precalculus

[summer ruin]

now apply that to that problem

[gentle talon]

The center in the problem in my question is (-5, 4)

My big problem is that

50/2 = 25

so the center is (-5, 4)

Do i add the 25 radius in the center?

[gentle talon]

Oh wait.........................

[gentle talon]

Yes I know but likeeee

OH SHET

I KNOW NOW THANK YOU LOL

Im legit so dumb :v

[gentle talon]

wait not gonna lie im still kinda confuse UHHHHHHHHHHHHHHH

From this one he added that equation to the center...

oke nvm

[hot bison]

need help with this

gotta find the intervals

@everyone need this quick, could someone help me out?

[summer ruin]

you just did everything backwards

it would be right if it wasn't backwards

and your second interval is literally backwards of what it should be

[limber patrol]

can anyone explain the steps in the second line (from sin 2c + sin c = 0)

[void matrix]

I took an AoPS course and it really helped. Intermediate Algebra is a good course to go along with it as the precalc course doesn't teach any new algebraic topics.

[void matrix]

So, the first part of the second is found by using the identity sin(2x) = 2sin(x)cos(x) and the second part of the second line comes from factoring out sin(c) from the expression on the LHS of the equation.

Does that help?

[limber patrol]

yes thanks

[void matrix]

yw

[gusty pasture]

can anyone help me with this

[summer ruin]

do you know the equation that describes circle?

[gusty pasture]

pythag?

or the one with k and h

[summer ruin]

that's only true for a circle centered at the origin

so I guess you do, now use that to solve the problem

[gusty pasture]

i tried it doesnt work

[summer ruin]

well explain what you tried

[gusty pasture]

i got 16/3,0

[gusty pasture]

i did distance to get the radius to write the formula but then i just got lost

my teacher never taught us x intercepts so i dont understand it

[summer ruin]

it's just a point at which this curve intercepts x-axis

so what does your equation for circle looks like after you've found the origin and radius?

[gusty pasture]

(x - 5)^2 + ( y - 1)^2 = 4

[summer ruin]

I guess you meant (y-1), not (x-1)

well that radius is wrong

[gusty pasture]

wait

is it

4

[summer ruin]

no

[gusty pasture]

wait

fuck

i mean

-2

[summer ruin]

radius can't be negative either and that's still far away from proper radius

[gusty pasture]

im doing this all in my head idk why this is in calc 3 work

2

?

[summer ruin]

no

[gusty pasture]

ok im dumb lmao

[summer ruin]

where did you get that radius squared is 4?

[gusty pasture]

Wait you wanted me to write the square root part?

I was just referring to the main equation

[summer ruin]

no, I'm just asking why do you think radius squared is 4 or same thing normal radius is 2

this doesn't come from anywhere

[gusty pasture]

Wait

Let me write this down

Because I’ve been doing this in my head the whole time

Is it this?

And then c is just 10

[summer ruin]

well you can check whether the circle with that radius indeed goes through that point you're given in the problem

[gusty pasture]

I’m confused on what to do after this part

Like, what do I do after writing the formula

(x - 5)^2 + ( y - 1)^2 = 10

[summer ruin]

well what can you say about points that lie on the x axis?

[gusty pasture]

Hold on I think I found smth

Is it (8,0) (2,0)

[summer ruin]

seems right

[gusty pasture]

Woohoo it is 😭

I was overthinking this LOL

Thanks for helping

[acoustic vault]

What do you think

If you're asking for help, you really should be more polite

[summer ruin]

the answer should be pretty obvious given that -1 =< sin(x) <= 1

[acoustic vault]

They're just a troll

[summer ruin]

since you don't care about getting help in solving this - plugging the question into wolfram alpha is much faster than waiting here

then don't ask for a solution I guess, but try to make it on your own with the hint I provided

you don't plug infinity anywhere

[narrow mason]

question! I have a line (lets call it L1), and its coordinates are 0,0 (P1) and 10,10 (P2). I'd like to draw a line at P1 that is perpendicular to L1, and a line at P2 that is also perpendicular to L1. I've got the slope-intercept equations for the perpendicular lines figured out. My question is, I'm trying to figure out how to draw the perpendicular lines from the origin points (P1 and P2) outwards by a certain number (lets call it length). I'm not sure where to fit length in my slope-intercept equation in order to offset the points correctly. Anyone have any ideas?

[summer ruin]

what does that mean to draw perpendicular lines outwards by a certain number?

[narrow mason]

somewhat similar to how a radius grows outward from a point

[summer ruin]

I still don't get it, you're comparing a non-linear curve to a line which behaves differently

moreover any line is infinitely long, so length doesn't really make sense here

[narrow mason]

Imagine I don't know the coordinates of point D or point C, but I do know that line2 is 5 units long. how can i calculate point D and C?

[summer ruin]

for points (0,0) and (10,10) you can easily find the line that goes through them in the form ax+by = 0 which turns into a+b = 0 and has solutions like a = 1, b = -1

then perpendicular line can be found by taking inner product of coefficients of the new line with existing one and equating that to zero, so $av_1 + bv_2 = v_1 - v_2 = 0 \rightarrow v_1 = v_2$, picking any $v_2$ (say - 1) we get $v_1$ which is also 1, so any line of the form $1x + 1y = c$ is orthogonal to the first one, varying the parameter $c$ allows to move the orthogonal line for whatever reason you'd want

[obsidian monolith]

Transparent_Elemental

[charred viper]

What rule is this?

Where can I read up on this

Seems a bit strange... Why can't I just find (f o g)(x) then determining the domain (D) and range (R)?

[charred viper]

Also I'm a bit dubious about R (f o g) like what does that even mean

Domain of F and Range of G?

[mental stirrup]

The domain range of f o g. As in, the domain of the function f o g (the composition of f and g) [typo]

[mental stirrup]

This is a way of expressing the domain and range of f o g in terms of the domains and ranges of f and g.

[charred viper]

Wait what if

I put (f o g)(x) together and I just evaluate the range/domain based on the resulting function

I had an example once but my answer was different from when they used the above

[mental stirrup]

Then you certainly made a mistake

[mental stirrup]

That's fine, but if you know D_f and R_g then you can avoid calculating a lot.

[charred viper]

It's as if they're saying if you have (f o g)(x) then to be acceptable then g(x) must have values that are the domain of f(x). But what if you redefine this function to be a completely new function, then you don't really need the domain to depend on the range of g(x), but rather only on the x-values of this function itself?

I guess what I'm really asking is, if I have (f o g)(x) = h(x), their domains and ranges might be different just because in the first one we have to evaluate the range of g(x) that will be acceptable to f(x). Why isn't it considered a new function h(x) where the domains and ranges depend solely on h(x)?

[mental stirrup]

I don't get what you mean, this is considered an entirely new function h(x).

It's just that we can calculate the domain and range of h(x) in terms of the domains and ranges of f(x) and g(x)

[...] if you have (f o g)(x) then to be acceptable then g(x) must have values that are the domain of f(x). [...]

This is simply true though, you need this to be the case for h(x) to be defined at that point.

That is the definition of h = f o g.

[charred viper]

,rotate

[charred viper]

This is what I mean I guess

[mental stirrup]

This is because you made an error here.

It is not true that:

$(\sqrt(x))^2 = x$

[obsidian monolith]

Boytjie (Plutonic Relations)

[mental stirrup]

It is only true when x is non-negative.

[charred viper]

Oh I see what you mean

Wait um but if x = -1 then isn't by definition i^2 = -1

sorry if i'm a bit slow

still trying to make sense of things

[willow bear]

@mental stirrup

[mental stirrup]

I noticed, I simply don't care

[willow bear]

$(\sqrt{x})^2 = x$

[obsidian monolith]

Ann

[mental stirrup]

Your lecturer is working only in the real numbers.

