use l hospitals rule to find the derivative of 1 is 0 and 0 divide anything is 0

[silent oasis]

I have just begun with limits. I haven't reached differentiation yet.

[queen shoal]

oh ok

the derivative of any constant is 0

so it doesnt include x or variable

[silent oasis]

$\frac{1}{1-\frac{3}{2*0}}$ after substitution of x with negative infinity

[queen shoal]

so if you see lim (something) c/x where c is a constant then the limit is 0

[obsidian monolith]

Nischal

[silent oasis]

is it a rule?

[queen shoal]

in l hospitals rule it is the derivative of the numerator divided by the denominator is the limit so if the numerator is a constant then its derivative is 0

[silent oasis]

can you tell me how to do without using L hospital's rule

[queen shoal]

idk how to do limits

i only know derivatives

and integration (maybe)

[silent oasis]

uhh ok. thanks

[queen shoal]

you can just remember that the derivative of any constant is 0

[silent oasis]

after that

[queen shoal]

wdym

[silent oasis]

i mean where do i use this?

[queen shoal]

if you have lets say lim > OO constant/x

OO is infinity

you know that the derivative of constant is 0

so using l hopsitals rule you get 0/derivative of x

which is 0

[silent oasis]

so 0/0?

[queen shoal]

i think if it is 0/0 then the limit is undefined?

im not too sure i only learnt this yesterday

[willow bear]

incorrect

you needed to substitute x with -∞, but you substituted it with 0 on the bottom.

[willow bear]

this is also incorrect

[queen shoal]

ok

[willow bear]

l'hôpital is way overkill for this limit

[queen shoal]

how would we find the answer

[willow bear]

$\lim_{x \to -\infty} \frac{ \sqrt{1 - \frac14 x^{-3}}}{1 + \frac{3}{2} x^{-2}}$

[obsidian monolith]

|Ann⟩

[queen shoal]

ok

[willow bear]

$x^{-3}$ and $x^{-2}$ both approach 0

[obsidian monolith]

|Ann⟩

[queen shoal]

i only know l hopsital

idk how to find limits without l hospital

[willow bear]

there are various algebraic tricks

[queen shoal]

ok

like what

[silent oasis]

how can you claim that?

[willow bear]

x goes to -∞, yes?

[silent oasis]

yes

[willow bear]

do you agree or disagree that $\lim_{x \to -\infty} \frac{1}{x^2} = 0$?

[obsidian monolith]

|Ann⟩

[willow bear]

DON'T bring l'hôpital into this.

[queen shoal]

ok fine

[willow bear]

this is a simple matter.

[silent oasis]

yes

[willow bear]

do you agree or disagree that $\lim_{x \to -\infty} \frac{1}{x^3} = 0$?

[obsidian monolith]

|Ann⟩

[silent oasis]

i agree

[willow bear]

ok then what caused your initial shock/disagreement/whatever?

[silent oasis]

this thing

[willow bear]

$\lim_{x \to \pm\infty} \frac{1}{x^n} = 0$ for any natural (or in fact positive) $n$

[obsidian monolith]

|Ann⟩

[willow bear]

(with the caveat that for n ∉ Z we don't speak of x -> -∞. but in our case that doesn't apply)

$\frac{1}{\infty} = 0$ is a convenient and semi-rigorous shorthand.

[obsidian monolith]

|Ann⟩

[silent oasis]

uhh ok

so is the ans 1?

[willow bear]

sure is.

[silent oasis]

i will revise this once again.

[hexed totem]

hi guys

someone can resolve this?

[willow bear]

!status

[tender quest]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[hexed totem]

1

[silent oasis]

Make a table try substituting values greater than zero like 0.1, 0.01 and so on

[hexed totem]

but isnt that an indermination?

[willow bear]

"indeterminate form" means only that direct substitution doesn't work, nothing else

[silent oasis]

don't put x=0.

[hexed totem]

?

[silent oasis]

you get 0/0 when you plug x=0

yes or no?

[hexed totem]

ya

[silent oasis]

so what does that + denote?

do you know?

[hexed totem]

ya ik

its like the values in right of zero

[silent oasis]

yeah. so you've to get values that are closer and closer to zero from the right.

for that plug 0.1,0.01,0.001 to the eqn and see what value the limit approaches

[hexed totem]

i didnt learned that form at school

[willow bear]

there is a better way than nischal's suggestion

[hexed totem]

i think its supposed to work with that

[willow bear]

but you need to turn the 3rd limit in that pic on its head a bit

lim[t->0] ln(1+t)/t = 1

[hushed sphinx]

On the other hand, if plugging in small values doesn't even make sense for Nina, then it's high time she learned about it.

