[\wrb{\int \f 1{\s[n]x + \s[n+1]x} ;\dd x,; n\in\N}]
#help-33
hidden plaza
elfin berryBOT
Pure
hidden plaza
yeah
orchid oracle
actually let me write it down
wait
okay so u = x^{n*(n+1)}
then du = the derivative of x^{n*(n+1)}
but thats way too long to be correct
🧐
ill try to solve the original problem then
hidden plaza
[\wrb{\dv x x^{n(n+1)} = n(n+1)x^{n^2+n-1}}]
elfin berryBOT
Pure
hidden plaza
but yeah thats fine
elfin berryBOT
Pure
hidden plaza
Sure
elfin berryBOT
Pure
orchid oracle
so i substitute u again
ok wait
some steps later...
now do u subst again
v substitution*
yep this seems to be correct
thanks guys
marsh citrusBOT
@orchid oracle Has your question been resolved?
marsh citrusBOT
marsh citrusBOT
Available help channel!
unborn agate
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rapid cosmos
The sequence is periodic (order 3) and k=-2
a_n = 2-4-1+2-4-1+2-4-1........
For part c why do you need to also subtract the 4 at the end instead of just adding the 2?
rapid cosmos
i ended up with -76
marsh citrusBOT
@rapid cosmos Has your question been resolved?
upper gust
hello
upper gust
$\lim_{p \to \infty} \|f\|_p = \|f\|_{\infty}$
tobi
upper gust
I just proved this
where the p norm of f is the L^p norm (integral) and the infinity norm of f is the essential supremum
now my task is to figure out what happens when p goes to negative infinity
I am not sure how to takle this problem
marsh citrusBOT
@upper gust Has your question been resolved?
marsh citrusBOT
@upper gust Has your question been resolved?
sour iris
how do i know if there is an ambiguous case for law of sines, is there one for law of cosines, and how do i find the answers if there is an ambiguous case
marsh citrusBOT
how do i know if there is an ambiguous case for law of sines, is there one for l...
@sour iris Has your question been resolved?
sour iris
<@&286206848099549185>
marsh citrusBOT
@sour iris Has your question been resolved?
still temple
can someone help me with some of my missing geomotry assignemnts
marsh citrusBOT
can someone help me with some of my missing geomotry assignemnts
still temple
yes bro thank you thank you
torpid owl
No way.
marsh citrusBOT
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chilly island
Can someone explain how the substitution here works? I'm not 100% on how you can use a substitution of -2xdx to cancel it.
marsh citrusBOT
Can someone explain how the substitution here works? I'm not 100% on how you can...
elfin berryBOT
THAT'S MY QUANT, MY QUANTITATIVE
chilly island
why is the x left on the u side?
chilly island
ahhhh, so you solve for u in terms of x after substituting, then plug that into the x left in that equation?
lyric crypt
you choose a substitution such that there is no x
it's just dependent on u
we have
u = 4 - x^2
which is
chilly island
I understand how they got u=4-x^2, and how du=-2xdx, but I'm not sure why you left the x in the du part of the equation.
elfin berryBOT
THAT'S MY QUANT, MY QUANTITATIVE
eager dagger
If g(x) = -2(3.5^x)
X is -1
How do I do the base and -1 if it’s a negative ??? How do I solve that
marsh citrusBOT
If g(x) = -2(3.5^x)
X is -1
How do I do the base and -1 if it’s a negative ??? ...
quaint elm
you're trying to find g(-1)?
eager dagger
Yes
quaint elm
$a^{-1} = \f1a$
elfin berryBOT
hayley
quaint elm
eager dagger
I have one more question
In those equations what is the domain and range?
The exponent base or initial value
quaint elm
domain is all of the valid values that x can be
range is all of the possible outputs of the function
so like for $j(x) = \sqrt x$, the domain is ``all positive numbers and zero''
elfin berryBOT
hayley
quaint elm
bc you can't take the square root of a negative number
quaint elm
for exponents like this, no; every value of x is valid
however, the range is restricted
it may be helpful to graph some of these
eager dagger
Okay I will. That won’t be cheating right ?
quaint elm
i don't think so
but you should look at the similarities between them, and the similarities in the graphs
eager dagger
Okay I will thanks
sturdy dawn
rotund pike
marsh citrusBOT
marsh citrusBOT
Available help channel!
vernal basin
marsh citrusBOT
vernal basin
i don't get vandermonde determinant
where does the upper index (j index) go between the second and third lines?
also, why does the formula $\prod_{i,j \leq n} (x_j - x_i)$ not have upper indices as well?
elfin berryBOT
omgatriple
long osprey
they're not upper indices
vernal basin
is there a symmetry in the matrix which i don't get
long osprey
it's powers
azure egret
marsh citrusBOT
@azure egret Has your question been resolved?
stone quartz
how do I go about solving this?
