help-0

[quick surge]

can u draw a line

[dusk smelt]

a is 20-12

Agree?

[quick surge]

at where the line is suppose to be

[dusk smelt]

From point c and down

[dusk smelt]

Yes

[quick surge]

i only have b^2

and thats height

[dusk smelt]

Ok

Now for the a

[quick surge]

a is?

[dusk smelt]

a is (22-12)/2

[quick surge]

the base or

[quick surge]

why

why did u /2

[dusk smelt]

22-12 gives you the extra width

[quick surge]

mhm

oh okkkk

[dusk smelt]

Which is split between the left and right side of the trapezoid

[quick surge]

okok

now i just

a^2 = 64 + 25
a^2 = 89

wat

[dusk smelt]

Ok

Take the square root of that

[quick surge]

a = 89 square root

i cant use a calculator when theres an exam

so like

hm

[dusk smelt]

Keep it as square root then

[quick surge]

ok

[dusk smelt]

Anyway I gtg now

Good luck

[quick surge]

oh

ok

bye

[shadow sparrow]

Aye, it seems like this room is clear?

Working on final review stuff, can someone help me understand this?

Brain is kinda blanking, I haven't done this stuff since the beginning of the semester.

I know that 2x-3 and 8-x are both their own function lines

[muted raft]

Did you graph it first?

You are asked to label four points.

[shadow sparrow]

nyet, from what I can remember, you just pick for instance 4 values greater than 3, then plug them in

and that is the x value, the equation it gives is the y value

[muted raft]

First two points that I would pick are 3 for both 2x-3 and 8-x

Label those points first

then add two more other points

Shouldn't matter which.

[shadow sparrow]

You can't pick 3 tho right?

Its x is greater than 3 for one.

You can for 8-x of course

[muted raft]

You can still use it just not shade it in

This will be useful when you are calculating the limit.

[shadow sparrow]

because it'll be from where they are not continuous?

[muted raft]

so kinda like this

See how the point is not shaded in

[shadow sparrow]

yeah

[muted raft]

that means it is undefined at that point.

So you could do this, or if not, pick another point.

It is up to you.

[shadow sparrow]

Okay so first equation set would be (3, 3), (4, 5), (5, 7), (6, 9)

Second would be (3, 5), (2, 6), (1, 7), (-1, 9)

[muted raft]

(3, 3) however is not a valid point, so make sure you don't shade that point in.

[shadow sparrow]

Yeah

So yeah

Got what you got

So now the question is the approaching limit thing

I know that x approaches 3- is from the right to left iirc

Then 3+ is vice versa

[muted raft]

does it?

[shadow sparrow]

I don't know

I'm not sure

lol

Thats my guess from what I remember

[muted raft]

You said second one was (3, 5)

what does that mean?

[shadow sparrow]

You mean that its a declining slope? or thats its endpoint? or

I'm not sure

[muted raft]

We are approaching the function f(x) from the left side.

Which function is used for left side of 3?

[shadow sparrow]

Second one I believe

[muted raft]

yeah y=8-x

What happens as we approach x=3 on y=8-x

[shadow sparrow]

Yeah because x < 3

As well right?

As another tell

[muted raft]

Yeah x<=3

[shadow sparrow]

The function decreases

[muted raft]

We don't care about that, we care what value are we approaching as x approaches 3

[shadow sparrow]

5

[muted raft]

yeah

so limx->3- is 5

[shadow sparrow]

So the limit as it approaches from the left is 5

[muted raft]

Now what about the right side.

What value are we approaching

[shadow sparrow]

Then limit approaching from the right is going to be before the undefined point right?

Because (3, 3) isn't defined

[muted raft]

Well, yeah it is undefined but we don't care about that.

[shadow sparrow]

So you're approaching 3?

[muted raft]

We simply care about what it looks like we are approaching

Yup

[shadow sparrow]

So a is 5, b is 3

[muted raft]

It looks like we are approaching 3, but we don't actually. But this doesnt matter since limit is all about approaching.

yeah

So now what is c)?

[shadow sparrow]

That one, I'm blanking on

lmao

[muted raft]

Well, for a limit to exist at a point C, then its left hand limit and right hand limit must be same.

Is it the same in this case?

[shadow sparrow]

No

so its undefined

?

[muted raft]

That means limit does not exist.

So that would be your final answer.

[shadow sparrow]

Because 5 and 3 don't match up

from the two

[muted raft]

Yeah

[shadow sparrow]

SO IF the limit approaching from both sides was continuous

Or well

I guess it wouldn't be continuous

Right?

Or would that be going into the approaching positive and negative infinity.