[hushed canyon]

Is sen supposed to be sin?

[hushed sphinx]

Even if there are better ways to comput the limit once you know what it means.

[willow bear]

yeah it's a Spanish thing iirc

[hushed canyon]

That seems more French imo

[hexed totem]

im so confused

[willow bear]

nope, it's Spanish

or maybe Portuguese

[hexed totem]

nope its portuguese

[hushed canyon]

What is "p" supposed to be in the 5th problem?

[willow bear]

it isn't a problem, but p is supposed to be any real number.

[hushed canyon]

I speak Spanish ToT

Ohhh ic

[hushed sphinx]

The formulas you posted a picture of are good to know in general, but they won't help you with this limit.

[willow bear]

the third one kinda sorta can

[hushed canyon]

It's really interesting seeing this kind of education in other countries

[hexed totem]

its real number

[willow bear]

but only with some preprocessing

[hexed totem]

this is the only thing I learned to resolve so idk

i will try with the third one

[hushed sphinx]

Did you learn a definition of "limit"?

[hexed totem]

of course

[hushed sphinx]

Can you tell us what it is?

[hexed totem]

for me its dificult to explain cause i need to do that in english

but iis like

when a fuction its getting closer to a number but never will be that number

[hushed sphinx]

I think it will be useful for you to break the definition down in small enough parts that you can translate them to English one by one.

Not for the sake of the English, but because it sounds like you haven't looked at the definition you have in enough detail to really understand what it means.

[hushed canyon]

Is it possible to factor and cancel out values from the fraction? 😭

I forgor the rule when it comes to limits lol

[silent oasis]

i don't think so

[hushed canyon]

Man

[hexed totem]

let f be a function defined on an open interval that contains the number A, except possibly on A itself. Then we say that the limit of f(x) as x tends to A is L.

this

is what i learned

[hushed sphinx]

This doesn't tell us anything about what L is at all.

[hexed totem]

is the result

[hushed sphinx]

Okay, you're missing half of the definition, and you have to learn and understand that half before you'll be able to make sense of the limit you started with.

You'll probably need help from someone who speaks Spanish Portuguese who can look at a picture from your textbook and walk you through it. And I don't.

[silent oasis]

did you solve your initial problem?

[hushed canyon]

Isn't her language Portuguese?

[silent oasis]

maybe both are quite similar, probably

[hexed totem]

but that is not the question cause i dont work with the definition a few years ago

[hushed canyon]

They're similar but not the same language

[hexed totem]

we need to know the notable limits and resolve

[hexed totem]

for us, portugueses is easy to understand spanish but for them no ahah

[hushed sphinx]

Apologies, I didn't see you had stated Portuguese.

[silent oasis]

yeap.

[hushed canyon]

Actually it's the same for Spanish, at least for me I can understand some Portuguese without a translator

Music from Brazil is so nice ngl

I rlly like Marcos Valle

[silent oasis]

is Spanish from Spain?

[hushed canyon]

Btw for ur problem, idk if you're using a text book but if you are, what topic is your problem under?

[hexed totem]

but my Portuguese is from Portugal

is completely diferent

[hushed canyon]

Yeah but people in most Latino countries also speak Spanish

[hushed canyon]

Ohhhh

I heard from my mom people from Portugal are very nice

I kinda wanna visit the country sometime

[hexed totem]

im not using, this exercise appeared in an exam of my professor last year

[hushed canyon]

Ohhh I see

[hushed canyon]

Man I wish I could help but idk much about natural logarithms

I'm unfamiliar

[hexed totem]

dw

its ok

i will ask some colleagues

[hushed canyon]

Piecewise? 💀

Is this the same problem?

[hexed totem]

yeah ahahahha

[hushed canyon]

I didn't have this in my precalculus class, this is new to me

[hexed totem]

ohh ok dw

here in portugal we have a better education than in other countries

so its normal

[viscid thistle]

bro at this point give up

[hushed canyon]

H o

[silent oasis]

better education!! what do you mean?

[hushed canyon]

Then I very kindly ask that u send ur math problems more often 👍👍👍

[hexed totem]

yeah its true

[hushed canyon]

I'll see if I can find a video on the topic

[hexed totem]

some students when go to other countries in erasmus need to do much more hours to get equivalence

[silent oasis]

what grade is that question from?

