Do you know the limit of 1?
#help-42
spark relic
tepid meteor
i know that the sin(1/x) is a bounded function but i don't know how the calculus are done
tepid meteor
Do you know the limit of 1?
of 1?
spark relic
I mean, the function when x = 0
spark relic
Yes
tepid meteor
and the limits?
spark relic
I think you have to study only the first function, the big one
tepid meteor
yeah, cause its 0+ and 0-
spark relic
$\lim_{x \to 0^{-}} \frac{\sin(\frac{1}{x})}{e^{\frac{1}{x}}+1}$
potent lotusBOT
Rub05
spark relic
And 0+ too
spark relic
I think you can say that sin(1/x) ≈ 1
spark relic
Technically yes, I think
tepid meteor
ok, then, when x->0+ the limt is 0
spark relic
Correct
tepid meteor
but how it is when x->0-?
spark relic
Undetermined
spark relic
Yes
spark relic
What happens if you multiply a number by 1?
tepid meteor
you get the multiplied number
spark relic
And what happens if you multiply that number with 0,5?
tepid meteor
you get the half of it
tepid meteor
so you don't know the exact result
spark relic
Yes
tepid meteor
okay i got it
spark relic
Perfecto
spark relic
errant vigil
tepid meteor
.close
tepid meteor
when x=0, in this case, how are the limits done?
calm coralBOT
when x=0, in this case, how are the limits done?
tepid meteor
the limit when x->0+ is done with the x^3-x function?
spark relic
0- is the first function, 0+ is the second one
alpine mason
is ths topic from Continuity?
ocean cypress
Yes
tepid meteor
is ths topic from Continuity?
yeah
tepid meteor
0- is the first function, 0+ is the second one
and for the first one, how is the limit?
alpine mason
hmmm the breaking points of the function here are x=0 and 1 check the continuity at those points
tepid meteor
because now we have 0*sin(1/x) ≈ 1
tepid meteor
hmmm the breaking points of the function here are x=0 and 1 check the continuity...
yeah i know, my problem is solving the limits
spark relic
and for the first one, how is the limit?
L'Hopital
tepid meteor
0/0 no?
tepid meteor
when doing L'Hopital the result is -cos(1/x)/0 no?
spark relic
Yes
spark relic
Doesn't exist
spark relic
when doing L'Hopital the result is -cos(1/x)/0 no?
If we have this in x -> 0, we can do the lateral limits
But we are in lateral limits actually
tepid meteor
then why doesnt exist?
spark relic
We cannot do lateral limits in a lateral limit
tepid meteor
oh, only by that?
spark relic
Yep
tepid meteor
wow, ok
spark relic
spark relic
Yes
tepid meteor
and when do i know that i have to do one-sided limits?
spark relic
When limit x -> a = ± ∞
tepid meteor
ok, thanks
and in this same exercise when x=1 the function is continious and can be derivated
how can i prove that is "derivable"?
spark relic
If lim x->1+ and x -> 1- is the same, then f'(x) is differentiable in x = 1
tepid meteor
is this well done, only the differentiable part (bottom one)?
i know that i miss to put some lim x->1
calm coralBOT
@tepid meteor Has your question been resolved?
rugged oasis
If I need to show x^4 -6x^2 -k = 0 has four roots, how can I do so?
rugged oasis
I can use b^2 - 4ac > 0 for k > -9 but I believe there is another constraint that k < 0 also
thick pond
See if you can solve for x^2 first
rugged oasis
What do you mean solve for x^2?
x^2 = (-x^4 + k)/6?
Or if you're meaning x^4 - 6x^2 - k = y^2 -6y - k then that is what gives k > -9 for two distinct roots to that
thick pond
Let a = x^2
a^2 - 6a - k = 0
Solve for a
The thing is you'll get 4 roots for all k except k = --9
rugged oasis
real roots** to specify
i think i see
so you end up with
$6 \pm \sqrt{36 + 4k} = 0$
potent lotusBOT
Alfie
thick pond
Well sometimes you'll get imaginary roots tho
rugged oasis
we're trying to find k for only real roots, forgot to clarify in the qn
so you need to disregard 6 + sqrt(36 + 4k) = 0 as that doesn't exist
so you have 6 - sqrt(36 + 4k) = 0
6 = sqrt(36 + 4k)
36 = 36 + 4k
4k = 0
k = 0
but then you lose k = 9 as the other part?
unless you need to solve and use the discriminant? to get both bounds
rugged oasis
sorry, i should better state the original question rereading it
thick pond
We need a > 0 I think
rugged oasis
trying to find all k in R such that the quartic has four real roots
thick pond
Ah okay
thick pond
trying to find all k in R such that the quartic has four real roots
4 distinct real roots?
