help-0

[mild oyster]

.close

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[gray umbra]

Hello! I need some help figuring out the solution set for
|x+3| / |x-5| ≤ 3
I was able to get ]-3, 3[ U [9, +∞[ by checking cases (x: ≤ -3, ≥3 ^ ≤5, ≥5).
Math motors like wolfram give the same answer but instead of "]-3, 3[", it is "]-∞, 3[".
Where am I going wrong? In the case I get different signs in denominator and numerator, I must have -3 ≤ x, otherwise numerator will go negative, right?? If so, why is it that the solution ranges from -∞?

[zinc haven]

your mistake lies here

what is the sign of x - 5 when x <= -3?

[gray umbra]

it would be negative

[zinc haven]

correct. so what happens when you multiply both side of an inequality by a negative number?

[gray umbra]

but I'm not multiplying though, I'm dividing the -1s from the numerator and the denominator

ohhhhh

I saw it

since x-5 is negative per sé I must flip the sign at the third step

aight ty so much 🫂🫂

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[zinc haven]

yes

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[tawny condor]

I have a piecewise function, let's say 0 for -pi < x < 0, and x/2 for 0 < x < pi
I want to make wolfram integrate this function times 2 between -pi and pi.
When I just do Integrate[Piecewise[{{0, -pi<x<0}, {x/2, 0<x<pi}}], {x, -pi, pi}] it works, but if I do:
Integrate[2 * Piecewise[{{0, -pi<x<0}, {x/2, 0<x<pi}}], {x, -pi, pi}] it doesnt understand my query.
I need to be able to take this piecewise function, multiply it by something and integrate.

[tawny condor]

I'm not looking to simplify it manually

[zenith pebble]

1v1 in math?

[tawny condor]

this channel is taken

[random gulch]

So the 2 would do: 2*x/2?

[tawny condor]

guys this channel is taken, I already sent a question here

take your discussion elsewhere please

[random gulch]

This is directed to you

[tawny condor]

Oh lmao

Yes

[random gulch]

Why not put the 2 into the piecewise?

[tawny condor]

But as I said, if possible I wouldn't want to manually place it in the piecewise definition

[random gulch]

I didn't read

[random gulch]

Why not tho?

[tawny condor]

I need to be able to multiply it by other things

This is just an example

[random gulch]

Oh I see

So I had a quick look at docs and couldn't find anything

I'm also a little busy so cannot help any further

I apologise

[tawny condor]

no worries

thanks for trying

[tight pier]

,w Integrate[PiecewiseExpand[Cos[x] * Piecewise[{{0, -Pi < x < 0}, {x/2, 0 <= x <= Pi}}]], {x, -Pi, Pi}]

[tawny condor]

okay but let's say the 2 isn't a constant

what happens if I replace it with cos(x) now

???

inside the integral

i want to integrate AFTER multiplying

[tight pier]

try it yourself

[tawny condor]

,w Integrate[cos(x) * Piecewise[{{0, -Pi < x < 0}, {(x/2), 0 < x < Pi}}], {x, -Pi, Pi}]

[ocean seal]

Results provided by WolframAlpha

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Requested by redstoneplayz09

[tawny condor]

see

[tight pier]

wtf

[tawny condor]

yeah wolfram is dumb

[tight pier]

what are you even trying

[tawny condor]

i guess it has something to do with how piecewise functions are interpreted

again, if possible without doing that

[random gulch]

@tawny condor

[ocean seal]

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Requested by adoniszgks

[random gulch]

I solved it

[tawny condor]

?

ok can you show me whats your prompt

[random gulch]

Use PiecewiseExpand[2*Piecewise[...]]

[tawny condor]

ok lets see

,w Integrate[PiecewiseExpand[cos(x) * Piecewise[{{0, -Pi < x < 0}, {(x/2), 0 < x < Pi}}]], {x, -Pi, Pi}]

[ocean seal]

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Requested by redstoneplayz09

[tawny condor]

that doesnt seem to work

,w Integrate[PiecewiseExpand[2 * Piecewise[{{0, -Pi < x < 0}, {(x/2), 0 < x < Pi}}]], {x, -Pi, Pi}]

[ocean seal]

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[random gulch]

It worked for me???

[tawny condor]

on the website?

[tight pier]

,w NIntegrate[Cos[x] * Piecewise[{{0, -Pi < x < 0}, {x/2, 0 <= x <= Pi}}], {x, -Pi, Pi}]

[tawny condor]

nice

that seems to work

[tight pier]

no

[tawny condor]

oh its numerical

[tight pier]

not that

[tawny condor]

?

[tight pier]

it does cos(piecewise) for some reason

[tawny condor]

I dont think it does

also it turns out that also doesnt work for me since I was planning to also have a parameter there

NIntegrate doesnt work with that

[tawny condor]

try it with bounds?

[tight pier]

bro xy-ing us

[tawny condor]

yeah kinda I know

but its a pain to explain

honestly if there is no good way Ill give up on this approach

[random gulch]

wdym

Without bounds on?

[tawny condor]

from -pi to pi

this channel is taken..

[karmic falcon]

oh

[random gulch]

@karmic falcon Goto #help-5

[tawny condor]

refer to #❓how-to-get-help

[karmic falcon]

my bad

[random gulch]

Sorry I don't understand wym

[tawny condor]

@random gulch try this

Integrate[PiecewiseExpand[Cos[nx] * Piecewise[{{0, -Pi < x < 0}, {x, 0 <= x <= Pi}}]], {x, -Pi, Pi}]

I have no idea why its working for u

hopefully this definite integral also will

[random gulch]

Sure

One sec

I can't copy paste on mobile

[tawny condor]

its Integrate[function, {x, -pi, pi}]

[tawny condor]

holy moly

what kind of sorcery is this

what app you using

[random gulch]

Wolfram cloud

Mobile app

[tawny condor]

this literally doesnt work whether I use the website or the wolfram api straight from code

like bruh

[random gulch]

Fr installed it for this help lmaoo

[random gulch]

That's so dumb

What did u send to api

Did u make sure u did n*x?