marsh citrusBOT
how do I go about solving this?
stone quartz
<@&286206848099549185>
main idol
If the slope of a function is negative, then the function is
increasing or decreasing?
stone quartz
if slope is negative
stone quartz
If the slope of a function is negative, then the function is
increasing or dec...
decreasing
main idol
is f(x) decreasing or increasing between x = 0 and 1?
main idol
but the graph shown is f'
That's all you need to answer the questions
stone quartz
f(0) is greater cos the function is decreasing so naturally, f(1) will be smaller than f(0)
right?
main idol
Yes
stone quartz
understood
marsh citrusBOT
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stone quartz
<@&286206848099549185>
marsh citrusBOT
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stone quartz
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mental kayak
Bro
mental kayak
What is so important
About finding the change
Of a function
Limits are absolutely FIRE
Integrals are basically geometry hackers
But what derivatives do
Nuthin
So i want someone to explain
how derivatives useful at all
what does + C do
why not - C
or A+
lucid zenith
I would argue derivatives are the most useful
mental kayak
I would argue derivatives are the most useful
How
lucid zenith
the only reason you can calculate most integrals at all is because of derivatives
mental kayak
the only reason you can calculate most integrals at all is because of derivative...
Ok but like
how is finding the change of a function
lucid zenith
derivatives give u information about how a function changes
lucid zenith
just graph it bruh
most real life applications of derivatives actually occur in differential equations, where you don't have a function you can simply graph
mental kayak
most real life applications of derivatives actually occur in differential equati...
wdym
lucid zenith
differential equations are what define the world
fervent rampart
derivatives are one of the really basic building blocks for building all sorts of useful mathematics on top of them. asking "why should i know derivatives" is like asking "why should i know algebra", you need algebra to learn calculus in the first place
mental kayak
derivatives are one of the really basic building blocks for building all sorts o...
ok whatevs
lucid zenith
wdym
a differential equation is an equation that doesn't involve the function itself, but it's derivatives
mental kayak
a differential equation is an equation that doesn't involve the function itself,...
Take the integral
lucid zenith
...
lucid zenith
no, but your question doesn't make sense then
lucid zenith
integration cannot exist without differentiation
mental kayak
integration cannot exist without differentiation
How so
lucid zenith
because they are inverse operations
it's like saying why do we use subtraction when we can just add a negative number
it doesn't make sense
lucid zenith
but the theory that makes you think integration is more useful than differentiation came after
integration doesn't necessarily represent "the area under a curve"
mental kayak
Area under a line
marsh citrusBOT
@mental kayak Has your question been resolved?
smoky frigate
Hello. I've learned these two equations in class. However, I'm unsure what the difference is. When to use which one?
marsh citrusBOT
Hello. I've learned these two equations in class. However, I'm unsure what the d...
frail orbit
level surface can be expressed (most of the time as) f(x,y)
quaint elm
aren't they equivalent?
frail orbit
I think the second is a more general version of the first
like...an implict version?
smoky frigate
The only difference I see is the +f(a,b) at the first one.
That would make them different right?
quasi vessel
The second one subsumes the first one. You can pick F(x,y,z) = f(x,y) - z and then picking a tangent plane to the level surface F(x,y,z) = 0 at some (a,b, f(a,b)), you obtain the tangent plane as described in the first one
You may in certain cases derive the second equation from the first one but it requires so called "implicit function theorem"
smoky frigate
Hmm, does this "F(x,y,z) = f(x,y) - z" go in general? Also, what is the difference between f and capital F?
Hmmmm, so in the first one, it "F(x,y,z) = f(x,y) - z", with z = 0? As it isn't a variable in the function?
So basically then if I'm correct z = fz(x,y) - z + fx (x,y) - x + fy (x,y) -y ?
= fz(x,y) + fx (x,y) - x + fy (x,y) -y
quasi vessel
It's just a way to make the graph of function f which is a function over two inputs into level curve of a function with three variables
smoky frigate
But it's not that you juse use the first equation for a function of 2 variables, and the second one for a function of 3?
quasi vessel
Hmm, does this "F(x,y,z) = f(x,y) - z" go in general? Also, what is the differen...
I am saying that you can derive the first one from the second one by setting F(x,y,z) = f(x,y) - z and considering the level curve F^-1(0)
quasi vessel
But it's not that you juse use the first equation for a function of 2 variables,...
Both of them are still computing tangent planes but now for different kinds of surfaces both in R^3
devout mauve
they are the same thing, just one for functions R^2->R and the second for R^3->R
quasi vessel
I am saying that you can derive the first one from the second one by setting F(x...
The solution set for F= 0 where F is defined as f(x,y) - z is exactly the points (x, y, f(x,y))
smoky frigate
Okay, cheers. So to conclude, the first one can be used for just equations with two variables. And the second one is the general one, which can be used for all equations of 3 variables.