[muted raft]

Well, if the limit on both sides were same, then the limit would exist at that point.

[shadow sparrow]

Yeah

Like in the case of x-1 and x+1

as functions of x=>1 and x<=1 iirc

[muted raft]

for example

[muted raft]

this would have limit

[shadow sparrow]

Let me give this one a shot rq to make sure I understand it clearly

yeah

[muted raft]

Go ahead, give it a try.

[shadow sparrow]

Because its undefined at one point right?

[muted raft]

It is undefined but if we just look at limit, it looks like it is approaching same value from both sides.

So in fact limit does exist.

However, the function itself is not continuous.

[shadow sparrow]

Yeah makes sense

Idk why but mathematics books have always been in my experience horrible at explaining concepts

At least higher level ones anyways

[muted raft]

Ah well,

[shadow sparrow]

Also so when it says label at least 4 points

That isn't really needed right?

In this case

[muted raft]

They just want you to graph it basically.

[shadow sparrow]

Like you could literally just graph 4

for both

and you'd have your answer

or 4 and 5 for one and 4 and 3 for the other

[muted raft]

Yeah, graphing usually gives you all the information.

And that is fine.

[shadow sparrow]

so

both equations are (4, 2)

With the one for 10-2x not being defined

but as its approaching, it would still be 2

so in this case, wouldn't all of the answers be 2?

[muted raft]

yeah

[shadow sparrow]

Yeah, that makes more sense now

I'll write it down that way where its just talking about the maximum y value as x approaches what number

This was more of a pre-question for the one I had, which is finding limits for more complicated limit equations

So like, if its as x approaches -2, if you think of it graphically, it would be when that equation is approaching -2

So would you set the equation to -2 and solve? or could you set x to -2?

[muted raft]

first factor out both numerator and denominator.

[shadow sparrow]

Yeah I'm horrible at this lmao

Like factoring out complicated fractions

[muted raft]

$x^2+3x+2 = x^2+x+2x+2$

[ocean seal]

zslya

[muted raft]

similar idea as well to the numerator

[shadow sparrow]

What confuses me is it looks like you just split 3x into x and 2x

[muted raft]

yes

Because I want to group them

[shadow sparrow]

factoring always confuses me because I don't know why that happens

[muted raft]

Practice more and it should come to you naturally.

[shadow sparrow]

For 2x^2 + 4x?

[muted raft]

2x^2+2x+2x+4

[shadow sparrow]

You mean group it with the numerator? Or

[dusk smelt]

@shadow sparrow think about which two numbers will yield a product of 2 but a sum of 3

[muted raft]

no no I meant factor the numerator lol.

[dusk smelt]

Then those are your factors

2 and 1

Then (x+2)(x+1)

That's it

[shadow sparrow]

I conceptually get that idea, its just always confusing how you pick what the factors are based off of

Because in this case its 3 and 2, but in some cases of more complex factoring, it gets confusing for me

[muted raft]

Like I said, practice.

[shadow sparrow]

Like I know you have to have two values from the equation and then find the product of one but a sum of the other

[dusk smelt]

Yeah u need to practice

[shadow sparrow]

Where would I practice factoring? Is there a good website for it?

I can do so after finals more specifically

[muted raft]

I would just google it and see if there are pdfs for it.

[dusk smelt]

^

[shadow sparrow]

So $x^2+4x+4$

[ocean seal]

Huntifer

[shadow sparrow]

factors of 4

2 and 2

so (x+2)(x+2)

So it would be

[dusk smelt]

Yes

[shadow sparrow]

$\frac{(x+2)(x+2)} {(x+2)(x+1)}$

[muted raft]

yeah

[dusk smelt]

Yes

[muted raft]

cancel out x+2 now

$$\frac{(x+2)(x+2)}{(x+2)(x+1)}$$

this is how you write it

[ocean seal]

zslya

[shadow sparrow]

Oh yeah hur dur.

[ocean seal]

Huntifer

[shadow sparrow]

So cross out two of the x+2

$\frac{(x+2)}{(x+1)}$

[ocean seal]

Huntifer

[muted raft]

yeah

notice how you can just apply the limit now

[shadow sparrow]

Yeah, so you can just apply -2 now?

!wolfram

.wolfram

[muted raft]

hold on

[shadow sparrow]

forgot the command

lmfao

[muted raft]

,w

[shadow sparrow]

,w ((-2+2)/(-2+1))

[ocean seal]

Results provided by WolframAlpha

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

Requested by Huntifer#4860

[shadow sparrow]

So undefined

[muted raft]

no

its 0

0 is not undefined.

[shadow sparrow]

Or wait

-inf?