[tawny condor]

yea

nx

just x

or just 2 *

nothing works

whenever I multiply it by something, it hates me

alright i give up for now but thanks for showing me that it works on the app

appreciate the help

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[random gulch]

,w Integrate[PiecewiseExpand[Cos[n*x] * Piecewise[{{0, -Pi < x < 0}, {x, 0 <= x <= Pi}}]], {x, -Pi, Pi}]

[ocean seal]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by lindenlinden

[random gulch]

Fuck

[tawny condor]

yup...

[random gulch]

How to do ≤

Idk can't be bothered

[tawny condor]

just <=

tho it doesnt matter

integral exists anyways

[lone heart]

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[zealous scaffold]

so, i'm studying math (Analysis) rn and i need a quick answer. if f is (strictly) convex <-> f' is (strictly) monotonically increasing based on the picture.
however, in the problem f is strictly convex, but strictly monotonically decreasing

[zealous scaffold]

this is the full one, konkav is just concave and fallend is decreasing

[vale crag]

f is convex and derivative of f is monotonically increasing

what's the issue ?

[zealous scaffold]

look at the problem (second picture)

f is convex, but it's monotonically decreasing

but the definition we got just before says it needs to be increasing :/

[vale crag]

yes and ?

no that's not what your definition says

f' isn't f you know

[zealous scaffold]

is f concave, then first derivate is increasing. if first derivative is increasing, then f is concave

[zealous scaffold]

f' is the first derivative of f

[vale crag]

yes i know

[zealous scaffold]

yes, so do i

[vale crag]

it looks like you're saying f can't be decreasing and concave

[zealous scaffold]

it looks like the definition is saying that

[vale crag]

how does that contradict your 'definition' in any way ?

[zealous scaffold]

or at least the given definition on the paper

[vale crag]

but first derivative increasing doesn't mean f itself has to be increasing

and vice versa

[vale wigeon]

-5 to -1 is also an increase

[vale wigeon]

as i understand, your issue is that the curve seems to be getting less steep scanning from left to right, which in your eyes is contradicting the statement that f' is increasing

do i understand your issue correctly?

[zealous scaffold]

oh wait now i understand

or wait, gimme a moment

yea i understand now

f is convex if the function f' is increasing, meaning we need f'' to determine if f' is increasing/decreasing. and if f'' > ∀x ∈ (0, inf) then it's increasing meaning it's convex. and vice verca

thanks @vale crag

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[quick fable]

b^2=a⋅c is formula in sequences, how to prove it is correct?

[mortal trellis]

I assume you mean that a,b,c are in a geometric progression?

if yes then just plug the formula for each of a,b,c in

[quick fable]

i have to prove that is truth for a, b, c other than zero

and then its geometric sequence

[mortal trellis]

remember that a geometric means that the common ratios are equal

what are those here

[quick fable]

yes

q is constant

[mortal trellis]

what are the common ratios of the sequence a,b,c

[quick fable]

b/a = c/b

[mortal trellis]

rearrange that

[quick fable]

oh okay

thanks

i got that i think

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[vapid shuttle]

If Selberg's inequality is a generalization of Bessels

Why does it not have the v_i term

I'd imagine it to be
$$\sum_{j=1}^{n}\frac{\langle x, y_j \rangle^{2}}{\sum_{k=1}^{n}|\langle y_j, y_k \rangle|}y_j\leq ||x||^2$$

[ocean seal]

Austin

[tacit arch]

did you read the proof

[vapid shuttle]

For which

Bessels yes

[tacit arch]

selberg

[vapid shuttle]

It uses Cauchy Schwarz

I am looking at it now, very short

[tacit arch]

in what way does the book say selberg is a generalization of bessel's inequality

[vapid shuttle]

"It can be seen as a generalization of Bessel's inequality"

I feel like what they mean is by letting the y_j be orthogonal in Selberg's you should be able to obtain Bessels

and it is so close to doing that

From the way its written in theorem 10 you get $$\sum_{j=1}^{n}\frac{\langle x, y_j\rangle^2}{|\langle y_j, y_j \rangle|}\leq ||x||^2$$

[ocean seal]

Austin

[vapid shuttle]

if they're orthogonal

which is kind of like Bessels?

[tacit arch]

original Bessels assumes orthonormal vectors
https://en.wikipedia.org/wiki/Bessel's_inequality

[vapid shuttle]

yeah it does

which is kind of my point, like if Selbergs is Bessels for non-orthogonal

oh i see you said orthonormal

if they're orthonormal you get original bessels

from selberg

awesome

that makes sense now

ty

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[vapid shuttle]

the bot is a she/they

[tacit arch]

this just follows from gram schmidt and triangle inequality

[vapid shuttle]

you can do a cooler proof using pythagoras

[tacit arch]

fax

[vapid shuttle]

x=(x-the sum)+(the sum)

they're orthogonal so you just win

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[proven marten]

hi everyone

[topaz sage]

hi

!da2a

[lone heart]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[proven marten]

Does anyone know of a bot that can solve any exercise?

[twin nimbus]

No, they do not exist

!nogpt

[lone heart]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[blissful whale]

Definitely not any.

[topaz sage]

If there existed one, then the Riemann Hypothesis would have been solved, and also the Navier-Strokes Equation

[blissful whale]

Wolfram Alpha can go a long way with computation stuff.

[twin nimbus]

There are problems for which we know that there is no way to even prove that they do or do not have a proof.

[blissful whale]

Also, yes, ChatGPT and the like, but it cannot be trusted. You must be able to verify it yourself. Ir will say a lot sound very eoquent but are wrong.

[proven marten]

so in theory no bot can solve the exercise? thank you

[blissful whale]

You asked for any, now the?

[amber fern]

What kind of exercise are you talking about?

[blissful whale]

When you are proofing a ststement with the forall quantifier but only existance was actually asked.

[proven marten]

like math, chemistry,....

[blissful whale]

Why?

For computaton and maybe chemestry wolfram alpha is not bad.

[proven marten]

I honestly don't even know what it is :))

[north rover]

What level?