I hope I got this correct now.
I don't need the full understanding of this problem, but atleast need to know which to use when.
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woven forge
marsh citrusBOT
woven forge
how can this 2x2 matrix be set up
i know how to do the process of finding the steady state vector
but i keep getting 6/7 as my answer
which isnt an option
marsh citrusBOT
@woven forge Has your question been resolved?
alpine olive
Why are null spaces always linearly independent sets?
marsh citrusBOT
Why are null spaces always linearly independent sets?
echo bough
wdym?
alpine olive
if you explicitly describe the null space as the span of some vectors, is it always linearly independent?
The vectors in that span
The free variables would be the weights
devout mauve
well if you wanna express it as a span of some vectors, then usually you choose those vectors precisely in a way such that they are linearly independent
if you can for example express it as the span of v1,v2 then you can also express it as the span of v1,v2,3v1-17v2
but the second choice is linearly dependent so you usually wouldnt do that
alpine olive
Hmm ok yeah in my book it mentionned how its always linearly indep when we find the null space of a matrix, but when we get the column space it may not always be linearly indep, so we would have to remove some of them to get the basis
Anyways I think I capische
Ty
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marsh citrusBOT
marsh citrusBOT
Available help channel!
still temple
marsh citrusBOT
still temple
can someone explain why n suddenly becomes n^2 + 2n / n + 2?
dull gust
can someone explain why n suddenly becomes n^2 + 2n / n + 2?
The LCM is multiplied to n, i.e, n+2. Making it n(n+2)/n+2
dull gust
oh you can just do that?
Yes since the LCM in your case will n+2
marsh citrusBOT
@still temple Has your question been resolved?
sonic yarrow
how many different ways can 3 juniors and 3 seniors stand in a line if the 3 seniors must stand next to each other
marsh citrusBOT
how many different ways can 3 juniors and 3 seniors stand in a line if the 3 sen...
modest forge
3!*2! ?
modest forge
3!* 2!* 3! ? maybe
sonic yarrow
Why the second 3 factorial?
cloud iron
If the three seniors must be next to each other, try treating them as one person
sonic yarrow
I did
That gave me 4 different ways to put seniors next to each other, disregarding the different orders of seniors it could contain
cloud iron
The seniors can stand next to each other in 3! different ways
normal venture
Answer should be 4!×3!
sonic yarrow
Answer should be 4!×3!
How?
cloud iron
And the juniors can as well, so so far we have 3! * 3!
If you treat the three seniors like one person, they can be in one of four spots among the juniors, so * 4
3! * 4 is 4!, so we have 4! * 3!
sonic yarrow
Oh
normal venture
How?
Because if we treat seniors as one single group we can arrange it in 4! Ways then 3 seniors can arrange themselves in 3! Ways
cloud iron
That leads to an interesting question I have
sonic yarrow
Yes?
cloud iron
Is it coincidence that 3! * 4! is (3*4)^2?
sonic yarrow
Hmmm
cloud iron
Doesn't work with 4 and 5, not sure what other pairs it does or doesn't work with
Just interesting thing I noticed there
cloud iron
That's kinda cool
deft matrix
Hi, im a little confused on how to set up the equation here. Any help or advice would be appreciated
marsh citrusBOT
Hi, im a little confused on how to set up the equation here. Any help or advice ...
bleak ibex
$\pi \int_{a}^{b}f(x)^2,dx$, right?
elfin berryBOT
Jelle
marsh citrusBOT
Available help channel!
warped moss
marsh citrusBOT
warped moss
Hello
I'm trying to solve this
the answer u see in the pic is the right
one
but i dont know how to get there
I know that $L(x) = f'(a)(x-a) + f(a)$
warped moss
in this case f is g
nova totem
a = 2
warped moss
yea
i have trouble with the derivative
its 1 i think
derivative at 2
is 1
so then
$L(x) = 1(x-2) + 10$
warped moss
so $L(x) = x+8$
warped moss
but thats not the right answer
pls help
opps i didnt do chain rule
.clsoe
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stuck marsh
marsh citrusBOT
stuck marsh
hello, if b = (4, 3, 3, 1), how do i solve Ax = b using PLU decomposition?
or QR, or cholesky?
<@&286206848099549185>
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pseudo viper
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serene scarab
how would I prove that the only critical point is (0, 0) of the following system:
marsh citrusBOT
how would I prove that the only critical point is (0, 0) of the following system...
serene scarab
The system of differential equations $\vec{x}' = A\vec{x}$, where A = $\bigl( \begin{smallmatrix}a & -b\ b & a\end{smallmatrix}\bigr)$ and $a, b > 0$
serene scarab
is this correct?
or is there a tool that lets me know if its correct?
