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[slow canopy]

hey all, I understand why |Xn+1| = 1 + |X1| + ... + |Xn|

but I don't know how to prove it

because the number of compositions of any given number equal the sum of the number of compositions of all lower positive integers + 1

like this

[charred flint]

you'll have to rephrase the compositions of n+1 into things with a size of |X1|, |X2|, ...

[slow canopy]

I'm not sure I understand

[charred flint]

like "composition of n+1 is the same as this, this, this, and this, which is 1, |x1|, |x2|,.... |xn| large

anyways let's see if we can find compositions of smaller integers in a bigger one

here is it for n=5

and for n=3

[slow canopy]

I know that the total number of compositions = 2^n-1

[charred flint]

and I can sort of see the composition of 3 within the 5's

[slow canopy]

Since my main problem here is pretty much total ignorance on how to structure a formal proof for this, could you elaborate on the "this, this, this, and this" part

[charred flint]

yeah, I'm trying to motivate what the this's are first

[slow canopy]

oh gotcha

yeah I see the 3-compositions nestled in the 5-compositions

[charred flint]

so you see you can split 5 into 2+(coposition of 3)

and logically you could do 3+(composition of 2), and other stuff like that

so we can split up 5's composition according to the starting number

in general, a composition of (n+1) will look like k plus (composition of n-k+1)

since the numbers have to add to the total

does that make sense? then we can start the formal proof

[slow canopy]

yeah I'm mostly following

[charred flint]

cool, so for every composition of 5, we can assign a lower composition, depending on the starting number

every lower composition also has it's own composition in 5, since you just add (n-k+1) in front and you'll get n+1 total

and that all means that |Xn+1| = |X1|+|X2|+...+|Xn|

(we forgot the 1 though, and that's when the composition is just the number itself, so there's no lower composition that shows up)

[slow canopy]

this part confuses me a bit

don't we do k + (n-k+1)

to get n+1

[charred flint]

oh I mixed up what k means yeah

if k is the first number in the composition sum, the lower composition is of (n-k+1) yeah

you're right

[slow canopy]

ok so how do you get from that

[slow canopy]

to this

also why did we leave out the 1

if that's what we're trying to prove

[charred flint]

yeah you have to explain that 1 needs to be there too

from the case that's left out, when (n+1) = (n+1)

this is a proof about a bijection between two sets

[charred flint]

this means that [compositions of n+1] is less than [sum of lower compositions + 1]

[charred flint]

and this means that [sum of lower compositions+1] is less than [compositions of n+1]

so the sets have equal size

I probably should've said "less than or equal"

[slow canopy]

aren't they always equal?

[charred flint]

I meant less than or equal

if you know about injections and surjections that's what's happening here

[slow canopy]

Kind of yeah

onto and 1-to-1

[charred flint]

like in general these proofs about equal sizes go "here's how to get from A to B, so A<=B, and here's how to get from B to A, so B<=A"

[slow canopy]

is that how I should go about this proof?

[charred flint]

yeah, assuming this is a discrete math class that needs formal proofs

[slow canopy]

I feel like I'm kind of asking for an "answer" here, but more like asking for an outline of how to structure the answer... I have the problem written down, and "PROOF:" below it, but I don't know what to literally write to get started

but yeah looking for a formal proof

[slow canopy]

Typically in formal proofs I couldn't just take 5 and use it as an example and assume that it works in all cases

[charred flint]

yeah the outline is
"We'll show that the # of compositions of n+1 is equal to the # of lower compositions plus one"
show how a composition of n+1 can be associated with a lower composition
show how a lower composition can be associated with a composition of n+1

right all the stuff with 5 and 3 won't show up in the proof

it's just the work you have to do to come up with the strategy

[slow canopy]

thank you, this helps

[lone heart]

@slow canopy Has your question been resolved?

[slow canopy]

still working

[slow canopy]

I think I got this down

how about part b?

is that something you can help me with?

for ease:

[charred flint]

sure, how are you seeing it

[slow canopy]

seeing it?

sorry had a phone call

[slow canopy]

since |Xn| is just the number of compositions of n, I get that it equals 2^n-1

not sure how to use induction to prove this

[charred flint]

that's what you're trying to prove

so with induction you show the base case and that n+1 works using n

[slow canopy]

I'll add that I know how inductive proofs work, prove it for n, then prove it for n+1 basically

[charred flint]

yeah so base case is easy and for the inductive step we need to show |Xn+1|=2^n

[slow canopy]