rugged oasis
yes
thick pond
So we know x^2 =6 +- sqrt(36 + 4k) / 2
thick pond
And x = +- sqrt ( (6 +- sqrt(36 + 4k) ) / 2 )
Now 6 +- sqrt(36 + 4k) must have 2 distinct real values
i.e. what is inside the sqrt is strictly > 0
rugged oasis
Sure
Which leads back to k > -9
we need to say also the outer square root must have 2 values
thick pond
Now 6 +- sqrt(36 + 4k) must have 2 distinct real values
*which are both positive
calm coralBOT
thick pond
🙂
void citrus
Which shape is correctly transformed?
calm coralBOT
Which shape is correctly transformed?
grave blaze
well not the blue one
void citrus
well not the blue one
Why?
grave blaze
translation is done by addition not multiplicatoin
so essentially you move all vertices one left and 3 down
void citrus
.close
astral breach
Hello I'm not quite sure how to get the x
calm coralBOT
Hello I'm not quite sure how to get the x
clear sapphire
Do you know that exterior angle of a triangle is equal to the sum of the interior opposite angles?
astral breach
yeah
clear sapphire
So can you apply that here?
astral breach
nope
valid needle
PAR and RAY should add to 180 degrees, if PAY is a straight line
valid needle
the three angles inside any triangle also add to 180 degrees
think you could solve it based on that system of eqs (variables to find are x and mPAR)
dull wagon
Why, not? Can you identify which two angles would add up to angle RAY here?
add up to RAY
glass ibex
calm coralBOT
dull wagon
you seemed to have forgotten about product rule
dull wagon
product rule
you have a product of functions
apply product rule first
then apply chain rule if needed for function compositions
dull wagon
yeh, and why's the t gone from the argument
glass ibex
derivative of pi t is pi
dull wagon
so?
that's no reason to erase the variable from the argment
chain rule doesn't state to replace the inner function with its derivative
glass ibex
is the cos part wrong too?
dull wagon
i mean despite the ignorance of product rule
remote mural
there is no rule to apply product rule first then chain rule
dull wagon
you actually differentiated sin(pi * t) itself correctly the first time
remote mural
u apply both together
dull wagon
there is no rule to apply product rule first then chain rule
here you can distinctly split the steps
dull wagon
no
glass ibex
what should it be?
glass ibex
cos(pi t) * (pi)
dull wagon
yes
dull wagon
yes
remote mural
good boy
remote mural
clean up the expressions
clean up with soap or wot
glass ibex
this works
so for this one, I get 2x - 14 (dy/dx) + (2dy/dx) = 14
not sure how to do implicit but I think its kinda like this
everytime I see a y, I need to differentiate and just put a dy/dx by it right?
dull wagon
why =14
dull wagon
differentiate both sides wrt x (or desired variable if you have something else)
i.e. apply d/dx to both sides of the equation
you didn't differentiate the right side here
and again you forgot about the product rule when diffentiating the 14xy
glass ibex
well what should happen to the 14xy?
dull wagon
apply product rule when diffentiating
glass ibex
so I will have 6 sets of ( ) total? since its 14 * x * y
glass ibex
well idk how I use product rule on something with 3 terms
14 x and y
f'(x) * g(x) * j(x) + f(x) * g'(x) * j(x) + f(x) * g(x) * j'(x)
dull wagon
well one of them is a constant which makes things easier
you don't need triple product rule for this
you could just apply constant multiple rule
glass ibex
0 * x * y + 14 * 1 * y + 14 * x * dy/dx
dull wagon
and factor the constant out of the thing you're differentating
or treat that as
14x * y
or x * 14y
depending on personal preference
d/dx (14xy) = 14* d/dx (xy)
would be the most convenient
glass ibex
so for this one, I get 2x - 14 (dy/dx) + (2dy/dx) = 14
you mean I had it right at first?
dull wagon
no
d/dx (xy) isn't dy/dx
and looking closer
d/dx y^2 isn't 2 dy/dx either
xy is the product of x and y
apply product rule
glass ibex
1y + xdy/dx
dull wagon
how exactly are you applying power rule...
glass ibex
x^2 = 2x
dull wagon
d/dx x^2 = 2x
glass ibex
y^2 = 2dy/dx
glass ibex
because its a y
dull wagon
so?