[charred viper]

ahh okay

thanks!

[mental stirrup]

No problem. Apologies for the interruption.

[shrewd prawn]

Hi

When we look at the graph of a the absolute function f(x) = |2x|, without looking at the equation itself, one would never know if thats a horizontal shrink or a vertical stretch right ?

[vapid plaza]

Well yeah because |2x| = 2|x| so it would be the same function anyways

[shrewd prawn]

sqrt(2x) = 2 sqrt(x) are these the same ?

[willow bear]

no

[shrewd prawn]

so that makes |2x| and 2|x| not the same

[willow bear]

no

|2x| and 2|x| are the same, but sqrt(2x) and 2sqrt(x) are not.

[shrewd prawn]

Can you elaborate a bit more ?

I'm talking not the same in the context of function transformations

[willow bear]

sqrt(2x) and 2 sqrt(x) are different as functions, but |2x| and 2|x| are not

the formulas f(x) = 2|x| and f(x) = |2x| are suggestive of different methods by which the graph of y = 2|x| may be obtained from that of y = |x|

[shrewd prawn]

yes

[willow bear]

sometimes two different transformations applied to the same graph can lead to the same result

nothing wrong with that

[shrewd prawn]

One acts inside the function |2x| and the other one on the outside 2|x|

one acts on the Horizontal level, other one on the vertical

that's how I see it

but looking at the graph after the transformation one would never know if that was |2x| or 2|x|

😄

hahaha

[fleet tendon]

Omg

[summer ruin]

you divided by zero twice

[fleet tendon]

Shut up 9=0

Point

[fleet tendon]

And proven

[willow bear]

[viscid thistle]

the only solution to x is 0🤨

[thorny sand]

You broke the universe

[unique mauve]

i can't figure out why c is 6, for a & b i got s(h) = h + 6 and 6.2 but for c s(3) = 3 + 6 = 9, not 6. when calculating IV do you always use 0 despite it asking for t=3?

[willow bear]

you do not want s(3), you want ds/dt(3)

what you calculated is s(3)

[unique mauve]

we haven't started learning derivatives yet, this is the equation im going by

[willow bear]

this is the derivative in all but name

also what you calculated in part a is not s(h) and should not be called that

[unique mauve]

oh ok

[summer ruin]

you should now write "@obsidian monolith tex" and then math

[mental stirrup]

As I stated in a follow up, the lecturer is clearly working in the real numbers alone.

[keen obsidian]

pretty sure this was multiposted ^^

[keen anchor]

Why would the parentheses stay after foiling?

[winged berry]

shut up nerds

[willow bear]

expanding (3x+z)^2 gives you a sum of terms, and it is this sum of terms that is being subtracted from x^2

if you did not put parentheses and just wrote x^2 - 9x^2 + 6xz + z^2 that would make only the 9x^2 be subtracted and the rest would be added instead

which is not what you want

[willow bear]

remove != erase blindly

[shrewd prawn]

You have a - sign before the parenthese, you have to foil that first then you can remove them

[sacred quartz]

how did this get 4 in the circle like im confused

[willow bear]

show the entire working uncropped

[sacred quartz]

kk

[willow bear]

ok

so they added 1 to both sides to complete the square on x and added 4 to both sides to complete the square on y

[sacred quartz]

OH OKAYYY

tysm

[crystal storm]

what do i do next?

do i graph

and if yes

how do i graph with sin?

[hushed sphinx]

You're being asked to "sketch", which means more or less "a drawing that has the correct shape for the graph, but probably drawn by hand, so we don't care about exact precision of the drawing as long as the key points are marked off correctly".

[crystal storm]

wait wait so the 2(sin 3x) -1 is correct?

and I have to graph that?

[hushed sphinx]

Yeah, that looks correct.

[crystal storm]

ohh okay okay, next question is how do I get those key points let say I want to use 1 as x, how do i solve the sin 3?

[hushed sphinx]

If you get the top and bottom y-cordinates of the wiggles right, and mark off the x-coordinates of a few of them, I'd say that qualifies as a sketch.

[crystal storm]

I'm more concerned about the points

[summer ruin]

why pick x = 1 when you can pick values of x for which this expression can be nicely evaluated?

[hushed sphinx]

You'd choose values for x where you know the sine is doing something interesting.

[crystal storm]

umm hehe what does that mean?

[hushed sphinx]

So for example x=pi/3 would be a point where you know the sine is 0.

pi/3 is about 1, and if you sketch your coordinate system freehand, that's as precise as you need to know that. Make ticks on your x-axis marked 0, pi/3, 2pi/3, pi, ... rather than ticks marked 0, 1, 2, ...

[summer ruin]

^ now you know the period which allows you to graph parts of sine properly

[crystal storm]

I uhh failed in trigonometry so i dont really get it

so lets say pi/3 where is that?

[summer ruin]

it's greater than 0 and less than pi/2

[crystal storm]

do I watch an organic chem video about trigo first?

[summer ruin]

and you don't need to know more than that

[hushed sphinx]

That suggestion looks like you might be forgetting that it's sin 3x rather than sin x?

[summer ruin]

no I mean to locate a point x = pi/3 you can just make regular intervals like 0, pi/2, pi, etc and mark a new point pi/3 between 0 and pi/2 for sin(3x)

[hushed sphinx]

You should definitely go back and revise some trigonometry if you cannot sketch sin(x) itself. Whether you should use a chemistry video for that sounds a bit specific.

Why would you mark off your regular intervals as multiples of pi/2 instead of multiples of pi/3 in the first place?

pi/3 is reasonably close to 1, which is the unit you need on the y-axis, so marking off a reasonably squareish grid by eye will be easier than trying to find pi/2.

[summer ruin]

you can go that way, I think it's just easier to think that way first

you don't have to draw anything accurately anyway, you can just mark pi/3 anywhere

like this

[hushed sphinx]

Right, but that is again sin(x), not the sin(3x) we're sketching here.

[summer ruin]

I know

[crystal storm]

how would I be able to identify the points that im going to use?

[summer ruin]

you need to know all the values of sine from the table

then you just plug in values of pi such that your sine can be evaluated using the table

[crystal storm]

are you reffering to this one?

[summer ruin]

yes

[crystal storm]

I dont understand with this, are you saying I'll just plug the values of the table above and then I graph my function?

[summer ruin]

no

since evaluating sine is the only tricky part here then you want to know for what x can you evaluate your sin(3x)

you know for what x sine is equal to 1/2 - that's x = pi/6

now solve for x the equation sin(3x) = 1/2

[crystal storm]

so 3pi/6?

oww wait pi/2?

[summer ruin]

sin(t) = 1/2 then t = pi/6

if t = 3x, then 3x = pi/6

but that's only one solution

you'd need to figure out multiple periodic solutions

and there's actually two families of solutions to sin(t) = 1/2

[summer ruin]

and now you know that for x = pi/18 your sine evaluates to 1/2

[crystal storm]

where did pi/18 come from?

[summer ruin]

from 3x = pi/6

[crystal storm]

im sorry but how? so sin (3x); where 3x=pi/6 how did we get from there to pi/18?

[summer ruin]

divide both sides by 3

[crystal storm]

ohhh okay okay, so whenever we do sin we try to eliminate the number besides x?

and that evaluates the sine to 1/2?

[summer ruin]

no, you're getting it backwards

first you look into the table and see for what values of the input you can evaluate sine

then you solve sin(3x) = 1/2 or whatever the number you want on the right hand side that comes from the table

because you know that there's a closed form solution to this from the table

sin(t) = 1/2 has solution t = pi/6, we know this from table

but if t was equal to 3x, then we get that 3x = pi/6

then solve for x and you're done

[crystal storm]

how do I identify what value I can and cannot input to evaluate the sine?