[north rover]

wolframalpha.com

[blissful whale]

https://www.wolframalpha.com

[proven marten]

high school

[proven marten]

tks

[north rover]

Yeah then wolfram will be fine

[proven marten]

thank you

[lone heart]

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[brave sentinel]

1+2+3+......+n-1=n(n-1)/2
please explain how

[vale wigeon]

have you heard the story of how Gauss figured out the computation of a sum like this when he was a schoolboy?

[brave sentinel]

nah, our schooling system brute forces these types of things

[vale wigeon]

well, the story goes that his teacher had the class do the sum of all numbers from 1 to 100 manually to keep them busy, but then little Gauss discovered a trick for how to do it.

[brave sentinel]

im left stuck at this

[vale wigeon]

wait, do you mean to say you already know the formula for the sum of an AP

[brave sentinel]

i highly appretiate the effort but i have my final exams in 8 hrs or so

[vale wigeon]

this is an AP with n-1 terms, first term 1, and last term n-1.

if you're going to bash the formula anyway, it's $S = \frac{n-1}{2} (1 + (n-1))$

[brave sentinel]

Sn=n/2 (a+l)

yes

[ocean seal]

Ann

[vale wigeon]

cancelling out the -1 in n-1 breaks the pattern of the arithmetic progression and so that does you no good

[brave sentinel]

i tried the other one and i think i got stuck somewhere too

lemme do it again

[vale wigeon]

the other what

[brave sentinel]

not cancelling out the 1's

i did 1+2+3...+n firat

then do the -1

[vale wigeon]

if you wanna do it that way, you need to subtract n off, not just 1.

[foggy dawn]

can someone help

[vale wigeon]

1 + 2 + ... + (n-1) means you stop the sum at n-1

[vale wigeon]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[brave sentinel]

ok what

[vale wigeon]

it is very likely that's what was meant

1 + 2 + ... + (n-1) as the sum of the first (n-1) natural numbers

rather than summing the first n natural numbers and then taking off 1

which would give you a different result!

[brave sentinel]

ok. makes a lot of sense

thats why i wasnt able to find it by googling it

[vale wigeon]

googling math expressions tends to go poorly tbh

[brave sentinel]

thanks

i got it now

[brave sentinel]

all other options are worse tbh

[last monolith]

the soft iron will become a permanent magnet and it will produce magnetic field lines like the bar magnet......

[brave sentinel]

gpt is on another level of braindead

[foggy dawn]

how will it look

[last monolith]

it will depend on the poles of the magnet.

[vale wigeon]

glad you realize that.

@last monolith exsu has their own channel now.

[last monolith]

the field lines will go from north pole to south.P form outside and South.p to North.p from inside.

[vale wigeon]

help them there.

not here.

don't hog this channel.

#help-11

[last monolith]

ok sry

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[zealous inlet]

can someone just go through and check/fix anything that's wrong. thanks big time <3

[broken pivot]

1 ✅

2 ❌

3 ⚠️ you didnt state the direction of the vertical translation

4 ✅

5 ✅

6 ✅

7 ✅

8 ✅

9 ⚠️ i highly recommend labelling which is A and which is B

@zealous inlet

[lone heart]

@zealous inlet Has your question been resolved?

[alpine sable]

Is it any GCSE question

[zealous inlet]

ty! what are the corrections?

[broken pivot]

Uh you need to figure that out

!noans

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The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

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[haughty jungle]

I dont understand where to begin to find the value at the bottom.

[vale wigeon]

well, for starters does $f^{-1}(3)$ exist?

[ocean seal]

Ann

[unkempt dove]

$$ f^{-1}(3) $$ is other way to ask "what is the value of x when y=3"

[ocean seal]

printf_hi

[haughty jungle]

at -3?

[unkempt dove]

The answer is of course x=-3

[haughty jungle]

okay cool so far

[unkempt dove]

So, we need dervative there

Or slope at that point

which will be the slope of the line segment on the left side

[haughty jungle]

so -1

which Is observable but in cases where it isnt then I would use the f(b)-f(a)/b-a

[unkempt dove]

yes

[haughty jungle]

awesome

thank you

.close

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[warm forge]

How can I solve this inequality according to the way my book shows it?

[charred flicker]

?

[thick lynx]

This channel is occupied, open a new one like #help-21｜아리스킨충1

[thick lynx]

I would document it by saying in which range t needs to lie in each case first

[unkempt dove]

@warm forge the way I use is get 0 on the RHS, it becomes easier as you have to deal with less number of cases

BTW, the step that you did by multiplying both sides by the denominator is wrong as denominator can be both -ve or +ve, and depending on the sign of denominator the sign of inequality will flip

[warm forge]

Could you maybe show an example? I have been trying to solve this since last night and to me it seems my book skips a big step.

[thick lynx]

He did two cases which correspond to that, it's just lacking some documentation

[warm forge]

Yeah I've wrote this out several times so I did just skip to the 0>t^2 + t - 6

[unkempt dove]

In both the cases, there will arise 2 subcases as well.

I will post my method, it is much easier

[thick lynx]

Not necessarily; view the denominator as a parabola

You will only get two cases

[warm forge]

So once i get to the 0 < t^2 + t - 6 do I factor this and come up with t = -2, 3?

[unkempt dove]

maybe because 2 out of the 4 cases are same

[warm forge]

my teacher had demonstrated a problem similar but not with a quadratic or factoring. Here is what he demonstrated. Which this was easier to understand than the part b of the problem which is what I am needing help with.

[unkempt dove]

$$ \frac{-36}{(t-6)(t+7)} > 1 $$
$$ \frac{-36-(t-6)(t+7)}{(t-6)(t+7)}>0 $$
$$ \frac{-36-t^2-t+42}{(t-6)(t+7)}>0 $$
$$ -\frac{t^2+t-6}{(t-6)(t+7)}>0 $$
$$ \frac{t^2+t-6}{(t-6)(t+7)}<0$$

[ocean seal]

printf_hi

[unkempt dove]

Then, it becomes easy if you know to use a shortcut

[warm forge]

But once you multiply the denominator to both sides it eliminates it no?