glass ibex
we are doing d/dx
glass ibex
ok 2y(dy/dx)
dull wagon
yes
if it wasn't clear enough, from chain rule
$$\dv{x} y^2 = \dv{y}y^2 \cdot \dv{y}{x}$$
potent lotusBOT
ℝamonov
glass ibex
ok
what happens next? x*y gave us y +x(dy/dx). y^2 gave us 2y(dy/dx)
x^2 is 2x
14 becomes 0
dull wagon
yes, be clearer next time
reef nimbus
just take the derivative of "14xy", not just "xy"
glass ibex
0 * x * y + 14 * 1 * y + 14 * x * dy/dx
like this?
reef nimbus
like this?
looks right, I don't know why you have 0 * x * y, though
glass ibex
derivative of 14
dull wagon
as mentioned earlier, triple product overcomplicating it
you "can" use it if you want
and it does work
just not needed
glass ibex
so what should happen if I didnt do long way
reef nimbus
yes. you can consider the products as "14x" and "y"
reef nimbus
you're subtracting "14xy" so the derivative of that term should all be subtracted
reef nimbus
yes that looks right
glass ibex
what comes next?
reef nimbus
what is the objective?
glass ibex
implicit differentiations]
glass ibex
.
reef nimbus
then you just "solve for" dy/dx
glass ibex
how do I do that
reef nimbus
the same way you would solve for anything
glass ibex
but this problem has x's y's and dy/dx
reef nimbus
you are only asked to find dy/dx though
glass ibex
so I put something on the other side of = sign
reef nimbus
well yes. think of "dy/dx" as being just another variable you need to solve for
glass ibex
so -14x and 2y have a dy/dx with them
dull wagon
dy/dx = stuff without dy/dx doesn't matter if there are y here
is what you want
glass ibex
so I will add 14x(dy/dx) and subtract 2y(dy/dx) both sides
reef nimbus
yes, whatever operations you would normally do
glass ibex
ok so now I have 2x-14y = 14x(dy/dx) - 2y(dy/dx)
reef nimbus
looks right
reef nimbus
where did 8/7 come from?
reef nimbus
Where are these numbers coming from?
Are you asked to calculate at a specific point?
reef nimbus
I still don't understand what you're doing
glass ibex
implicit differentiation
reef nimbus
Denominator from what?
glass ibex
middle
reef nimbus
I still don't understand. Just solve for dy/dx
glass ibex
i did. divide by 14x and -2y both sides
reef nimbus
As I said before, consider it as a variable. Call it "z" if that helps
You can't just get rid of the other variables, either
glass ibex
ok so now I have 2x-14y = 14x(dy/dx) - 2y(dy/dx)
.
we need dy/dx by its self
so I divide by 14x and -2y
reef nimbus
Yes. Like I said, call it "z". Then solve for z using typical rules of algebra
glass ibex
what do you mean
reef nimbus
$2x - 14y = 14xz - 2yz$. solve for "z"
potent lotusBOT
cwatson
glass ibex
yes, so I divide by 14x and -2y
reef nimbus
no
if you divided by "14x" you would then get $\frac{2x - 14y}{14x} = z - \frac{2y}{14x}z$
potent lotusBOT
cwatson
glass ibex
so I should bring the regular 2x or -14y over to the dy/dx side?
reef nimbus
no. solve for it like you would solve for any variable. this is just algebra now
glass ibex
can you give me a hint
reef nimbus
simplify the RHS
glass ibex
but 14x and -2y have different variables
glass ibex
2z(7x-y)
reef nimbus
well you want to solve for z, not 2z. but yes
glass ibex
so thats wrong?
reef nimbus
well it's right, but don't factor out a 2. so z(14x - 2y) = 2x - 14y
glass ibex
so now divide both sides by 14x - 2y
reef nimbus
yes
glass ibex
so I divide by 14x and -2y
so I had it right the whole time?
reef nimbus
no, you said you put 8/7 as the answer
reef nimbus
re-check your arithmetic. the first line is right
glass ibex
I don't know where it went wrong
reef nimbus
look carefully. the first line on the right is correct
random stratus
f a function such as f(1) != 0
i have to find an equivalent of
random stratus
$\int^1_0 x^n f(x) dx$
random stratus
no idea how to do that
tropic isle
what is f(x)?
random stratus
f a function such as f(1) != 0
i have to find an equivalent of
^
tropic isle
Right but if f(x) can be anything than the integral is kinda meaningless?
random stratus
we don't know f ye
tropic isle
So what are you trying to find?
jovial pebble
calm coralBOT
calm coralBOT
@jovial pebble Has your question been resolved?
tropic isle
Think worst case senario, whats the most amount of books you could take without getting a math book
calm coralBOT
calm coralBOT
Available help channel!
gloomy cargo
I have a question about integration: a ball starts rolling at 8m/s and ends at 0m/s. It has a deceleration of 2m/s^2. How much distance does the ball travel.