[summer ruin]

you've got a table

now you know for what theta you can evaluate sine

[crystal storm]

ohhh, but let say in my question here, it didnt specify what sin is equal to

so how do i know which value to use?

its only sin 3x

[summer ruin]

there's no need to "know", don't you see how you can pick arbitrary value for sine from the table and still get a solution the same way?

pick sin(3x) = sqrt(3)/2

well then 3x = pi/3

or x = pi/9

so for x = pi/9 sine(3x) evaluates to sqrt(3)/2

the point of this is that you pick many different points and evaluate sine at them, get yourself a table with (x, sin(3x) and then draw them on paper, then you can make a smooth curve that passes through them to get your graph

[crystal storm]

so it just comes to the matter of finding x for each value of sin and then plotting those points to the graph?

[summer ruin]

yes

[crystal storm]

ohhh icic

[summer ruin]

then you also know that since you've got a sin(3x) your period is 3 times as much as regular sine, so values of x repeat faster

to plot more than given in the table you need to use periodicity of sin(3x) and that's why there's multiple solutions to sin(3x) = 1/2 or sin(3x) = 0, etc

[crystal storm]

why do you call it solution instead of a point in the graph?

[summer ruin]

a value of x that satisfies this equation is a solution to this equation

here you can see that there's infinitely many points of intersection as well and next solution is 5pi/18

[misty sundial]

all i would do for this is make the x values negative and simplify it as usual correct?

[summer ruin]

yes

[misty sundial]

so the bottom is how it would be written out?

[summer ruin]

it's (-x)^4, not -x^4

[misty sundial]

ohhh okay

but the other side of the equation is right or do i put parentheses on that x as well

[summer ruin]

if you want to you can

[misty sundial]

thank you !

[misty sundial]

how does the equation change if the f in f(x) is negative so -f(x)?

with this, would i multiply everything by -n

[crystal storm]

do i conjugate this to cube root of (1+3x^2) + 1?

[misty sundial]

i’m very confused on how to graph this, like i don’t even have an idea, do i just use 3x and -3x as a starting point or something

[summer ruin]

it's equal to 3x for x < 1 and -3x for x>=1

[misty sundial]

so it would be written like
f(x<1)=3x

f(x_>1)=-3x

[summer ruin]

that doesn't make any sense, x<1 isn't a valid input for a function

[misty sundial]

oh

what do i use as the input for the function then because i still don’t get it and my notes aren’t helping me much

[summer ruin]

If I asked you to graph f(x)=3x for all x that are less than 1, what would you do?

[raw hill]

It's a line that passes through the origin, so I would graph that

[obsidian monolith]

messy circle creation

[misty sundial]

does that look right? i’m pretty positive with the increasing and constants but i’m still really confused on whether or not it would be decreasing from the x axis or y axis in interval notation

nvm

[misty sundial]

can someone just tell me what i would just search in google to find examples and notes of this

[prisma aurora]

I mean, technically this seem to be arithmetic operations with functions

(f-g)(x) = f(x) - g(x)

like you were multiplying

[misty sundial]

damn wish i would’ve saw this when i woke up 😭

could anyone give me any ideas on what i’m doing wrong cause i’ve rewritten this problem like 4 times and i don’t understand, and i’m following the process in my textbook as well

[summer ruin]

you're doing multiplication wrong

[misty sundial]

how, i thought that it just automatically multiplied if it was next to a square root? or do you mean in the step before that

[summer ruin]

clearly $(x^2 +6x) \cdot \sqrt{9-x} \neq x^2 + 6x \cdot \sqrt{9-x}$

[obsidian monolith]

Transparent_Elemental

[misty sundial]

wow

i am dumb💀

thank you 🙏🏻

[viscid thistle]

can someone please help with this

[oblique gazelle]

Well, what do you think it is?

Remember that the double bracket notation indicates that you take the value to the next highest integer

[hushed sphinx]

Huh, I would understand "the greatest integer function" to mean the floor function.

As in "floor(x) is the greatest integer n such that n <= x".

[oblique gazelle]

You're correct, my bad

[keen vault]

does anyone here know the difference between math 222 and math 225

was told math 225 didnt meet the prereqs for taking ab calc but math 222 did

i do know some students who took 225 and got into bc, but idk ab it

[keen obsidian]

Numbers of courses differ extremely based on what the school provides

[keen vault]

i see

nah i want to take ab

but i was taken out of ab since i didnt take math 222

since it didnt set up some infrastructure or smth

i was just confused on what said infrastructural difference is. i googled it and on the college class page for 225, it said it was just 222 mixed with 130 or smth

i took 225 over the summer since the dean told me to

[keen obsidian]

Best way to check is to check the course descriptions ig

[keen vault]

ill pull it up

math 225 "Equivalent to MATH 130 and 222. This course combines the topics of Trigonometry and Pre-Calculus..."

💀

[keen obsidian]

How can equivalent classes have different requirements and stuff

[keen vault]

no cluew

and i was literally told to take it after we asked ab AP calc

[keen obsidian]

ask your counselor or somebody

[keen vault]

was really insulting. they kept saying "youll get a D, this is like going from spanish 1 to french 3"

parents sent an email

i sent an email to my professor

[keen obsidian]

darn

[keen vault]

whats worse was that i was in the class for a whole 2 weeks, then they email that they mistakenly put me into AB calc AS im in the ab calc class

[misty sundial]

wouldn’t it just be the same thing? my notes always have f(x+h) in the examples so i’m a little confused

[keen vault]

everything i see says math 225 is more difficult than 222, but maybe 222 does a tiny bit more infrastructural things?

[keen obsidian]

[keen obsidian]

same thing as what?

[misty sundial]

as f(x)=-9x+2

[keen obsidian]

It wouldn’t be the same thing, what have you tried?

[keen vault]

wouldnt?

dont u just plug it in

[misty sundial]

this

[keen vault]

dont u just plug that into the formula?

[keen obsidian]

Pretty much

Dang my autocorrect is murdering me

[keen vault]

ah

[keen obsidian]

Or if you know the rules you could apply those

[misty sundial]

here i’ll send a pic of one of the example problems from a previous problem

[keen obsidian]

ok

[keen obsidian]

the result is not -2x+5

[keen vault]

^

well obviously not that

but its not -9x+2

remember ur subtracting and are distributing the negative

[misty sundial]

i got it🦾

[viscid thistle]

-9?

or is it 0

[oblique gazelle]

Neither

If I floor -0.4, what do I get?

[raw hill]

,w floor(-0.4)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by messy circle creation#1465

[raw hill]

b/c -0.4 rounds down to -1

[oblique gazelle]

Correct

And so, 6 + 3(-1) gives us 3

[raw hill]

Wait srry 💀

ok lemme just leave lol

[oblique gazelle]

Oh

That's alright, hopefully the person comes back and sees the reasoning

[viscid thistle]

oh, that's right

sorry, my professor didn't go over this yet. Appreciate the help, guys

[oblique gazelle]

No problem

[misty sundial]

ignore the answer in there, but this equals -x^2+2x right?

nvm i got it

[lost roost]

idk if this is the right place to ask, also sorry that my question is a photo but i can't type maths in discord

[willow bear]

if you define $S_n$ as the sum of the first $n$ terms in the sequence ${a_n}$ and then want to say that $a_n = S_n - S_{n-1}$ then you have to say $S_0 = 0$

[obsidian monolith]

Ann

[lost roost]

the sequence starts from element 1, at least in my book, so it's logical that S0 would be 0. I've seen other sources label the first element as a0, do you use that method?

[willow bear]

is that "you" as in generic-you or "you" as in me specifically

[lost roost]

I'm asking you personally bc i wanted to clarify this to avoid misunderstandings

[willow bear]

whether i start the indexing of a sequence at 0 or at 1 or at some other point is entirely a matter of context-dependent convenience

it's not something that really matters in the grand scheme of things

[lost roost]

oh ok

[lost roost]

was this the answer to my question? bc if it was i kinda didn't understand, could you explain further?