[unkempt dove]

sorry, i couldn't understand

you don't mutiply the denominator in this case

[thick lynx]

I will do the same one your book did to not take this one away from you. \ \hr \ We want to find all $t$ with $$\frac{-30}{(t + 3)(t - 8)} > 1.$$ Consider the denominator as a parabola that is opened upwards with zeros at $-3$ and $8$. Then we have two cases: \begin{enumerate}[{Case}] \item 1. $t \in (-\infty, -3) \cup (8, \infty)$. Multiplying by the denominator doesn't change the sign. $$-30 > (t+3)(t - 8) \iff -30 > t^2 - 5t - 24 $$$$\iff 0 > t^2 - 5t + 6 = (t-3)(t-2)$$ and again, since this is a parabola, we get $t \in (2, 3)$. Now intersecting this with the intervals from the case, we get \underline{$I_1 = \varnothing$}. \item 2. $t \in (-3, 8)$. Multiplying by the denominator changes the sign here. We get $$-30 < (t+3)(t-8) \iff \cdots \iff 0 < (t-3)(t - 2).$$ This is fulfilled for $t \in (- \infty, 2) \cup (3, \infty)$ and so we need to intersect the interval from the case. This gives us \underline{$I_2 = (-3, 2) \cup (3, 8)$} \end{enumerate} Now we union the two possible intervals, giving us $I_2$.

[warm forge]

So we haven't done these as parabolas and only on number lines and thats what he wants us to demonstrate

[thick lynx]

I'm not really doing this as parabola, it's just good to imagine it as that

I'm doing it the same way it's described in your notes

[unkempt dove]

Continuing my answer:
$$ \frac{t^2+3t-2t-6}{(t-6)(t+7)} < 0$$
$$ \frac{(t-3)(t+2)}{(t-6)(t+7)} < 0$$

[warm forge]

Something like this but okay thank you I will look at these ways and try and come up with something.

[mental python]

,rccw

[thick lynx]

This is for a linear denominator

[ocean seal]

printf_hi

[thick lynx]

You have a quadratic denominator in the one in the notes

[warm forge]

yes i know he never demonstrated an inequality with a quadtratic denominator which is why I needed help.

[unkempt dove]

Actually, my method without the shortcut trick is not useful. And the shortcut may not be useful for you as you have to explain each step

[thick lynx]

I think the method used in #help-0 message would be the most direct one; you just need two cases, one where multiplying by the denominator changes the sign and one where it doesn't. For a better imagination of where this happens, you can imagine it as parabola, but you don't need to. You can also just imagine it as two linear factors

That's also exactly what's done in the notes, just expressed in a clearer way

[warm forge]

Thank you Kepe I am looking at your method and attempting to plug it in with the numbers in my hw question. After the quadtratic is that supposed to be an equal sign or < >?

[thick lynx]

It's an equal sign

I factored the quadratic

You could also use the quadratic formula to deduce the two zeros, I saw the factors right away

[thick lynx]

[The method printf used also works, it's getting everything to one side and then it's just a matter of checking signs; but I feel it's a bit more effort if not everything factors nicely. If everything does factor nicely, perhaps it's a tad quicker if you practiced it]

[warm forge]

Thank you to you both if I were to look up some more examples what would this type of problem be called? Inequalities with quadtratics in the denominator?

I like Khan academy and some of the other YT math teachers not that you guys didn't help I just learn better with demonstrations,

[thick lynx]

Rational inequality, probably

[unkempt dove]

you will still have to show the two subcases each in my opinion. Like the following:
If, $$ 0>(t-3)(t-2) $$,
then, $$ 0>t-3 $$ and $$ 0>t-2 $$ (both -ve)
or, $$ 0<t-3 $$ and $$ 0<t-2 $$ (both +ve)

[warm forge]

I'm lost what is +ve and -ve?

[thick lynx]

He means positive and negative

[ocean seal]

printf_hi

[thick lynx]

[This is referring to #help-0 message]

[unkempt dove]

@thick lynx the method is posted is definitely correct, but I hadn't earlier seen that one. Maybe different ways to say the same thing

In the end, @warm forge can use the one his teacher has taught him.

[thick lynx]

Yeah, and that one is what I posted

With the exception of viewing the denominator as an opened-upward parabola instead of thinking about when the linear factors are positive and negative

[unkempt dove]

Ah, I understand it now

[warm forge]

He never demonstrated a rational inequality like this that was our 7th learning objective and he stated, "The book is unclear on the case-by-case basis. You must make the GOOD-BAD graphs as we did in class. Then line up the common GOOD sections."

[unkempt dove]

actually, u are using the shortcut i was talking about

[thick lynx]

"good-bad graphs" must refer to when both linear factors are positive or negative and one is positive, the other negative, getting the two cases when the sign changes

[unkempt dove]

I just instead of using "good" or "bad", I use "+ve" "-ve"

[warm forge]

I posted the photo of what the book uses as the demonstration for us and that was the problem you had solved Kepe. It just seems it skips some big steps after getting -30>(t+8)(t-3) how did they get the -3<t<8?

it was actually -30>(t-8)(t+3)

[thick lynx]

-30 > (t-8)(t + 3) <==> -30 > t^2 - 5t - 24 <==> 0 > t^2 - 5t + 6 <==> 0 > (t - 3)(t - 2)

[warm forge]

how are we finding that -3 is in fact less than t?

[thick lynx]

This gives us 2 < t < 3

[warm forge]

Ohhhhh now I see that!

[thick lynx]

You get "t < -3 or 8 < t" from the condition of the case right at the start

You made this condition because like this, when multiplying by the denominator, the sign doesn't change

Because you multiply by something positive

Now you just need to make sure both of your inequalities that you obtained are fulfilled

(t < -3 or 8 < t) AND (2 < t < 3)

[warm forge]

So then for my problem its 0>(t + 3)(t - 2)

and ill do those each separately so 0>(t+3) and 0>(t-2)

[thick lynx]

And this will never be fulfilled, so you get no solution (this is all for your first case)

Then you go on to your second case, where the direction of the inequality does change and do this all over again

[warm forge]

to get t > -3 and t > 2

[thick lynx]

You miss out on cases with that though!