[hexed totem]

12

last year of secondary school

but we learned that last year without logaritms

[silent oasis]

ohh! I see

[hushed canyon]

Regarding this, is it required to graph the equation?

[hexed totem]

no?

in the first one its to study the continuity

[hushed canyon]

Yey oki

[viscid thistle]

Can you translate it?

[hushed canyon]

So in the x > 0 part I could use any positive number?

[hexed totem]

the second one
horizontal asymptotes

[hushed canyon]

A h

[hexed totem]

no

you do the right and left limit

and needs to be equal a -1

[hushed canyon]

Ohhhhhhhhh

I think I get it now

[hexed totem]

good 🙂

[hushed canyon]

Oh not the answer just like what I'm supposed to do

I'm sorry

[hexed totem]

right dont worry i just give up

[hushed canyon]

:(

[hexed totem]

guys

i got it

[little sonnet]

the answer is 0, obviously. 🙄

i was fucking around

was that actually the answer

[vocal vapor]

no

[little sonnet]

liar

h = 0

[hexed totem]

Wut?

[charred epoch]

How do I determine if 2x+7 is a factor of 8x³+16x²+18x-602?

I thought I was supposed to use synthetic division or long polynomial division but I keep getting the wrong answer

[hushed sphinx]

The quickest is just to check if x=-3½ is a root of the cubic,

[winter comet]

where did you get that number from

oh

2x+7 = 0

gotcha

[hushed sphinx]

That's the root of 2x+7

[winter comet]

i see

and then you plug it in

and see if it still 0

i've never thought of that 💀

[hushed sphinx]

Yeah,

Synthetic division essentially makes the same computation, but with extra bells and whistles such that it can also tell you the quotient polynomial.

[charred epoch]

The question also asks for a remainder if it's not a factor @hushed sphinx

So it wants me to do division I guess but I must be doing it wrong

[winter comet]

oh, then yea you'd have to use synthetic division

or just long division i don't really know the difference too much

synthetic is just without the variables in like a matrix sort of way right 💀

[hushed sphinx]

If you just want the remainder and don't care about the quotient, then the value of the cubic at x=-3.5 is exactly that remainder!

[winter comet]

wait a minute 💀

why does that work?

[hushed sphinx]

Let p(x) = 8x^3+...-602.

[winter comet]

mk

[hushed sphinx]

Polynomial division then gives you a polynomial q and constant c such that
p(x) = (2x-7)q(x) + c
for all x.

If you plug in an x such that 2x-7 = 0, then the (2x-7)q(x) term disappears and all that is left is
p(x) = c
for that particular x.

[winter comet]

oh i see

nice proof XD

[charred epoch]

What if it goes up to power 4?

Oh I figured it out

I just evaluate f(6) = polynomial

I guess I was overcomplicating it

[half pollen]

yeah f(6) gives you the remainder

if the remainder is 0 then it's a factor

[wind bridge]

guys can soemone help me pls

[raw hill]

[wind bridge]

well its this question

[raw hill]

Do you mind sending a picture?

I'm not comfortable w/ downloading random files

[tranquil inlet]

Can someone explain what the dot product of 2 vectors actually gives us

I’ve heard it give you a scalar/number but I don’t know what that number really represents

[hushed sphinx]

If the two vectors are perpendicular, it gives 0.
If they are parallel, it gives the product of their lengths.

If they are neither parallel or peroendicular, you get something in between.

"In between" can be specifid either as (v+w)·u = v·u + w·u -- which fixes the result if you can write on of the vectors as the sum of something parallel to the other and something perpendicular to it.
Or by saying it's the product of their lengths and the cosine of the angle between them.
(These two descriptions lead to the same behavior).

[daring tapir]

Ig they were asking what that number actually signifies instead of how it is calculated or defined

Anyhow this doesnt belong in precalc atleast

[hushed sphinx]

85% of what it "actually signifies" is that it is 0 iff the vectors are perpendicular -- everything else can be seen as just incidental consequences of the easiest-to-work-with computation that achieves that.

[silent oasis]

can anyone check my work

[willow bear]

you made an even number of mistakes which cancelled each other out

(and in this case the mistake count is also positive)

[silent oasis]

Where did i make mistakes

[willow bear]

rewriting $\sqrt{x^2}$ as $x$ instead of as $-x$, twice: first in line 3, then in line 5 (though the second is a little less direct)

[obsidian monolith]

|Ann⟩

[willow bear]

you also took a few more steps than necessary imo, but that is less important

[silent oasis]

isn't $\sqrt{x^2}=x$

[willow bear]

otherwise you successfully applied the ideas i taught you last time (get rid of infinitely-big factors & 1/infty = 0), so good job on that

[obsidian monolith]

Nischal

[willow bear]

$\sqrt{x^2} = |x|$ generally, not $x$.