I actually don't know how to start with base case for this

[charred flint]

first step is to use (a), |Xn+1| = 1+|X1|+...+|Xn|

then you can use the inductive hypothesis and it'll work out

so the base case is n=1

|Xn| = 2^(1-1) = 1

to prove that you just write out all the compositions of 1

[slow canopy]

kk got the base case down

[charred flint]

inductive step is show |Xn+1| = 2^n

you're basically allowed to use these 3 facts:
formula for |Xk| from (a)
formula for |Xk+1| from (a)
|Xk|=2^k-1 (inductive hypothesis)

[slow canopy]

so I might be a little rustier with how to structure these inductive proofs than I thought

don't I need to replace n with k at some point

[charred flint]

oh I guess people usually use k in the proof itself, I'll just replace the n's

[slow canopy]

could you walk me through the inductive step a little more

[charred flint]

in general or for this problem

[slow canopy]

for this problem

[charred flint]

so you start with |Xk+1| = 1+|X1|+...+|Xk-1|+|Xk|

you can simplify this a lot because the formula for |Xk| is basically the same thing

do you see how or should I keep going

[slow canopy]

You should keep going if you don't mind

[charred flint]

|Xk| is the same except the last term

so plug that in to get |Xk+1| = |Xk|+|Xk|

[slow canopy]

the plugging in process

[slow canopy]

ok

so I get why |Xk+1| = |Xk| + |Xk|

not sure where to go from here

[charred flint]

we used 2 of these, all that's left is the inductive hypothesis

use that and it'll be finished

[slow canopy]

hmm

so isn't that saying that

|Xk+1| = 2^k-1 + 2^k-1

[charred flint]

yup

[slow canopy]

but that's not what I'm looking for right

[charred flint]

well a + instead of *

it's what you want

[slow canopy]

so why is that what I want to show in the inductive step?

[charred flint]

that the theorem or whatever works for k+1

here it's |Xk+1| = 2^k

[slow canopy]

2^k = 2^k-1 + 2^k-1 ?

[charred flint]

yeah

you're adding the same thing right, so it's two of that thing

2* 2^k-1

and you combine exponents to make it 2^k

[slow canopy]

hmm ok

sorry, my algebra is waaaaay rusty

alright I got it, thanks!

.close

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[limber lava]

how tf do I solve for t here:

[crisp grove]

by noting that denominator is not zero

[limber lava]

?

stupid high school math making life hard 😠

[alpine sable]

🙃

[limber lava]

so how do I solve it

Pls Help 🙏

🪄

Wtf is this answer sheet

[gleaming granite]

to simplify try asking yourself what will make the left side of the equation equal 0?

[lone heart]

@limber lava Has your question been resolved?

[teal turtle]

Multiply both sides by t(t-1), noting that t(t-1) isn't equal to zero because your denominator isnt zero.

You'll get 2(1 - 2t) = 0. From there you know 1 - 2t = 0 and you should be able to see what t is

[lone heart]

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[digital cloud]

hi, show that there exists a nonzero differentiable function f such that xf'(x) = 5f(x)

[alpine sable]

$f(x)=(x^0+5)e^x$

[ocean seal]

hvna

[alpine sable]

This works 😄

I'm sure that's not what you're looking for though

[digital cloud]

wait wuh.

its supposed to be like a proof

[alpine sable]

I know

I'm messing around

This is wrong anyway

It doesn't work

[digital cloud]

oh lmao

[alpine sable]

I just always throw e^x at these problems

And it usually works

I know some professors do not like that though

It's a trivial solution

[digital cloud]

does it actually work tho thats pretty funny

[alpine sable]

It doesn't because f'(x) isn't xf'(x)

That's an even more trivial solution

[digital cloud]

anyways lol um i think im supposed to prove logical equivalances but idk

[teal turtle]

Logical equivalences?

[digital cloud]

its like to prove 'p if and only q' we need to prove p->q and q->p

or like how else would u do it

coz im lost

[teal turtle]

But to show there exists a function f you can just think of a random solutuon right? (Or find a general solution if you need to look for that)

[digital cloud]

yea a general solution

wait actually yea maybe i shhould just input a value that works

sike figured it out its x^5

.close

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[quaint pine]

how du approach a problem like this

[quaint pine]

i have bad handwriting so if u don’t understand just ask

[alpine sable]

You can just 'simplify' the limit

[ocean seal]

hvna

hvna

[alpine sable]

Because the constants in the numerator and denominator aren't going to have an effect as x -> infinity

[quaint pine]

that makes so much sense ty so much ❤️

.close

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[lost dew]

why 0! = 1 since isnt ! same as multiplying all numbers till the one spesified?