[willow bear]

i was answering your question, yes. but in a somewhat obtuse manner.

what i meant is that S_1 as calculated from the explicit formula for S_n and S_1 as calculated by virtue of equalling a_1 are always the same,

and that to make the formula $a_n = S_n - S_{n-1}$ work for $n=1$, you need to state explicitly that $S_0 = 0$

[obsidian monolith]

Ann

[lost roost]

give my brain a sec to process that

[willow bear]

||this can also be said more succinctly as "the sum of nothing is 0"||

[lost roost]

ok so, if the a_1 from S_1 and a_1 from a_n end up not being the same, then there is no sequence for which (this specific) S_n would be a sum of all elements from a_1 to a_n?

and what about the last question, does it tell anything about the sequence beyond that it exists? I'm asking bc in all examples that I've done, when both a_1 's were equal, the sequence was arithmetic

[lost roost]

are you advanced in maths? we usually omit these statements in class bc they are kinda obvious, so that seems like some advanced stuff

[willow bear]

depending on your definition of "advanced", i probably am.

but also yes you're right in that "the sum of nothing is 0" is more of an obviousness than some deep, paper-worthy, meaningful fact

[misty sundial]

what would the vertex form of this quadratic function be

[summer ruin]

don't you know how to complete the square?

[misty sundial]

completing the square method? i’ve heard of it but i haven’t done it before personally

[raw hill]

'Divide' should say 'factor out a', but here

[summer ruin]

or just use the fact that you already know how to expand (x+a)^2 and do the process in reverse

[raw hill]

^

[spring acorn]

im dying someone help my cahnnel hahahha

[charred viper]

I know $0<|x-2|<\delta$, and this shows that $|x-2|<\frac{\epsilon}5$ but why do you decide $\delta = \frac{\epsilon}5$. The explanation here just seems very minimal, like "oh, they look similar so let's just slap an equal sign and see where it goes?" I want to know why they can be considered equal.

[obsidian monolith]

bluepianist

[charred viper]

Another example is equally as ambiguous :

[hollow lance]

Well in reality a lot of deltas work. But the definition requires us to find a single delta for which the inequalities hold

[charred viper]

So we take the extreme case?

[hollow lance]

In some cases yea, sometimes you can be more lenient, depends on the function really

[charred viper]

Gotcha, thanks

[spark scroll]

this is a photo my teacher has on our classroom. i understand how y' = 8x -4, but how does it end up as 2? (he wrote therefore ... =2 bottom right)

idk if im asking in the right place but the document i get my questions from are called precal

[summer ruin]

it doesn't

the red text explains it

[spark scroll]

so wtf is he writing bottom left

righth

oh hes explaning how he got the derivative of inner function

[summer ruin]

bottom left is the answer, bottom right is the explanation

[spark scroll]

yea ok thanks

is this precal? cuz alot of the questions im seeing above look much harder

[summer ruin]

well some schools cover this, some don't

[hushed sphinx]

If you're computing a derivative, then there's nothing "pre-" about it. It's just calculus.

[spark scroll]

is y' = 4x

sorry if im asking thihngs that seem basic or something but i dont hhave anywhere to check and i dont wanna get tmy basics wrong

[summer ruin]

the derivative of the sum is equal to the sum of derivatives

[willow bear]

the derivative of x is 1, not 0

[spark scroll]

oh yea i missed a step there

ty

y'' would be 4 then righht

[willow bear]

yes

[spark scroll]

im stuck on this. it can be rewritten as (x+5)^1/2, from here i went to 1/2(x+5)^-1/2, but now im stuck.

1/2(x+5)^-1/2 can be rewritten as

1/2 / (x+5)^1/2

but then thats a fraction on the top which dosent look right. am i taking the righht step or am i supposed to rewrite it just as 0.5

if theres anything im wrong ab lmk

[summer ruin]

"supposed to" is the wrong term to use

if the answer is correct then who cares what it's written like as long as it can be easily understood

if it's wrong well then even more so

[spark scroll]

so uh is my way of solving it correct? or am i taking wrong steps in processing the question

[summer ruin]

well that's one way to solve it

[spark scroll]

1/2(x+5)^-1/2 is y' right

[summer ruin]

1/2 * (x+5)^-1/2, yes

[spark scroll]

• is meant to symbolize multiplication?

[summer ruin]

yes

[spark scroll]

ok ty

[summer ruin]

1/2(x+5)^-1/2 is ambiguous

[spark scroll]

what's the less ambiguous way of writing my answer then

[summer ruin]

it must be written such that the order of operations cannot be interpreted in two or more ways

[spark scroll]

so 1/2 / (x+5)^1/2 is better?

[summer ruin]

just use the multiplication for complicated fractions

[spark scroll]

i dont understand

[summer ruin]

$1/2 / (x+5)^{1/2}$ could be $\frac{1}{\frac{2}{\sqrt{x+5}}}$

[obsidian monolith]

Transparent_Elemental

[spark scroll]

oh shit im stupid

1/2 is just = to root i forgot that

so yeah its just 1/2 / root x+5

[summer ruin]

you're missing the point

then I could just write $\frac{1}{\frac{2}{(x+5)^{\frac{1}{2}}}}$

[obsidian monolith]

Transparent_Elemental

[summer ruin]

and that's not at all what you mean anyway

hence writing 1/2 / (x+5)^1/2 is ambiguous

[spark scroll]

can u show me how u would write it

[hushed sphinx]

If you want to have two divisions, you need parentheses to show which of them produces the input to the other.

But what one usually does is put on a single fraction bar.

[summer ruin]

tl;dr manage your fractions properly $\frac{\frac{1}{2}}{\sqrt{x+5}} = \frac{1}{2} \cdot \frac{1}{\sqrt{x+5}} = \frac{1}{2 \sqrt{x+5}}$

[obsidian monolith]

Transparent_Elemental

[spark scroll]

ok yeah i understand this now. took me a few mins

[spark scroll]

my y'= is 9(8x^3-1)^8 * 8

but on the internet i used a calculator and it says

216x^2 (8x^3-1)^8

where does the 216x^2 come from

[willow bear]

chain rule

[spark scroll]

so i should be using chain rule for questions when the parensethes have a power higher than 2?

if its something just like (2x-1)^2, i can just make it 2(2x-1) * 2 and be done with it since the power is just 1 right?

[hushed sphinx]

That's also using the chain rule.

[spark scroll]

so how come when in my 2nd example i dont get an x infront of my 2 at the start or a power

[hushed sphinx]

What do you mean by that? Applying the chain rule to (2x-1)^2 gives you an outer derivative of 2(2x-1) and an inner derivative of 2.

[spark scroll]

yeah so how come when using the chain rule on the image above, the outer derivative has an x^2 and not just 216

[hushed sphinx]

(For a second power you could also use the product rule on (2x-1)·(2x-1) to get 2(2x-1)+(2x-1)2 and simplify that to 2·2(2x-1)).

It's not the outer derivative that has an x^2. It's the inner one.

What is the derivative of 8x^3-1?

[spark scroll]

just 8 no?

[hushed sphinx]

How do you get that?

[spark scroll]

only been taught in class that its the number infront of the x

does it change when the x has a power?

im currently going ahead so my teacher hasnt taught this part yet

[hushed sphinx]

So what you're missing is the power rule.

That ought to be somewhere in your textbook.

[spark scroll]

well my teacher uploaded a 80 page doc with what we needa learn. havent seen power rule in it

[hushed sphinx]

(And again, this is by no means PREcalculus, it is actual #calculus).

[spark scroll]

sorry ill switch to there after this, or would u like to take our converstion there now?

[hushed sphinx]

We can continue, but please ask further questions there.