[obsidian monolith]

|Ann⟩

[willow bear]

in your case, x was approaching -∞ and so was negative.

so you needed sqrt(x^2) = -x.

of course, if you had taken the shorter route, you could have factored out $\sqrt{x^4}$ as $x^2$ --- and then the sign is no longer a concern since $x^2$ is positive anyway.

[obsidian monolith]

|Ann⟩

[silent oasis]

yeah! that would have been shorter.

[silent oasis]

What about this?

[willow bear]

looks fine to me

[queen shoal]

e = lim n->infinity (1+1/n)^n
but if we substitute infinity for n we get
(1+1/infinity)^infinity
1/infinity = 0 so we have
(1+0)^infinity = 1^infinity = 1

where did i go wrong

[willow bear]

$1^{\infty} \neq 1$

[obsidian monolith]

|Ann⟩

[hushed sphinx]

That's not even wrong.

[narrow briar]

Hi! One question: if I upload an image of an excersise that is wrong but I dont know where is the mistake, would you help me to find the mistake? I would appreciate it, thanks.

[willow skiff]

ok

[narrow briar]

its a simple integral

ok, im sending it

i sometimes forget putting dx on the integrals, sorry about that

i used the integration by parts method

[willow skiff]

Why are you using integration by parts when u-sub will do

u = sin x
du = cos x dx

[narrow briar]

ik, but i wanted to try it on parts too

when i tried using substitution it went right, but now im trying with parts and it isnt the right solution

sorry about my english

[willow skiff]

Your English is fine btw

[narrow briar]

thanks

[willow skiff]

But yes the integral of sin^4 x cos x is 1/5 * sin^5 x

• c

[narrow briar]

when i tried with substitution it was sin^5 x +c

[narrow briar]

and the solutionarie says its this

[willow skiff]

The original equation is to integrate 5 sin^4 x cos x right

[narrow briar]

yes

[willow skiff]

Yes then you multiply this by 5

[narrow briar]

hmm

but the 5 was multiplying the integral already when i did the equation

idk if im explaining

[narrow briar]

when i arrived to the same integral, i put the 5 multiplying the integral on the left side of the equation

[agile sparrow]

put*

not putted

[narrow briar]

thanks

[agile sparrow]

Anyway what's the integral

[willow skiff]

If u = sin^4 x, then dv = cos x
So du = 4 sin^3 x * cos x, v = sin x

[narrow briar]

this

[willow skiff]

Yeah you got to the right step but your u and v and stuff is all mixed up

[agile sparrow]

sin^4x. (cosx or cotx)

Which one is it

[willow skiff]

So yes if $I = \int 5 \sin^4 x \cos x \ dx$

You get $I = 5 \sin^5 x - 4I$

[obsidian monolith]

south

[willow skiff]

cos x

[agile sparrow]

sinx = u

[willow skiff]

Cause -20 = -4 * 5

[willow skiff]

They want to do it by parts for some reason

[narrow briar]

hahah yes

sorry about that

but im trying it

[willow skiff]

So do you agree with this step?

[narrow briar]

so when i try to solve an integral by using parts, i have to name u and dv

[narrow briar]

and then calculate du and v

[willow skiff]

I know

[agile sparrow]

Ahahahahah

My type of guy

Complicating simple stuff

[narrow briar]

yes hahaha

[agile sparrow]

and being annoying

Gotta love it

[narrow briar]

i already solved it on substitution

but then i wanted with parts

hahaha

[agile sparrow]

Integrate cosx and keep differentiating sin^4x

[queen shoal]

why

[willow skiff]

You really have to test it out for yourself

Try calculating (1 + 1/10)^10, (1 + 1/100)^100, and (1 + 1/10^6)^(10^6)

Now try calculating (1 + 1/10)^20, (1 + 1/100)^200, and (1 + 1/10^6)^(2 * 10^6)

You can get closer to 1 and you can approach infinity at different rates

[queen shoal]

ok

[silent oasis]

is there any way to do this quicker without making tables and finding right and left hand limits

[willow skiff]

Yes, 3 sin(x - 2) approaches 3 * (x - 2) as x approaches 2

Basically you have a situation where 3 sin(x - 2) gets really small around x = 2 yet the numerator stays constant

So you just have to worry about whether g(x) goes to positive or negative infinity on each side

[silent oasis]

where did sin go?

