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#help-0
lone heartBOT
@vivid heath Has your question been resolved?
vivid heath
you are confusing dot with cross
so no k?
vale wigeon
of course no k...
alpine sable
.help
lone heartBOT
Commands:
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help: .help
Type .help <command name> for more info on a command.
knotty steeple
Cpukd I ha e help how to calculate this please with the figures provided?
lone heartBOT
Cpukd I ha e help how to calculate this please with the figures provided?
@knotty steeple Has your question been resolved?
teal ore
should i use the quadractic formula?
lone heartBOT
should i use the quadractic formula?
teal ore
median oar
I don’t think that’s the point
teal ore
so i just solve for area?
teal ore
how will i find it after?
fallow bridge
Make the given expression in the question equal to the expression you get from solving the area
median oar
Show your work
lone heartBOT
@teal ore Has your question been resolved?
lone heartBOT
@teal ore Has your question been resolved?
teal ore
.close
spiral grove
i just need a little help is a number example 22.22 the bolding 22 are reperting numbers so then i should 2 zeors
vale wigeon
did you not post a very similar question just twenty or so minutes ago?
you know the method for these
so tell us what is troubling you in this one
spiral grove
i want to confirm that i am doing it right
spiral grove
i meant there are 2 reperting numbers and should i put 2 zeros
vale wigeon
yeah, sure.
spiral grove
if there are 3 reperting number and 1 number normal number then i put 3 000 example 1.1222 bold is the reperting number]
hello
inner rune
spiral grove
wt
vale wigeon
if there are 3 reperting number and 1 number normal number then i put 3 000 exam...
this looks a lot like word salad
but yes, if you have a 3-digit repeating bit, then you will want to multiply by 1000.
spiral grove
o k
vale wigeon
no matter if there's some non-repeating digits before it or not.
lone heartBOT
lone heartBOT
Available help channel!
dire sonnet
taylor expansion?
lone heartBOT
taylor expansion?
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
1
surreal meadow
what is your question
lone heartBOT
@dire sonnet Has your question been resolved?
dire sonnet
what is your question
it asks to write it explicitly
I don't know how it works with two variables
lapis valley
what does a quotient set mean in the context of an equivalence relation
lapis valley
like is it a set of all equivalence relation?
knotty steeple
Cpukd I ha e help how to calculate this please with the figures provided?
Any help with this answer?
tawny condor
what does a quotient set mean in the context of an equivalence relation
I think a quotient set is just the set of equivalence classes
maybe someone who knows more about this should help tho
marsh rapids
where x ~ y iff phi(x) = phi(y) ?
marsh rapids
a classic bijection you'll use very often
lapis valley
oh
marsh rapids
if you were to assign a class to a value of phi(X), which one should it be ?
lapis valley
sorry i dont understand
marsh rapids
what function would make sense, to make a bijection ?
marsh rapids
like phi (x) = phi (y) and we prove thy are both injective and surjective?
that doesn't define a function between classes and images
lapis valley
oh
so we need to find a bijective function to show that there is a bijective map ?
but i dont understand what the function is supposed to be
marsh rapids
take phi(x) = x mod 2, so that N is split into the classes of even and odd number
an example should make it easier
lapis valley
ok
marsh rapids
then you want to assign to each of these two classes, a value in the image set of phi, i.e. 0 or 1 mod 2
anything coming to mind ?
lapis valley
sorry but i dont think i understand this stuff
not ur fault tho
i dont know where to start with these stuff cuz i dont really understand my prof during class
marsh rapids
then you want to assign to each of these two classes, a value in the image set o...
you'd want to map the even numbers to 0, because they're all 0 mod 2
And all the odds to 1 mod 2
right ?
lapis valley
yea
marsh rapids
in this case that's your bijection
marsh rapids
you'd want to map the even numbers to 0, because they're all 0 mod 2
And all the...
you got 2 classes in your input set
lapis valley
yea
wait can u also explain what this is saying
in this method are we trying to prove that its bijective by proving its injective and its surjective?
marsh rapids
as they state yes
lapis valley
ok
devout summit
But, before all that they showed the map is well-defined
marsh rapids
yes ofc
lapis valley
this part?
devout summit
That capital phi map works in the first place because given a P, phi(P) is uniquely defined
lapis valley
ok
upbeat hornet
im trying to solve this q but i dont know where to start
is this the first isomorphism theorem?