[spark scroll]

ok

so to understand where im going wrong i gotta go search up power rule ye?

[hushed sphinx]

Yeah.

The derivative of x^n is n·x^(n-1).

[viscid thistle]

Just started pre calculus this year

[unique portal]

it sucks lol

[keen anchor]

It's easy, the only thing that might throw you off is logarithms and sequences

Precalculus is just Alegbra 2 on caffeine.

[unique portal]

best comment LMAO

[turbid lark]

My precalculus was a semester of algebra then a semester of trigonometry. I’m pretty sure it’s newer trend to teach these classes combined into 1 course to help students prepare for calc

you guys have my condolences I think it’s better to take your time and learn things a bit slower rather than rush the fundamentals

[potent acorn]

Can anyone help me answer this 😭😭😭

[willow bear]

did you crop off part of the problem? what's "it"?

and what are we asked to do?

[remote spoke]

There’s a formula for that find the mid points of those points

[gritty smelt]

Hey if I need to find an equation where every real number is a fixed point, does f(x) = 0 work?

[hushed sphinx]

Not every real number equals 0.

What does "is a fixed point" mean?

[viscid thistle]

is this correct?

If not, how would i plug -4 into 2? there is no variable

and 0, and 3

[hushed sphinx]

When there's no variable nothing happens when you replace it with a value.

[viscid thistle]

oh, ok

so i just replace it with those values?

[hushed sphinx]

Um, that may or may not be right; I don't completely understand what you're proposing.

[viscid thistle]

I'm saying do i have to replace -4 for 2 since -4 is less than 3, we use the first piece in the piecewise function, is that correct?

same thing with 0 and 3, since that's what the rule of the function is

[hushed sphinx]

If it makes you feel better, you can also plug into the \emph{entire} definition, getting facts like
$$ f(-4) = \begin{cases}2&\text{if }-4 \le 3\3\cdot(-4)-4&\text{if }-4 >3 \end{cases}$$
$$ f(3) = \begin{cases}2&\text{if }3 \le 3\3\cdot3-4&\text{if }3 >3 \end{cases}$$
$$ f(6) = \begin{cases}2&\text{if }6 \le 3\3\cdot6-4&\text{if }6 >3 \end{cases}$$

[obsidian monolith]

Troposphere

[hushed sphinx]

Since -4 is indeed less than or equal to 3, we see that f(-4) = 2.

[viscid thistle]

oh, ok so what i'm understanding that nothing changes when there is no variable

like you stated a few minutes ago

[hushed sphinx]

Right.

[viscid thistle]

Appreciate it

[strange echo]

Basically for every x value that is less than or equal to 3, the y value is 2

[weak kettle]

how do i find the range of a piece wise function

[hushed sphinx]

If you are 11, then you're not old enough to use Discord legally.

[rustic bison]

How is that possible though child genius's are not even at that level yet also what troposphere said. But If you telling truth my self esteem gonna get hit hard 😅

[viscid thistle]

Someone ban this dude

[robust elk]

how do you do this? 🙇‍♀️

[hushed sphinx]

Do you know something about how the complex roots of a polynomial with real coefficients are distributed?

[robust elk]

no, not really

[hushed sphinx]

They come in conjugate pairs.

[robust elk]

Oh! Thank you, I get it now

[viscid thistle]

Well, I am also Indian but I live in America. I am in high school. As far as I know about the Indian Curriculum, you are only able to take trigonometry in 10th grade. So I am interested to know more. May you please just elaborate on how you are able to and how you are managing to take this class at 11 years old( 5th grade)?

[tall lily]

anyone know how to graph a piece wise function

[summer ruin]

just graph the pieces separately for those x for which the piece is defined

[tall lily]

uh

[sick quartz]

I'm not sure they expect you to know conjugate pairs yet? otherwise you would have known immediately.

If you just use some prior knowledge you can get to the answer. They say it's degree 5, and give 3 terms. If we write it out like:

(2 - x)(3i - x)((3 + i) - x) = 0

you can distribute the complex terms, and create a quadratic. now you can set that equal to 0, and you'll find the remaining 2 roots with some clever algebra

[hidden quiver]

hi! can anyone help me with this? im rly stuck

[sick quartz]

based rotring pencil user

[hidden quiver]

lol

i was intending to cover my name

[sick quartz]

what part are you stuck on

[hidden quiver]

the first part

[sick quartz]

write out what they tell you using the function they give. so at sea level h = 0, and evaluating W(h) at h = 0 gives 140

[hidden quiver]

ok ill try with that

omg it worked it was a lot easier than i thought but thanks so much

[sick quartz]

no worries. you got a value for w right?

[hidden quiver]

yup

[sick quartz]

alright the rest of part 1 is just algebra. see if you can go from here

[hidden quiver]

yea im trying number 2 now

i think i got itt

[sick quartz]

nice. the last 3 questions are especially important in calculus

[proud jungle]

very basic question, but I want to know if the answer is C or D

if we choose -2 as our k value, once the graph gets reflected the range becomes y <= 2

now I want to know if that would be y <= -k as we defined k to be -2 in the question or it will be y <= 2 as the value of k changes to 2. thanks

[sick quartz]

every f(x) = y before the reflection, become f(x) = -y after the reflection. if the function reaches y = k as the end of the range, after the reflection -y = k should be true. so y = -k. your conclusion is correct

[proud jungle]

I argued with 2 friends, and a math teacher. they said its C but I still argue the answer is D as k was set to be a negative number originally. which is why I had no other choice but to ask here

they told me because it's a transformation the value of k becomes 2

[sick quartz]

oh mb i didn't notice. you used the wrong symbol for your domain

you got the end of the domain right

but if you have a porabola that goes up to infinity, and has a minimum at k.

the reflected porabola has a maximum at k

which flips the inequality sign if you use that in the algebra instead of just the = sign

[proud jungle]

x-axis reflection moves the k value

[sick quartz]

the range in what you sent is correct

so you chose k = 2?

[proud jungle]

-2

[proud jungle]

the question defines it to be less than 0

[sick quartz]

those are the values before it is reflected

[proud jungle]

Yes

but we are writing the range in terms of k

and k was set to be -2

and the range is to be y <= 2
which in my opinion is y <= -k

[sick quartz]

it's not an opinion it's how it is, you're correct either way.

[proud jungle]

thank you for your confirmation. I need to find a way to prove the answer to the people who argue the answer is C

[sick quartz]

k < 0 <=> range: y => k before the reflection

if you choose a negative number as k, the inequality satisfies itself without you defining k to be negative

-y => k
y <= k

[sick quartz]

it isn't. you defined k as negative and chose a negative number. you shouldn't define k as negative, because the number you chose is negative

you satisfy k < 0 by choosing a negative number. so choosing k = -2 would flip the range and the sign of the end value. so y=> -2 becomes y<= 2

so it's C

[proud jungle]

I dont quite get what you mean here, as the question said k is a negative number, and the range of the original graph before the reflection is y => k, once it gets reflected it becomes y <= -k as the value of k and a has changed. if the value of k remained the same the answer would be y <= k and if the value of a stayed the same the answer would be y>= -k

original graph:

y >= k

reflected graph:

a flips the sign
k moves to -k

y <= -k

[sick quartz]

pay attention to the inequality sign. if something is a maximum before the reflection, its opposite value becomes a minimum after the reflection

y >= k before the reflection, yes.
now reflected:
-y >= k
y <= -k

get it? just look at the graphs. -k = 2 . which is what we expect all the y-values to be below in the 2nd graph, as is

[proud jungle]

didn't you just prove D is the answer

[sick quartz]

i mistakenly did because i looked at the maximum value, not the range as a whole. y = k becomes y = -k. but range also flips

