precalculus

[arctic galleon]

56(b)

[viscid thistle]

hey man what program is this that you're using, would be greatly appreciated if you told me 😄

[kindred grove]

good notes

5

good notes 5 i think

[viscid thistle]

is it possible to add 24 functions at once?

[patent beacon]

Sure

[viscid thistle]

I don't find any examples on the internet, other than adding only two functions ;_;

[restive hound]

Hello, what should I do to get rid of negative 12? Because if I divide everything by -12 there will still be a negative number even if I multiply it everything by -1/12

[glacial lantern]

how do i write the equation of a square root graph that goes up wards

https://gyazo.com/5e6c8f670fddeb668d662c3550cf6ea2

[steady imp]

can anyone help me for my pre calculus?

[restive hound]

If it is about circle, ellipses, and parabola, I can try

[willow bear]

@steady imp don't ask to ask, just ask

i.e. post your problems so that any potential helpers can know if they can help you

[steady imp]

hereeeee

[steady imp]

yeahhhh exactly

[restive hound]

On number 2, I think number 4 is a parabola. Because on my pre-calculus book on chapter 1 it has a picture of an antenna with the description saying “A parabolic antenna has the capability of receiving and reflecting radio waves to one focal point etc…”

[steady imp]

wow , we don't have book :(((

only in online class

[restive hound]

Ah yes, I had online class too and we didn’t use our books that much

[steady imp]

i see. Help me please

[restive hound]

Okay, I will try. From the first part, where you have to identify what type of conic section that each general equation will produce. First, I think you have to be familiar with the standard form of a circle, ellipse, parabola, and a hyperbola. Because in ellipse, there are two right? Like if the smaller number is under x squared and the bigger number is under y squared then you will expect a vertical ellipse, and it’ll be a horizontal one if the bigger one is under x and the smaller one under y. And if you were given an equation of an ellipse in standard form, you will know it is an ellipse based on the type of standard form it is. You know what I mean? In the standard form of a circle, its standard form is (x-h)^2 - (y-k)^2 = r^2. So if you get this kind of answer with this form then you’ll know it is a circle. If you get the answer with either of these form (x-h)^2/a^2 - (y-k)^2/b^2 =1 or (x-h)^2/b^2 - (y-k)/a^2 = 1, you know it’s an ellipse. So I think you should be really familiar with the standard form of each conic section. From what I know, since you’re given the general equation. Turning them into standard form is quite similar for all the conic sections. In letter a, rearrange the equation. Put the +1 on the right side and everything on the left, rearrange them, place all the x and y together. So x^2-2x+y^2-4y= -1. Here, I think you’ll either expect a circle or an ellipse @steady imp

@steady imp

With the architectural designs and where you have to identify what conic section it is, remember the type of graph that each conic section will produce.

Visualize

[steady imp]

can u do the solution? A , B , C and D?

you can send me on private message

[viscid thistle]

thanks man really appreciate it 🙏

[restive hound]

Hello, how can I add x^2 to 1/4x^2?

[uncut mulch]

$x^2 + \frac14x^2$
?

[obsidian monolith]

ℝamonov

[restive hound]

What does it mean?

[uncut mulch]

is that your question?

[restive hound]

Yes

[uncut mulch]

have you ever combined like terms before?

[restive hound]

Oh yes! But I don’t know with fractions 😅

[uncut mulch]

then this seems like a case of expressing
$$1 + \frac14$$
as an improper fraction

[obsidian monolith]

ℝamonov

[uncut mulch]

would you be able to do that?

[restive hound]

Ah okay so will it be 6/4x^2? Because there’s 1 in x^2, so it’s like adding 1x^2 to 1/4x^2?

6/4 and improper fraction

[uncut mulch]

no

how are you getting 6/4

[restive hound]

I supposed that x^2 has a coefficient of 1, so if I combine like terms then I would add 1x^2 to 1/4x^2 and when I calculated in the calculator I did 1 + 1/4

[uncut mulch]

what exactly did you to with
1 + 1/4 to get 6/4
because 1 + 1/4 isn't 6/4

[restive hound]

Ah! It’s 5/4x^2

Isn’t it? Lol I made a mistake

Thanks a lot @uncut mulch !

[craggy furnace]

Hello, I am struggling with an online homework, I believe my answer is right but its marking it wrong. Can someone tell me if they think my answer should be different? I should be writing the Domain in the top text box, in interval notation, and the range in the bottom text box

[uncut mulch]

the graph looks cut off

is that what they have you

[craggy furnace]

Sorry that was just my screenshot

It still doesnt appear to have an end point

[uncut mulch]

was any other information given

[viscid thistle]

How do i solve this

[craggy furnace]

No just *If your graph continues to infinity, drag the edge of your rectangle to the end of the graph paper

[uncut mulch]

where's -3.44 coming from

[viscid thistle]

I know how to address 4/3/x but not 4/(3/x)-1

[craggy furnace]

the orange point on the x axis

[uncut mulch]

how do you know it's -3.44

[craggy furnace]

the endpoint

[uncut mulch]

are you told that it's -3.44?

[craggy furnace]

yes

[uncut mulch]

I asked you if any other information was given and you said no...

[craggy furnace]

Ahh sorry!

[uncut mulch]

is there anything else you aren't telling me

[craggy furnace]

The orange end points show -3.44 and 3

The green dot at the end of the green shaded area shows -3.22

That is all I know

[uncut mulch]

then the values you gave seem alright

[craggy furnace]

Ok, Ill just accept this one is marked wrong and move on

[viscid thistle]

can someone help me plz

[uncut mulch]

what have you tried?

[viscid thistle]

i dont even know how to approach it

Wait a minute

[viscid thistle]

Can i solve for x in the denominator and use that value when dividing with the numerator

i think i figured out how to solve the problem

Yes i did

I did

By “solving for x” I meant making the denominator equal to 0

And i didnt realize that i would be dividing by 0 so disregard that “dividing with the numerator”

What would the domain and range be for: ln(-x-1) and sqrt((ln(1-x^2)+2)

what do you know about the behaviour of the function ln?

@viscid thistle

[jovial bridge]

1. Domain of any log function is a>0, in this case a=-x-1
   Range of any log function is always positive

2. domain of any sqrt function is positive numbers. And since log function is always positive we would just need the domain of the inside log. Inside the log 1-x^2 > 0, which is -1 < x < 1
   Range of the sqrt function is always positive

[viscid thistle]

Ohh so then 1. D: X > -1, R: x>0 and 2. D: (-1,1), R: x>0?

Ln can’t be negative, so x>-1

@jovial bridge @viscid thistle

[jovial bridge]

correct

[viscid thistle]

no

[viscid thistle]

this is very incorrect in a lot of places.

[viscid thistle]

and no to this too.

-x-1>0 isn't x>-1. recall that you have a -x not an x on the left hand side.