lapis valley
i dont know ;-;
devout summit
For sets
lapis valley
its just some practice q that prof gave but its hard to understand at this stage ;-;
devout summit
It's not that difficult of a concept in itself and appears regularly enough for us to recognise and give it a name
lapis valley
ok
lone heartBOT
@lapis valley Has your question been resolved?
somber charm
ive square rooted both sides and turned into cosecx = +-sqrt2
granite badger
ive square rooted both sides and turned into cosecx = +-sqrt2
hi
so like cosec is 1/sin right
try 1/sin^2x = 2
sin^2x = 1/2
and have a go from there
rugged bone
Let $a_1, a_2, \ldots, a_n$ be positive real numbers such that $n(a_1 + a_2 + \cdots + a_n) = a_1a_2 + a_2a3 + \cdots + a{n-1}a_n + a_na_1$. Prove that for all positive integers $n \geq 3$, we have:
$\sqrt[n]{a_1a_2 \cdots a_na_1} + \sqrt[n]{a_2a_3 \cdots a_1a_2} + \cdots + \sqrt[n]{a_na1 \cdots a{n-1}a_n} \geq \sqrt[n^3]{n^3 + a_1^3 + a_2^3 + \cdots + a_n^3}.$
ocean sealBOT
MrRobert
rugged bone
<@&286206848099549185>
near apex
How are terms in LHS different from one another in inequality ?
They all seem just product of ai's.
lone heartBOT
@rugged bone Has your question been resolved?
strange fractal
hi
lone heartBOT
@strange fractal Has your question been resolved?
strange fractal
.close
royal bay
When a side BC of a triangle ABC is produced to an exterior point D, answer
- Write down the relation between <angle>ABCD and <angle>ABC+<angle>BAC
- Verify experimentally that the relationship between <angle>ACD and <angle>ABC + <angle> BAC
- If the triangle ABC is an isoscaled right angled triangle right angled at B, find the ratio of measurements of <angle> ABC and <angle>ACD
lone heartBOT
@royal bay Has your question been resolved?
kind patio
Having trouble figuring out how to write this integral
lone heartBOT
Having trouble figuring out how to write this integral
alpine sable
integrate both cases separately
kind patio
integrate both cases separately
So would I be doing the integral of 4 from -4 to 0 + the integral of 16-x^2 from 0 to 4?
The answer I got was 58.66 which it says is wrong
alpine sable
odd?
alpine sable
oh yeah probably because you are approximating
lone heartBOT
@kind patio Has your question been resolved?
warped topaz
lone heartBOT
https://i.imgur.com/ts7hZqx.png
warped topaz
I got pretty far but im stuck on something
I got the derivative and then subbed in 0 for t
So I have 18cos(1) and 20cos(-2) for x and y
Do I just sub these in for a and b in ax + by = c? How do I get c?
tardy stag
I think you want a to be 1/(18cos1)
you have a parametric line equation of x = (18cos1)t + x_0 and similar for y
so solving for t you'd get it in fractional form
anyway you know it's tangent at t=0 which means it's coincident there, you can use the point of the curve C at t=0
warped topaz
I think you want a to be 1/(18cos1)
you have a parametric line equation of x = (...
But why do we solve for t?
We have t as 0 right?
tardy stag
we're trying to make an equation in x and y right?
so we have x = (18cos1)t + x0
and y = (20cos-2)t + y0
and we're trying to unite these to make an eqn out of x and y
easiest way to do that is to solve for t
warped topaz
so we have x = (18cos1)t + x0
and y = (20cos-2)t + y0
Where did you get these sry?
tardy stag
that's just the derivative of C at t=0
warped topaz
Why is there a t if t=0?
tardy stag
let me try again
that's the tangent line to C at t=0
tardy stag
imagine a particle traveling along the path C, then at t=0 remove whatever's influencing it and it'll go in a straight line
every unit of time its x coordinate will grow by 18cos1, and its y coordinate will grow by 20cos(-2)
warped topaz
Yes im with you so far...
tardy stag
so the equations that describe those are x = (18cos1)t + x0 and y = (20cos-2)t + y0
where x0 and y0 are the positions of the particle at t=0
tardy stag
imagine a particle traveling along the path C, then at t=0 remove whatever's inf...
you can write this as dx/dt = 18cos1 and then integrate both sides to see this
warped topaz
Why not just (18cos1)t and (20cos-2)t sry? Why are we adding the positions of the particle at t=0?