[silent oasis]

this is the full question.

[willow skiff]

It's based on sin(x)/x going to 1 as x approaches 0

But actually you don't need to use that in the question

[silent oasis]

how do I do it without making table?

[willow skiff]

So 3 sin(x - 2) is negative for x < 2
And x will always be positive in fact
So (negative) * (positive) / (negative) = positive

[silent oasis]

yeap! makes sense. thanks

[willow skiff]

No worries

[urban stream]

i got these solutions for the inverse of $3x^2 + 9x -7=y \ f^{-1}(x)\rightarrow\frac{-9+\sqrt{165+12x}}{6}$

the other one is $f^{-1}(x)\rightarrow\sqrt{\frac{x+\frac{55}{4}}{3}} - \frac{3}{2}$

[obsidian monolith]

uh

[urban stream]

which one is correct

if not both

[willow skiff]

https://www.desmos.com/calculator/h5ceyh0kpw

[obsidian monolith]

uh

[willow skiff]

They're both correct by graphing

[urban stream]

ah

[willow skiff]

if you wanted to find the top branch of the inverse

[urban stream]

yeah

[willow skiff]

cool

[urban stream]

thanks

[willow skiff]

np

[shut monolith]

wow I didn’t know AP Precalculus was a thing

i thought just AP calculus

[edgy edge]

AP Precalc is just a scam

It’s an OKAY starter AP class

But AP gov would be better

[crisp summit]

i've just watched Michael Porinchak's playlist about how to study for the precalc exam and its actually pretty helpful

i took basic notes over the chapter summaries and the 4 "how to crush" videos so dm me if you want them

[arctic sundial]

In what way

That it is easy or not worth it

[edgy edge]

Not worth the exam

[arctic sundial]

Ye barely any colleges accept credit

[edgy edge]

Most colleges don't even accept it

Yeah

Plus, it'll most likely be overshadowed if you were to take AB/BC calc

[arctic sundial]

I’m still taking it lol but I got 100% on the mocks I took

It’s so easy

I’m taking ab exam also

[edgy edge]

Isn't 2 sections algebra recaps?

[arctic sundial]

Yep

Unit 1 and 2

3 is trig

But it’s not that bad

And 4 isn’t even on the test

Lmao

[crisp summit]

im taking it cause i signed up for honors precalc/trig and at the start of the year my teacher said that we can do the ap exam if we wanted to so i agreed to do it

plus its free for my school

so why not

but yeah its prolly useless cause im planning on taking bc next year so if i do good on that it will just make precalc useless

[chrome ether]

did they drop unit 4 from AP precalc ?

im far too old for this but just curious as to what’s going on lmao

[crisp summit]

unit 4 is in it

but its not tested on

so basically 3 units

[willow bear]

conjugate bullshit

[crisp summit]

I did that too

Good luck today

[charred epoch]

Why does the leading coefficients of the numerator and denominator of a rational function determine the horizontal asymptote if the highest degrees of both are eaual?

[hidden lagoon]

because

if they’re the same degree they increase at the same rate so the leading coefficients determine what they will approach

[willow bear]

divide the numerator and denominator by x^n (where n is the shared highest degree)

then your numerator and denominator will become sums of their respective leading coefficients and other shit that goes to 0 as x -> infty

[charred epoch]

Increase? What if the lines go in opposite directions?

[hidden lagoon]

wdym

[charred epoch]

Like here they go in opposite ways

[hidden lagoon]

they have the same horizontal asymptote

[charred epoch]

I feel so dumb when I can't get good sleep. I gotta talk to a doctor about my sleep.

I guess I'm just gonna memorize the procedure for this stuff and not try to understand the "why"

[charred epoch]

@grave oxide here's my course description

[grave oxide]

oh damn

thank you :3

[elfin patrol]

Is my precalcus just for babies?

Cuz wtf is that

[winter comet]

what kind of precalc does urs have

[urban stream]

which grade are you in

why did my school make us do all this in 8th grade wtf

[charred epoch]

I'm in college. Dropped out of highschool when I was 16.

[urban stream]

that course description was ur college's?

[winter comet]

"precalculus" could be literally anything before calculus except it could be anything with calculus as well lol

completely depends on the class

[real crypt]

so are limits considered pre-calculus or calculus?