[viscid thistle]

and for example the domain of 2 isn't correct either. that is wrong. recall that you are dealing with sqrt(log(1-x^2)+2) and not just log(1-x^2)

[flat goblet]

hey people I have a hw that needs to be done

9 questions

[viscid thistle]

@flat goblet this channel is occupied at this moment.

please use an unoccupied one. and read #❓how-to-get-help

[flat goblet]

thanks

[viscid thistle]

I’m still here, just seeing that you’re typing so I will wait to see what you’re typing

[viscid thistle]

and now onto this, regarding "the range of any log function is always positive", is wrong by itself, take a simple counterexample f(x)=log(x), the range of f is R.

[jovial bridge]

can you elaborate?

oh

shooot

i meant the domain

and base

[viscid thistle]

as well as "any sqrt is positive numbers", non-negative numbers, since 0 is also included

[jovial bridge]

fair enough

uh

well i need to tell him now

[jovial bridge]

Hey, this is actually wrong I went wrong with my concepts. The domains are correct, but the range for 1 should be all numbers, and the range of 2 should be all positive numbers including 0

is that right?

[viscid thistle]

no, it isn't still

[viscid thistle]

look at what i wrote here

[jovial bridge]

its x<-1

wait

yeah x<-1

i must of mis wrote

hey platypus

[viscid thistle]

but the bigger issue comes with the "since log function is always positive we would need the domain of the inside log", that is wrong as well, you do need to investigate on solving log(1-x^2)+2>=0.

[jovial bridge]

i messed up

[viscid thistle]

So what the interval notation? (-1, infinity) for domain for 1?

[viscid thistle]

so nothing of this is correct.

[jovial bridge]

im gonna rework this

[viscid thistle]

Wait that can’t be

Has to be non negative

[viscid thistle]

no, non-negative is for sqrt. ln can't take values like 0

unlike sqrt(0)=0

so essentially, you want to solve -x-1>=0, as you want the argument to be non-negative

[jovial bridge]

1. D: x<-1 R: all real numbers

[viscid thistle]

also avoid just giving away the answers. it goes against the purpose of the server

[jovial bridge]

yeah but

i already messed it up

and plus we were explianing that ealier

[viscid thistle]

it was explained incorrectly

[jovial bridge]

but our convo fixed it

[viscid thistle]

But how can we get a value for x that doesn’t give a negative value since it’s -x - 1

[jovial bridge]

• • (-21525120) = positive

2 negatives

[viscid thistle]

that doesn't guarantee that they understand it

Ohhh

[jovial bridge]

im going to explin things in one piece afterword

so do you understand the 1st one?

the domain and range?

[viscid thistle]

So for domain for 1. Is x>-1

[jovial bridge]

no

x<-1

less than

[viscid thistle]

Right

Two negatives

[jovial bridge]

yup

[viscid thistle]

Then the range.. (0, infinity)?

[jovial bridge]

it can be any number

let me elaborate

[viscid thistle]

+- infinity

[jovial bridge]

yes

because a is the out come of our log

and exponents can be any number

2. $D: -\sqrt{1-1/e^2}<=x<=\sqrt{1-1/e^2}$

[viscid thistle]

Okay and now 2.

[obsidian monolith]

DR

[jovial bridge]

can you re write your original question

incase i wrote the question wrong

but i believe that that is the domain

[viscid thistle]

So I’m trying to understand the rules/limitations for domain and range for sqrt((ln(1-x^2)+2)

[jovial bridge]

ok

so the first thing we need to do is evaluate the domain of a sqrt function

as we know the domain of a sqrt(x) is x>=0

[viscid thistle]

Yes

[jovial bridge]

x is what is inside of the sqrt function

but we have a another function inside of the sqrt function

so we have to say $\ln({1-x^2})+2>=0$

[viscid thistle]

Oh okay

[obsidian monolith]

DR

[jovial bridge]

after that

we need to solve for x

im not sure if you learned this but we have to raise e to both powers

so then we could get rid of the log function

[viscid thistle]

No

[jovial bridge]

oh

let me show you what it looks like

$e^{\ln({1-x^2})}>=e^{-2}$

[obsidian monolith]

DR

[jovial bridge]

we do this because one of the propeties of the logs

the top part of the image explains what are are doing

where a = e

after that

we would simplify

$1-x^2>=1/e^2$

[obsidian monolith]

DR

[jovial bridge]

$x^2<=1-1/e^2$

[obsidian monolith]

DR

[viscid thistle]

Because e^-2 is 1/e^2

\frac{}{} for fractions.

[jovial bridge]

yup

[jovial bridge]

oh i know but i feel that 1/e^2 isn't neccasary for that

ok so

once we have that thing simplified

we just need to find this domain

which in return would give us

2. $D: -\sqrt{1-1/e^2}<=x<=\sqrt{1-1/e^2}$

[obsidian monolith]

DR

[viscid thistle]

\leq as well for <=.

[jovial bridge]

oh yes thanks for that

i forgot that

[viscid thistle]

Ohh because x is +- what we solved for

[jovial bridge]

yeah

[viscid thistle]

Okay that makes sense 🙂

And now the range

[jovial bridge]

ok so the range of would be pretty simple

can you tell me the range of a sqrt function is?

any sqrt function

[viscid thistle]

Non negative

[jovial bridge]

you are missing a small detail

[viscid thistle]

0 to infinity

[jovial bridge]

but you are right on that

yes

correct

we did it 🥳

[viscid thistle]

Yay! Thank you both! Lol I will be back again soon

[jovial bridge]

no problemo, thank you to al3dium for correcting me xD

[viscid thistle]

if by this you mean that the range of 2 is (0,inf), it's once again not correct either.

[kind arrow]

how does linear factorization theorem apply to perfect square trinomials?

[jovial bridge]

;-; it would include 0

[viscid thistle]

that's not correct either.

you keep having the missundestanding of "if one function of sqrt has range [0,inf), then every sqrt function has range [0,inf)".

and that's not true.

how do i find the domain

as long as a function is above the x axis its considered positive right?

like if part of the function is at the top left that is positive or is it negative since thats the negative quadrant

[dusk elm]

The domain is all the possible x values. Since it is a rational function, you have all real numbers as the domain other than the values in which the denominator equals 0 because you can’t decide by 0

[dusk elm]

Yes the function takes on positive output values above the x axis. In other words, f(x)>0

[dusk elm]

Try solving for y

[jovial bridge]

once you solve for y, you can act like the amplitude became 1/2

[viscid thistle]

how do you know if a function is not even or odd

[sick steppe]

if $f(-x)\neq \pm f(x)$ then it's neither even nor odd

[obsidian monolith]

Mosh

[restive hound]

Hello, how can I get the square root of y^2 = 2x^2+1?

[steel venture]

is there more to the question @restive hound

[warm forum]

wdym

[steel venture]

sorry for the late response

[restive hound]

It’s okay, wait I’ll show you

I’m solving systems of nonlinear equations and this is the question I have. I’m planning to do the substitution method because I don’t think elimination method would work. So I chose to isolate y^2 in y^2 - 2x^2 = 1. But I don’t think I could substitute y^2 = 2x^2 + 1 to the y^2 in the first equation @steel venture @warm forum

[steel venture]

why couldn't you?

[restive hound]

What do you mean? So I could substitute y^2 = 2x + 1 in the y^2 on the first equation?