[ebon condor]

I mean you can make an argument

N! = prod 1 to n

And going backwards

[lost dew]

oddly 1! = 1 also

[alpine sable]

The factorial also represents the number of ways that a specific data can be arranged. And since 0! can only be arranged one way ->>

[ebon condor]

N-1! = n!/n

Do you agree?

[lost dew]

that seems reasonable

[ebon condor]

So 0! =1!/1

[lost dew]

seems reasonable

[ebon condor]

It’s hard to even extrapolate upwards if 0! Was equal to 0

Cause the product of n times 0 is…

[lost dew]

yeah it just seems illogical that both 0 and 1 ! are 1

[ebon condor]

Well, there is no guarantee that it’s one to one

A lot of functions aren’t one to one

[lost dew]

isnt this exponential, anyway i got my answer, thanks

[ebon condor]

I hope it makes sense

[lone heart]

@lost dew Has your question been resolved?

[lone heart]

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[alpine sable]

[(x^2 + 4x + 4)/(2x^2 -x-1)] > 0, find x.

[alpine sable]

My answer is coming out to be x belons to (- infinity, -2) U (-2,-1) U (2, infinity)

I just want to confirm whether it's right or not

[severe sluice]

uh... no.

[alpine sable]

so what's the correct solution to this

[severe sluice]

wait a sec
what course are you taking?

[alpine sable]

have done schooling this year

[severe sluice]

eh

nvm

i think you can just find the signs

[alpine sable]

that wud be equivalent to completing calc 1

am revising for college entrance

[severe sluice]

can you... uh, show your work, please?

[alpine sable]

ohh sorry I can't as I m on my PC so can't take a pic of my notebook

[severe sluice]

anyways

i think we can find the signs of the numerator and the denominator

and thus figure out the sign of the expression itself

[alpine sable]

I did made sign chart and after simplifying LHS

but still...

[severe sluice]

well

i think you can factor the numerator first

=(x+2)^2

and then for the denominator

you basically cant

so the numerator would be positive

so we only have the denominator

which is a parabola that opens upwards

now i think you can just find the roots

(sry i'm multitasking)

[alpine sable]

ok

.close

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[alpine sable]

What do you need @lavish birch

On x or y?

@lavish birch

[dusk pecan]

@lavish birch the value for slope (or gradient) is measured as rise divided by run

[alpine sable]

@dusk pecan jojo fan sup

[dusk pecan]

part 6 december 💯

[lone heart]

@lavish birch Has your question been resolved?

[lone heart]

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[junior thicket]

I keep finding 18 but the answer is 19, i thought that T is a range between 9 and 25 (both are included) and the only integer in set G i could find was 4

[junior thicket]

U is supposed to be G in the image

I just realized i don't actually need to find the value of x, since the image of the function is between sqrt(32) and 4, 5 also needs to be a possible value.

.close

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[hollow siren]

Hi

[hollow siren]

Is there a way to calculate indefinate integrals in a graphing calculator

[placid zinc]

Google your calculator type

[lone heart]

@hollow siren Has your question been resolved?

[alpine sable]

Unless you have a calc that does algebraic

[lone heart]

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[whole wadi]

f(x) is a linear function whose graph passes through the point (0, 4) and intersects the graph to g (x) = -x-2 at a right angle. What is f(x)?

[ocean seal]

Theophania

[next pulsar]

$\frac{1}{x}$ is not a linear term

[ocean seal]

Joy!

[next pulsar]

go on desmos and type your graph in, you are looking for a straight line, is f(x) = 1/x a straight line?

[strong hornet]

well you know how to get perpendicular lines right

the slope of your equation * the slope of g(x) =-1

[whole wadi]

Yes

[ocean seal]

Theophania

Theophania

[strong hornet]

well then it should be easy

it is not -x

it's the slope itself

not including the x

it would be

-1 * the new slope = -1

[whole wadi]

ok sec

$-1\cdot \frac{1}{1}=-1$

Like that?

[ocean seal]

Theophania

[strong hornet]

yup

[whole wadi]

Okay, so we have

[strong hornet]

now just plug in (0,4) to y=x

[whole wadi]

$f\left(x\right)=x+m$

[ocean seal]

Theophania

[strong hornet]

yep

plug in the point it passes through

and

[whole wadi]

$f\left(0\right)=0+m=4$

[ocean seal]

Theophania

[strong hornet]

find hte value of m

[whole wadi]

$\implies m = 4$

[ocean seal]

Theophania

[strong hornet]

exactly

[whole wadi]

$f\left(x\right)=x+4$

[ocean seal]

Theophania

[whole wadi]

Okay thanks for the help

[strong hornet]

right

no problem

[whole wadi]

.close

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[dreamy latch]

Can someone help me with this question about making x the subject?