[hexed tangle]

can be either, usually taught at the end of some pre-calc courses and at the beginning of calculus

[winter comet]

i consider it calculus, but its taught in a lot of precalc too

[urban stream]

yeah

but damn the diff is huge

[winter comet]

i mean...

i don't really think its that big of a gap

they're just different subjects

sometimes its more of the prerequisites for calculus and sometimes its just the basic concepts that you apply when you do parts of calculus

idk

[arctic sundial]

Lmao the ap precalc test was a joke

So easy

[echo axle]

should I do AP pre-calc? I've heard some not-so-great things about it, so I'm not sure if it's a generally suggested course.
Take note that I am fairly comftorable with algebra and proofs, and am planning on taking AP Calculus Senior year

[winter comet]

do you know trig?

[echo axle]

I think so

I know the basic functions, but I am comfortable working with the unit circle and have a gist of the identities

(I am in geometry, doing alg 2 over the summer, so I haven't done alg 2 formally yet)

[winter comet]

alr

uhh

[river drift]

you'll need to take some sort of precalc course anyway, so i don't think there's any harm in taking the ap course. don't necessarily plan to rely on getting college credit for the exam (like you would for other ap courses), though

[winter comet]

really?

[river drift]

some course called "precalculus" is usually required in my experience. always good to check your own school's requirements, though

[winter comet]

you mean as a prerequisite for calc?

i always thought it was skippable

i never took it lol

[swift oak]

help me 😭

[cobalt atlas]

can the average person learn all of the trigonometric identities solutions/rules (including double angle/ sum and difference problems) within an hour or two?

[winter comet]

the...average person?

idk about the average person

[cobalt atlas]

okay slightly above average individual who is taking AP Physics 12

[winter comet]

i would say one could memorize most that information in an hour or two, relatively

[cobalt atlas]

thanks big man needed the motivation/assurance

[winter comet]

👍

XD

[tranquil inlet]

Can someone explain the thought process behind solving these problems

[tranquil inlet]

edit: i finished 48-51, i only need help on 52

[viscid thistle]

!status

[tender quest]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[viscid thistle]

3

[willow bear]

you don't need to invoke the !status command on yourself lmao you can just say "status 3"

[viscid thistle]

My working

[willow bear]

anyway, show your work and the source that told you your answer is wrong

[viscid thistle]

I don't remember 1,2,3,...

[willow bear]

k

[viscid thistle]

The same paper as yesterday

[willow bear]

ok first off

you need to write $\infty$ in a way that doesn't look like 8.

[obsidian monolith]

|Ann⟩

[willow bear]

cause right now it's tilted upward way too much

[viscid thistle]

I was told that so many times

What nest ?

Next

[willow bear]

this looks suspicious to me

like i cannot follow your logic here at all

except for "take the limit in one sub-expression and somehow keep the rest as-is even though all of it depends on n"

which is bullshit

[viscid thistle]

An and a(n+1) are the same when n tends to infinity right ?

[viscid thistle]

That's the third time

[willow bear]

sure they are, and i can concede that a_n - a_{n+1} does tend to zero -- assuming that lim a_n is finite, and only then.

(which i don't think is even known)

[viscid thistle]

So they both are the same only when an is finite

[willow bear]

i said lim a_n, not a_n itself. don't conflate a sequence with its limit.

[viscid thistle]

My bad

[willow bear]

even then, (n+1) (a_n - a_n-1) is an ∞*0 situation

so even taking the limit of this won't do you any good

so this is not only bullshit but it is also unsalvageable

give me a couple minutes...

[viscid thistle]

I have the right solution

I just didn't understand why mine is wrong

Thank you

Infinity times 0 is not 0?

[willow bear]

no, it is not

it is an indeterminate form

[viscid thistle]

okay

👍

[viscid thistle]

Status 1

[willow bear]

imagine cutting the conical surface along AQ and unrolling it onto a flat plane

then the path from A to B will be a straight line

[viscid thistle]

Is the image clear enough ?

[willow bear]

the image is fine but i am describing something not directly shown in it

[viscid thistle]

Okay

[viscid thistle]

What next ?

One min, what do you get whe you cut it ?

[willow bear]

a circular sector, what else

[viscid thistle]

I am not very good with 3D structures

Next .

?

[willow bear]

the path from A to B will be a straight line

height up the mountain translates to distance from the center of the conical sector

seriously just make the drawing lmao

i recommend drawing the circular sector with angle less than 180° for illustration purposes

[viscid thistle]

That helps

[viscid thistle]

Distance between the centre and ?