tardy stag
well, because at t=0 it might not be at the origin
tardy stag
I think in this case x0 is 9sin1
if you just write down those eqns with x0 and y0 filled in then you can just eliminate t and you'll have your target form
warped topaz
How do I get x0 and y0
warped topaz
I think in this case x0 is 9sin1
How did you get this
tardy stag
by evaluating your original curve at t=0
tardy stag
$L(t) = C'(0)t + C(0)$
ocean sealBOT
Hayley
tardy stag
L being the linear path the particle takes
tardy stag
we're trying to relate x and y by eliminating t from the component equations
tardy stag
the ones for x and y that we've been computing
tardy stag
so the equations that describe those are x = (18cos1)t + x0 and y = (20cos-2)t +...
these
warped topaz
okok
So we have x = (18cos1)t + 9sin(1) and y = (20cos-2)t + 10sin(-2)
How do I eliminate t?
tardy stag
solve for t in both equations and then equate them to each other
warped topaz
x and y are zero?
tardy stag
no
but you can solve for t in both equations
like the first one would be $\frac{x - 9\sin(1)}{18\cos(1)} = t$
ocean sealBOT
Hayley
ocean sealBOT
odietje
warped topaz
and then set them equal
warped topaz
Ahh oke
I think I can manage that
I will leave this thread open in case I have trouble but I doubt it
alpine sable
lone heartBOT
Please read #❓how-to-get-help
rare grotto
how to solve this
lone heartBOT
how to solve this
vale wigeon
think about how many digits the number 4444^4444 would contain if written out in full. at least approximately.
rare grotto
how would one do that?
vale wigeon
well
what happens to the digit length of a number when you square it?
again, approximately
rare grotto
increases?
vale wigeon
sure it increases, but by how much?
rare grotto
there's no pattern tho right?
vale wigeon
there's no easily describable strict pattern.
but i'm trying to steer you towards intuition on this.
if i told you to square a 5-digit number, how long would you expect the result to be?
vale wigeon
yeah, about 10 digits long.
what if i had you square a 7-digit number? about how long would that be?
rare grotto
13 ,14 ,15
vale wigeon
about 14 digits.
rare grotto
ok.....
vale wigeon
my point is: when you raise a number to the 2nd power, its length approximately doubles.
more generally when you multiply two long numbers together, their lengths tend to add.
rare grotto
mhm makes sense
vale wigeon
so raising a number to the 4444th power will make its length grow approximately 4444 times.
rare grotto
mhm
rare grotto
so raising a number to the 4444th power will make its length grow approximately ...
by isnt that a big approximation?
rare grotto
go ahead , Imma try to follow
vale wigeon
4444 < 10000
therefore 4444^4444 < 10000^4444 = 10^(4*4444)
therefore 4444^4444 < 10^17776 and so 4444^4444 has at most 17776 digits.
rare grotto
mhm
vale wigeon
this lets you put an upper bound on the sum of these digits
rare grotto
correct....
west girder
how to solve this
Bound the sum of the digits and take mod 9
rare grotto
Bound the sum of the digits and take mod 9
ok.... but why 9?
vale wigeon
the mod 9 will come in much later
rare grotto
alr
vale wigeon
for now we just want to establish that ds(ds(ds(4444^4444))) is small enough that we can actually use mod productively
(using ds() here as the notation for the digit-sum function)
right now i want you to fill in the blank here:
rare grotto
k
vale wigeon
4444^4444 has at most 17776 digits, therefore ds(4444^4444) ≤ _____
rare grotto
28?
vale wigeon
twenty-eight??
you've got all these thousands of digits, and you claim they add up to at most twenty-eight?
are you sure that's what you wanted to say?
rare grotto
159984
rare grotto
Right.
vale wigeon
we can even loosen up a bit and say it's less than 160000 -- it won't matter that much
we know ds(4444^4444) is a 6 digit number or shorter
this lets us put a very tight bound on ds(ds(4444^4444))
rare grotto
mhm
vale wigeon
in the same fashion
do you see how to continue with this idea?
mind you, you won't be done just yet, but you will have gone a long way already
rare grotto
ok we get 9*6
vale wigeon
wording
yeah, sure, we get ds(ds(4444^4444)) ≤ 54.
we can do a little better though.
we know that if there is a 6th digit then it can only be 1. so in fact ds(ds(4444^4444)) cannot exceed 1+9*5, or 46
rare grotto
we know that if there is a 6th digit then it can only be 1. so in fact ds(ds(444...