[steel venture]

yeah, i don't see why not

[restive hound]

Oh okay

Also, how could I solve if the given two equations are a standard form of an ellipse and a circle?

Still systems of nonlinear equations

Here

I made them both to general form but looks like I’m wrong, then I tried removing the denominator on the 1st equation and did elimination but I can’t because the (y-k) aren’t the same

[warm forum]

$y^2=1+2x^2\implies 9x^2+2(1+2x^2)=54$

[obsidian monolith]

Muzan Jackson

[restive hound]

Ohhh okayy

So it is possible

[warm forum]

yep

[restive hound]

Thank you guys

[restive hound]

How about this one? 😭 I’m not sure what to do

I turned them both to general form but when I did elimination it looks wrong, then when I removed the denominator for the first equation and did elimination I can’t eliminate the y because their value of k aren’t the same, only x

[somber sigil]

(x+3)²=16{1-[(y-4)²/9]}

16{1-[(y-4)²/9]}+(y-7)²=25

[restive hound]

Thank you very much @somber sigil but how did it became like that? 😮

[manic lantern]

Hello my fellow smart people

[viscid thistle]

https://gyazo.com/40bed4b911f2551c6f9ccfd883bfe15e

can someone explain why this is from the left now

like how does it become x->6 from the left

[sick steppe]

x^2 is slightly more than 9, so 15-x^2 is slightly less than 6

[small wyvern]

does anyone have basic questions on local maxima, local minima, global minima and global maxima

[dusk elm]

Like example problems?

[merry chasm]

Is this true?

[mild swan]

No it is not

[merry chasm]

Damn

Thought so

[mild swan]

Square both sides and see what you get

You'll have to expand the right side out using foil

[merry chasm]

Ah I see

[gleaming nebula]

I still don't understand how to solve this😵

[mild swan]

Not enough context for that

[gleaming nebula]

That's just it.

Ngl

[mild swan]

You stopped it mid sentence

[gleaming nebula]

Not really that's just it

It says for the measurement of length: 1km ± 50m and expected value is 1100m, find the relative error.

[mild swan]

🤨

Relative error - not sure I'm familiar with that term

[gleaming nebula]

Damn, no one's been able to solve this so far.

[mild swan]

What did Google come up with when you looked up relative error?

[gleaming nebula]

Wait

[mild swan]

Not sure what the ± 50 m indicates now knowing the definition of relative error

[gleaming nebula]

Basically 50 metre

And the plus minus sign

[mild swan]

Yes I get that

But I don't see its purpose in the problem

You can probably just do |1000 - 1100|/1100

[raven kindle]

How to graph this one?

[mild swan]

Just start with a diagram - step by step

[raven kindle]

Shoul i start on 10?

To 50?

Should*

[mild swan]

Once you get your first graphical piece of information, put that down into a diagram

[raven kindle]

okay thank you

[small wyvern]

yep about 10 probs

[mild sluice]

getting confused on this one, wondering which step im messing up

could anyone help real quick?

im normally super good with inverses and theyre never a problem, but this one is tripping me up

[hollow lance]

What did you try so far?

[mild sluice]

i’m writing out a way to do it that i think may be right rn

i may solve it in a minute i’ll get back to you

i hope you can somewhat follow what i was getting at, its hard for me to not do these sorts of problems in my head

just realized i put both as root 5, the inside is supposed to be 3

[hollow lance]

the inside or the outside one

think about it carefully

[mild sluice]

like that

i think ¿

[hollow lance]

you should probably write it again more cleanly and step by step, it's reverse actually

[mild sluice]

oh

hmmm

so the outside is 3 and inside is 5?

[hollow lance]

and by reverse I mean the 5th and 3rd root are swapped

yes

[mild sluice]

ah i see

everything else is right?

[hollow lance]

Yeah, my personal preference is to put positive numbers inside a root first so I'd write it as 8 - (x)^(1/5)

but it is fine like that as well

[mild sluice]

woop woop

[viscid thistle]

what does it mean when someone asks you to "state the finite differences that will be constant"

for example if im given f(x) = x2 +3x -1

[rancid jolt]

do you know first differences

second differesnces

etc.

[viscid thistle]

Guys

[restive hound]

Hello, I’m learning about Mathematical Induction. I was just wondering how he factored out k^2(k+1)^2 + 4(k+1)^3 ? He said he’ll take the GCF, how?

[uncut mulch]

a^3 = a^2 * a

[restive hound]

How? 😅 is it when I simplify 4(k+1)^3?

[uncut mulch]

(k+1)^3 = (k+1)^2(k+1)

[restive hound]

ah ok so from what I understood, that's when you factor (k+1)^3? but what will happen to the 4 next to it? and the k^2 next to (k+1)^2? because he said that he will get the GCF and factor the (k+1)

[lucid flame]

hey y'all I know how to find the Domain and Range, but I'm not entirely sure what the other values are for or what I'm supposed to do with them

ah sorry it's sideways

[dusk elm]

The domain is just all the possible x values for the function and the range is all the possible y values. For example f(x)=x^2 has a domain of all real numbers with a range of [0, infinity)

[lucid flame]

I know that I'm just a bit confused about the other values that are given from a-d on each problem

[dusk elm]

That is just asking what the function is equal to at those x values. So f(-1) is just asking what is the y value corresponding to x=-1 on the graph

[lucid flame]

ahhh okay I was overthinking that super hard

thanks brother

[silent cave]

i dont know how to do this but my hw asks for the zeros of f(x) =(3x-5)/(x-4)

[kind arrow]

what makes f(x) = 0

that's what it means to find the zeros

[silent cave]

i got 5=3x

does that seem right

[viscid thistle]

I don't know what I just saw but today I was doing this problem

Did as it asked me to

The fun part starts here

So I was playing around a bit with the numbers

I took points from a line
convert them into ln
And then draw a graph of those points
Everything in this graph looks like they line up

Went to wolfram alpha to check if they are actually all on a line

So anyway, I did it again but for y=x^2

And then tried it for x^3 and it works too

What I want to know is why do they form a line?

Nevermind found the answer

https://mathbench.umd.edu/modules/misc_scaling/page11.htm

[undone girder]

Hi

[willow bear]

kahree appears to have left.

[warm forum]

This is what I did xD

[trim hemlock]

oh wow, thats quite some dedication. Erm I dont know how i should put this but i should have said that the problem is solved

but erm i think i might look through your solution

tho the intended solution is much much shorter

[warm forum]

that's fine

[warm forum]

I also use derivative but it ended up useless :v

[trim hemlock]

hmm it looks like you are following the direction of a cardano-like formula?

heres the intended direction if you are curious

so basically they acquired the function x^3-9x^2+ax-b from the given information

and then proceed to take the derivative

[warm forum]

yep

[trim hemlock]

what was the special thing of this solution is that they used the behaviour of cubic equations

so they conclude that in order for f(x) to have 3 roots in [-1,5] then f'(x) must have 2 roots in that interval

because as we know a cubic function can only have 3 roots when it has 2 critical points

[warm forum]

yes

[trim hemlock]

so that was their catch which was pretty cool

the rest is just solving so yeah

i mean i was quite close to the answer too, what i messed up is that i thought they meant 3 distinct roots

whereas it was never mentioned in the question

gotta note that to myself and not make that mistake again

[warm forum]

so it uses inequalities?