How would I go about solving that? I have this so far:
V1 = V0 + a*t
Solving that gives me t =4s
But that's not using integration so I'm trying to get v(t) as a formula so I can integrate that to get s(t)
gloomy cargo
I know but since it's in my book in the chapter of integration I'd rather do it that way, since my teacher explained it badly and my book is unnecessarily complicated i have to practice it a bit
night portal
Ohk makes sense
calm coralBOT
@gloomy cargo Has your question been resolved?
calm coralBOT
calm coralBOT
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hasty sun
(a) f(x) = 2x^3 + 3x^2 + 2x + 8
= (x+2)(2x^2 -x + 4)
From this, one real solution is x = -2
(b) Using answer to part (a) ,hence solve x^3 + 3x^2 +4x + 32 = 0
hasty sun
Question (a) is solved, how do I relate (b) to (a)?
gray leaf
Question (a) is solved, how do I relate (b) to (a)?
think
gray leaf
i know
dull wagon
did you copy the question down correctly
hasty sun
did you copy the question down correctly
yes
remote mural
hint: 32/8 = 4
hasty sun
So I multiply first equation by 4 and compare the two equations
since the constant 32 will be the same?
gray leaf
wrong way
hasty sun
Yeah i tried and cant seem to relate it to part a
I just factorise normally and got x=-4 as the only real root
remote mural
Yeah i tried and cant seem to relate it to part a
once you multiply by 4, try and find a substitution that makes a look like b
hasty sun
once you multiply by 4, try and find a substitution that makes a look like b
If answer is x = -4 as only real root then I should replace x with x+2
gray leaf
I just factorise normally and got x=-4 as the only real root
you'll get this
gray leaf
If answer is x = -4 as only real root then I should replace x with x+2
nope
think not just about addition but multiplication too
hasty sun
Cant see the link sir, please help
I am compare part(a) 8x^3 + 12x^2 + 8x +32 = 4(x+2)(2x^2 - x +4)
vs part (b)
x^3+3x^2+4x+32 = (x+4)(x^2-x+8)
remote mural
compare the coefficients of part a and b
8/1 = 8
12/3 = 4
etc
what do you notice?
hasty sun
Its 8,4,2,1
I never learn this before
decreasing by factors of 1/2
Could u link me any video that teaches this concept
gray leaf
Could u link me any video that teaches this concept
what if you replace x by x/2?
calm coralBOT
@hasty sun Has your question been resolved?
hasty sun
what if you replace x by x/2?
Oh yes, that works!
Thank u so much
I shall be more aware of multiplication and division too in the future
calm coralBOT
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sudden nebula
Conceptual Question, there's a thereom that states that if f(x) is differentiable, then it is also continous. Is it possible for a function to be differentiable but not continuous?
slender yoke
I think not cuz a function needs to be continuous to be differentiable
velvet osprey
there is a theorem that states all differentiable functions are continuous. do there exist functions that satisfy the hypotheses of the theorem but not the conclusion?
what do you think?
fathom shuttle
there is a very specific statement in MVT which requires a function to be continuous on [a,b] and differentiable on (a,b) because if you do not include the underlined statement it could be discontinuous at the endpoint
kinda dumb huh
sudden nebula
what do you _think_?
I don't know. I think the examples I've worked with happen to satisfy the theorem.
I can't think of an example that doesn't.
Most of the functions are trig, poly, exponential, rational, and such. Which are continous about their domain.
velvet osprey
i mean heres the thing
if there were a counterexample
the theorem would not be a theorem
sudden nebula
TRUE
velvet osprey
like literally thats the whole point of a theorem
sudden nebula
I remember that from my proof writing book
velvet osprey
so you're just casting doubt on logic
still echo
through (3,1) with direction vector (2,-1) . I get as rico 2 but the solution says (-1/2)
calm coralBOT
through (3,1) with direction vector (2,-1) . I get as rico 2 but the solution sa...