why does it need to be 1 again?
vale wigeon
what's "it"
rare grotto
the 6th digit
vale wigeon
it doesn't "need to" be 1, it CAN ONLY be 1.
since ds(4444^4444) ≤ 160000
putting a 2 or higher there would already make you exceed that
rare grotto
i see.....
vale wigeon
likewise, once we know ds(ds(4444^4444)) ≤ 46, we know that ds(ds(ds(4444^4444))) ≤ 4+9 = 13.
do you understand why
rare grotto
no.
rare grotto
we know that if there is a 6th digit then it can only be 1. so in fact ds(ds(444...
did you mean first digit here? first digit can only be one if there is a sixth digit/
lone heartBOT
Available help channel!
alpine sable
Did I do something wrong?
lone heartBOT
Did I do something wrong?
fallen verge
Whys your sum going from 6 to 3
fallen verge
Then you would go n=6 on the bottom and 9 on top
ocean sealBOT
fallen verge
Its probably because of the asymptote
alpine sable
What about it
lone heartBOT
@alpine sable Has your question been resolved?
lone heartBOT
@alpine sable Has your question been resolved?
void tinsel
Can someone explain if it matters by the product rule in partial intergration which part you set as f(x) and g'(x). For example: Intergrate g(x) = x * e^-x
void tinsel
i don't get the same answer if i set x to f(x) or set e^-x to f(x)
vapid shuttle
can you show your work?
worn fox
It doesn't matter in theory, but one choice is going to be a lot easier than the other, or not terminate if you keep making the "wrong" choice for each integral that shows up
vapid shuttle
, rotate
ocean sealBOT
void tinsel
this is the answer i should have got
vapid shuttle
how did you integrate $\int{-e^{-x}\cdot \frac{1}{2}x^2 dx}$ in the second line?
ocean sealBOT
austinu
vapid shuttle
seems to me you just ignored the e^(-x) and integrated the x^2 part
which is incorrect
also, you can tell that you chose the wrong f(x) and g(x) for your integration by parts easily, because when you compare the integral in your second line to the first line, you can see it just became more complicated rather than more simple
void tinsel
i intergrated -e^-x as e^-x
void tinsel
yeah i see that, but it should not matter right?
vapid shuttle
It doesn't matter in *theory*, but one choice is going to be a lot easier than t...
as for it not mattering, ZAC had a good response above.
I suggest you read it if you haven't
void tinsel
yeah i read it, but can you make mine right? just for me to understand what i did wrong
besides making the wrong choice
vapid shuttle
yours won't work, unless you apply integration by parts a second time and possibly more, and make better choices these next times around
vapid shuttle
no problem
lone heartBOT
lone heartBOT
Available help channel!
cerulean herald
help
lone heartBOT
@cerulean herald Has your question been resolved?
austere lagoon
and in the middle of -5 and 3 its negative so shoulnt it be 1
what do u mean
cerulean herald
what do u mean
shouldnt the answer be the + areas
austere lagoon
it shows for less than 0
so u gtta take the negative area
u can confirm using testpoints
austere lagoon
question?
lone heartBOT
@cerulean herald Has your question been resolved?
alpine sable
lone heartBOT
tacit arch
Maybe raise both sides to the 6th power?
alpine sable
Thai one
tacit arch
Maybe raise both sides to the 6th power?
Did you see this @alpine sable
rugged bone
.reopen
lone heartBOT
@alpine sable Has your question been resolved?
tacit arch
<@456226577798135808> Has your question been resolved?
@alpine sable what questions do you have still
tacit arch
correct
tacit arch
no, you have to do more thinking
alpine sable
2^15=2^8
Almost
Those are the numbers
Idk but why did we do that
But no equal sign
Y=2^15 x=2^8
alpine sable
But no equal sign
But why did we do that
tacit arch
Couldn't we just put 2
put 2 for what
alpine sable
Instead of multiplying by 6
alpine sable
put 2 for what
X and y since its bigger than 1
ocean sealBOT
Result:
32768
,calc 2^8
ocean sealBOT
Result:
256
No
tacit arch
do those two look equal?
alpine sable
No
tacit arch
Couldn't we just put 2
does that answer this question
alpine sable
Idk
You can think of it in the following way:
The RHS is an 8th power so the LHS has to be an 8th power too. So every prime number should be a multiple of 8 times in the primefactorization. Because gcd(8,15)=1 this means that y has to be an 8th power. Same wat you can conclude that x has to be a 15th power. You’re looking for the smallest possible values so this gives y=2^8 and x=2^15
Oooh OK I get it
Now
Ty
tacit arch
You can think of it in the following way:
The RHS is an 8th power so the LHS has...