P = 100(y-x) / x

Im stuck on y-x/x = P/100

[next pulsar]

what happens when you multiply both sides of the equation by x?

[dreamy latch]

y-x^2 = Px/100?

[next pulsar]

why x^2?

[dreamy latch]

x multiply x = x^2 right?

oh

y-x = Px/100

?

[next pulsar]

yep

now what is stopping you from taking all the terms with x on one side of the equation?

[dreamy latch]

oh figured out the answer now

thank you for the help

.close

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[queen quiver]

Yoo wassup

[queen quiver]

One sec let me get the question

How exactly do you do binomial expansion with a fractional power?

[bleak ridge]

Do you have a formula?

[queen quiver]

That's the issue I'm not sure what the formula is

[bleak ridge]

Theres some like ridiculously long formula to do it

[queen quiver]

I have this but it doesn't work for fractions I think

[bleak ridge]

Yeah that one

[queen quiver]

or am I just not using it correctly

[bleak ridge]

That should work with fractional powers

[queen quiver]

ohh

but how does calculation of the nCr combinations thing work with fractions

is what im confused about

[bleak ridge]

Gamma function

Horrible

[queen quiver]

huh

all i can find is the coefficient for the first term is 1

how do factorials work for fractions ;-;

[bleak ridge]

Factorial is actually a continuous function

,w plot x!

[ocean seal]

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[bleak ridge]

Idk if thats what they expect you to use

[queen quiver]

wott

according to my teacher the first 3 terms end up as 1, -1/3 x, and -1/9 x^2

(1/3) / [2! (1/3 - 2)!]

can this be calculated using a graph like that?

[bleak ridge]

Yeah I think so

[queen quiver]

ima try using my calculator to find ncr actually

bruh it says domain on my calculator :(

[bleak ridge]

Hmm

Well wolfram can help out a bit

,w (1/3) / (2! (1/3 - 2)!)

[ocean seal]

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[bleak ridge]

Uhhhhhh

[queen quiver]

yea

[bleak ridge]

,w (1/3) choose 1

[queen quiver]

its sposd to be -1/9

[ocean seal]

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[queen quiver]

ohh awesome

[bleak ridge]

,w (1/3) choose 2

[queen quiver]

,w (1/3) choose 2

[ocean seal]

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[queen quiver]

oh ur faster

[bleak ridge]

Oo

[queen quiver]

yay it works

[bleak ridge]

Les go

[queen quiver]

huh maybe i messed up using the formula or something

[bleak ridge]

Perhaps

[queen quiver]

but this substitution is correct tho right?

(1/3) / (2! (1/3 - 2)!)

[bleak ridge]

Oh you missed the factorial on the top

[queen quiver]

OHH

yea that makes sense

,w (1/3) choose 2

[ocean seal]

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[queen quiver]

????????????????/

why is this different from my calc

[bleak ridge]

Hmm

[queen quiver]

im pretty sure it's possible to find the answer by hand tho

cause this questions gonna be on paper 1 which is no calc

[bleak ridge]

Im not sure

[queen quiver]

;-;

[bleak ridge]

Its not really easy to calculate ncr of decimals by hand

He prob wouldnt have such a nasty equation?

I hope

[queen quiver]

let me show the method he gave

this mans is a singaporean menace to society i tell you

(no racism one of my best friends is singaporean)

but he literallt teaches nothing

[bleak ridge]

Reasonable

[queen quiver]

how does this calculation work

[bleak ridge]

I mean that seems to be what we did

[queen quiver]

i'm confused as to why the second and third term coefficients end up like that

[bleak ridge]

,rotate

[queen quiver]

thx

why is 1/3 choose 1 = 1/3, and 1/3 choose 2 = -1/9

[bleak ridge]

Ive honestly got no clue

[queen quiver]

darn

u wanna know how he explained it in class

he didnt bruh this is what one of the students wrote on the board and i just copied it down

;-;

especially confused as to how 1/3 choose 2 can be expressed in this way

[bleak ridge]

Hmm

[queen quiver]

well i understand why my calculator thinks its 0.333 for 1/3 choose 2 now

[bleak ridge]

Oh

[queen quiver]

bruh google gets it correct

[bleak ridge]

Bruh

[queen quiver]

yes

[bleak ridge]

Just add parens ig

[queen quiver]

it doesnt work with parens on my calc

[bleak ridge]

It must just not be able to calculate fractional ncr

[queen quiver]

yea

it starts screamin domain

uhh i think this applies right

bruh mans never told us about falling factorials this is news to me

[bleak ridge]

Thats prob what he used

Ive never heard of that either

[queen quiver]

i guess i get it now

seems very specific

[bleak ridge]

Yeah

[queen quiver]

thx for your support chief

i was boutta have a breakdown cause this needs to be submitted soon

[lone heart]

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[queen quiver]

yes thank u bot

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[viral python]

The cost of tickets for a play is $3.00 for adults and $1.00 for children . In total $1100 were collected. 50 fewer children tickets were sold than adults.