[willow bear]

and a point on the path

like you clearly are interested in the point at which the path switches from uphill to downhill

i.e. when your altitude is highest

[viscid thistle]

Any simpler method?

For a high schooler

That is me

[willow bear]

i cannot think of any method that is any simpler than this

[viscid thistle]

Way too complex, I will leave it
Thank you

In which channel should I ask integration questions?

[willow bear]

#calculus, obviously.

[viscid thistle]

Okay ty

[jovial hare]

At which points is the function discontinuous and at which is it continuous? In the answers it says that it is discontinued only at x=0, but when I try to find their limits at x => 0 in 1. function I find 2 and 2. - 0. What does this mean? Shouldn't they be equal?

[willow bear]

the fact that they aren't equal is precisely the reason why f is discontinuous at x=0.

@jovial hare

[jovial hare]

But how will I know if they are continuous at the other points?

[willow bear]

you should know the continuity of some basic functions

such as polynomials, roots and trig

and also that the sum and difference of continuous functions is continuous

and that the quotient of continuous functions is also continuous wherever the denominator is not zero

[jovial hare]

I didn't know I needed to know them.

Thanks

[primal river]

what is the required knowledge to solve this?

[viscid thistle]

The paper says the question is from applications of derivatives

High-school ig

[primal river]

I wish kahn academy had more exercises like this one

or maybe I just haven't seen them yet

[agile sparrow]

Indian

[honest osprey]

hello

i am working on a college class and we are covering function notation rn

and i need help on it

[naive sluice]

Can I take a look?

[honest osprey]

yes

i have been so pissed off today i can't really even think straight rn

[naive sluice]

Question b why is it orange did you get one wrong?

[honest osprey]

its orange because i got partal credit on it

[naive sluice]

Ah I don't know what that means but ok

[honest osprey]

ok

[honest osprey]

so ?

[honest osprey]

damn

[tired stone]

is there a difference between 2cosx and cos^2(x)?

[hushed sphinx]

Yes.

[tired stone]

damn

i found a yt video going over the problem but the number its slgithly different

[hushed sphinx]

One is "take the cosine and multiply it by 2", the other is "take the cosine and multiply it by itself".

[tired stone]

damn. if im horrible at trig can I still pass calc 1?

[hushed sphinx]

Depends on what exactly "calc 1" is in your location and how high the bar for passing is, but in most places I think it would be a challenge.

[tired stone]

canada at uni

is it mostly trig?

[hushed sphinx]

(That is, unless you manage to stop being horrible, of course -- that's not a static property).

[tired stone]

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

[golden cairn]

Question is asking to use laws of logarithms to write ln(f(x)) as an expression of natural logarithms

How do I do that?

[winter comet]

try writing
f(x) = (x^3(x+1))^(1/2)

does that help

[golden cairn]

I already did that step but what do I do next

[winter comet]

did you take the natural log of both sides

[golden cairn]

No

[winter comet]

try it

[golden cairn]

Ok

Now what’s next

[winter comet]

technically you're done 💀 but do you know the exponent property of logs?

[golden cairn]

The answer key has some completely different answer

It doesn’t even make sense

I will send it

[winter comet]

ok

i think i know what they want you to do

they probably want you to use the exponent property of logs

[winter comet]

yeah

[golden cairn]

They brought it in the front

[winter comet]

yeah

you can do that with ln

[golden cairn]

Oh and they multiplied

Or expanded I meant

[winter comet]

yes

and then they moved it to the front again for ln(x^3) = 3ln(x)

[golden cairn]

Yeah

But what happens to the ln infront of f(x)

When we multiplied on both sides

[winter comet]

you're trying to find ln(f(x))

its still there

[golden cairn]

Oh yeah

Okay

[winter comet]

thats what u were solving foryeh

[golden cairn]

Is x^3(x+1) one term?

[winter comet]

...

difficult question

technically

i guess technically its 1 term

but if you expand it then its 2

or if you take the log and use the property

then its 2

[golden cairn]

Because I was confused how the ln multiplied and made two seperate terms

[winter comet]

ln(ab) = ln(a) + ln(b)

its just a log rule

and

ln(a^b) = bln(a)

[golden cairn]

Oh yeah you’re right

[winter comet]

also a log rule

[golden cairn]

I forgot about that

[winter comet]

yeh

they just used those

[golden cairn]

Okay thanks

[winter comet]

technically you would be done at ln( (x^3(x+1))^1/2 ) 💀 but i think its assumed that ur supposed to expand it

because like

you did what they asked

i don't think one is more "simplified" than the other

lol

i think the question was a bit odd

[golden cairn]

I think I actually remember the teacher saying that the answer key just shows unnecessary extra steps

[golden cairn]

Yeah

[winter comet]

yea

or unclear

[golden cairn]

Yeah

This question is asking to solve for x but I am getting confused because there is ln in the exponent

[winter comet]

e^ln(u) = u

it cancels out

[golden cairn]

What is u?