!nosols
lone heartBOT
You can think of it in the following way:
The RHS is an 8th power so the LHS has...
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
alpine sable
Oh ok
tacit arch
<#486128019652476939> and <#525440994242461702>
you too @alpine sable
alpine sable
But ty @alpine sable
lone heartBOT
@ruby flume Has your question been resolved?
loud skiff
loud skiff
help on this, is this right? the original point doesn't even lie on the surface
lone heartBOT
@loud skiff Has your question been resolved?
loud skiff
or screenshot / take a picture of the problem
tacit arch
,calc atan(pi + -pi) + 2 *sin(pi) - 3 * (2^2) + 3
ocean sealBOT
Result:
-9
huh interesting
pale kestrel
tacit arch
https://cdn.discordapp.com/attachments/903481101744484362/1130630076518649876/IM...
this looks right
tacit arch
help on this, is this right? the original point doesn't even lie on the surface
you're right the point isn't on the surface. you should tell that to your teacher and hope for extra credit
loud skiff
you're right the point isn't on the surface. you should tell that to your teache...
this was the solution after i submitted lol
can u help me understand why they used dz/dx = -Fx/Fz
tacit arch
this was the solution after i submitted lol
oh i guess we were supposed to use chain rule
loud skiff
i meant like
loud skiff
oh i guess we were supposed to use chain rule
how was i supposed to see that lol
like yea with the solution i get how they get the answer, but how was i supposed to tell before
tacit arch
how was i supposed to see that lol
no idea, i can't read your teacher's mind
lone heartBOT
@loud skiff Has your question been resolved?
reef delta
lone heartBOT
vapid shuttle
What are you trying to foil?
$a(b+c)=ab+ac$ you distribute. Nothing is being squared here, and no binomials are being multiplied
ocean sealBOT
austinu
vapid shuttle
yes
reef delta
can this be further simplified?
i assumed it was fully simplified cause no like terms
sour dove
yep thats about as simple as it gets
reef delta
awesome thank you
quick followup question
for something like this do i simplify 115
or enter as is
wait thats supposed to be 144 not 124
135
sour dove
just just remember PEMDAS 🙂
sour dove
bingo
sour dove
okay so with that info:
6(2)^2 = 24
7(2) = 14
So we have 24 -14 + 5 = 15 🙂
reef delta
thank u again
sour dove
of course!
reef delta
.close
safe cape
how to integrate (x^4-2x^2)^2
lone heartBOT
how to integrate (x^4-2x^2)^2
sour dove
FOIL it out. No fancy tricks needed for this one
safe cape
what's foil
proven leaf
what's foil
$(x+a)(x+b)=(x+a)x+(x+a)b=x^2+(a+b)x+ab$
First, Outer, Inner, Last
i.e. expanding
ocean sealBOT
XxMrFancyu2xX
sour dove
we make a good tag team
proven leaf
FOIL it out. No fancy tricks needed for this one
oh I didn't even notice the pfp change Mellow, looking classy!
safe cape
I'm so fucked
sour dove
lmao thanks
sour dove
I'm so fucked
nah you got this fam
safe cape
pls guys send prayers that my exam give the most normal looking question
it's in an hour
sour dove
$(x^4 - 2x^2)(x^4 - 2x^2)$
ocean sealBOT
mellowdramallama
proven leaf
its mostly algebra that gets people, just make sure your not rusty on that :) (I've seen so many people I teach drop negatives everywhere 
)
sour dove
$= x^8 - 2x^6 - 2x^6 + 4x^4 = x^8 - 4x^6 + 4x^4$
ocean sealBOT
mellowdramallama
sour dove
So then you have $\int (x^8 - 4x^6 + 4x^4)dx$
ocean sealBOT
mellowdramallama
safe cape
its mostly algebra that gets people, just make sure your not rusty on that :) (I...
in order to get rusty at something, you have to be good at it at one point
sour dove
now just use the power rule
safe cape
my algebra was always pretty mediocre
proven leaf
now just use the power rule
and linearity!