[trim hemlock]

not really

[warm forum]

hmmm

[trim hemlock]

i mean kind of

well of course in order to get a minimum value of a we must have some kind of inequality in terms of a

[warm forum]

I also did this but I used inequalities a lot

[trim hemlock]

but the main thing is the conclusion where the derivative must have two roots in [-1,5]

[warm forum]

yeah

[trim hemlock]

yeah i have noticed

that was actually pretty nice too

all in all that was quite tricky i have to admit

[viscid thistle]

$$\boxed{\cos 2x = \cos ^2 x - \sin ^2 x}$$
$$\boxed{\sin 2x = 2\sin x \cos x}$$

[obsidian monolith]

Spica

[viscid thistle]

it'll sort out

[unique sierra]

is somebody available to help me understand better Quadratic Functions and Applications on a vc?

[mild swan]

Just ask a question in chat and people will do their best to help you

[dreamy siren]

^

[restive hound]

Hello, is it possible if I move the 1 next to secant and the tan to the right hand side? So it could be sec^2 data - 1 = tan^2 data?

[past meadow]

yes

although its theta

not data

[restive hound]

Ah okay, thank you !

[steel venture]

make sure you understand that you're not just "moving" stuff, but you're subtracting 1 from both sides, and adding tan^2(theta) to both sides

[restive hound]

Ah yes okay, what do you mean by adding tan^2 (theta) to both sides? So if I add it on the left side it would cancel out and on the right side, the tan^2 (theta) would be positive. And at the same you said I should subtract 1 from both sides. So on the right hand side, it’ll cancel out and on the left side I’ll have -1. Like that?

[steel venture]

exactly like that

sin^2(x) - tan^2(x) = 1
sin^2(x) - tan^2(x) + tan^2(x) = 1 + tan^2(x)
sin^2(x) - 0 = 1 + tan^2(x)

that's one example

then -1 on both sides will yield the result you wanted

[restive hound]

Ah okay, thank you very much

Oh yeah, if I multiply cos^2 (theta) and sec^2(theta) that would give me 1 right? Because cos multiplied by sec is 1?

[steel venture]

yup

and that's because $\sec = \frac{1}{\cos} \implies \sec \cdot \cos = \frac{1}{\cos} \cdot \frac{\cos}{1}= \frac{\cos}{\cos} = 1$

[obsidian monolith]

maximo

[restive hound]

Ahh okay! thank you very very much!

[restive hound]

Thank you too for reminding me this

[steel venture]

no problem. so many teachers gloss over it that a ton of people just don't recognize what's going on

[restive hound]

Yeah, it is important to know

[restive hound]

What will happen if I were to add sec(theta) and sec(theta) together? Would it just be sec(theta)+sec(theta)?

[past meadow]

you could write 2sec(theta) if you want

or 2/cos(theta)

[restive hound]

Ah okay, but what if I would add a different one like cos(theta)? And what if it was subtraction?

[civic furnace]

247how do I prove that if a and b are of the same sign then
a/b +b/a >= 2
I have
(a^2 +b^2)/ab -2 >= 0, so I'd have to prove that
(a^2 +b^2)/ab >= 2

[civic furnace]

'An integer indivisible by 3 is of the form 3n +1 or 3n-1'
say n = 1
3n+1 = 4
3n-1 = 2
1 is also not divisible by 3, yet it is not included?
likewise 5 isn't divisible by 3, and it is also not included

thus, wouldn't be a form of an integer indivisible by 3 be
3n+1 or 3n-1 or 3n+2, or 3n -2

likewise, for all integers a number indivisible by an integer x is of the form
xn +- x-(x-1) = xn +- (x-x+1) = xn +- 1
or
xn +- x-(x-2) = xn +- (x-x+2) =xn +- 2
and so on until for y = x,
xn +- x-(x-y) = xn +- (x-x+y) = xn +- y = x(n +- 1), which is divisible by x

Am I making some mistake or is the solution to this problem false:
"Show that the square of an integer indivisible by 3 during the division by 3 gives the remainder 1"
The solution: (3n+-1)^2 is of the form 3k +1
???

[dense saffron]

I think you need to put n=0,n=2 over there for example at n=0 3n+1=1 and at n=2 3n-1=5. Hope that helps

[civic furnace]

oh right

what's wrong with

likewise, for all integers a number indivisible by an integer x is of the form
xn +- x-(x-1) = xn +- (x-x+1) = xn +- 1
or
xn +- x-(x-2) = xn +- (x-x+2) =xn +- 2
and so on until for y = x,
xn +- x-(x-y) = xn +- (x-x+y) = xn +- y = x(n +- 1), which is divisible by x
?

what's the form of all integers indivisible by x if not this

[willow bear]

i think you're overcomplicating this

a lot

if you're allergic to the form nx + r where n ∈ Z and 1 ≤ r ≤ |x|-1, or if you really want that plus-minus symmetry, you could instead do nx ± s for 1 ≤ s ≤ |x|/2

[dense saffron]

I don't think this is the pattern cause if you try putting 7 into the equation xn-+1 it would yield only 6 and 8 as indivisible

I think you should follow remainder theorem basically that @willow bear suggested

[civic furnace]

i meant that all the number indivisible are of either of these forms

[willow bear]

you meant what?

[civic furnace]

for 7, are all the number indivisible by it
7n+-1
7n+-2
...
7n+-6

[willow bear]

some of those are redundant

you only need to go up to 7n ± 3

any number of the form 7n + 5 can also be expressed as 7n' - 2 for example

[civic furnace]

makes intuitive sense, but what's the proof

[willow bear]

the proof for what

[dense saffron]

Dd = DQ+R and max remainder = divisor-1
And for 5/7 either remainder can be 5 or -2 but generally I think 5 is preferred because division theorem says remainder can't be negative but half remainder theorem states that remainder can be negative

https://proofwiki.org/wiki/Division_Theorem

[willow bear]

consider also this: https://www.wikiwand.com/en/Euclidean_division

even though you chose not to help me know what exactly it was that you wanted proof of

[civic furnace]

my question has been answered

[shut totem]

multiply and then bring the 2ab to the other side

[uncut mulch]

what's the question asking for

[mild sluice]

need help with 23

not sure how the 2x would affect the process

in D(x)

[mild sluice]

nvm figured it out with long division, still having trouble with synthetic tho

[tidal ferry]

need some help heheh

from calculus

I always mess up the domain and range

I feel like for 1, there should be no value for which it's undefined right? so range -inf, inf, and domain R?

idk

[dreamy siren]

Correct, #1 has domain of -inf, inf

[tidal ferry]

wait that's its domain?

man I ALWAYS mess that up

[dreamy siren]

Domain is asking “what x value doesn’t break the function?”