Show your work, and if possible, explain where you are stuck.
still echo
wanton wave
It looks like you calculated the slope between the points (3,1) and (2,-1), but that's not what the problem says. (2,-1) is the direction vector, not a point on the line
The direction vector gives the slope explicitly
(2,-1) means 2 right, 1 down
which is slope -1/2
@still echo
calm coralBOT
wanton wave
np 👍
cunning veldt
use log on it, then put everything on the same side, it's a polynomial of degree 2
get the roots, and then the solutions
you can but not necessarily
the properties of the log will suffice
no
you forgot parenthesis
and also bad use of properties I think
log(5^x/sqrt(5)) = xlog(5) - log(5)/2
log(2(6^(x²-x)) = log(2) + (x²-x) log(6)
log(sqrt(5)) = log(5)/2
quadratic fomula
it's a polynomial of degree 2
no
use quadratic formula
leaden thunder
group terms together and put it in the form
ax^2 + bx + c = 0
-log(5/2) - log(2) can be grouped together
do the same for log(5) and log(6) for the x term
you can try plotting in desmos to compare
keen palm
how do i get the partial integral of y''-y+ xe^x when i use y= (ax+bx^2)e^x
leaden thunder
is that y'' or y'
oh you're given y?
do you just plug in y and find integral (y'' - y + xe^x) dx ?
exotic falcon
Are you trying to find the general solution?
calm coralBOT
@keen palm Has your question been resolved?
calm coralBOT
calm coralBOT
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little kiln
leaden thunder
Closed your other channel
leaden thunder
You should show your work from there
proven vortex
@little kiln show your work
calm coralBOT
@little kiln Has your question been resolved?
calm coralBOT
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brazen kraken
calm coralBOT
brazen kraken
thats the only work they provided, how did they get it?
specifically how did they get value of f(x)?
main marlin
$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}$
potent lotusBOT
MrFancy
brazen kraken
yes i know thaat
main marlin
simply find f(-2) and g(-2) and do the division
main marlin
also one more question
fire away!
main marlin
the domain are the numbers that make the function defined
this function is undefined at x=-3,3
main marlin
hence they are not part of the domain
brazen kraken
3^2 = 9
9-3=6
main marlin
should be $\pm\sqrt{3}$ :)
potent lotusBOT
MrFancy
keen inlet
Solid of revolution?
calm coralBOT
potent lotusBOT
slippingcircle
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
warm wren
hey isn't this the usamts
calm coralBOT
warm wren
lmao
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Available help channel!
remote mural
calm coralBOT
remote mural
so i believe its 40/7 because U would do 4/7 x 10/1 which is 40/7
but mathway says 640/7 and idk why
im assuming it has to do with the cube root
pliant hedge
but mathway says 640/7 and idk why
640/7 isnt even one of the options...
pliant hedge
so i believe its 40/7 because U would do 4/7 x 10/1 which is 40/7
it should be 40/7 yup
pliant hedge
let me check actually
remote mural
alr
remote mural
could u show that with the bot thing
pliant hedge
sure
$\lim_{x\to0}\frac{10\sin^3(4x)}{7x^3}=10\lim_{x\to0}\left(\frac{\sin(4x)}{7x}\cdot\frac{\sin(4x)}{x}\cdot\frac{\sin(4x)}{x}\right)$
potent lotusBOT
light
pliant hedge
$$\lim_{x\to0}\frac{\sin(ax)}{bx}=\frac ab$$
potent lotusBOT
light
pliant hedge
yup
desert marsh
help idk what to do im vonfused
calm coralBOT
help idk what to do im vonfused
desert marsh
what do i do idk
sweet tapir
help idk what to do im vonfused
that's not a question. Can you formulate an actual questions? What have you tried?
desert marsh
wdym
sweet tapir
you did not ask a question. Do you have an actual question about the problem you posted?
desert marsh
im confused on what to do ive tryed plugging in 0 but i dont think it works
im confused on what i put as x
sweet tapir
I'm not trying to be a dick, it will be helpful for you to take your confusion and formulate it as a question
and you still haven't asked a question
sweet tapir
why do you think you should plug something in for x?
desert marsh
because in class when we were learning squeeze theorem we would plug in a vallue to x (the number x aprovhes )
sweet tapir
I think there's a conceptual step you're missing there
what does the squeeze theorem say?
sweet tapir
used for what?
desert marsh
its used to find where the x is approvhing
when its stuck in the middle of 2 functionsd
sweet tapir
that's right, it's used to compute limits
but it's only helpful when the middle function is "squeezed" in between the outer functions
can you give an explicit description of how to tell when the squeeze theorem applies?
desert marsh
ummm well my twacher didnt explain that so i dont know when it is applied thats all i really know about it
sweet tapir
okay, well you will have no hope of solving problems about the squeeze theorem if you don't even know what it says
I guarantee it's in your textbook
go take a look
desert marsh
we dont use our textbook...