[proud jungle]

the parabola opens up in the original graph
which makes the inequality sign >=

the parabola opens down in the reflected graph which makes the inequality sign <=

however the value of k also gets reflected. it doesn't stay k. it becomes -k

which is why I believe the answer is y <= -k

[sick quartz]

ye

everything checks out. how come they said c?

ask if the k < 0 restriction still holds. if so then the range will always be restricted to the value of k in relation to that

[proud jungle]

no idea, they agree with the range and -2 proving part but they tell me because its a transformation the value of k becomes 2 so y<=k

[sick quartz]

yeah, y<=k so that it satisfies the k < 0

you chose k = -2 first, reflected becomes k = 2. this doesn't satisfy k < 0. y always has to be greater than or equal to k before the reflection. after, y always has to be less than or equal to k

[proud jungle]

the k < 0 restriction is only for the initial graph

After the transformation it doesn't have to keep the restriction

[sick quartz]

yeah the graph doesn't lie

it's probably in how they deal with the inequality restrictions then. but the first explanation was right given how you defined k, and the reflection you drew

[proud jungle]

my best bet is to find the original source of the question and the answer key

but it's from an old diploma exam

but I do really appreciate your help

[sick quartz]

no bother, you figured it out yourself beforehand. maybe someone else can weigh in to make sure

[cold oar]

Hmm, anyone know what to do within this situation?

f: {(1,1), (2,3), (3,0), (4,5)} ; f(2)

Is it like F as the numerator and the bottom as denominators, and just divide by f(2)?

[proud jungle]

👍

[sick quartz]

think about what the notation means. it tells you that those pairs of numbers, coordinates, belong to f

[cold oar]

So, I'm guessing this is Domain and Range at this point.

[sick quartz]

the next statement asks something about f. if you know functions, you know what it's asking.

[cold oar]

Yes yes.

[sick quartz]

yes and no

[cold oar]

But those functions don't coordinate to anything.

[sick quartz]

it gives information about the domain, which is required for the next part to exist

they aren't functions

think of a curve or a line

actually, think of a bunch of points scattered in a graph

[cold oar]

Oh, I see I see.

Am I suppose to find where the dot and line meets eachother?

[sick quartz]

we call f the set that makes up all those points. assuming x,y plane, a coordinates looks like (x, y)

the next part asks what f is at x = 2, or f(2) = ?

[sick quartz]

no

don't think about any lines or division. just think about these:

(1,1), (2,3), (3,0), (4,5)

coordinates

on a graph

[cold oar]

Oh okay okay.

Well, it's only that, so I feel like it is just Domain and Range.

[sick quartz]

see those points? f is the set (or collection of those points)

[cold oar]

Right right.

[sick quartz]

the function f(x) relates the x-value of a coordinate to its corresponding y-value. the same coordinates (or points) that make up f on the graph.

[cold oar]

Right.

x,y

[sick quartz]

so f(2) asks what the corresponding y-value for the point at x = 2 is

[cold oar]

Ohh, I see.

k(x) = 4x + 3 ; 3f(-2), do you know what this would be?

Also, I found the value within the graphing.

Helped a lot with the graph solution.

[sick quartz]

it's good that you thought about domain and range. because for the function f(2), or any f(a), to exist, the point has to exist

[sick quartz]

where did you find that?

[cold oar]

Right, that's what I was figuring out while doing it.

[sick quartz]

what's the full problem?

[cold oar]

My math textbook 😭

That is the full problem

That can't be a thing like 4 doesn't equal towards k, right?

[sick quartz]

yeah it's just fancy notation for a table with two columns, like you're used to seeing in other places

[cold oar]

Right, that's true.

[sick quartz]

we've all been there

i meant if you could send a picture

[cold oar]

Oops, sorry, I actually got it, don't worry about that one now xd

[viscid thistle]

We talked in DM's and that person said that they were homeschooled so when his/her parent's decided to let them go to real school, they combined some grade levels such as 7 and 8 and 10 and 11.

[fleet tendon]

I’ve seen people demostrate this by the quadratic formula, i know that, but anyone has any way to demostrate like, since the beginning?

[willow bear]

where did you get this

the derivative of the function ax^2 + bx + c is certainly not ±sqrt(b^2 - 4ac)

[fleet tendon]

Its just to find x

[hushed sphinx]

So you're saying that at the x where the polynomial is 0, the derivative of the polynomial is such-and-such.

But the derivative is different at any other x.

[fleet tendon]

If want to find in other image just move it to polynomial

[hushed sphinx]

I don't understand what you're saying there.

[fleet tendon]

d=ax^2+bx+c

d-d=ax^2+bx+c-d

0=ax^2+bx+c-d

And start all over

[hushed sphinx]

I have no idea what you're even trying to say now.

[fleet tendon]

lmao

3x^2+2x-4=7

3x^2+2x-4-7=7-7

3x^2+2x-4-7=0

3x^2+2x-11=0

3=a
2=b
-11=c

But yea

Only when the polynomial is 0

If not, have to do that to get different c

[willow bear]

and what exactly is the point of all this?

to overcomplicate the otherwise simple process of getting the derivative of a polynomial function of degree 2?

[fleet tendon]

Why does it happens

But in a linear demostration

From the start all in the same line

Is it possible?

[willow bear]

I have no idea what you're even trying to say now.

[fleet tendon]

.

[viscid thistle]

do you know what the +-sqrt{D} means?

[fleet tendon]

Yup

[viscid thistle]

nvm you did it in the pic you sent

[proud jungle]

@sick quartz I asked another teacher and they told me the correct answer is D as I thought. So there's that

[cerulean flower]

very quick question: do "factor the polynomial completely" and "find all real solutions of the equation" mean the same thing? (both are about polynomials)

[zinc fulcrum]

I'd say no, not really

connected, but not really the same thing

[cerulean flower]

thanks! i realized that after i got help in another channel 😅

[hushed sphinx]

(Please don't crosspost questions in multiple channels like that).

[zinc fulcrum]

the equations 2(x^2 - 1) = 0 and x^2 - 1 = 0 have the same solutions, but the polynomials 2(x^2 - 1) and x^2 - 1 have different factorizations

[zinc fulcrum]

and oh lol, well there's my answer anyway

[fleet tendon]

DID IT!

[viscid thistle]

Math!

[fleet tendon]

Well, at the end its not exactly well written but u get the idea

[vapid plaza]

This only makes sense when x is a root of ax^2 + bx + c

So no

I guess you found an expression for the derivative of a quadratic at its two roots which is pretty cool ig

(not actually that cool imo)

[hot sigil]

how do you do this

[sick quartz]

great

[hushed sphinx]

First you'll need to find out what the exercise means when it says "level of difference". It's not a standard term and might be invented specifically for the particular course you're following.

(E.g. it's not like the second differences of 0,12,10,0,-12,-20 look particularly distinguished).

[fleet tendon]

You are literally getting roots of polynomial

So kinda yes

Also x from the derivate is the same as the x which is in the polynomial

[vapid plaza]

???

[fleet tendon]

ax^2+bx+c

x in that polynomial, is the same x as in ax+b

[vapid plaza]

Yeah what is your point

[fleet tendon]

That x is always the roots?????

[vapid plaza]

Nah bruh do you get what I’m saying
“2ax + b = +-sqrt(…)” is only true whenever x is a root

[hushed hollow]

Wtf is that method of completing the squarw

[hushed hollow]

@fleet tendon

[fleet tendon]

d-d is 0

ab^3+tg-wr^2 = h

ab^3+tg-wr^2**+d-d** = h

Is that your question?

[viscid thistle]

@viscid thistle hes a college grad that left and said aw screw that

[viscid thistle]

https://cdn.discordapp.com/attachments/955196822702403714/1018345228043817051/IMG_5951.png

May somebody please help?