[tidal ferry]

I really can't tell the difference between domain and range, it seems they're the same at times

[dreamy siren]

Typically something / x will break it because then you can’t have x=0 for example

[tidal ferry]

yee

[dreamy siren]

Range is asking “what values of y are possible for the entire function”

[tidal ferry]

but if it's defined for all x's, what would the range be? cuz I feel like that too is -inf, to inf

AHa o

hmmmmmmm

[dreamy siren]

If you look at the graph of the function, it’ll be easier to explain

[tidal ferry]

it's a parabola, so it can't be -inf to inf, but something to inf

aight holup

[dreamy siren]

Exactly. Because it’s not a downward parabola

[tidal ferry]

yuss got it up

ok so it can't be -inf

1,inf?

[dreamy siren]

Exactly

[tidal ferry]

domain R, range (1,inf)? ^^

O yayyy

[dreamy siren]

[1, inf)

[tidal ferry]

but 3 is messier

OH RIGHT

[dreamy siren]

Because you can include 1

[tidal ferry]

u think I'm allowed to use a graph calculator for calculus?

like idk, it doesn't say so idk if I'm doing it correctly hahah

[dreamy siren]

Idk ask your teacher lol

[tidal ferry]

oh wait oh wait

so [0,2] means 0 and 2 and all inbetween

whereas (0,2) means values between 0,2 but not 0 and 2

[dreamy siren]

Looking at graphs is helpful in the beginning to build your understanding. But it isn’t necessary once you get the hang of it

[dreamy siren]

Correct

[tidal ferry]

yehh not very used to it yet

gee thanks! ^^

[dreamy siren]

So for #3

What values of x will ‘break’ the function?

[tidal ferry]

its a half parabola

laying down LOL

it's a weird one

lemmme see

4

and anything negative right?

[dreamy siren]

What about x=5?

[tidal ferry]

wouldn't be allowed

[dreamy siren]

Why?

[tidal ferry]

cuz it'd become negative and the graph doesn't go under y=0

[dreamy siren]

Yeah

So I like to look at these problems and think “hm, what can break?”

“Oh, I see a square root. I know that will break if I take the square root of something negative. When will the inside be negative?”

[tidal ferry]

hmmm

but the domain is a lil weird for me still

[dreamy siren]

Then you can begin digging into the answer

[tidal ferry]

ik x<5

[dreamy siren]

Domain is just x

All x values that don’t break it

[tidal ferry]

but it seems to expand to -inf

wait no

yes

(-inf, 4)

[dreamy siren]

So follow my thought process

You agree it breaks if the inside of the sqrt is negative right?

[tidal ferry]

yeah

[dreamy siren]

So use that knowledge to make an inequality to find when it’s negative

(8-2x) < 0

[tidal ferry]

o smart

wait I can only think of 4 still lol holup

and anything above

[dreamy siren]

Solve that inequality and tell me what you get

[tidal ferry]

it's -inf<x<4

[dreamy siren]

8-2x < 0
-2x < -8
x > 4

That’s how you solve an inequality

[tidal ferry]

but

how did the arrow switch lol

[dreamy siren]

Because I divided by a negative

[tidal ferry]

ahaa

oh right

[dreamy siren]

So with that, we know it breaks if x is bigger than 4

[tidal ferry]

yee ik it's undefined for all x>4

[dreamy siren]

Now to check the boundary, will it break if x Equals 4?

[tidal ferry]

yeh

so x=>4

[dreamy siren]

Plug x=4 into the function and tell me what you get

[tidal ferry]

sqrt(0)

[dreamy siren]

Does that break it?

What is the sqrt of zero?

[tidal ferry]

I mean I think, it's not allowed

[dreamy siren]

Try it on your calculator

[tidal ferry]

it's either not allowed or 0 lol

ah got 0

[dreamy siren]

So it’s allowed

Doesn’t break it

[tidal ferry]

O

right I get a horizontal line

cuz no x

[dreamy siren]

So as long as it isn’t bigger than 4, the function is okay

So our domain is (-inf, 4]

[tidal ferry]

AH

[dreamy siren]

Because 4 is okay

We can include it

[tidal ferry]

I thought sqrt(0) was the same as x/0

[dreamy siren]

No

[tidal ferry]

ahh

okok ye I follow u

[dreamy siren]

Cool

So that’s domain

Now range, that’s the y values

[tidal ferry]

(-inf, 4], and range is all positive y's

so R_+? or [0,inf)

[dreamy siren]

Since the function is sqrt(something) then yes we will never get an answer that’s negative

Otherwise it’s okay

[0, inf) is exactly right

[tidal ferry]

YAYYY

I FINALLY GET IT

domain is for x range is for y

[dreamy siren]

Yes

[tidal ferry]

duude tysm

[dreamy siren]

Yw

[tidal ferry]

thx! have a good day ^^

[dreamy siren]

Take care

[keen beacon]

can someone explain how i got this wrong

[fresh hearth]

what is

-2i * 2i

would it be

(-2)(2)(1)

so like

-4

?

or -5

or 5

?

my calculator says 4

but it's 5

[visual hinge]

how so?

its 4

wait no

its negative 4

hold on

you cant get the subgroups of a negative sqrt so I am wrong

hold on

I have all these different ideas

but they dont work

the negatives when you multiply them cancel out'

[hardy mango]

can someone help solve this

[slow marten]

Hi can someone tell me what the domain and range is for this graph

[visual hinge]

is domain basically y intercept

my brain incinerated half of everything I know in math overnight os I wouldn't know

[uncut mulch]

is domain basically y intercept

no

[visual hinge]

ok

[uncut mulch]

far from it

[visual hinge]

good to know

[foggy thunder]

domain is all the values x can take while allowing the equation to be able to plotted on the cartesian plane (for example, plotting -1 for x into f(x)=sqrt x would not make sense, so -1 is not in f's domain)

[restive bridge]

I don't quite understand the rationale behind this proof. Currently reviewing geometric sequences:

https://i.imgur.com/sHR2vD6.png

Why do they substract Sn - rSn?

Wouldn't that lead to a negative output?

[willow bear]

it might lead to a negative output, so what?

@restive bridge

[restive bridge]

How does subtracting rSn from Sn of eventually lead you to knowing the sum of the geometric sequence?

Let's say that r = 2 and the first term of the Geometric Sequence was 2.

If we add up to the Nth term, that would be a 14.

14 - 28 = -14

So then how would combine these two factors allow you to un--

I get it now

[trim hemlock]

Translation: Given f(x) such that its derivative f'(x)=(given below). How many extremas are there?

so just one thing im really curious, what is a good way to make sure that some of the roots of f'(x) isnt an inflection point

i mean i could do the second derivative and test to see if its an inflection point or not, but that is way too rigorous

[sick steppe]

extrema occur when the 1st derivative changes sign

by 1st derivative test.

[trim hemlock]

oh right...

[sick steppe]

also it's a multiple choice question, who's telling you to be rigorous?

[trim hemlock]

ehhh i was too hot-tempered with the question then

[trim hemlock]

i mean since its a multiple choice question, i would always try to go with the fastest route, you know to save time. This is in the form of a national test, so im trying to have my mindset in that position.

tho yeah, i was thinking too complicated, 1st derivative test should just do it right

[sick steppe]

yeah, you just see which roots have odd multiplicity in f'

[trim hemlock]

yep

[ionic summit]

for the third step

i was confuse if we do sin^2(...) does it give x^2 or sinx

[willow bear]

x^2

[ionic summit]

is it cuz we should not see the 2 as sin(sin(f(x)))?

but it's (sin(f(x)))^2

but is there times where sin^2(x) would be sin(sin(x))?