[viral python]

please help

i had this questions

so basically A3 + C1 = 1100 and A = C - 50

3(C-50) + C1 = 1100
3C - 150 + C1 = 1100
C2 = 950
C = 475

oh it's acually C+ 50
3(C+50) + C
C2 = 1250
C = 625

[lone heart]

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@viral python Has your question been resolved?

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[dreamy otter]

when i checked the answer they did y2- x^3+2x and I was wondering why the sign changed?

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@lean stump Has your question been resolved?

[slim bone]

i say just list it out
134256
134265
134562
135642
156342

and say them back

but keep 34 together

dont move them round

[lean stump]

there’s 720 possible lists lol

i can’t list out each combo

[oak chasm]

@lean stump Well, first get the relative positionings of 2, 3, and 4.

Like:

2 3 4
2 4 3
3 4 2
4 3 2

Does that make sense so far?

Then you place a positioning in the final sequence.

There are 3 items to place in 6 places.

So, maybe with the positioning 2 3 4, you'll get

_ _ _ 2 3 4
_ _ 2 _ 3 4
_ _ 2 3 _ 4
_ _ 2 3 4 _
⋮

Then there are three numbers to randomly place in the remaining places.

So, you multiply together:

· the number of acceptable relative positionings of 2, 3, and 4
· the number of ways to put each relative positioning into the final sequence
· the number of ways to rearrange the last three digits

[lean stump]

nah

i tried that

doesn’t work

i’m getting a number bigger than 6! when i count everything up

[oak chasm]

What do you get for each part?

What is the number of acceptable relative positionings of 2, 3, and 4?

What is the number of ways to put each relative positioning into the final sequence?

What is the number of ways to rearrange the last three digits?

@lean stump

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[hazy ingot]

is a=0,c=2,b=5,m=2 a counter example to this

[hushed pasture]

no that's not true

$ac \equiv bc [m] \implies m/c(a-b)$

[ocean seal]

Salah

[hushed pasture]

then if $m \wedge c = 1$ then we can use Gauss' Th and say that $m/a-b$

[ocean seal]

Salah

[hazy ingot]

but doesnt my solution disprove that example

because 0 x 2 = 5 x 2 mod 2 is 0=0 then 0≠5mod2

[hushed pasture]

I didn't see your solution, i talked only about the photo

[hazy ingot]

oh my bad

isnt the counter example basically proving the problem false while following its universe of discourse

[hushed pasture]

no wrong

a shouldn't equal to 0

[hazy ingot]

why

isnt 0 a integer

[hushed pasture]

otherwise it's like you saying that 0 = k x m + 5

in this case m = 2

then you said the 5 = 2k and k in Z

[hushed pasture]

is this logical?

[hazy ingot]

where are you getting k and Z from

[hushed pasture]

the definition of congruence

$a \equiv b [m] \iff \exists k \in \bZ: a = k\cdot m + b$

[ocean seal]

Salah

[hazy ingot]

so the a has to be greater than 1

or does it just have to satisfy the congruence definition

[hushed pasture]

take for example take m = 2 and c = 2 as well

[hushed pasture]

because if gcd(c,m) = 1 then this is correct

you got that?\

[hazy ingot]

yeah

[hushed pasture]

so i'll give a counter example

c = m = 2, a = 4, b = 1

$2 \times 4 \equiv 2 \times 1 [2] \implies 4 \equiv 1 [2]$ which is wrong

[ocean seal]

Salah

[hazy ingot]

yeah because 8≠2mod2

[hushed pasture]

no the opposite

[chrome kite]

8 is cong to 2 mod2

[hushed pasture]

8 = 2 [2]

and 2 = 0 [2]

[hazy ingot]

what does the [m] mean

[hushed pasture]

then 8 = 0 [2]

[chrome kite]

[m] = mod m i think

[hazy ingot]

ohh

[hushed pasture]

just type error

[hushed pasture]

so you get that example

because 8 and 4 = 0 [2]

[hazy ingot]

is 0[2] 0 modulo 2

[hushed pasture]

what is this

[hazy ingot]

is 0[2] the same as saying 0 modulo 2

[hushed pasture]

don't make this confuse you always turn that modulo into normal equality

$a \equiv b [n] \iff \exists k \in \bZ: a = k\cdot n + b$

[ocean seal]

Salah

[lone heart]

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[icy trail]

how would i do these 2?