<@&286206848099549185> ?
marsh citrusBOT
@serene scarab Has your question been resolved?
solemn loom
is this 1 or 2
marsh citrusBOT
is this 1 or 2
copper raven
solemn loom
ok yea i did (A - 3I) and then found the RREF and found that the bottom 2 rows were all zeros
marsh citrusBOT
@solemn loom Has your question been resolved?
solemn loom
Would PD^100P^-1 be a valid approach for this?
marsh citrusBOT
Would PD^100P^-1 be a valid approach for this?
stoic saddle
indeed it would
analog sun
How can I get the last term (marked with red) through the previous term?
marsh citrusBOT
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analog sun
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surreal hull
Investigate whether the relation R⊂C×C defined by the relationship
z1Rz2⇐⇒ℜez1=ℜez2,
where ℜez denotes the real number part of z, is an equivalence relation. If so, determine the equivalence classes relative to this relation
marsh citrusBOT
> Investigate whether the relation R⊂C×C defined by the relationship
> z1Rz2⇐⇒ℜ...
surreal hull
no idea how to do this
marsh citrusBOT
@surreal hull Has your question been resolved?
forest kayak
So what is an equivalence relation, first of all
forest kayak
@surreal hull
marsh citrusBOT
@surreal hull Has your question been resolved?
solar heron
Hello @surreal hull did you still need some help?
surreal hull
Hello <@981586502154866708> did you still need some help?
yep
surreal hull
okay
surreal hull
<:FubukiThis:939197591361581096>
equivalecne relation
is
a relation
for which
3 other relations are true
yyyy
reflexivity, transitivity
and
symmetricity
yes, it ought to be symmetrciity
soo, we have a relation between
z_1 and z_2
which is equivalent to
the real number part of z_1 being equal to the real number part of z_2
to determine whether it's an equivalence relation
we see that
- it is reflexive because
ℜez1=ℜez1,
- it is symmetric because ℜez1=ℜez2 <=> ℜez2=ℜez1,
- it is transcitive because
ℜez1=ℜez2, ℜez2=ℜez3 -> ℜez1=ℜez3
so it's an equivalence relation
but idk how to do determine equivalence classes 😦
glass silo
Well, of course, the equivalence classes are basically going to be formed of complex numbers that have the same real part
Can you think of how they'd look like, if you sketched them out on an Argand diagram or something?
surreal hull
no idea what that is
glass silo
if you sketched out every complex number with the real parts corresponding to the x coordinate and the imaginary parts the y coordinates
forest kayak
Okay so
There is a definition
Let's go back to that alright ? @surreal hull
I just translated the equivalence relation from its definition
Now these z' will be all of the complex numbers
Something good would be to express z' wrt what you have already
marsh citrusBOT
@surreal hull Has your question been resolved?
solar heron
I believe they would be vertical lines in the Re/Im plane.
narrow token
Is my ans right?
marsh citrusBOT
Is my ans right?
narrow token
rapid cosmos
If u wrote y= 5x+1 then yea
marsh citrusBOT
@narrow token Has your question been resolved?
marsh citrusBOT
native frigate
its looking like a quotient rule/product rule to me
how are you going to take the derivative of (x^2 - 1)(x^2 + x + 1)?
oh you could do that too
id just use product rule personally but same difference
yes
i have no idea what this means, sorry
why did you do division?
do you know what the product rule is?
its not just multiplying the derivaties together?
and you cant do it for division like that
okay
so you still have a question?
you use product rule when you take the derivative of just the bottom part
you have to use it within quotient rule
twilit grove
I guess if you multiply everything in the denominator, maybe
because they are in the denominator. you can apply the product rule with 1 / (x^2 + 1) and 1 / (x^3 - x + 1) if you want
twilit grove
1 plus 1?
<@&268886789983436800> trollspam
what would be?
the derivative of that is not 0/2x. you need to review the quotient rule
let f(x) = 1 / (x^2 + 1), and g(x) = 1 / (x^3 - x + 1)
marsh citrusBOT
@still temple Has your question been resolved?
cerulean oxide
so
cerulean oxide
i am asked to show that v * (v x x) = 0
that if you take the cross product of v and x and times that by the same vector v you get 0
this is what i've done, but it doesnt add up
long osprey
you have a3b3 in the first coordinate of your very first calculation
instead of a3b2
cerulean oxide
nice
long osprey
oh that's you writing the definition
cerulean oxide
thanks for telling me
cerulean oxide
someone explain this to me like i am a child:
Show that the composition of orientation-perserving linear maps is orientation perserving.
marsh citrusBOT
someone explain this to me like i am a child:
Show that the composition of orien...
cerulean oxide
I am not sure what it means
sullen ice
im not entirely sure what most of that means
but it sounds like its saying applying orientation perserving linear maps on to each other creates another orientation perserving linear map that is still orientation perserving
thats a huge shot in the dark though
only word i recognise there really is composition
long osprey
can you give your definition of orientation preserving?