[hushed sphinx]

Anything.

[golden cairn]

Okay

[torn kestrel]

bro just felt like using a quirky variable

[golden cairn]

So ln(1-x) equals 1?

[hushed sphinx]

What? No?

[golden cairn]

I don’t get it

Does it equal 1-x?

[hushed sphinx]

What is your definition of "ln"?

[golden cairn]

I meant e^ln(1-x) equals to 1-x?

Natural

Log

[hushed sphinx]

Yes,

[golden cairn]

Okay

[hexed tangle]

if ln is natural log then e^ln(1-x) = 1-x

[golden cairn]

How though

[hushed sphinx]

Okay, let me rephrase the question: What is your definition of "natural logarithm"?

[torn kestrel]

isn’t it the log rule?

[hexed tangle]

ln is inverse of e^x

[golden cairn]

I don’t even know what it means

[hexed tangle]

so ln(e^x) = x

and e^ln(x) = x

[golden cairn]

Ok

[winter comet]

ln is log base e

ln(x) = log_e(x)

[golden cairn]

Okay

[winter comet]

log_e(x) = y
x = e^y

they are inverse functions

e^x and ln(x) are inverse functions

so if you apply one to the other, they cancel out

[golden cairn]

What if I write e^ln(1-x) as ln(1-x)e. What would I do then?

[winter comet]

uhh

what

no

;-;

[golden cairn]

Why it as caps

Put

Oops

I don’t know what I’m even doing

[winter comet]

do you know what logarithms are

like log base b of a = y

[golden cairn]

Yeah

[winter comet]

a = b^y, right?

[golden cairn]

What?

Oh yeah

Yeah I get it

[winter comet]

yea

[golden cairn]

Yes

[winter comet]

log base e

is natural log

[golden cairn]

Okay

[winter comet]

log_e(x) = y
ln(x) = y

so what is x

[golden cairn]

Im not sure

[winter comet]

log_b(x) = y
x = b^y
log_e(x) = y
btw e is a number, x is a variable

[golden cairn]

Oh I get it

I understand that lne is equal to 1

But

Okay so just tell me if e^ln is equal to lne

Are they both 1?

[hexed tangle]

$$e^\ln(x) = x$$

[golden cairn]

Like how is e^ln(1-x) equal to 1-x

That’s what I don’t understand

So is x=1/3 in the question I asked earlier?

[hexed tangle]

let y = 1-x, e^ln(1-x) = e^ln(y) = y = 1-x

[fallow field]

Im stuck here

[hushed sphinx]

e^ln does not make sense in the first place.

[golden cairn]

Yeah I realized that afterwards

After I typed it into my calculator

So would e^ln(x)=x, like is that a rule?

[hexed tangle]

yes beacuse e^x and ln(x) are inverses

[golden cairn]

Okay

I get it

[hushed sphinx]

It's hard to guess exactly what is confusing you, but have you tried plugging in some values for x and comparing the two sides as actual numbers?

[golden cairn]

I get it now though

[hexed tangle]

you need to do 1 0 0 0 -2 in the first row because x^4 - 2 = x^4 + 0x^3 + 0x^2 + 0x - 2

there has to be one column for each power

[fallow field]

there will always be 4 number for synthetic right?

[hexed tangle]

what do you mean

[fallow field]

hold on let me try and solve it the way you put it

[fallow field]

How is it now?

[hexed tangle]

good just note that the 1 is unnecessary

[fallow field]

Yeah I know

I think my graph is wrong?

[winter comet]

you didn't do the 9 - 2x if x > 2 part correctly

[fallow field]

Is it suppose to be a diagonal line?

[winter comet]

yeah

but it should be going down

[fallow field]

Down to up

Or 9 units down

[winter comet]

i mean

the line itself starts at (0,9) and the y values decrease as the x values increase

but for your piecewise function

you only draw the part of that line AFTER x = 2

[fallow field]

?

Oh

?

[winter comet]

um

not exactly

the 9-2x line is wrong

its not 2x - 9

its 9 - 2x

so you start at positive 9, and go down by 2x