[icy trail]

im completly lost

i factored the first one 😅 but no idea where to go from there

[glass lichen]

you can only take log of positive stuff

[icy trail]

ohhh

ok

so i solve (x-5)(x+1)>0 for domain

what about the range?

i think its just all reals?

[lone heart]

@icy trail Has your question been resolved?

[raven rover]

Correct

[raven rover]

For the range, think about what are all possible outputs for logarithms?

Make sure you keep in mind the domain restrictions. In other words, be careful of what you can't put into the function, because whatever values those are, you can't get a corresponding output.

[icy trail]

the log is just the exponent of the 3 that gets x^2-4x-5 right

[remote heron]

did you figure it out leaf?

or still need help

[icy trail]

not quite

:/

[remote heron]

oof sorry

@icy trail which one to start with

ping me if youre back, sorry i was takin a break

[icy trail]

all good

im still tryna find the range of the first one

[remote heron]

oh

$2 \log _3 (x^2 - 4x - 5)$

[ocean seal]

jan Niku

[remote heron]

so

i guess we can ask like

log is definitely one of our "problem functions" wrt to domain right

so what are problem values for log

inputs

,w graph log x

[ocean seal]

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[remote heron]

where is log not defined

[icy trail]

wrt means?

[remote heron]

with regard to

[icy trail]

ah

you put "to" twice then, but go on 😅

[remote heron]

oh

[icy trail]

[remote heron]

so where are problem values for log

[icy trail]

problem values?

[remote heron]

yea

let me throw out a few

[icy trail]

like values that arent in domain?

[remote heron]

log(1)

log(0)

log(-1)

yea

[icy trail]

log(-1)

[remote heron]

how about log 0

[icy trail]

log(0)

[remote heron]

log 1?

[icy trail]

log(1)

[remote heron]

yup

so can you summarize

what kinda values do we not let be inside logs

[icy trail]

nonposibitves

[remote heron]

yup

[icy trail]

[remote heron]

so, just remember, well shelve it for a moment

log domain is only positive numbers

then lets look at the argument to the log

x^2-4x-5

what kinda shape is this?

what sorta object

you there?

[icy trail]

oh yea

sorry

its a parabola

[remote heron]

yup

is it up or down

[icy trail]

so you solve x^2-4x-5>=0

[remote heron]

well, sorta

we do care about the roots

so maybe this is not a bad way to do it

[icy trail]

its up

[remote heron]

its pretty easy to factor, yea

[icy trail]

ye

(x-5)(x+1) ? 0

[remote heron]

lemme draw but just for funsies

[icy trail]

,w x^2-4x-5>=0

[ocean seal]

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[icy trail]

is it inclusive?

its not right

[remote heron]

is it?

[icy trail]

dont think so

[remote heron]

.

[icy trail]

since 0 doesnt work

[remote heron]

alright so

we have this up parabola

its 0 at -5 and 1

and in between those points, its...

[icy trail]

below

[remote heron]

positive, negative?

[icy trail]

negative

[remote heron]

yea

[icy trail]

[remote heron]

so 0 at 1 and -5

negative between

[icy trail]

which doesnt work

[remote heron]

but its inside a log

[icy trail]

so between -1 and 5 doesnt work

[remote heron]

inclusive, yea

okay, lets bring everything together

[icy trail]

x<-1 or x>5 ?

[remote heron]

$f(x)=2 \log _3 (x^2 -4x -5) = 2 \log _3 (x-5) + 2 \log _3 (x+1)$

[ocean seal]

jan Niku

[icy trail]

oh?

it seperates liek that?

[remote heron]

yea

$\log (ab) = \log (a) + \log (b)$

[ocean seal]

jan Niku

[icy trail]

oh okay

[remote heron]

so we need the whole thing to be defined for the function to have a value

we already know were throwing out values between -5 and 1, inclusive

whats happening just after 1, moving more positive?

maybe lets check an easy number....like....the base is probably a good choice

[icy trail]

3-5 is negative

[remote heron]

oh, good point

maybe factoring it this way is a bad choice

oh, wait

i screw it up

[icy trail]

yikes

[remote heron]

$2 \log _3 (x+5) + 2 \log _3 (x-1)$

[ocean seal]

jan Niku

[remote heron]

whats a good value, you think

just to the right of 1

[icy trail]

1.1?

2?