[willow bear]

yes

if you want sin(sin(x)) then write sin(sin(x))

the different meanings of ^2 and ^-1 are the result of history and bad conventions

if you ask me this is why you should write arcsin and not sin^-1

[ionic summit]

oh okay i see

so for trigo the power is abit different

thanks!

[narrow lion]

Domain: Reals \ {-3}
Range: y>1

[jolly raven]

so how do you even notate the domain and range of int(x) or floor(x)?

[willow bear]

Z

[jolly raven]

oh ok

cool

so just like s(x) E Z

E being the element of sign

bruh it said the domain of int(x) is (-inf, inf)

[uncut hollow]

oh the domain is R

or (-inf,inf)

but range is Z

[jolly raven]

i guess all points do end up being included

on the domain

[willow bear]

just say "in" lol

[jolly raven]

Oh ok

[ionic summit]

hi can i ask if all absolute extremas are considered local extremas?

for the domain [a, b], is point 1 considered absolute max and local max, point 2 considered abs min and local min?

[willow bear]

yes all absolute extrema are also local extrema

also it's extremum in the singular

one extremum, two extrema.

[high hornet]

help pls i dont get why it's 10,-10

also how to know which side is shaded?

[fresh hearth]

Hi guys

I looking for precalc study buddy

<3

[viscid thistle]

how is f(x) = -7x ^4 +2x^3-3x^2+4 from quadrants 2 - 4

the degree is even but the leading co is negative

scrap that i was looking at the wrong answer key

hbdsjkal;;'

SORRY

[thin dune]

yooow help me plsss find the seats only

[ionic summit]

oh ok thx

[sharp hollow]

I'm not sure if I did this right?

is mapping notation correct??

[restive bridge]

To prove that Log is an inverse of an exponential function, do I need to know how to use ln?

[willow bear]

no, to prove that log_a is the inverse of x ↦ a^x you need only know the definition of log_a

because that's literally it

[restive bridge]

Yeah, remembering the definition is simple enough. It's more about really just making sure my understanding is solid.

https://i.imgur.com/0LJsDbi.png

To test if a function is an inverse, we would normally be able to do: f^-1(f(x)) = x. No? @willow bear

What's bugging me is that I don't know how to use that method to subsequent get x.

[willow bear]

...

log_a(x) IS the inverse of a^x

like

thats the definition of log_a(x)

[restive bridge]

I understand that,

Lol

[willow bear]

so

theres nothing else to it

don't overthink it

[restive bridge]

I just don't get why they would use that format. It bugs me that I just have to take the definition and run with it. Granted it does make sense. Y = 3^4 ---> (4, 81). so it makes sense that the inverse is (81, 4). But gosh, that bugs me.

Oh well, I'll just take the definition and go with it.

[restive hound]

Hello, where did that minus sign come from?

[willow bear]

expanding (csc(θ)+cot(θ))(csc(θ)-cot(θ))

[restive hound]

Ohhh okay, thank you

If I multiply sin(theta) to sin(theta)/cos(theta), that would give me sin^2(theta)/cos(theta) right?

[willow bear]

of course

[restive hound]

Ah but what if I multiply sec^2(theta) to sec^2(theta), that would be sec^4(theta) right? 😅

[willow bear]

yes

[restive hound]

Where did the cos^2(theta) in sin^2(theta) - cos^2(theta) x sin^2(theta) / cos^2(theta) come from? And was sin^2(theta) there because it was multiplied by sin^2(theta) / 1?

[willow bear]

parentheses!!! and also do not use x for multiplication.

[echo wagon]

^ it comes from creating a common denominator for the two terms

So you are writing sin^2( θ) as sin^2(θ)cos^2(θ)/cos^2(θ)

[restive hound]

Ah so sin^2(theta) was multiplied by cos^2(theta)/cos^2(theta)?

[echo wagon]

Yes

[restive hound]

Oh okay thank you very much! Thanks for the reminder too Ann !

Oh wait, did it become (1-cos^2(theta)) because it is equal to sin^2(theta)? If so, what will happen to the two sin^2(theta)?

[echo wagon]

The 1 -cos^2(theta) came from taking out sin^2(theta) as a common factor

[restive hound]

Ah so the numerator was factored out?

Like this? @echo wagon

[echo wagon]

Brackets

And it should all still be over sin^2(theta)

I mean, it's kind of exactly like they wrote it

Wait

[restive hound]

Ah okay

[echo wagon]

It should be over cos^2(theta)

Not sin^2(theta)

[restive hound]

Ohh okay, thank you very much!

[timber bear]

sorry my mouse double clicks

[restive hound]

Is it possible to separate 1+sec(theta)/1-sec^2(theta) to [1+sec(theta)/1-sec(theta)] [1+sec(theta)/1-sec(theta)]?

[willow bear]

please i am begging you

[restive hound]

Ah not possible then?

Sorry, I’m practicing proving trigonometric identities

[viscid thistle]

(sin(x)+cos(x))^2 = 1 + sin(2x)

[willow bear]

no gain that's not what made me cry just now

what made me cry is your continued lack of parentheses where they belong

[restive hound]

Ahhh whoops… where should it be then? 😅

[willow bear]

you meant $\frac{1-\sec(\theta)}{1-\sec^2(\theta)}$, yes?

[obsidian monolith]

Ann

[willow bear]

that's (1-sec(θ))/(1-sec^2(θ)).

[viscid thistle]

the denominator, can be factored in difference of squares

a^2-b^2 = (a-b)(a+b)

sec(theta)=1/cos(theta)

[restive hound]

Oh yes but the numerator is plus not minus

Ah so if the denominator can be factored in difference of squares then it will be 1-sec(theta) and 1+sec(theta)?

[willow bear]

the denominator here factors into (1-sec)(1+sec)

[restive hound]

Ohh okay thank you guys very much!

[viscid thistle]

hello, can someone help me with this problem?
2x (x + 9) = x (x – 2)
i need to transform it to a quadratic

[civic sierra]

What have you tried?

[viscid thistle]

tbh i dont know where to start

im trying to distribute the variables outside the parentheses

[civic sierra]

God start

What'd you get?

[viscid thistle]

2x^2 + 18x = x^2 - 2x

[civic sierra]

Yep

Now get all the terms on one side

[viscid thistle]

i should transpose it

2x^2 + 18x - x^2 + 2x = 0

x^2 + 20 = 0

[civic sierra]

yep

[viscid thistle]

then what technique should i use to get the roots?

[civic sierra]

Whichever one you want

here you can just take the square roor

root

though it does give you complex solutions

[viscid thistle]

umm using the zero property i got 0 and -20, is that correct?

im betting its correct. thx for the guidance

[civic sierra]

Its not correct

it is correct nvm

you missed an x after 20

[viscid thistle]

oh yea thx

[restive hound]

What does conjugate mean when proving trigonometric identities?