long osprey
oh
cerulean oxide
also something to do with its direction
long osprey
were you not given a definition in notes, class, lectures?
cerulean oxide
yh but you're supposed to get an intuition for why this works
long osprey
oh
cerulean oxide
not that theroem 459 tells me its true
long osprey
in that case: if you have a map that keeps right handed right handed, and then do another map that does the same thing, overall you've still kept handedness the same (and then maybe another, and another, ...)
is what it's asking you to show
cerulean oxide
the matrix of the composition of the linear maps is the product of the two matrices
long osprey
yes
marsh citrusBOT
@cerulean oxide Has your question been resolved?
raw crown
what 😭
raw crown
for 1 i was going to start like suppose y is an element of A
since A is a subset of S then
yeah no i have no idea
brave marsh
Go with the definitions. If x is in A, then what can you say about f(x)?
raw crown
its in f(A) ?
raw crown
if i use preimage do the fs like
cancel out
so then ill get x is in f'(f(A))
which is also A
brave marsh
It's just the definiton though right?
if f(x) is in f(A), then by definition x is in f-1(f(A))
Because x is mapped inside f(A).
raw crown
oh okay
brave marsh
f(x) is in f(A)
so x is in f-1(f(A))
It's just convoluted because of the f(A), but you could let f(A) = C and you would have the same
f(x) in C, so x in f-1(C).
raw crown
when u say it like that it sounds so simple but the way she explained it in class was so
overwhelming
raw crown
so just to double check
preimage: if f(x) is in something then x is in the invers
image: if x is in y then f(x) is in f(y)
whats the whole thing with the equal sign in the first definition
brave marsh
y is in f(A) iff there is some x such that f(x) = y.
In other words, the elements of f(A) are those that mapped to by f
You could imagine that some functions might not map to every element of their codomain (think x^2 : R -> R, it only maps to positive R)
raw crown
so basically its like x is in A so then f(x) is in f(A)
and then f(x) is just y
thank you sm 😭
im like in love w u or something 😭
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still temple
Unsure of how this works
marsh citrusBOT
Unsure of how this works
dim schooner
answer is -3 <= x < -2 or x>2
marsh citrusBOT
answer is -3 <= x < -2 or x>2
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.
lilac siren
Unsure of how this works
how'd you start
dim schooner
dono just taped it in my algebric calculator
lilac siren
@still temple are thee there
dim schooner
i guest your supposed to multiply ( 1 / x+2 ) on both side so that you can resolve on one side only
lilac siren
i guest your supposed to multiply ( 1 / x+2 ) on both side so that you can ...
nah you'd solve for x case-wise, yielding different inequalities
vagrant trail
You should solve linear inequality, also take into account where denominator is zero
lilac siren
thx for an attempt, but on one side pls no direct proof solutions and on the other you get wrong cases
dim schooner
nice
lilac siren
thx for an attempt, but on one side pls no direct proof solutions and on the oth...
since multiplying by x-2 for instance doesn't necessarily preserve inequality
likewise for x+2
still temple
<@456226577798135808> are thee there
yes
lilac siren
yes
k, you'd start by removing the fractions, meaning you can first multiply by (x-2)
however careful, at this point you already need to split into cases
since if (x-2) is positive, inequality holds
if (x-2) is negative, inequality flips
still temple
so you would multiply the 5 by (x+2) and the 1 by (x-2)
lilac siren
meaning
for x>2: 5 >= (x-2)/(x+2)
for x<2: 5 <= (x-2)/(x+2)
still temple
wouldnt you get to a point where it is basically 5x+10-x+2 is greater than or equal to 0
wait hold on
lilac siren
wouldnt you get to a point where it is basically 5x+10-x+2 is greater than or eq...
hm, not sure whether that is true
lilac siren
meaning
for x>2: 5 >= (x-2)/(x+2)
for x<2: 5 <= (x-2)/(x+2)
but this part is clear right
the case-splitting is important when solving inequalities
likewise if you now multiply by x+2
you again get two cases: x+2<0 and x+2>0
meaning
for x>2 and x<-2: 5*(x+2) <= (x-2)
for x>2 and x>-2: 5*(x+2) >= (x-2)
for x<2 and x<-2: 5*(x+2) >= (x-2)
for x<2 and x>-2: 5*(x+2) <= (x-2)
however here you can quickly see that some cases become redundant: x>2 and x<-2 isn't possible
and x>2 and x>-2 is the same as x>-2
lilac siren
once i have gotten to x is greater than or equal to -3
that's correct
but doesn't exclude wrong cases
it only limits the right ones
for instance x = -1 is wrong, yet -1 >= -3
still temple
but doesn't exclude wrong cases
how do i exclude wrong cases then
lilac siren
meaning
```
for x>2 and x<-2: 5*(x+2) <= (x-2)
for x>2 and x>-2: 5*(x+2)...