sweet tapir
but you can
you can't expect to just absorb all of calculus from a 50 minute lecture a few times a week
desert marsh
r u tlking about the thing where we can only use the theorem if all f(X) = L as x aproches c
and g(x) as x aproches c equal h(x) as x aproches c = L
sweet tapir
that second sentence is borderline gibberish, but yes
how does that apply to your problem?
desert marsh
well because im assuming f(X) is being sqweezed by them
sweet tapir
how can you tell whether a function is squeezed between 2 others? As you say, that's what the problem is asking you to decide
desert marsh
do we graph it to find out if its being squeezed?
sweet tapir
that's one way, but graphing can be tricky
it can be hard to tell whether the functions actually squeeze, or if you just have a bad drawing
desert marsh
sweet tapir
can you compute the limits and compare?
desert marsh
u can find out if its beeing squeexed if the values or h(x) and g(x) are the same number
sweet tapir
yes if the limits of h and g as x approaches a exist and are equal, then the squeeze theorem says that the limit of f as x approaches a exists and is also equal
so in your problem
when are those outer limits equal?
desert marsh
im not sure
desert marsh
the limmit is 0 right>
sweet tapir
which limit? there are many limits to compute in this problem
desert marsh
i assumed for all of them
sweet tapir
many (most?) of those limits are not equal to 0
desert marsh
i thought it was because it says which of the inequalities can be used with the squeeze theorem to find the limit of the function as x approaches 0 ?
sweet tapir
yes, you need to take a limit as x approaches 0
that's not what you originally said, but that's correct
desert marsh
to find the limmit we would need to plug in numbers closs to 0?
sweet tapir
let's pick an example
1/|x|
plugging in 0 doesn't make sense, you can't divide by 0
how do I compute the limit as x -> 0?
desert marsh
would u put a number close to 0?
sweet tapir
what happens if you plug in a number close to 0?
desert marsh
the number would get bigger and bigger as it gets closer to 0
desert marsh
+00
desert marsh
no
sweet tapir
cool, why not?
desert marsh
becauxe not enough info is given
sweet tapir
I want to believe you're on the right track, but I don't know what you mean by that
desert marsh
oh\
um
well we were told that it can only be used when the h(x) and g(x) are equal otherwise u say u dont have enough info to determin the limit with sqeeeze theorem
sweet tapir
that's a weird way to phrase it, but yes I'm on board
the two functions don't squeeze because they have different limits as we approach 0
so that answers 1 of the 3
desert marsh
so it cant be 3
for 1 i think it cant be it either becaused the values as x aproch 0 is 2 dif values
desert marsh
i was just confused i never seen it like this
i assumed we plugged in 0 beccause thats what our teacher said or thats how i understood it
sweet tapir
it's a good skill to learn to be able to think critically when things are presented in new ways
you definitely knew all the things
calm coralBOT
calm coralBOT
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bright mulch
Hey so I was wondering if I did this correctly?
calm coralBOT
Hey so I was wondering if I did this correctly?
bright mulch
We’re supposed to find the derivative
merry python
looks right
calm coralBOT
calm coralBOT
Available help channel!
naive estuary
Uh.
naive estuary
How to find
cursive linden
yeah?
calm coralBOT
@naive estuary Has your question been resolved?
leaden thunder
.close
brisk pelican
calm coralBOT
brisk pelican
i dont know where the mistake is
potent shoal
i dont know where the mistake is
It seems good to me, how do you know there was a mistake?
brisk pelican
well from there i get sin^3a / 3
and then i think thats just [(x^2-1)^(3/2)]/3
but according to wolfram alpha theres an extra x^3 in the denominator and i have no idea where it comes from
omg
yes i do
im just blind
.close
calm coralBOT
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regal remnant
calm coralBOT
regal remnant
New to induction confused on what to do
merry python
So you need to start with the base case
n = 1
You need to show that the expression is true for n = 1
merry python
Good
Now comes the key part
You want to assume that it works
So let us say it works for some N
You want to show that it will still work if you consider the next case, N+1
regal remnant
So let us say it works for some N
So this would be N/N+1 ?
merry python
So you know that for n = N, the formula gives N / N+1
that's for n= N
I want you to replace N with N+1
regal remnant
Ahh right
merry python
DOn't overthink this
regal remnant
So just N+1/ (N+1) + 1
merry python
Yes!