sour dove
my algebra was always pretty mediocre
algebra is just practice and repetition. The more you do it, the easier it becomes
proven leaf
my algebra was always pretty mediocre
there's always Khan Academy or Organic Chem Tutor on YouTube for cramming too! :)
I learned a lot from the latter
safe cape
now just use the power rule
wdym
ocean sealBOT
XxMrFancyu2xX
sour dove
yeah that
safe cape
the answer I get when doing that is not the same as your initial method though
or am I wrong
proven leaf
can you show your work? :)
safe cape
this pretty much
ocean sealBOT
XxMrFancyu2xX
safe cape
**mellowdramallama**
then what was this for
proven leaf
,w int (x^4+3x^2)^2
ocean sealBOT
proven leaf
then what was this for
$(x^4+2x^2)^2=(x^4+2x^2)(x^4+2x^2)$
ocean sealBOT
XxMrFancyu2xX
proven leaf
thus nothing is rotten in the Kingdom of Denmark 
proven leaf
,
safe cape
,w integrate 4x^8-4x^6+4x^4
ocean sealBOT
safe cape
wait
proven leaf
no 4 in front of the x^8
lone heartBOT
proven leaf
ty for helping!
lone heartBOT
Available help channel!
merry pelican
still struggling with this actually haha, I don't really get it
merry pelican
I know that sin is negative... so I guess that means cos is negative. But why? I don't much about these functions outside of plugging in theta to get an answer
wary stream
notice one thing about tan being positive
lone heartBOT
Please read #❓how-to-get-help
sour dove
I know that sin is negative... so I guess that means cos is negative. But why? I...
if tan is positive, then either we're in Quad 1 or Quad 3
since sin < 0, then we're in Quad 3
and in quad 3, both sin < 0 and cos < 0
merry pelican
What is the relation between the functions and the quadrants? Also, a corrolary question: is this sketch of the graph right?
sour dove
sketch is perfect! but you have negative x values here
merry pelican
oh I see, that explains why 3 would be negative then
but it's confusing to me because
wary stream
What is the relation between the functions and the quadrants? Also, a corrolary ...
Well tan = sin/cos. If sin is neg that means tan = -/? and the only way that tan would be positive is if cos is neg
sour dove
here's a simple answer. Recall that tan = sin/cos. For tan to be postive, either both sin and cos have to be positive OR sin and cos both have to be negative
merry pelican
I guess I didn't have enough knowledge about these functions. I know tan(theta) is opposite over adjacent, and since it's 4/3 it seems like it's saying that the adjacent is positive 3, always
but since I've now been given the relation of tan, sin, and cos, it makes things clearer
I think I'm going to go watch some videos on this and then come back to the problem. Thank you guys for your help
.close
stable jacinth
Hey I am pretty stuck
lone heartBOT
Hey I am pretty stuck
stable jacinth
I understand the basic concepts of sign charts
but reading them always throws me off
tawny tapir
I understand the basic concepts of sign charts
hello
where do u have doubt in the question?
stable jacinth
where do u have doubt in the question?
in identifying the crit points\
obviously there is a sign change at 3 from + to -
meaning its a max I believe
and at 6 from - to + making it a min correct?
tawny tapir
it does not work like this
stable jacinth
huh?
tawny tapir
the crit point are 0, 3 and 6
stable jacinth
why 0?
tawny tapir
or as per the question, it will be 3 and 6
lone heartBOT
Show your work, and if possible, explain where you are stuck.
stable jacinth
I need another opinion
lone heartBOT
Show your work, and if possible, explain where you are stuck.
stable jacinth
theres no real work to show
where the lines are, are where I believe the signs switch
making them critical points
tawny tapir
yes
stable jacinth
and when you go from + to - I believe its a max
tawny tapir
f(x)= x^3/3 -9x^2/2 +18x +3
stable jacinth
where the point at the top is 3
tawny tapir
let us take this example
stable jacinth
f(x)= x^3/3 -9x^2/2 +18x +3
we arent looking for an equation
stable jacinth
I dont want to find the point of that, I need to use the sign chart
not and equation
tawny tapir
i made this equation using the graph
stable jacinth
how?
tawny tapir
sign chart*
actually sorry if u think it inconvinient
but i think it will be easier to clear from example
stable jacinth
x = 0:
Since f'(x) changes sign from positive to negative at x = 0, it indicates a local maximum. The function increases as we move from the left of x = 0 and decreases as we move to the right of x = 0.