[remote heron]

wait

we gotta track a sign error

(x+5)(x-1) = x^2-4x-5

no

(x-5)(x+1) = x^2-4x-5

yes

[icy trail]

x-5, x+1

[remote heron]

then our roots are -1 and 5

we throw away values between -1 and 5, inclusive

[icy trail]

[remote heron]

lets look at a value just to the right of 5

maybe 6

[icy trail]

6 is fine

[remote heron]

,calc 2log _3 (11) + 2 log _3 (5)

[ocean seal]

The following error occured while calculating:
Error: Unexpected type of argument in function multiplyScalar (expected: number or Unit or string or boolean or BigNumber or Complex or Fraction, actual: function, index: 1)

[icy trail]

oh

[remote heron]

oh, doesnt like logs

thats okay

its 3.54

what do you think this number is gonna do as we go further and further to the right

[icy trail]

increase

[remote heron]

looks like f(10) = 7.3

is it ever gonna stop increasing?

[icy trail]

doesnt look like it

[remote heron]

yea

logs just kinda increase forever

[icy trail]

[remote heron]

anyways

were just playin around at this point

i think we answered the question

but now you should hopefully sorta know the shape?

[icy trail]

so the domain is x<-1 or x > 5

[remote heron]

lets check i guess

,w (-2)^2 - 4*(-2)-5

[ocean seal]

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[remote heron]

hmm

[icy trail]

hmmmmmm

[remote heron]

splitting the function in the way we did

we lose a side

but thats okay

[icy trail]

[remote heron]

you should have another side

so, yea

youre right

and the range should be everything

it should hit all numbers

[icy trail]

oki

the other question now?

[remote heron]

alright

[icy trail]

ye so domain would be everything right

[remote heron]

we can check

you know your normal problem functions

roots

logs

division

i see division

division problem is 0 in the bottom

can 2+3^x be zero?

[icy trail]

oh

no

[remote heron]

why not

[icy trail]

cuz 0 in denominator bad

[remote heron]

no i mean

is it possible for 2+3^x to be zero

just in general

usually youd like

[icy trail]

hmm

[remote heron]

so we know its division, yea?

[icy trail]

i dont think so

[remote heron]

we dont want the bottom to be zero

[icy trail]

[remote heron]

so we need to check if it can be zero

[icy trail]

2+3^x cant be zero

[remote heron]

and if it can, we need to throw out those numbers from the domain

[icy trail]

cuz that means 3^x = -2

which is not possible

[remote heron]

yup

[icy trail]

ok

so it is all real numbers

[remote heron]

yea, no other place for problems to come from really

[icy trail]

[remote heron]

exponentials are pretty nice functions

so domains everything

[icy trail]

and for range

[remote heron]

can 5 ever be negative?

[icy trail]

you can do the swap x and y thing right

[remote heron]

can 5 ever be 0

[icy trail]

wdym can 5 be negative

[remote heron]

can 5 be negative

[icy trail]

no

[remote heron]

is there any x that can make 5 negative

okay

so positive number in the top

can 2+3^x ever be negative?

[icy trail]

no

[remote heron]

nice

so we got 2 positive numbers

that means our fraction is always positive

make sense?

[icy trail]

yes

[remote heron]

can 5 ever be zero?

[icy trail]

no

[remote heron]

the only way to get 0 out of a fraction is if the numerator is 0

yup

[icy trail]

mhm

[remote heron]

so what range can we already lob off our range

its a pretty big interval

5/(2+3^x) has to be ____

[icy trail]

positive

[remote heron]

yup 😄

[icy trail]

😄

[remote heron]

okay

heres some stuff you should know about fraction behavior

especially where the bottom is changing

but the top isnt

[icy trail]

[remote heron]

you have two general types of behaviors

if the bottom becomes smaller and smaller

then the fraction overall shoots off to infinity

[icy trail]

makes sense

[remote heron]

if the bottom gets really really big

the opposite happens

it flattens down to 0

[icy trail]

also makes sense

[remote heron]

'dividing a pie into more and more pieces'

so which one do we have here

[icy trail]

both?

[remote heron]

whats gonna be the end behavior of this thing, as x gets really big

well, for really big x values

[icy trail]

if x gets really big then it flattens

[remote heron]

to what?

[icy trail]

0

[remote heron]

yup

[icy trail]

wait

[remote heron]

?