for x>2: 5*(x+2) >= (x-2)
for x<-2: 5*(x+2) >= (x-2)
for -2<x<2: 5*(x+2) <= (x-2)
for x>2: 4x+12 >= 0
for x<-2: 4x+12 >= 0
for -2<x<2: 4x+12 <= 0
for x>2: x >= -3
for x<-2: x >= -3
for -2<x<2: x <= -3
first inequality holds for any x>2
second inequality holds for any -3<=x<-2
third inequality is impossible
tada, that's how you solve for x
if any steps are unclear mention
lilac siren
shall we go through from the start once more?
still temple
```
for x>2: 5*(x+2) >= (x-2)
for x<-2: 5*(x+2) >= (x-2)
for -2<x<2...
how can the greater than/less than symbols differ over here despite the fact that the equations remain unchanged
lilac siren
how can the greater than/less than symbols differ over here despite the fact tha...
because they can differ case-wise
that's why you can't simply modify the inequality and get one equation
still temple
case wise meaning?
lilac siren
for instance:
5/x > 3
if you multiply by x here
then x can either be negative or positive
if it's positive, then the sign remains unchanged: 5 > 3x
however if x is negative, then the sign flips
because if you multiply both sides by a negative number, the greater side becomes the smaller side
5 < 3x
still temple
wait so
lilac siren
meaning you now have two different inequalities for two different cases
still temple
i have found the critical x values
lilac siren
an inequality can have any number of cases
still temple
x=2, x=-2, x=-3
how would i go about testing these?
just plug the numbers in?
and see if it works?
still temple
theres a thing that says tests on the side
still temple
but it asks for it
lilac siren
test that they are critical?
k
so we have
for x = 2, left side is undefined
==> critical
for x = -2, right side is undefined
==> critical
for x = -3, you receive 5/(-5) >= 1/(-3+2) which is -1 >= -1
same values on both sides
==> critical
still temple
so those are your proofs
lilac siren
yeah, that's how you'd show they're critical
still temple
so all im doing is just plugging in
lilac siren
yes
still temple
2, -2, and -3
lilac siren
a value for an inequality is critical if one of two cases occur:
still temple
idk though on the answer key numbers like
lilac siren
the inequality is undefined or holds equality.
lilac siren
-4, 0, -2.5, 3?
if we plug in -4, we get -5/6 >= -1/2
so it's not a critical value
still temple
so what, are we just proving that those randomly selected values are not critical values?
lilac siren
ofc not, plugging in -4, 0 and -2.5 is useless
the only purpose it could serve
is that it would show the inequality is defined on all intervals except {-2, 2}
but that should be obvious from the start
therefore testing these values is rather redundant
you only need to test -3, -2 and 2 for showing critical values
but are the inequalities from before using cases clearer now or should we go through them? @still temple
still temple
but are the inequalities from before using cases clearer now or should we go thr...
well if you are talking about finding critical values, i understand that
but i dont understand the test part
or the solutions
lilac siren
ok so the inequalities one receives were:
which are -3 <= x < -2 and x > 2
what that means is that
lilac siren
do you mean from here to -3 <= x < -2 and x > 2?
or showing the inequalities in the first place?
still temple
from here
still temple
yes but how do you even get those lines
lilac siren
via case-wise solving for x
still temple
solving for x i only found that x >= 3
lilac siren
let's just do the case-wise solving again, then it should become clearer
so we start with
and want to solve for x. as you did, we can start by multiplying both sides by one of the denominators
still temple
yes
lilac siren
so let's choose to start with multiplying by x-2
still temple
5(x+2) >= x-2
lilac siren
5(x+2) >= x-2
now here it's already crucial that you don't do that
because then you ignore all cases
still temple
so you only multiply by one at a time?
lilac siren
so back to multiplying by x-2. If x-2 is positive, then we multiply both sides by a positive number, meaning the inequality remains the same
so IF x-2 > 0: then 5 >= (x-2)/(x+2)
and IF x-2 < 0: then 5 <= (x-2)/(x+2) (since if x-2 is negative, the sign flips)
that part needs to become clear
following so far?
still temple
yeah sort of
lilac siren
this splitting into cases means that we can already end up with two inequalities as our solutions
that's the weird part about inequalities
still temple
wait so just to be clear
so at first i multiply by (x+2)
then separately
by (x-2)
lilac siren
you don't necessarily just get something in the form of x >= ..., but instead possibly ... <= x < ... OR ... < x OR ... < x < ...
still temple
and get two different inequalities?
lilac siren
so at first i multiply by (x+2)
nono so far we've only multiplied by x-2
we haven't multiplied by x+2 yet
only x-2
that very multiplication by x-2 is what splits it into two cases already
because you either multiply by a negative or by a positive number
lilac siren
no, the left side becomes 5
since 5/(x-2) * (x-2) is 5
and right side becomes 1/(x+2) * (x-2) = (x-2)/(x+2)
here if you prefer latex
that is the crucial part
if you got that then you can apply this principle repeatedly
and get all solutions for x.