[hollow lance]

A useful identity you'll come across a ton is (a+b)(a-b) = a^2 - b^2

Whenever you have a fraction that looks something along the following lines

[obsidian monolith]

peaceGiant

[restive hound]

Ah thank you very much @hollow lance !

Btw, since we can manipulate pythagorean identities.. if I were to manipulate sin^2(theta) + cos^2(theta) = 1, I could get sin^2(theta) = 1-cos^2(theta) and cos^2(theta) = 1-sin^2(theta). But is it not possible for me to have 1+cos^2(theta)? Or possible?

[hollow lance]

From that identity alone, the best you could do is 1 + cos^2(x) = sin^2(x) + 2cos^2(x) which might not be very useful, instead you can consider cos^2(x) = 1/2 + 1/2 * cos(2x) though it really depends on the question

[restive hound]

Oh okay, I was wondering because I’m not sure how to prove this

[willow bear]

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

[obsidian monolith]

Ann

[restive hound]

Ah (a-b)(a+b) was used here?

[willow bear]

yes, i used the difference-of-squares identity.

[restive hound]

Ah okay then foil method was used in the denominator on the middle right? Then combine like terms?

[willow bear]

"foil method"...

i mean

maybe you have to resort to that dreaded FOIL mnemonic...

[restive hound]

What do you mean?

[willow bear]

what i mean is idk about you but i feel no need to include any intermediate step between (1-t)(1+t) and 1 - t^2

if you do not see why these are the same then you can of course do the expansion the long way

i.e. using the distributive property

but i do recommend practicing some normal algebra for some time until things like this are automatic

it's kind of like the multiplication table

it's good to get really practiced with algebraic moves like this so that you don't have to slow down every time you see one of them being made

[restive hound]

Okay, I’ll do. Thank you very much, I’ve learned a lot today

[glacial monolith]

Pain

[mild swan]

Damn - those remainders are so stupid lmao

[fathom seal]

Hi precalc

[viscid thistle]

cam someonewa

hepl

i need help with my algebra 2 homework

[full pagoda]

#rules

dont ask just send

welp he left the server sucks to suck

[restive hound]

Hello, I was wondering if what I did was right? So I could prove that the right side is equal to the left? I stopped at sin^2(theta)sec(theta)+sin(theta). I’m wondering what I could do from there, should I get one of their quotient identities or reciprocal? Also I made the denominator as one because when I multiplied sec(theta) to cos(theta) it equals to 1

[restive hound]

<@&286206848099549185>

[rough parrot]

im not sure about what you did, but here's how i approached this. i hope it helps:

1. Expand both sides:
sin0 + sin0tan0 = tan0sin0 + tan0cos0
sin0 + sin0tan0 = sin0tan0 + sin0

2. Subtract sin0tan0 from both sides: (Not really needed haha)
sin0 = sin0

Hence proved

i used 0 to represent theta

[uncut mulch]

use t instead

(or x) if you don't have access to greek symbols

[restive hound]

Oh okay but I need to make tanx(sinx+cosx) to sinx(1+tanx) though but thank you, it’ll help !

[rough parrot]

noted!

[restive hound]

Is it possible to combine these denominators? Or it has to be cot^2x too?

[trim hemlock]

it has to be cot^2(x) of course

they must share a common denominator

[placid knot]

please be respectful to others

openly making others uncomfortable can lead to consequences (muting, banning) depending on the intent

[willow bear]

the intent is made very clear by considering the message he replied to is from 2 years ago.

[restive hound]

Thank you @trim hemlock !

[vivid moth]

@miyorumi#5370 willing to help you can Dm.

[restive hound]

If sec(theta) multiplied by cot(theta) is csc(theta), then if I multiplied cot^2(theta) to sec(theta), does that give me csc^2(theta)?

[uncut mulch]

do you mean
multiplied cot^2(theta) to sec**^2**(theta)

[willow bear]

$\frac{1}{\cos(\theta)} \cdot \frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}$

[obsidian monolith]

Ann

[restive hound]

Ah nope, I mean cot^2(theta) to sec(theta). The secant doesn’t have a square @uncut mulch

Wait, if 1 is multiplied to cos(theta) how did cos(theta) multiplied by sin(theta) equal to sin?

[uncut mulch]

$\frac{\cos(\theta)}{\cos(\theta)} = , ?$

[willow bear]

\cos

[obsidian monolith]

ℝamonov

[willow bear]

it's basic fraction simplification. since you're clearly having trouble with it, you would do well to practice doing it without trigonometric shit getting in the way.

remind yourself of how to multiply fractions.

seriously, i've seen you stumble on basic algebraic things here a lot

it's a bit concerning that you keep trying to do things with trig when algebra is clearly where your gap in knowledge is

[restive hound]

I understand now why it was 1/sin(theta). And yep you’re right, I actually forget but practicing could make up for it. Thanks for reminding me, can you give me advice on what algebra methods or ways I could use when proving trigonometric identities? Aside from factoring, foil, the difference of squares identity

[willow bear]

just about everything algebraic

there really isn't much you learn in algebra that doesn't also apply to trig

[restive hound]

Okay thank you very much, I’ll try to always keep that in mind

[manic lantern]

hello

how do I find m

[harsh oxide]

To solve for m, you have to first get rid of the square root. What's the reverse of a square root?

[manic lantern]

squared

[harsh oxide]

Good. The first thing to do is square both sides to get rid of the square root

[still comet]

Is this correct?

[harsh oxide]

Not quite. There's also a negative branch to consider. For example, when x = -5, you get 25>4, which is definitely true

[manic lantern]

100= m/100

[harsh oxide]

Not quite. The 10 in the denominator is also in the square root, so you don't have to square it to 100

Well, not just don't have to, you shouldn't

[manic lantern]

1000?

[harsh oxide]

Is that your answer for m or...?

[manic lantern]

uhh

[dark ingot]

By squaring both sides you get:
10^2 = m/10

If you multiply again by 10, you get:
10^3 = m

or

m = 1000

——

This is basic algebra and does not belong in the pre-calculus channel. If you are attempting to learn calculus or even trig for that matter, fundamentals in algebra is a necessity.

[restive hound]

Hello, I was wondering if what I did was right? In number one, I used the pythagorean identity for sin^2x and combined like terms for the two -cos^2x. In number two, since there’s 2cos^2x, I separated them and made it cos^2x + cos^2x. Because from what I know, when you combine these two, it’ll be 2cos^2x. Then for sin^2x, I used the pythagorean identity so (1-cos^2x). Then I cancelled out cos^2x and -cos^2x.

[trim hemlock]

yes both are correct, but for the second one, you might want to erase the second line because when presented like that, it creates confusion

if you do want to have that intermediate step, you should write it out as an equation, because you started with an equation

[restive hound]

Oh okay, on what part? which line? Also, if I got cos^2x+1 that’s just the same as 1+cos^2x right? I could just rearrange it?