[icy trail]

but what about the 2

oh

nvmnvm

its the denominator as a whole

[remote heron]

yup

alright, so for really really big x, this thing just flattens out to 0

[icy trail]

yes

[remote heron]

how about around x=0

[icy trail]

but never reaches 0 right

[remote heron]

whats this thing look like

nope, since 5 can never be zero :p

[icy trail]

x=0,then its 5

bottom is 5

[remote heron]

whys that

yup

[icy trail]

2+3^0

[remote heron]

wait

no

bottom isnt 5

[icy trail]

3

its 3

[remote heron]

okay

thats the bottom

whats the fraction as a whole

[icy trail]

5/3

[remote heron]

alright so

this thing crosses from the left where it does something

passes through 5/3

and then flattens out to 0

lets look at negative numbers now

[icy trail]

okok

[remote heron]

as we go further and further to the left

what happens to the denominator

2+3^x

[icy trail]

so thats a negative exponent

[remote heron]

yup

[icy trail]

which is the same as rooting it?

[remote heron]

so it looks something like $2+\frac{1}{3^x}$

[ocean seal]

jan Niku

[icy trail]

or is that exponents between 0 and 1

[remote heron]

fractional

[icy trail]

wait

what do

negative exponenets do again

[remote heron]

they move it to the other side of the fraction

[icy trail]

oh

inverse

yesyes

[remote heron]

yea

well here

[icy trail]

well 1/3^x gets really small

[remote heron]

,w graph 5^x

[ocean seal]

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[remote heron]

heres a nice toolkit shape

you may want to memorize

key points:

at x=1, you always get your base (5 here)

[icy trail]

mhm

[remote heron]

at x=0, you always get 1 (always)

then it explodes to the right

and flattens out to the left

since negative exponents will make it in the bottom of a fraction instead

[icy trail]

[remote heron]

so

2+3^x

as we go to more and more negativer x values

what does this thing start to look like

[icy trail]

it approcahes 2?

wait

[remote heron]

yea, it approaches 2

so what does our fraction approach overall

5/(2+3^x) as you go to more and more negativer x values

[icy trail]

5/2?

[remote heron]

yup

from underneath? or above

[icy trail]

from under

since x=0 is 5/3?

[remote heron]

yea, exactly

[icy trail]

ok

[remote heron]

so were stuck

[icy trail]

[remote heron]

were stuck in between 5/2 and 0

[icy trail]

which is the range

[remote heron]

we start out, at the way left end of x

really close to 5/2

then we shrink and shrink and shrink

passing through 5/3 at x=0

then get closer and closer to 0

yup, the range

[icy trail]

[remote heron]

so fractions can be really screwy

if you get a domain that has 'holes'

which is really common for like

fractions sometimes have 0's in the bottom

so instead of losing whole intervals like we did in the first problem

you lose just points

so that process we did for the range, youll want to do something like that

but once for each continuous domain

each one between the holes

so if you have 2 holes (a quadratic in the denominator), youll need to do the range process 3 times

[icy trail]

[remote heron]

as you approach the problem x values

something like uhh

,w graph 1/(x^2-4x-5) from x = -10 to x=10

[ocean seal]

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[remote heron]

see, is gonna have really weird behavior near problem values

[icy trail]

isnt there a \ thing that means exclude

[remote heron]

yea, but its not usually used for arbitrary values i think?

usually just 0

unless im crazy

just by convention

[icy trail]

idk

i think i was taught that you could just remove any

[remote heron]

normally for these , its appropriate to use interval notation

set notation is not as uhh

conventional

interval notation keeps a sense of order i think

which is nice

rather than set which just has you plucking out random values

[icy trail]

yea that makes sense

well tysm for the help, you explained very well

.close

[lone heart]

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[remote heron]

np, good luck

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[paper holly]

.

[paper holly]

I've tried it without the square brackets too

[spring kayak]

its because its 3 to infinity

[paper holly]

nope wrong answer

last attempt left 👀

[smoky cloud]

did u try taking the [ of 0 ?

not sure but hv u tried it ?

[snow nest]

i think that's maybe it

[paper holly]

yeah I tried that too

[snow nest]

huh

that's weird

[spring kayak]

omfg im sorry

i thought it was 3 to infinity because it doesn't decrease starting at 5

[paper holly]

it's cool

figured it

it was to 7 not infinity lol

how do i close this

[snow nest]

what

that's so weird

.close

i think

[paper holly]

ikr

.close

[lone heart]

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[alpine sable]

hi

[alpine sable]

.close

[lone heart]

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[tight locust]

how exactly does Bessel's correction work?

[tight locust]

something about degrees of freedom but wtf does that actually mean

<@&286206848099549185> !

right that makes sense intuitively

but why n-1 in particular?

haha yeah. just doing the formulas is easy but as soon as you go a little deeper there's a lot of complicated math

just try finding the pdf for the t dist haha

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[tight locust]

no

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