<@&286206848099549185>

[sick quartz]

represent the numerator as a series

you should get one of these:

and the rest is just the rest

[limpid crystal]

I’ve been stuck on a problem from a worksheet that goes like this:

If f(x)=6-3x+k and f(f(k))=16, determine the value of f(-6).

I’m not sure on how to handle the multiple variables since we haven’t taken it up in class

To clarify I mean x and k

[hollow lance]

So you want to find f(-6). When you plug in x=-6 into f(x)=6-3x+k what do you get?

[limpid crystal]

f(-6)=24+k?

[hollow lance]

Yup yup, so now we need to find k

The only other information we haven't used is f(f(k)) = 16

There is a lot going on with the nested f(f(...)) so let's simplify, what is f(k)?

[limpid crystal]

f at k?

[hollow lance]

yup, f(x) evaluated at x=k

[limpid crystal]

Does this mean I can simplify f(f(k)) into f(k)? and set it equal to the f(-6) equation?

[hollow lance]

No no, I want to work with f(k) first just to simplify the calculation of f(f(k)).
When you find f(k), you can plug that in in f( f(k) )

f(k) = 6 - 2k right?

[limpid crystal]

How do you get that equation?

[limpid crystal]

Oh wait nvm I understand

[limpid crystal]

Ok so do we go from this to f(f(k))=6-3x+(6-2k)?

[hollow lance]

When we want to compute f(f(k)) we replace f(k) first.
f(f(k)) = f(6-2k)
Now what is f(6-2k). It's f(x) = 6-3x+k evaluated at 6-2k.
Or f(6-2k) = 6-3(6-2k)+k

With this, you've successfully rewritten f(f(k)) into something more manageable.
Thus you go back to f(f(k)) = 16 or 6-3(6-2k)+k = 16 and solve for k, and then go back to f(-6) = 24+k and plug in that k

[tropic elm]

So how do you learn this?

[limpid crystal]

I get it now, thank you so much!

I was very confused with how I was supposed to work through the 2 functions

This makes it clear

🙏

[viscid thistle]

Thank you.

[rigid topaz]

U solved it

[rigid topaz]

?

[viscid thistle]

?

[fleet tendon]

?

[cyan island]

I am having problems with this one question
Determine the interval(s) on which the function is increasing and decreasing.
g(x)=(3x+1)^2 +2
I can't use derivatives because my teacher told us not to and I also have not reached that point yet

[viscid thistle]

?

bruh that is like 4 question marks. 4 is a perfect square of 2 and the square root of 4 is 2.

[fleet tendon]

IM NOT SAYING THIS IS CORRECT

But something must must be wrong, in which part and why? I would say you cant power a negative power to a fraction, so (-1)^1/2 is the error, what do u guys think? (Or know)?

[fringe charm]

(-1)^(1/2) = (-1)^(2/4) = sqrt(i)² = i

[shell brook]

Idk if this is the right channel but if someone can help me see if this is the right way of solving fraction modulus

6/5 mod 7
5x = 6 mod 7 //6 -> 13 -> 20
5x = 20 mod 7
X = 4 mod 7 = 4

[willow bear]

@wind python do you still need help with this

[sleek saddle]

@ everyone heyy guyss heyy what does the d in dx/dt signify pleasee tell????

[hushed sphinx]

This is parabola in vertex form. Do you know how to sketch it and see its monotonicity intervals on the graph?

[sick quartz]

ah that was stupid.

just see the end-behavior for the numerator and the denominator to see if it converges or not

since they're both polynomials. you just see what their degrees are, and if like what the leading coefficient is

the correct answer should be 0 then

[elder coyote]

Need math tutor

<@&286206848099549185>

[limber patrol]

Can anyone explain what happened between these two pictures

[summer ruin]

just multiplied by 1

it's a very weird looking one, but $\frac{-2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{-2}$ is indeed equal to 1

[obsidian monolith]

Transparent_Elemental

[limber patrol]

ok, but how did we get the $\frac{-2}{\sqrt{5}}$ on the left and the $\frac{\sqrt{5}}{-2}$ on the right, in the first place (this might be a dumb question because my book doesn't explain this section very well)

[obsidian monolith]

mstf

[summer ruin]

you're multiplying

what's the difference anyway if $a \cdot b = b \cdot a$

[obsidian monolith]

Transparent_Elemental

[sturdy bobcat]

How is this the same as 3^-3?

[summer ruin]

it's not?

$(x^{-1}y^{-1})^{-3} = (\frac{1}{xy})^{-3} = (xy)^3 = x^3y^3$

[obsidian monolith]

Transparent_Elemental

[sturdy bobcat]

my math book says it is but im not sure either

yes i know , after calculating the parenthesis and applying exponent rule i get x^3 y^3 but my book says that its also the same as 3^-3

[summer ruin]

most likely you're misunderstanding your math book then or missing the context in which it is the same

with no context it is not the same, but with context it could be

[sturdy bobcat]

not sure how i could misunderstand this

[summer ruin]

I suppose this is from the answers section?

[sturdy bobcat]

ye

[summer ruin]

well what does the question itself says

[summer ruin]

no, no, the question

it says complete the something

[sturdy bobcat]

complete the following statements:

[summer ruin]

what's after the arrow?

[sturdy bobcat]

x^3 y^3 = ... and then just where im supposed to complete

[summer ruin]

well then that's true

you're given that x^-1 * y^-1 = 3 and asked to find x^3 * y^3

so in that context it is the same, the context being that 1/(xy) is equal to 3

[sturdy bobcat]

oh yeah of course, i was just looking at answers section so i forgot the actual question

[lean pendant]

precalc looks easier than algebra 2

i know ite not

[hushed cape]

Hi all, I'm reviewing precalc basics, like real basics and I've been watching freecodecamp's precalc video. There's one thing im a bit confused by and it's probably quite trival and im overthinking probs but

[lean pendant]

the domain

so

[hushed cape]

So the person in the videos find the domain by taking the denominatgor

[lean pendant]

x to x

[hushed cape]

So like, what about the rest of the equation

[lean pendant]

wdym

[hushed cape]

the x numerator

[summer ruin]

the domain of the function are all values of x such that the function is defined

numerator changes nothing

[lean pendant]

rn ur probably just figureing out the bottom

[hushed cape]

So if I wanted to find the domain in a equation with like x^2/x^2-4x+3, I just take the denominator>?

[summer ruin]

you would care about numerator if it was something like $\frac{\frac{1+3x}{x}}{x^2-4x+3}$ because now in that case the numerator defined only for some x, but not for all

[obsidian monolith]

Transparent_Elemental

[hushed cape]

So how can I tell if a numerator defines some aspects of x?

[summer ruin]

at what point the function $\frac{1+3x}{x}$ is undefined?

[obsidian monolith]

Transparent_Elemental

[hushed cape]

When it reaches 0?

when x is 0?

[summer ruin]

yes

so domain can't contain the 0 for that function

therefore the domain of original function doesn't contain 0 either

it's the same thing really for all the other functions and their combinations like $\frac{\log (x)}{x}$, $\sin (\frac{1}{x})$, $\log (|x|)$ etc

[obsidian monolith]

Transparent_Elemental

[summer ruin]

as long as you know for what values of x each individual function is defined

[hushed cape]

ahhhh I see

Sorry just looked on desmos

I see, I don't need the care about the numerator function unless they interfere with the values of the domain of the denominator

[rustic swallow]

I’d like some help with this please

[summer kiln]

It’s a composition of functions

I think

What does fog mean

So like for the first one you’d plug in the function g into the function f

[split holly]

what does the E mean in between n and z+

is it equal?

[uncut mulch]

element of / in

[hollow summit]

E means belongs to

[limber patrol]

You can get the distance using integration for the given function, and the second function takes absolute value because time can't be negative

[sturdy bobcat]

why is (z/xy)^6 the same as (xy/z) ^-6?