In fact, it won't even be true

You say you want 0 > t + 3 and 0 > t - 2

So both of the factors are negative

But negative * negative = positive

So you don't even have 0 > (t+3)(t-2)

[warm forge]

So once i get here how should i move t to the left hand side? 0 > (t+3)(t-2)

If i shouldnt handle them separately?

[thick lynx]

You need to think about when this is fulfilled. There are two methods you could go for

1. By doing cases.

You want one of the two factors to be negative, the other positive

Only then will the product be negative

So (t + 3 > 0 and t - 2 < 0) OR (t + 3 < 0 and t - 2 > 0)

2. By imagining it as parabola.

You just need to find the zeros in this case (which you read off as -3 and 2) and then you need to imagine in which regions this parabola is negative

Well, that will be at (-3, 2)

[warm forge]

So (t + 3 > 0 and t - 2 < 0) OR (t + 3 < 0 and t - 2 > 0) This is for both cases correct? Case 1 OR case 2?

[thick lynx]

This is all for case 1, this

We didn't talk about case 2 at all right now

[thick lynx]

I'm just arguing this: Think about what it means if $0 > a \cdot b$.

[ocean seal]

Kepe

[thick lynx]

There are two possibilities this can be achieved: a is positive, b is negative

or a is negative, b is positive

Nothing else works (both positive gives positive, both negative gives positive too)

[warm forge]

So then I got t > -3 and t > 2 but as you said theres no solution as t cannot be greater than -3 but less than 2?

wait yes it can

[thick lynx]

You get -3 < t < 2 from your 0 > (t+3)(t-2) inequality

[warm forge]

So (t + 3 > 0 and t - 2 < 0) why did the sign flip when doing t - 2 < 0?

[thick lynx]

Because one of the factors need to be negative

The other positive

Only then is the product negative

If they are both positive we get something positive

(t + 3 > 0 and t - 2 < 0) OR (t + 3 < 0 and t - 2 > 0) gives you

(t > -3 and t < 2) OR (t < -3 and t > 2) gives you

(-3 < t < 2) OR (FALSE)

So you get -3 < t < 2

[warm forge]

Okay and then this is not a no solution answer as t can in fact be in between -3 and 2?

[thick lynx]

Yes, now you just compare it to what your condition for the first case was

Which is t < -3 or 8 < t

And -3 < t < 2 lies exactly in the middle of that!

So no intersection

=> No solution for case 1

[warm forge]

Well this is the problem I am comparing and working out. Not the demonstration from the book.

Ohh i drew that inequality on a number line and that is correct that is no solution

[thick lynx]

[Ah, you went to the second case with this, we need to compare it to the interval from the second case then of course]

I was still at the first case

[thick lynx]

Then as before, make two cases, when the inequality sign changes and when not

After multiplying by the denominator

[warm forge]

so flip this to -36 < t^2 + t -42. Then subtract to get 0 on the left then factor the quadtratic

And that will lead me to no solution

[thick lynx]

What case are you on?

You need to document this so later on you know

[warm forge]

right now I am working -36 > t^2 + t - 42

[thick lynx]

What allowed you to multiply and not change the sign?

[warm forge]

So the way the book has it wrote is id yield this for case 1 -36 > t^2 + t - 42 and this for case 2 -36 < t^2 + t - 42

Originally this -36 > (t-6)(t+7) and -36 < (t-6)(t+7

[thick lynx]

Exactly, now just document what case 1 and case 2 stand for. Case 1 stands for (t-6)(t + 7) >0, so t < -7 or t > 6

Case 2 stands for (t -6)(t+7) < 0, so -7 < t < 6

[warm forge]

I am just plugging in the numbers in my given problem and solving

[warm forge]

Thank you

They give us the answer key but obviously doing them and knowing the work is best

This is question 3b btw

[thick lynx]

[I will be heading to sleep now, I just want to say that I really think you should imagine this entire 'when is the denominator positive or negative' as just 'when is the parabola over the x-axis']

For (x-6)(x-7) for example you have this

,w plot (x-6)(x-7)

[ocean seal]

Results provided by WolframAlpha

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

Requested by kepe_5235

[thick lynx]

Between 6 and 7, it's below the x-axis

Everywhere else, it's above

So you could right away answer (x-6)(x-7) > 0 with x < 6 or x > 7 with this visualization in mind

And similarly (x-6)(x-7) < 0 with 6 < x < 7

[warm forge]

Thank you Kepe I apologize this was so difficult for me to understand

[thick lynx]

No that's normal when first learning anything

[warm forge]

I will close this out now and save all of your screenshots and such have a goodnight

[thick lynx]

Thanks

[warm forge]

.close

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[lone heart]

@shell narwhal Has your question been resolved?

[jagged cobalt]

you realise nobody is going to do anything if you dont even add a single word of contribution
you havent even said what youre having a problem with

[broken pivot]

!status @shell narwhal

[lone heart]

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

[broken pivot]

"If you're not willing to do the work [to solve your problem], why should we?"

[shell narwhal]

We can find the potential energy lost due to the mass falling

Is the angular momentum conserved?

[broken pivot]

Wait i just realized i have no clue how to solve this

Sry

[lone heart]

@shell narwhal Has your question been resolved?

[lone heart]

@shell narwhal Has your question been resolved?

[charred flint]

not really since the cart is fixed in the y-direction. linear momentum + energy conservation should solve for everything

hm maybe if you do angular momentum around the center of mass

[shell narwhal]

But we don’t know where the center of mass is right?

[charred flint]

halfway on the rod right?

it's clear at theta=0 and 180

[shell narwhal]

Well, that seems like an approximation

Let me try what you suggested, i’ll post my results

[lone heart]

@shell narwhal Has your question been resolved?

[lone heart]

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[finite fern]

Am I tweaking or is this NOT the definition of logarithm

[finite fern]

I thought logarithms are the power to which a number must be raised to get some other values?