[trim hemlock]

from the first line to the second line of problem 2

[trim hemlock]

yes, those are equivalent

[restive hound]

Ah! Okay so you said that if I want to have that intermediate step, I should write it out as an equation because I started out with an equation which is 2cos^2x+sin^2x? So I could write it as 2cos^2x + (1-cos^2x) and cancel out the -cos^2x from (1-cos^2x) and the cos^2x from 2cos^2x, which will leave me with cos^2x + 1 which is the same?

[trim hemlock]

no what i meant is, basically you should have an equation in the second line, i.e 1+cos^2(x)=cos^2(x) + cos^2(x) + (1 - cos^2(x) )

or if you want to write that as a note or sth, then you should write it to the side of the question, or write it out clearly that this is a note

the reason you should not put the second line there right below the question is because readers will infer that the second line is equivalent to the question, which it is not since the question is an equation.

i mean i might be too pedant here but thats just representation-wise

[restive hound]

Ah! Okay, I see. Thank you !

[somber sigil]

u have to consider the -2, so x²-4>0, factorise so (x+2)(x-2)>0. use sign analysis or a graph to get the range of values of x

[trim hemlock]

Translation: Given f(x)=(in picture) satisfying (given conditions in picture). How many extremas are there in y=|f(x)-2019|?

so i have to say, im stuck here

tho i have a picture of what i should be doing, so basically i should first find how many extrema f(x)-2019 can have, and then i have to find how many roots f(x)-2019 can have, and then just add them together and i should have the amount of extremas for |f(x)-2019|

but the problem arises when trying to find the roots and extremas of f(x)-2019

i have also tried to use the given conditions, but im not sure how i should apply it

[hollow lance]

Do you perhaps have the answer key?

||I suspect it is 5, but I might be wrong||

||You need to use the inequalities, along with the argument that the cubic has a leading coefficient of 1. It means to the left (-inf) it goes down, and to the right it goes up, so it certainly crosses the x-axis once, the inequalities should tell you that it must cross it 3 times, from there it's easy||

^ hint

[trim hemlock]

i dont, but i should have it soon enough, ill take a look at the hint

[hollow lance]

And by extremes I think you are referring to local extremes, since there would be only one global one

[trim hemlock]

hmmm why does a+b+c-2018<0 implies that f(x)-2019 should have 3 roots?

[trim hemlock]

yeah, local ones

the question just speaks in general and we just all assume so

[hollow lance]

I don't want to spoil most of it away, however those inequalities have something to do with f, perhaps maybe for some specific x the inequality happens

[trim hemlock]

hmmm

[hollow lance]

Or another way to think of it, rewrite the inequalities in terms of f

[trim hemlock]

its just that im having a bit of trouble to connect the a and b in the inequality to the a and b in the function

lemme just wrap my head around that for a bit

[hollow lance]

Yeah, think about what you need to do to f, so you can work with a and b.

I'll be afk for a bit, I'll return in 15 minutes or so

[trim hemlock]

sure

thanks for the hint btw

ohhhhh

i see where we are going with this

ah alright, this problem is solved. Channel is open

[hollow lance]

Great, yeah the problem boils down to showing that the new function f(x)-2019 is positive at x=0 and negative at x=1, so going from left to right we have three roots on the interval
(-inf, 0), (0, 1), (1, inf), We make a quick sketch and use the absolute value

[trim hemlock]

yep, thats such an elegant solve

thank you very much

[hollow lance]

Of course, anytime

[ocean spire]

Anyone wanna help?? Im so incredibly confused

[obsidian monolith]

codemonkey

[viscid thistle]

this is from problem The length is twice as long as the width. The height is 2 inches greater than the width. The volume is 192 cubic inches.

but not sure how to approach it at all

$(x^3+2x^2) = 96$

[obsidian monolith]

codemonkey

[viscid thistle]

$2(x^3+2x^2) = 96$

[obsidian monolith]

codemonkey

[viscid thistle]

meant to put this my compupter is laggy sorry

[dusk edge]

Do you need to solve for x?

[cedar bramble]

how do i factor this?

$(x^2-4)(x+3)^1/2 - (x^2-4)(x+3)^3/2

$(x^2-4)(x+3)^1/2-(x^2-4)(x+3)^3/2

[viscid thistle]

$(x^2-4)(x+3)^1/2 - (x^2-4)(x+3)^3/2

uh

$(x^2-4)(x+3)^1/2-(x^2-4)(x+3)^3/2

hm

[cedar bramble]

plz wtf?

i’m so confused

[uncut mulch]

identify common factors in both terms

also consider that (x+3)^(3/2) = (x+3)*(x+3)^(1/2)

[forest otter]

anyone know how to discern if the function is continuous?

[pine oracle]

How can I find Sn?

[mild swan]

That's a telescoping series

So write out the first few terms and see if you can recognize a pattern

[visual brook]

My teacher says when taking limits of composite functions you keep taking limits no matter if the functions are continuous at the previous functions' limit??

Is this untrue

This is what he did

Isn't this wrong

[hollow lance]

f isn't defined at x=1, but that doesn't mean the limit (x->1) f(x) doesn't exist

so in this case (the limit from left and right is the same) we can compute that limit

[visual brook]

Some definitions I've seen online point out that f should be continuous at lim(x->0) g(x) in this case in order for the limit to exist

Oh nvm I just understood

[visual brook]

Thank you

[ornate island]

can someone explain the frequency of a sinusoidal function

[thin dune]

isn't a is the logest segment in hyperbola?

[dusk elm]

Polynomial functions are continuous every. Since this is a piecewise function you just need to test the point x=0 to ensure that the function isn’t discontinuous at that point

[dusk elm]

Depends on what you are defining each variable to be

[dusk elm]

You have no variables defined from the graph. What is a, b, c, and d?

[thin dune]

yeah thats y m confused

[dusk elm]

I would recommend asking your professor for clarification or consult your notes

Like this?

[proud raven]

multiply by the lefthand denominator

then if you dont have eagle eyes and a brain of steel you have to distribute and start the equating

[dusk elm]

Yeah what he said, after you distribute I would collect all the like terms and you will get 4 equations in which you can solve for A, B, C, and D

[proud raven]

i can give you a big hint

turn your paper sidewise so you dont have to use multiline equations in a few steps

[viscid thistle]

how do you do the question, "for each polynomial, determine the value of K ig the remainder is 3" (kx^2 +3x +1) / (x+2)

[dusk elm]

try out some long/synthetic division

[viscid thistle]

do i not use factortheorum?

[dusk elm]

i would recommend distributing the right side then collect like terms. This will give you 4 equations

4 equations and 4 variables means you can solve it

https://www.storyofmathematics.com/multiplying-polynomials this might help

[ornate island]

<@&286206848099549185>

[sharp fiber]

can someone show me how to do 33

then i can do the rest by myself

[valid badger]

plug in 5

[wet knoll]

Is it differentiation

[dusk elm]

(4(x+h)-4x)/h

You should see some really nice cancellation and simplification

[fresh hearth]

HIiiiiiii

How would I go about this problem

I'm stuck on this part

So the part where I would do

7^-x = 2401

How would I evaluate that??

<@&286206848099549185>

[proud raven]

just in the same

[fresh hearth]

so like