does something remain uncertain about this operation?
still temple
this is what the key says
idk if the methods used are the same or not
it looks like the key does something different
lilac siren
they come to the same conclusion, it's a slightly different approach
instead of splitting into cases they refactor the fractions
to get (4x+12) / (x+2)*(x-2) >= 0
and then in order for that to be >= 0, either the top and bottom are positive
or the top and bottom of the fraction are negative
but that also yields cases
either way, you receive cases, the solution doesn't properly highlight that though
still temple
i have no idea where 4x+12 comes from though
lilac siren
let's stick to the first solution approach then I can show you the one they've shown as well
but I'd recommend writing it more elaborately than how it's shown there because a few mid-steps are missing in the key
ah and btw the tests section doesn't refer to testing critical values
testing critical values is just insertion
the tests refer to testing the defined intervals of the inequality
that's why they inserted these seemingly arbitrary numbers
for each interval they just chose some random number within that interval
-4 is in (-inf., -3)
-2.5 is in (-3, -2)
0 is in (-2, 2)
and 3 is in (2, inf.)
lilac siren
ok, but firstly back to showing the inequalities, is that part clear? @still temple
still temple
yeah i guess so
lilac siren
ok. because cases can become hard to keep track off, it makes sense to write all cases under each other to not forget some
so after multiplying by x-2 we get:
for x>2: 5 >= (x-2)/(x+2)
for x<2: 5 <= (x-2)/(x+2)
for x>2 is the same as if x>2 btw.
lilac siren
these two lines are just the cases from the image
since IF x-2 > 0 is the same as IF x > 2
and IF x-2 < 0 is the same as IF x < 2
lilac siren
currently there's no 3
still temple
earlier one of the x values was -3
lilac siren
now we can multiply by x+2
lilac siren
earlier one of the x values was -3
well we still need to solve for x
so far we've only multiplied by x-2, we haven't yet solved for x in each case
now, if we multiply by x+2, again, that can either be positive or negative
since we multiply by x+2 in both current cases, these two cases again split into two new cases each
one where x+2 > 0 and one where x+2 < 0
if x+2 > 0, then the sign remains unchanged
if x+2 < 0, then the sign flips
meaning we get:
for x>2 and x<-2: 5*(x+2) <= (x-2)
for x>2 and x>-2: 5*(x+2) >= (x-2)
for x<2 and x<-2: 5*(x+2) >= (x-2)
for x<2 and x>-2: 5*(x+2) <= (x-2)
ok, and after that?
lilac siren
now we can solve for x in each case
lilac siren
yes, let's look at the uppermost
5*(x+2) <= x-2
is the same as
5x+10 <= x-2
is the same as
4x+12 <= 0
is the same as
x+3 <= 0
is the same as
x <= -3
that's where the 3 comes in
now if we solve for x in each case, you get:
for x>2 and x<-2: x <= -3
for x>2 and x>-2: x >= -3
for x<2 and x<-2: x >= -3
for x<2 and x>-2: x <= -3
that part clear?
still temple
yes
lilac siren
k, now since we've solved for x in each case, these are our solutions
now we only need to clean them up a bit
as they're a little bloated
and some are redundant
the first case says x > 2 AND x < -2 AND x <= -3
however x can't be greater than 2 and smaller than -2
so that case falls out
for x>2 and x>-2: x >= -3
for x<2 and x<-2: x >= -3
for x<2 and x>-2: x <= -3
the second case says x > 2 AND x > -2 and x >= -3
which is the same as x > 2
x > 2
for x<2 and x<-2: x >= -3
for x<2 and x>-2: x <= -3
the third case says x < 2 AND x < -2 and x >= -3
which is the same as -3 <= x < -2
x > 2
-3 <= x < -2
for x<2 and x>-2: x <= -3
the last case says x < 2 AND x > -2 AND x <= -3
but that's also not possible, since x can't be smaller than -3 and greater than 2
x > 2
-3 <= x < -2
🦇
these are the solution
still temple
ok ok that matches with the answer key
lilac siren
this approach works always
lilac siren
you can always split an inequality into cases this way
regardless of whether you have roots, logarithms, etc.
you'll always get the correct answers
lilac siren
do u know how the other one works
yes, I'll do that one in latex to show how they change the inequality, one sec
still temple
yes, I'll do that one in latex to show how they change the inequality, one sec
ok thanks
lilac siren
the first step they do is alter the two fractions
by expanding them
the fractions themselves remain unchanged, since the numerator and denominator are multiplied by the same values
on the left in the fraction's top and bottom are multiplied by x+2
and on the right the fraction's top and bottom are multiplied by x-2
the intent behind this is that now both fractions have the same denominator
both have (x-2)*(x+2)
still temple
so LCD is (x+2)(x-2)
lilac siren
which becomes useful afterwards
lilac siren
so LCD is (x+2)(x-2)
ys
next step:
just subtracting the right fraction
on both sides
now, since both fractions have the same denominator
you can combine them
by subtracting their numerators:
and now they simplify the numerator:
still temple
ok i understand that part
lilac siren
ok, now in order for that to be >= 0
what must be the case such that the fraction is >= 0?