How is it a type of average in the first place

[charred jewel]

sounds like the geometric mean to me

[lavish cave]

your teacher messed up, yes

[lilac dome]

same!!
but i was a bit confused because this is such a weird practice sheet

[finite fern]

Darn. Thanks for the clarification though

[lone heart]

@finite fern Has your question been resolved?

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[tight nest]

Something is sesriously wrong with my calculation😭😭

Where did i do wrong??

[raw jetty]

ln|e^x|=e^x?

[tough relic]

Can't you just break the fractions here?

[raw jetty]

although im not even sure if you can just evaluate the integral like that

[tight nest]

isnt it correct?

[mental python]

$\int\frac{xe^x+1}{x}\dd x$

[ocean seal]

Bonk

[mental python]

this is the integral, correct?

[tight nest]

yea im just curious what happens this way

[tight nest]

yesyes

[mental python]

so then you sub in u=ex+1, and you get $\int\frac{xu}{xe^x}\dd u$

[ocean seal]

Bonk

[mental python]

however, xe^x+1 is not equal to x(e^x+1)

xe^x+1=x(e^x+1/x)

[alpine sable]

anyone here knows something about determine each roots of quadratic equations? i forgot and don't want to get cooked

[mental python]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[alpine sable]

oh mb

[tight nest]

x(e^(x)+1) = xe^(x) + x

no??

[mental python]

yes

but whats not whats written

$\int\frac{xe^x+1}{x}\dd x\neq\int\frac{xu}{xe^x}\dd u$

[ocean seal]

Bonk

[mental python]

with u=e^x+1

[tight nest]

wait why is that??

ohh wait

[mental python]

$xu=x(e^x+1)=xe^x+x\neq xe^x+1$

[ocean seal]

Bonk

[tight nest]

ahh I get it

so it was wrong to set u as e^x + 1

[mental python]

yes

you dont even need a u-sub actually

[tight nest]

yea😭

Ill just spilit the fractions and do them

thanks anyways!

.close

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[fringe kite]

is $\sqrt{49+64} = to \sqrt{49}+\sqrt{64}?$

[ocean seal]

neo

[vale wigeon]

no

[fringe kite]

oh wait yeah duh

[vale wigeon]

the word to is not necessary there either, but no, square roots DON'T distribute over addition

[fringe kite]

gotcha thank you

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[wispy timber]

uh I got to the step |e^y+3| = Ae^1/4ln|2x^2+4x+1| and than I got confused

[wispy timber]

nvm solved

.close

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[alpine sable]

how to do this again?

like with the cube

[lament wyvern]

well, fortunately your cubic equation has no constant term

so you can take a factor of x out of it all

[alpine sable]

uhh how do i do that

[lament wyvern]

$4x^3 - x^2 - 3x = (4xx^2) - (x * x) - (3 * x) = x(4x^2 - x - 3)$

[ocean seal]

PrettyPrincessKitty FS

文也

[atomic prawn]

$sin(252525)$

[ocean seal]

文也

[sullen ocean]

?

[atomic prawn]

wut

[alpine sable]

whats that?

[cloud stump]

#latex-testing

[lament wyvern]

I have no idea, but it's not relevant to your question

[atomic prawn]

ight thx

[alpine sable]

is that the answer? or do i still need to do smth with it

[lament wyvern]

well no that's just the first step

but you don't need to worry about trying to factorise cubic (x^3) or quartic (x^4) equations

[alpine sable]

do i need to distribute the x? To the 4x^2 whatsoever

[lament wyvern]

it's exactly the same process as the last question was

[alpine sable]

ok ill try to do that for the others

this?

[lament wyvern]

reading it

[alpine sable]

okayy take ur timee

[lament wyvern]

factorisation looks good, but you don't need to write out the (4.x.x^2) - ... if you don't want to. it was just to make it clearer to explain. if it helps you then of course leave it in

[alpine sable]

ah okays

[lament wyvern]

you can do some factoring of that $x^4 - x^3$ too

[ocean seal]

PrettyPrincessKitty FS

[alpine sable]

so x?

wait nvm

how do i factor that

[lament wyvern]

well what terms do they have in common

if I write it out - $(x * x * x * x) - (x * x * x)$

[ocean seal]

PrettyPrincessKitty FS

[alpine sable]

x

so its x?

[lament wyvern]

well no, because we are factoring

we want to find a common term that we multiply everything by, and only multiply by it once

$4x^3 - x^2 - 3x$ for example - we discovered we are multiplying each individual term by $x$. so we factor that out and only multiply the whole thing by x instead - $x(4x^2 - x - 3)$

[ocean seal]

PrettyPrincessKitty FS

[alpine sable]

so like x^3(x-1)?

ok what do i do now

[lament wyvern]

yep!

so now we have two nicely factored fractions we want to multiply

[alpine sable]

yeah

[lament wyvern]

we can make them into one big fraction and cancel

[alpine sable]

ok wait

[lament wyvern]

hint: -(stuff)(more stuff) = -1*(stuff)(more stuff)

[alpine sable]

like thiss?

[lament wyvern]

be careful - you've written this like you've got a subtraction in the numerators

I'd either put the -1 at the start (so it becomes -x) or write (-1)

[alpine sable]

so i switch the places?

[lament wyvern]

also we can factorise $4x^2 -x - 3$ 🙂

[ocean seal]

PrettyPrincessKitty FS

[lament wyvern]

yup

[alpine sable]

(4x+3)(x-1)

[lament wyvern]

looks good

[lament wyvern]

time to cancel!

[alpine sable]

i cancel the x-5?

and the x-1?

[lament wyvern]

yup, but there's some more you can do too

[alpine sable]

what else can i cancel?

the x?

one of the x’s i mean

[lament wyvern]

I mean $x^3 = x . x . x$

[ocean seal]

PrettyPrincessKitty FS

[alpine sable]

like thiss?

[lavish sage]

Yes

[lament wyvern]

yup!

[lavish sage]

But that 5 kinda looks like a s

[lament wyvern]

you can also write it as - <big fraction> or -(x+2)(4x+3) but this is good

[alpine sable]

lmfaoo

ok

wait so this is the answer?