x = 3:
Since f'(x) changes sign from positive to negative at x = 3, it indicates a local maximum. The function increases as we move from the left of x = 3 and decreases as we move to the right of x = 3.
x = 6:
Since f'(x) changes sign from negative to positive at x = 6, it indicates a local minimum. The function decreases as we move from the left of x = 6 and increases as we move to the right of x = 6.
tawny tapir
yes...this is part i wanted to tell u
stable jacinth
actually I think I messed up 0
tawny tapir
actually I think I messed up 0
yes
stable jacinth
0 isnt a max or min
tawny tapir
becuse in general, chart is extended from -infinity to + infinity
stable jacinth
but because the funtion is from 0 to inf, 0 is a crit point because its an endpoint right?
stable jacinth
Ok
so since its an endpoint but no sign change is it neither a max or min?
or because the slope is positive between 0 and 3 is it a local min?
tawny tapir
so since its an endpoint but no sign change is it neither a max or min?
u cant tell it is end point
tawny tapir
becuse in general, chart is extended from -infinity to + infinity
this happenns
stable jacinth
for this specific problem though it bounds in our domain?
) Let f(x) be continuous on [0,∞) and differentiable on (0,∞).
stable jacinth
the bracket on [0 indicates its a solid point
tawny tapir
there is the domain
tawny tapir
so because the slope is positive from 0 going to 3
.
grizzled ermine
can someone check my work?
tawny tapir
@stable jacinth likewise, 3 is max, and 6 is min
stable jacinth
<@798658094045462578> likewise, 3 is max, and 6 is min
yes
we are on the same page now
tawny tapir
sorry to take a while, i was writing it
stable jacinth
and to know the global max or min, I would need the parent function right?
tawny tapir
yes
stable jacinth
not right now
grizzled ermine
already taken channel
sorry!
short blaze
does anyone know the answer to this
lone heartBOT
does anyone know the answer to this
remote heron
write the numerator as a fifth root
short blaze
would it be 4^5a^5b^5c^9 to the 5th root
remote heron
$4=2^2=2^{2 \cdot 5 \cdot \frac15} = = \sqrt[5]{2^{10}}$
ocean sealBOT
janniku
short blaze
wait isnt that the same as 4 to the 5th to the 5th root
remote heron
you can combine the fraction under the radical
remote heron
wait isnt that the same as 4 to the 5th to the 5th root
it is
short blaze
so why do we turn it into 2
remote heron
no reason
short blaze
remote heron
remote heron
how do i do that
law of power of quotient
short blaze
is that right
fossil latch
c²⁰, not c⁹
ocean sealBOT
1pkame
short blaze
okay
short blaze
how do i simplify this
can i just make a5 into a3
and get rid of the a2 in the bottom
and the same for the other variables
fossil latch
Yeah since they are in the same root
ocean sealBOT
1pkame
short blaze
im not sure how that applies
ocean sealBOT
1pkame
fossil latch
Yeah
loud brook
How to get help
short blaze
get help
loud brook
Here I ask my question?
fossil latch
**1pkame**
So $ \sqrt[5]{a} \sqrt[5]{b} = \sqrt[5]{ab}$
Cmon why is bot not working
lone heartBOT
How to get help
Please read #❓how-to-get-help
fossil latch
To make things short, $ \frac {\sqrt[5]{1024}}{\sqrt[5]{48}} = \sqrt[5]{\frac{1024}{48}}$
short blaze
uh
fossil latch
Bot isn't working sorry
short blaze
any way to explain it without the bot?
ok wait so ur tryna say its just the 5th root of 1024/48?