[woeful pulsar]

0 is a number

0/0 is not a number

[shadow sparrow]

Yeah

[dusk smelt]

It's 0

[woeful pulsar]

0/1 is a number

[shadow sparrow]

Sorry, brain was being dumb

So in the case of basically all of these

would you pretty much just factor?

Unless it was simplified enough to just test the limit

[muted raft]

try it out.

See if they are factorable or not.

[shadow sparrow]

Okay so b

https://cdn.discordapp.com/attachments/490557019623915520/841132078232895508/unknown.png

[woeful pulsar]

if it's 0/0 when you plug in then you need to do something

[shadow sparrow]

Factor numerator first

-8 and 7

[dusk smelt]

No

[lapis harness]

in all of them

you need to factor

and cancel

that's it

[dusk smelt]

-1 and -7

[shadow sparrow]

-1 and -7?

For x^2-8x+7?

[dusk smelt]

They are easy

[woeful pulsar]

not find zeroes

factor

[dusk smelt]

(x-7)(x-1)

[woeful pulsar]

okay

how about the denominator

[shadow sparrow]

[woeful pulsar]

wait

urgh

[shadow sparrow]

lol

[woeful pulsar]

please don't give answers directly (if you are)

[shadow sparrow]

When I'm looking at factoring, I'm thinking of looking at the numbers in the equation

so in the denominators case, -6 and -7

[woeful pulsar]

the numerator/denominators are quadratic

you need to write them as a product of linear terms

that's what we mean by factor

[shadow sparrow]

uh

Okay

mm

So I know the leading term is x^2

arent you looking for the roots?

in the case of a quadratic

That being what two numbers when multiplied will equal one term, and added will equal the other.

[muted raft]

what are you factoring?

[muted raft]

So factor the numerator first.

[shadow sparrow]

-8 and 7 right?

[muted raft]

Break apart 8x like I did on previous questions

[shadow sparrow]

I'm not saying -8 and 7 are the roots

I'm saying you see what multiplies to -8 and adds to 7

I think that was what they misunderstood before

[dusk smelt]

Opposite

Adds to -8

[shadow sparrow]

So the first number will be addition

the second will be multiplication

[dusk smelt]

Yes

[shadow sparrow]

so yeah, -1 and -7

that makes sense.

Because -1 and -1 added is -8, multiplied is 7.

so (x-1)(x-7) is the numerator

added to -6 and multiplied to -7 would be uhh

1 and -7

so (x+1)(x-7)

Or wait no

other way around.

Sorry lmao

[woeful pulsar]

what do you mean other way around

[shadow sparrow]

Right?

Wouldn't -1 be (x+1)?

Thought the signs flipped when doing that

[woeful pulsar]

eh

at least it's correct now

so what do we do after factoring

[dusk smelt]

We multiply them back again

[woeful pulsar]

?

[dusk smelt]

Then factor again

For fun

[shadow sparrow]

lol

[woeful pulsar]

urgh

[dusk smelt]

Lol

[shadow sparrow]

So wait

[woeful pulsar]

don't troll plz

[shadow sparrow]

Would it be (x+1)(x+7)?

or (x-1)(x-7)

To clarify

In relation to the numerator

[dusk smelt]

Which ones give u -8 when u add them?

If you're unsure you could always expand them by multiplying and see if it matches

[shadow sparrow]

What I'm asking basically is do the signs not flip?

Thats what I had remembered

[dusk smelt]

It's the second one

[shadow sparrow]

Okay

So in the case of it it was 4 and -8

It would be (x+4)(x-8)

Just clarifying so I know

[dusk smelt]

Which question is that

[shadow sparrow]

It isn't one

it was random values I thought up to test it.

[dusk smelt]

Yeah

Correct

[shadow sparrow]

So in the case before it'll now be

[woeful pulsar]

if you are unsure of your factoring you can multiply to check

[shadow sparrow]

$\frac{(x-1)(x-7)}{(x+1)(x-7)}$

[ocean seal]

Huntifer

[shadow sparrow]

Cancel out the x-7 iirc

then fill in the approaching limit which would be 7

[obsidian badge]

how to solve for the top part

[shadow sparrow]

@obsidian badge If this is another question, channel is in use, please read #❓how-to-get-help

[obsidian badge]

k

[shadow sparrow]

Make sure to check the message time to see if its recent when asking for help

thats what I do

Anyways, so

$\frac{((7)-1)}{((7)+1)}$

[ocean seal]

Huntifer

[shadow sparrow]

?

Which is just 6/8 or 3/4ths

So b) 3/4

[dusk smelt]

Yes

[shadow sparrow]

Okay, so I next one, this one is different

Approaching 0 from the left side

So negative to positive

Factor x^2-2?