[lament wyvern]

all I can think of is if by "standard form" your teacher wants you to expand the factors

so write it like ax^2 + bx + c

[alpine sable]

xd how do i do that bruh

[lament wyvern]

reverse factoring 🙂

$(x+2)(4x+3) = (x*4x) + (x * 3) + (2 * 4x) + (2 * 3)$

[ocean seal]

PrettyPrincessKitty FS

[lavish sage]

AKA expanding

[lament wyvern]

yes

[alpine sable]

so dats the answer?

wait i solve that right

[lament wyvern]

well no because you can simplify it

you can make it look like ax^2 + bx + c, and the same with the denominator

and you'll get something that looks like the fractions you started with

[alpine sable]

so like for the one u gave, i solve it?

[lament wyvern]

simplify it, yeah

[alpine sable]

like for the first one, 4x^2

okk thx

[lament wyvern]

solve means "we have found value(s) for all the unknowns (the x)", whereas simplify is "we have made the formula look nice and tidy and neat"

[alpine sable]

?

[lament wyvern]

well 3x + 8x can be simplified

[alpine sable]

11x

[lament wyvern]

yep

that's the numerator in standard form, and you can do the same to the denominator

[alpine sable]

ohh uhh

how do i do that

like ax+by=c?

[lament wyvern]

nono, you've got $x^2(x-5)$

[ocean seal]

PrettyPrincessKitty FS

[alpine sable]

ohh ok

[lament wyvern]

also just a warning, remember that your numerator has a -1 at the start 🙂

so you actually have -1(4x^2 +11x + 6)

[alpine sable]

okk

what do i do next?

do i distribute the -1?

[lament wyvern]

so we can multiply -1 and the (4x^2 ...) polynomial

yep

and we can distribute the x^2 over the (x-5)

[lament wyvern]

looks good to me!

[alpine sable]

thats the answerr?

[lament wyvern]

well I ran it through wolframalpha to check and would you look at that, you've solved it 😉

[alpine sable]

yayy thanks

ok next

uhh

bro im so cooked 😭

[lavish sage]

OOF

You're not cooked

It'll just take a bit

[alpine sable]

yep

Uhhhhhhhh damn bro

[lavish sage]

I can already see the answer flashing in front of me

[lament wyvern]

break it down into little pieces and you've got this

[lavish sage]

^

[alpine sable]

huh how

[alpine sable]

ill try ok

[lavish sage]

Start by factorising

[alpine sable]

do i do the first 2 first?

[lament wyvern]

...though I really hate using both / and the division symbol because is this saying div (polynomial * 14) or 14 div polynomial

[lavish sage]

My teacher assigned heaps of homework on factorising

[lament wyvern]

I assume it's meant to be the first version

as in, this is 14(x+4)/(x-8)

[alpine sable]

oh

[alpine sable]

huhh?

[lavish sage]

Don't forget to take the reciprocal lol

[lavish sage]

The last 2 fractions

[lament wyvern]

factorise the first fraction. see if you can cancel anything
factorise the second fraction. see if you can cancel anything
combine the two into one big fraction, see...
take the reciprocal of that last fraction, because division
combine the two sides, cancel anything

[alpine sable]

ok

ahhh okayzz

tysmm ill try

i combine it then cancel?

[lavish sage]

You can factorise 2x^2-128

[alpine sable]

how tho

[lavish sage]

Factor out 2 first

[alpine sable]

2(x^2-64)?

[lavish sage]

Yes

You can still factorise that

[alpine sable]

ohh ok thxx

oh

how?

[lavish sage]

Difference of 2 squares

[alpine sable]

i forgot 😭

[lavish sage]

2(x+8)(x-8)

[alpine sable]

ahh ok

[lavish sage]

The rest should be a piece of cake

[alpine sable]

?

[lavish sage]

Where did you get x+8 on the denominator from?

[alpine sable]

oh wait

what do i do next

[wet tusk]

Distriboot.

[alpine sable]

the 2?

[wet tusk]

2 to x and -8

[alpine sable]

wha next

[wet tusk]

at this point you just asking for answers is pretty straight forward what to do now

[alpine sable]

-2/7

?

guys what do i do now

<@&286206848099549185>

[lament wyvern]

okay, so you've fully expanded the top. can you distribute/expand the bottom now?

[alpine sable]

which do i distribute lol

[lament wyvern]

you have 2(x-8)

you would like it to be in ax^2 + bx + c

[alpine sable]

2x-16x+2?

idk

[lament wyvern]

$2(x-8) = 2x - 28$

[ocean seal]

PrettyPrincessKitty FS

[alpine sable]

2x-16

[lament wyvern]

yup

[alpine sable]

ok

what nextt

[lament wyvern]

what do you have? what do you need to end up with? do you have what you needed?

[alpine sable]

so 2x-16 is rhe answer?

[lament wyvern]

that is what you get if you expand the brackets of the denominator.

[alpine sable]

oh okay

then i solve for it?

4/x-16?

[lilac dome]

this is just an expression
its in its most simple form
if it was in an equation form that is if it were equal to something, we could've solved it

e.g
you can solve
x+100= 23
but you cant solve x+100

[alpine sable]

ahh okay

[alpine sable]

what do i do with this?

[lilac dome]

cuddle it

thats your answer

[alpine sable]

ohhh ok tysm

.close

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[prisma night]

how do I generalize it further for x1 x2 x3...xn and t1, t2 t3..tn ?

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@prisma night Has your question been resolved?

[dire plover]

do induction, as in
t2x2+t3x3+... tnxn = TX for T = t1+t2+.... and X = t2x2+t3x3+..._tnxn

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[alpine sable]

How do I prove this using epsilon delta idk which value of f(x) to use

[lone heart]

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[alpine sable]

.close

.close

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[clever tundra]

I don't understand how to solve D and E, please run me through it

[clever tundra]

I have to find the truth value (true or false)

and the domain is the set of all real numbers

So it says for every x there is a y that makes every z true

right?