[gusty frost]

when you have a negative exponent, you flip the fraction

so pretend its like (x^-6 / 1) that would be the same as (1/x^6)

it switches the negative to positive on the exponent and also flips the fraction

@sturdy bobcat

[sturdy bobcat]

thx

[gusty frost]

yep

[torn path]

can someone pls help with this , i just started precalculus 1 week ago

[frozen schooner]

can someone guide me through? my proctor did a "you're on your own" and im literally pissed abt it so idk what to do since its due in a few hours and yt vids aint helping

[limber patrol]

can anyone explain how to solve this (i have the solution i just need the explanation)

[limber patrol]

https://www.mvcc.edu/learning-commons/pdf/trigonometric-identities-sept-2017.pdf
I'd recommend memorizing and understanding this then try again with the questions

[silent lynx]

ill be doing integrals all up to pre-uni tmr 🙂

not that you need to know 😅

[limber patrol]

thanks i understand it now

[viscid thistle]

I like the pfp of Srinivasa Ramanujan.

[wraith flume]

okey bro

[wraith flume]

Ah,yes.

[limber patrol]

quick question: can we simplify this as $\frac{9}{2} (ln 9)$

[obsidian monolith]

mstf

[uncut mulch]

no

the sqrt(x) in the numerator appears to be the power of 9

[limber patrol]

thanks

[rich grotto]

Can someone help me with this please

[summer ruin]

$u(x) = p^{-1}(q(x))$

[obsidian monolith]

Transparent_Elemental

[worn sequoia]

How can I calculate zeros vs x intercepts when using a quadratic equation? The equation being:
x^2+x-30

[summer ruin]

zeros and x-intercept are the same thing?

you just set it equal to zero and solve for x

[worn sequoia]

Hm

Okay

Thank you

[fading wedge]

Is x+y^2=2 a function ?

[lost grove]

it's a relationship

but if you subtract y^2 from both sides it's just x=f(y)

so yes

a function

kind of

[fading wedge]

I mean like does it pass vertical line test typa function

[lost grove]

what?

[fading wedge]

Is there just 1 y value for each x value

[lost grove]

no, since this is a function of y

not a function of x

[fading wedge]

But i plugged it into the graphing calc and it passed vertical line test

[lost grove]

did it pass it for x values or y values

if it passes it for x values you graphed it wrong

[fading wedge]

Wdym there was only one point at each x value

Oh my god

God damn it

[lost grove]

what

[fading wedge]

I forgot to do plus or minus the square root

[lost grove]

??

just subtract both sides by y^2

this isn't a quadratic

[fading wedge]

No i made it y = root of 2 - x

[lost grove]

what?

why

that doesn't even work

nvm

it does

but it's way more complicated

[winter crater]

Yo anyone can help me with my pre_ clac shit?

[winter crater]

Would be nice if anyone can Dm me

[viscid thistle]

can anyone help me with conic sections?

please dm me

[limber patrol]

hi do we just ignore 2x is that how natural logarithms work?

[willow bear]

no, we do not "ignore the 2x"

what is done here could be called a "chain rule in-reverse" or a "silent u-substitution" perhaps

but the point is that the derivative of $\ln(x^2+9)$ is $\frac{\dv{x} (x^2+9)}{x^2+9}$.

[obsidian monolith]

Ann

[limber patrol]

thanks for the help

[obsidian monolith]

mstf

[limber patrol]

can anyone explain how to solve this

[fiery saffron]

Try turning a cot ^2(5x) into (csc^2(5x)-1)

No promises

Hm

Should work if you know the antiderivative of cot

[limber patrol]

actually solved it did what you said then converted one of them to cos 5x/sin 5x and used ln then got correct answer but anyways thanks for helping

[fiery saffron]

Not that it matters but it should be #calculus youre def passed precalc

[limber patrol]

the category says pre university this is grade 12 for me

but i'll put it in calculus next time

[fiery saffron]

Ah gotcha. It doesn't really matter

[keen anchor]

Statistics question:
There are 18 unique filling options at chipotle. How many possibile combinations of 2 or more (ex. Lettuce+rice, lettuce+beans, lettuce+beans+cream) are there?

[wraith flume]

one second...

[wraith flume]

You undarstand bro?

[willow bear]

@wraith flume don't give out answers.

[viscid thistle]

Erdos is a fan of Srinivasa Ramanujan and so am I.

[rigid cliff]

the answers just come to him

[limber patrol]

Yes

[limber patrol]

I said in an earlier message that I solved it

[wraith flume]

okey

[atomic rune]

the cross product of two unit vectors is a unit vector right?

but then why is the cross product of these two unit vectors not a unit vector?

the answer is

which is NOT a unit vector

[viscid thistle]

the cross product of a unit vector with itself is zero

[atomic rune]

i'm talking about the cross product of two distinct unit vectors

not a unit vector with itself

[viscid thistle]

the cross product of any two distinct unit vectors is not a unit vector

[atomic rune]

oh ok nvm

[knotty zephyr]

Hello! I need help, how can I solve this?

1. Find the inverse and domain and range of f(x) and f^-1(x)

a.) y = 3x^2-5x+4

[slender bobcat]

When |a+b|=3 and |a|+|b|=11
Find ranges of a and b

Please help

[frigid python]

Guys need helppp

[viscid thistle]

on what?

[slender ferry]

Can someone solve this n let me know what they get ?

[opal tree]

Are you solving for d?

[slender ferry]

Yes

[opal tree]

What have you tried

I have an answer

[slender ferry]

My answer is 0.98

But the teacher’s is 0.9118 something ?

idk what I did wrong

[opal tree]

Yours is closer

0.98 is literally just about 0.01 off from the actual value

[slender ferry]

It’s my first work of pre calc 😭😭

[opal tree]

I'm wondering if it's because they rounded π to 3

But 0.98 is literally close enough

[slender ferry]

Okayy I’m just vv worried cuz it’s gonna be on a test on Monday

[opal tree]

I mean if you wanna know

$$\frac{9.8}{π^2} ≈ 0.992$$

[obsidian monolith]

Umbraleviathan

[opal tree]

Which is the answer

So your teacher made a dum dum moment somewhere

[slender ferry]

ahhh I see lol thank u so much

[rich grotto]

Can someone help me please

[viscid thistle]

i fw ur pfp

[slender ferry]

@viscid thistle thank uuu, u know the artist ?

[viscid thistle]

destroy lonely

[slender ferry]

u cool asf

[viscid thistle]

nah you cooler bra

[dull rain]

x+cy=6
cy=6-x
y=(6/c) - (x/c)
y=(-1/c)(x) + (6/c)
This is of the form y = mx + y0
where m is the slope
Here, m=(-1/c)
In question, it is given slope = m = 2
So, substituting m,
2 = (-1/c)
c = (-1/2) = -0.5

There are 2 quadratic polynomials P(x) and Q(x) having roots R1, R2 and r1, r2, respectively. If I sum them up, then can the roots of the new quadratic polynomial R(x) = P(x) + Q(x) be calculated using R1, R2, r1, r2?

[willow bear]

don't think so

x^2 - 1 + 2(x^2 - 2) has different roots than x^2 - 1 + 10(x^2 - 2), surely

yet in both cases the quadruple (R1, R2, r1, r2) would be the same

[dull rain]

Wdym by this?

[willow bear]

take $P(x) = x^2 - 1$ and $Q(x) = 2(x^2 - 2)$

[obsidian monolith]

Ann

[willow bear]

now take $P_1(x) = x^2 - 1$ and $Q_1(x) = 10(x^2 - 2)$

[obsidian monolith]

Ann

[willow bear]

if you took these polynomials and distilled them to their roots, you'd get the same four numbers in both cases

so, if a formula of the kind you seek were to exist, it'd give the same result for these two pairs of P and Q

do you agree?