Perfect
Which is (N+1) / (N+2)
Now you want to show that
$\sum_{n=1}^{N+1} \frac{1}{n(n+1)}$
potent lotusBOT
TooManyCooks
merry python
gives you that
all I did here is change the upper index of the sum
You have $1 + \cdots + \frac{1}{N(N+1)} + \frac{1}{(N+1)(N+2)}$
potent lotusBOT
TooManyCooks
merry python
Still with me?
regal remnant
regal remnant
😭
merry python
Let's take a step back
You want to show that it will hold for N+1
You have $1 + \cdots + \frac{1}{N(N+1)} + \frac{1}{(N+1)(N+2)}$
potent lotusBOT
TooManyCooks
merry python
I want you to add that up using the thing you assumed
You assumed $1 + \cdots + \frac{1}{N(N+1)} = \frac{N}{N+1}$
potent lotusBOT
TooManyCooks
regal remnant
Wait where did the 1/ (n+1) (n+2) come from?
merry python
That's the (N+1)th term in the summation
The inductive step requires you to show that your assumption will still hold true if you consider the N+1 case
regal remnant
Ah alrightt
merry python
So you ahve the N+1 sum
And you have the assumption as well
Use those two bits of information to simplify $1 + \cdots + \frac{1}{N(N+1)} + \frac{1}{(N+1)(N+2)}$
potent lotusBOT
TooManyCooks
merry python
Wait why do you still have small n
regal remnant
Ah mb that should b big N
merry python
Also, you can't have that on the right hand side
That's what you're trying to prove
You don't know if that equality is true (yet)
merry python
I would do that much
but your professor may want something more elaborate, so I'll let you decide
remote mural
calm coralBOT
remote mural
Stuck on what error I made
leaden thunder
,rotate
potent lotusBOT
regal remnant
Does this look good?
calm coralBOT
@remote mural Has your question been resolved?
remote mural
<@&286206848099549185>
calm coralBOT
@remote mural Has your question been resolved?
broken rover
anyone have like a doc of all the polynomial function rules
calm coralBOT
anyone have like a doc of all the polynomial function rules
broken rover
like if "two subtraction signs are next to each other it becomes a positive"
and how would i even start
pure kayak
you just sub 2 into f
exotic falcon
yeah g(x) is unrelated here
it only is asking you to evaluate f(x) at x=2
so substitute x=2 into f(x)
broken rover
so substitute x=2 into f(x)
so would it be
2(2)^2 -5(2)?
exotic falcon
yes
broken rover
i gotta start locking in
been aceing the hw but getting 60s on tests and quizzes
def gotta sleep before 11:30
grade in math is a 72 😵
broken rover
<:kekw:752021756650455061>
am i doing smth wrong
i got -18
but calculator said -2
exotic falcon
8-10=-2
broken rover
YOOOOOOOOOOOOO
💀 💀 💀 💀 💀 💀 💀 💀 💀 💀 💀 💀 💀 💀 💀 💀 💀 💀
IM TRIPPIN
CANT DO SIMPLE 4TH GRADE MATH
exotic falcon
guess what
yknow how f(2) you plug in 2 for x
for f(x-2)
you plug in x-2 for x
groundbreaking
try that out
broken rover
uhhhhhhhhhhhhhhhhhhhhhhhh
broken rover
(x-2)^2 use box method
i figured out everything besides where the 8x comes in
got stuck so i put it into a calculator
confuzzled
calm coralBOT
@broken rover Has your question been resolved?
broken rover
@exotic falcon ser!!
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Available help channel!
lone sparrow
calm coralBOT
lone sparrow
I believe my answer is wrong. can someone walk me through this?
pure kayak
on the second line, how did those e^x cancel?
lone sparrow
I divided them
pure kayak
how?
lone sparrow
I just did
pure kayak
$\frac{1}{e^x} \cdot \frac{1}{e^x} \neq 1$
potent lotusBOT
AℤØ
pure kayak
it would probably be easier to expand the numerator then split the fraction
remote mural
if a and b are real, then 1/(a+bi)=1/a + 1i/b
calm coralBOT
if a and b are real, then 1/(a+bi)=1/a + 1i/b
velvet osprey
no
remote mural
if a and b are real, then 1/(a+bi)=1/a + 1i/b
How,
velvet osprey
what if you take one of them as 0
remote mural
yeah false 1/(a+bi)=(a-bi)/(a²+b²) ?
velvet osprey
yeah false 1/(a+bi)=(a-bi)/(a²+b²) ?
parenthesis crimes!
remote mural
yeah false 1/(a+bi)=(a-bi)/(a²+b²) ?
Numerator too
velvet osprey
missing around a-bi as well
calm coralBOT
calm coralBOT
Available help channel!
shy iron
Why when using reverse chain rule. For the first one I wouldn’t write y=4x^3-5 but the bottom one I do?
calm coralBOT
Why when using reverse chain rule. For the first one I wouldn’t write y=4x^3-5 b...