[vale wigeon]

look at just the statement $\forall z (z = \frac13(x-y))$ and wonder how it could possibly be true

[ocean seal]

Ann

[vale wigeon]

(i mean, whatever (x-y)/3 is, surely it can't equal both 420 and 69 at once?)

[clever tundra]

you're saying for x and z?

[vale wigeon]

no i am telling you to temporarily close your eyes to the quantifiers on x and y

[clever tundra]

okay

so you're saying there can't be multiple values for z

so it can't be true

if the quantifier was existential then it would be true

[vale wigeon]

yes but don't try to do more than what the problem asks of you

[clever tundra]

So what you're saying is I can't plug in 420 or 69 for z

because x-y/3 can only be 1 answer?

so it's false because the universal quantifer says every value for z

am I understanding correctly?

[lament wyvern]

you can pick z=420 and then for all x, there does indeed exist some y such that x-y/3 = 420

but because it's all, then you are saying that for the same choice (of any) x and the specific choice of y, z also equals 419 and 421 and 69 and 0

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[obsidian girder]

im

confused

on how to

get started

[vale wigeon]

if the curve was confined to a sphere of a particular radius, say c, centered at the origin

what could you then say about r(t) ?

[keen idol]

i spent some time thinking about the problem and i would also consider the dot product between the two vectors and how the magnitude of the position vector behaves wrt time

[obsidian girder]

ahhh i dont get it its so confusing

[vale wigeon]

ok lets dial it back a bit

way back

[obsidian girder]

yes yes

[vale wigeon]

let's say that a point with position vector r lies on the sphere centered at the origin with radius 1

what can you then say about r

[obsidian girder]

what can i say in the sense

i dont know 😭

[vale wigeon]

ok lets go further back to fundamentals

what is a sphere?

[obsidian girder]

its

like

a ball

circle but 3d

[vale wigeon]

what's the formal definition of a sphere?

or, if you don't know, what's the formal definition of a circle?

[obsidian girder]

like

it has a fixed center

no edges or vertices

distance between any point on the circle and the center is

the radius

[haughty lake]

Hi guys. Can I get assistance on this please. To calculate Income it's Income=Demand×price but I was given the demand formula only, where to I get the price one?

[keen idol]

i have a solution but i really dont want to help u brush through the fundamentals

ill try guiding you

[obsidian girder]

alright

[keen idol]

what can you say about the dot product between the tangent and position vector?

[keen idol]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[obsidian girder]

it'll be 0 because its perpendicular

[keen idol]

okay good

now do you know how to express
|r(t)|^2 in terms of dot product?

basically magnitudee of position vector squared

[obsidian girder]

one second

ah i dont know

[keen idol]

how do you find the dot product of two vectors

[obsidian girder]

length of a multiplied by length of b

[keen idol]

aaaand?

its multiplied by one more thing

[obsidian girder]

cos theta?

yes cos theta

NO DOUBTING MYSELF

(if its wrong itll be embarrassing)

[keen idol]

u r right

okay |a||b|cos(x) = a dot b

now what happens when a = b

(think of what happens to cos(x))

[obsidian girder]

it'll be 0

[keen idol]

think again

the angle between the two vectors is 0, yes

but what about cos(x)

[obsidian girder]

im not sure

[keen idol]

cos(0)?

[obsidian girder]

yeah

thats what i

meant

[keen idol]

cos(0) = 1

[obsidian girder]

yeah

[keen idol]

so if a = b then we have a dot a = |a|^2

[keen idol]

now we go back to this question

[obsidian girder]

its the same thing

[keen idol]

write it out

[obsidian girder]

in terms of r?

[keen idol]

yes

[obsidian girder]

r dot r = |r(t)|^2 cos(0)

[keen idol]

correct!

[obsidian girder]

yay

[keen idol]

now take the time derivative of r(t) dot r(t)

(basically product rule)

[obsidian girder]

left hand side?

[keen idol]

yes

[obsidian girder]

my brain is not really working rn

[keen idol]

do u know the product rule?

[obsidian girder]

yes yeah

im just very exhausted how do i derivate r dot r with respect to time?😭

[keen idol]

(this is not rigorous and if someone in this server sees this, i will be exiled and sentenced to death so lets keep this between us)

think of the dot as a multiplication

so you are taking d/dt (r(t)*r(t)) basically

you can see why product rule works for dot products later

its getting late here and i want to sleep soon

[obsidian girder]

yeah yeah same

[obsidian girder]

alrighty

[keen idol]

taking time derivative of r(t) dot r(t) gives
r'(t) dot r(t) + r(t) dot r'(t) = 2r(t) dot r'(t)

[obsidian girder]

yes

thank you

[keen idol]

2 [r(t) dot r'(t)]

notice something about this part?

[obsidian girder]

IT IS

ZERO

BECAUSE

[keen idol]

yes

[obsidian girder]

YEAH

you are THE SMART mia

[keen idol]

so we know
|r(t)|^2 does not change wrt time since its derivative is 0

this means that |r(t)| does not change wrt time either

[obsidian girder]

yes

[keen idol]

now the magnitude of the position vector doesnt change

this means the distance from the origin to every point on the curve is always constant

[obsidian girder]

yes

[keen idol]

hence all points are at a fixed distance from the origin

[keen idol]

this brings us to this point

if the vectors are in R^3, it becomes a sphere

[obsidian girder]

yes

[keen idol]

there you have it

[obsidian girder]

wow

thank you so much

i have my math midsem tomorrow and this server helped like crazy today and i joined just today

OKAY goodnight gang

.close

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[long spade]

Guys can I have help with ex4 question b

[proper field]

my neck hurts and is this french

[long spade]

Show that (z1*z2 in IR —> z1 + z2 = 2i)

Yes this is french

My head hurts cuz I have an exam tomorrow

@forest marsh
Yo bro can you help with this if you can

Nvm I solved it

.close

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[long spade]

Hello

[long spade]

Can anyone help with the last question

For anyone who can speak french

[high rapids]

,rotate

[lone heart]

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[alpine sable]

im depressed

[azure needle]

Type .close
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