like what condition must the numerator and denominator fulfill
still temple
one of them must equal 0
lilac siren
hm not quite
if the denominator is 0, then the fraction is undefined
as you divide by 0
they must fulfill that either top & bottom are positive or top & bottom are negative
because +/+ = +
and -/- = +
where "+" = positive and "-" = negative
still temple
but the + or - solution may not be 0
lilac siren
exactly, that's why I asked for >= 0
meaning when is the fraction positive
that's the tricky part of this second approach
seeing that either top & bottom are positive
or both are negative
which means, as before, we get 2 cases 🥳
still temple
oh sorry i didnt see the =
lilac siren
oh sorry i didnt see the =
, np
lilac siren
no worries :)
so if we write the two cases out:
4x+12 >= 0 AND (x-2)*(x+2) > 0
4x+12 <= 0 AND (x-2)*(x+2) < 0
for the denominator it's > 0 instead of >= 0, since if it's 0 we divide by 0
likewise < 0 instead of <= 0 in line 2
now that we got these two cases
we can solve for x again
x >= -3 AND x² > 4
x <= -3 AND x² < 4
that part that makes this approach less appealing is that we now have a quadratic term x²
so far clear?
(x-2)*(x+2) = x²-4 btw
still temple
im not following
lilac siren
k which part made you unfollow
still temple
you have 4x+12/(x-2)(x+2) >= 0
lilac siren
yes
still temple
and you want a value equal or greater than 0
and you want to find x
how do you find x with that inequality
lilac siren
yes, now that fraction can only be >= 0 if the fraction has the form +/+ or -/- as said before
which means you get two cases:
- the fraction is +/+
- the fraction is -/-
still temple
ok understood
lilac siren
that's what I've written out here:
on the top, both terms are positive (+)
on the bottom, both terms are negative (-)
lilac siren
well now we again have cases which we can try to directly solve for x
in the first line, if you look at the left
4x+12 >= 0 is the same as x+3 >= 0 is the same as x >= -3
and on the left bottom that term becomes x <= -3
x >= -3 AND (x-2)*(x+2) > 0
x <= -3 AND (x-2)*(x+2) < 0
alright i understand that
lilac siren
k and now the two inequalities on the right
let's multiply out: (x-2)*(x+2)
= x² + 2x - 2x - 4 = x²-4
lilac siren
so x² > 4 and x² < 4
for the two lines
x >= -3 AND x² > 4
x <= -3 AND x² < 4
now, the bottom case is already impossible
because if x <= -3
then the square is >= 9
since (-3)² = 9
(-4)² = 16
(-5)² = 25
etc.
it strictly grows
still temple
ok
lilac siren
which means we can eliminate the bottom case
and only have
x >= -3 AND x² > 4
left side is solved for x, though how about the right
now we can apply another new rule for cases regarding inequalities:
what I've shown you before is that if you multiply both sides
you split into two cases
now the new rule is that if you take the root on both sides
you also split into two cases
never never never just do this:
x² > 4 --> x > 2
because it ignores the second case
still temple
ok
lilac siren
when taking the root of an inequality, the sign doesn't flip for the positive case and flips for the negative case
what do I mean by negative and positive case here? Well if you look at x² = 4 for example
then x is either 2 or -2
2² = 4 = (-2)²
here the positive case would be x = 2 and the negative case x = -2
but we have an inequality
so we get:
x² > 4 ---> x > 2 OR x < -2
still temple
got it
lilac siren
`x² > 4` ---> `x > 2` OR `x < -2`
ok that means we, ...again..., get two cases
so that case splits into two
x >= -3 AND x > 2
x >= -3 AND x < -2
now the upper one can be simplified to x > 2
x > 2
x >= -3 AND x < -2
🦇
same solution as before
still temple
alright
lilac siren
but to be frank more tedious
lilac siren
np, is the part with the tests clear as well or should I briefly go over it too
still temple
i understand that for the most part
lilac siren
k, just to generalise that section:
once you have any critical values
in this case -3, -2 and 2
you split all real numbers into different intervals
from -inf. to -3
from -3 to -2
from -2 to 2
and from 2 to inf.
and theoretically you could also prove (or test) whether these intervals are even defined
you'd do so by putting in a value
for each of those intervals
for the first interval for instance
(-inf., -3)
we can e.g. choose -4
because -4 is in that interval
plug that into the inequality and look whether that throws any mathematical errors