@shy iron Has your question been resolved?
Available help channel!
neat gorge
I'm trying to understand fractions at a more intuitive level. The analogy I've "created" (I'm sure it's been used many times before) is one using measuring cups. It's the only time it's ever made sense to me.
For example:
18 / 1/2 makes sense when you think of volume. We can either keep pouring in half a liter into an 18 liter bucket and then tallying up how many times we added 1/2 liter to fill it up -- or, we can first combined two 1/2 liters to make a full liter, and then multiply it by 2 (since, for every "pour," we're adding 2 half liters). This gives us the multiplicative inverse 1/2 * 2/1 (i.e., two steps to get to 1 whole unit), and as such, if we multiply 18 times the multiplicative inverse of what we're dividing, we will know how many half liters we poured. 18 * 2/1
My head makes sense of this pretty easily, but for some reason I can't visualize and expand the analogy to less-easy fractions. I can easily think of 3/2 as "a bit more than a full unit" and I can think of 15/8 as being "a lot more than a full unit." Since if we added one more 8th, we'd have two full 8/8s. So, 15/8 must necessarily be really close to 2. Likewise, 48/7 is nearly 7.
My issue is visualizing something like 4/3 % 3/2. I know that I multiply it by the multiplicative inverse and end up with 4/3 * 3/2. And I also know that 2/3 * 3/2 = 1. And I know that 6/6 = 3/3; 2/2 = 80/80, and so on.
But continuing with the measuring cup metaphor:
Suppose we have a measuring cup which holds exactly 3 liters. We add 2 liters. We've now gotten ourselves 2/3 liters. If we wanted a full liter, we'd obviously just add another 1/3 liter. We'd have our 3/3 liters and be very happy.
neat gorge
But similarly, if we multiply 2/3 by 3/2, we end up with 6/6 = 1. In other words, multiplying 2/3 by 3/2 is the same as 2/3 + 1/3. In other words, 3/2 == 1/3. But this doesn't make intuitive sense to me in the analogy.
Since 3/2 is "a bit more than 1," how can it be that multiplying a bit more than one full unit by something already partially filled gives us a full unit of the same size?
If we have our 3 liter measuring cup and put in 2/3 liters. How can we multiply (repeatedly add) an entire 1.5 units without overflowing the cup? If we multiply both the numerator and denominator, the cup must get larger. Since the base is now going to be 6 instead of 3.
I'm sure this is incredibly stupid, but I'm just trying to understand and build conceptual knowledge. Full disclaimer, it's obvious that the math works and I'm not doubting it, I can do it 100/100 times, I'm just confused when thinking about it analogously.
If anyone does get around to helping, can you continue with the measuring cup analogy?
oblique current
well initially your analogy was for dividing a number by another number, but in your second message it looks like you're trying to use the same interpretation for a multiplication?
if it were (2/3) / (2/3), a division like you had before then your analogy would clearly tell you this is 1
calm coralBOT
@neat gorge Has your question been resolved?
clear delta
i think one of the issues here is that unlike with addition and subtraction, where you start with two things of the same unit type (volume, mass, time, etc) and end up with something of that same type, the result in multiplication is a different type than what you started with. If you're thinking of multiplying two measuring cups partially full of water and arriving at a new measuring cup, then yeah that doesn't make any physical sense.
but if you think about a measuring cup that is 2/3 of a liter, and filling it up one and a half times, then it might make more sense that you end up with exactly one liter. (the units here are liters / cup, and cups)
calm coralBOT
leaden junco
yo can someone explain about a common factor by grouping of terms??
leaden junco
ik is easy but i forgot it
astral mural
!status
calm coralBOT
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
!show
calm coralBOT
Show your work, and if possible, explain where you are stuck.
leaden junco
6ax + 3a +1 + 2x
leaden junco
```
What step are you on?
1. I don't know where to begin
2. I have begun but got...
1
<@&286206848099549185>
merry python
You can factor the first two terms. They both share a 3a
merry python
No
You're only factoring the first two terms
You're still adding the last two
So if I had a b + ac + d + e. I asked you to do a (b+c) + d + e
chilly gorge
e^-n is the same as 1/(e^n)
rare drum
Yup
chilly gorge
which you can write as (1/e)^n, since 1^n = 1
rare drum
Is this the only time its possible to move the exponent from denominator to cover the whole fraction, is when the numerator is 1?
leaden thunder
Is this the only time its possible to move the exponent from denominator to cove...
,tex .exp rules