but im not sure how that helps since 1024/48 doesnt work out evenly
but 1024 to the 5th root is 4
if that helps
$ \frac {\sqrt[5]{1024}}{\sqrt[5]{48}} = \sqrt[5]{\frac{1024}{48}}$
$ \frac {\sqrt[5]{1024}}{\sqrt[5]{48}}$ = \sqrt[5]{\frac{1024}{48}}$
$\frac {\sqrt[5]{1024}}{\sqrt[5]{48}}$
ocean sealBOT
icecream
short blaze
$\frac {\sqrt[5]{1024}}{\sqrt[5]{48}}$ = $\sqrt[5]{\frac{1024}{48}}$
ocean sealBOT
icecream
short blaze
@fossil latch
short blaze
1024 to the fifth is 4
fossil latch
Hmm could you show the answer in fractions
fossil latch
No I mean in complex fractions
short blaze
what
fossil latch
64/3 = 2⁶/3
remote heron
uh i searched it up but i dont really get it
omg im sorry i spaced 
short blaze
its ok
$sqrt[5]{\frac{64}{3}}$ = $(\frac{64}{3})^(\frac15)$ = $\frac{2^6^{\frac15}}{3^{\frac15}}$
ocean sealBOT
icecream
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
remote heron
space for \ at start
fossil latch
So $\sqrt[5]{\frac{64}{3}} = (\frac{64}{3})^(\frac15) = \frac{2^{\frac65}}}{3^{\frac15}}$
ocean sealBOT
1pkame
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
remote heron
?
short blaze
wait what
fossil latch
Sorry was complaining for the complexity of latex
short blaze
64/3 to the fifth equals 64/3 times 1/5?
remote heron
So $\sqrt[5]{\frac{64}{3}} = \qty(\frac{64}{3}\qty)^{\frac15} = \frac{2^{\frac65}}}{3^{\frac15}}$
fossil latch
Where's 2^(6/5) part gone
ocean sealBOT
janniku
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
fossil latch
This is why I hate using latex
remote heron
fossil latch
So uhh do we have paper or something
remote heron
$\sqrt[5]{\frac{64}{3}} = \sqrt[5]{\frac{2^6}{3}} = \frac{\qty(2^6)^{\frac15}}{3^{\frac15}}$
short blaze
oh ok
so u converted 64 to 2 to the 6th
and then changed the radical to exponent
how does that help tho
ocean sealBOT
janniku
remote heron
@fossil latch
remote heron
oh ok
ill be home in like
20
if no one else comes
then i can write
and we can talk abt it
latex sucks for this stuff
short blaze
plugged it in mathway and got 2^6/5 over 3 ^ 1/5
damn i hate math
<@&286206848099549185>
remote heron
alright I'm home
lets make sure were on the same page with the letters first
@short blaze were you able to simplify the letters?
short blaze
it probably is
remote heron
i think its a 2
short blaze
this course is so shit
remote heron
if you right click where the text is, does it give you any options
short blaze
oh yeah its a 2
remote heron
okay
lets work on just the numerator here
are you okay to just do the letters?
actually, i think you got the idea there
short blaze
short blaze
yeah
short blaze
i made it the same root on the top as the bottom
remote heron
the letters are gonna end up being $\sqrt[5]{a^3b^2c^9}$
short blaze
its 9 instead of 19
remote heron
you can believe me
ocean sealBOT
janniku
remote heron
yes
short blaze
okay
remote heron
alright
we can still simplify this a tiny a bit more
you can rewrite that c piece a little bit
remember that we can write like $d^5 = d^2 d^3$
ocean sealBOT
janniku
remote heron
yea, that works
short blaze
so we can get rid of the c5
remote heron
so $\sqrt[5]{a^3b^2c^4c^5}$
ocean sealBOT
janniku
remote heron
get rid?
short blaze
turn it into just c
remote heron
yup
okay so we get to the letters being simplified, i think all the way
$c\sqrt[5]{a^3b^2c^4}$
ocean sealBOT
janniku
ocean sealBOT
janniku
remote heron
howd you rewrite that 4?
remote heron
okay
short blaze
whichh is 1024 to the fifth root
remote heron
i just mean, write the root over the entire fraction
short blaze
1024/48 to the fifth
remote heron
do we change 48
it works being easiest to write the numbers in their prime factorization
$\sqrt[5]{ \frac{ 1024 }{ 48 } }$
ocean sealBOT
janniku
short blaze
ok
remote heron
well, i guess you dont have to entirely factor them into primes
but it is easiest
you said 1024 = 4^5, right?
short blaze
you said 1024 = 4^5, right?
yes
remote heron
,w simplify 1024/48
ocean sealBOT
ocean sealBOT
janniku
remote heron
ocean sealBOT
janniku
short blaze
yeah
remote heron
and the letters are $c \sqrt[5]{a^3b^2c^4}$
ocean sealBOT
janniku
short blaze
yes
remote heron
short blaze
remote heron
you know
i think i see it
but you have to do something kind of unusual
try to match the powers on the letters in the answer
then see if you can figure out how they did the rewriting
short blaze
a^3b^2c^4
remote heron
yup