[lapis harness]

no point

cancel x/x^2

and plug it in

[shadow sparrow]

Remind me, do you mean expand the numerator or cancel it with the denominator?

[lapis harness]

cancel with denominator

so you get

$\frac{x^2-2}{x}$

[ocean seal]

PBj4FPxrZF7ovZ

[shadow sparrow]

So you're canceling one of the x from the numerator and denom

kk just wanted to make sure I was understanding right.

Same thing you do with another older equation iirc where its like n(n^2-8x)/n or whatever it was

then you cancel the numerator with the demoninator for n^2-8x

So its as the limit approaches 0 from the left..

So would you set it to 0? or like -1?

[muted raft]

you can't just substitute 0 yet.

see if you can figure out a work around for this.

[shadow sparrow]

By adding something to the equation and the limit?

[dusk smelt]

x²/x - 2/x

[lapis harness]

why not ?

plug in 0-

[muted raft]

Division by 0?

[shadow sparrow]

Yeah it would just equal 0

[lapis harness]

it aprroaches 0

wouldnt it be + or - inf ?

[shadow sparrow]

I meant like if you did that

uhh

• infinitity

[muted raft]

Yes it is inf.

[shadow sparrow]

if its moving towards the right?

[muted raft]

this.

[shadow sparrow]

or positive values

Oh

[lapis harness]

so whats the problem with plugging it in

i dont get it

[shadow sparrow]

$\frac{x^2}{x} - \frac{2}{x}$

[ocean seal]

Huntifer

[shadow sparrow]

splitting the fraction

So would I plug in 0 when I split it?

Or would I do something else with it in this case

[dusk smelt]

Simplify x²/x

[shadow sparrow]

x

[woeful pulsar]

can the limit be undefined?

[muted raft]

I guess its got to do with "formal" way of showing it?

I mean, I would also usually just plug it in and call it end of the day

[woeful pulsar]

oh it's 0 from the negatives

[shadow sparrow]

yeah if its 0 approaching from the left

It would just be the opposite infinity?

So if its heading towards the right, its +inf, to the left, its -inf

[dusk smelt]

Yes

[shadow sparrow]

How would you know though at what point you could just call it infinite?

If that makes sense

[woeful pulsar]

there's a formal definition for it

[dusk smelt]

But it's -2/x

[woeful pulsar]

that involves epsilon delta

[dusk smelt]

So negative negative inf

[woeful pulsar]

,calc -2/(-0.0000000001)

[ocean seal]

Result:

2e+10

[shadow sparrow]

I guess its just mainly

like how far should I go before I know its infinite when reducing

Or actually, maybe this will help explain it mentally for me

[jade spear]

There’s no memes channel? What about all the spicy math puns

[jade spear]

Good meme

Thanks

[shadow sparrow]

@jade spear #chill

Pretty sure #math-discussion would be fine with them too

So for e

factor the top and buttom

Top one would be finding what adds to equal 1 and multiplies to equal -5

Denom would be adds to equal -2 and multiplies to 1?

well the second one would be -1 and -1

the first one my brain isn't figuring out for some reason

Actually, I don't think you can factor the numerator

with rational numbers

[woeful pulsar]

no, it's not suppsed to be factoring

try dividing the top can bottom by the fastest growing term you see

x^3

[shadow sparrow]

Because of the as it approaches -inf?

Uhh

$\frac{x^2}{x^3} = \frac{1}{x}$

[ocean seal]

Huntifer

[shadow sparrow]

?

oh also thought I'd take a shot at d

Let me write up what I did

$$\frac{\sqrt{x}-4}{x-16}$$
$$\frac{x^2-4}{x-16}$$
$$\frac{x-4}{-16}$$
$$\frac{16-4}{-16}$$

Kk works

1min

[woeful pulsar]

how did you do that?

something looks off

is it a variable substitution?

[ocean seal]

Huntifer

[shadow sparrow]

Square root is equivalent iirc to a square

in respect to fractions

thats how my brain did it anyways

[woeful pulsar]

what do you mean?

that looks wrong

[shadow sparrow]

Probably is

[woeful pulsar]

there are many ways to do this

one way is to substitute x=y^2

[shadow sparrow]

don't you have to get rid of the sqrt?

[woeful pulsar]

yeah, that will help to get rid of the square root

another way is multiplying by sqrt(x)+4 on numerator and denominator

[dusk smelt]

@shadow sparrow factor the denominator to (√x-4)(√x+4)

[alpine sable]

how come we dont cancel out the 2x at the bottom?

[shadow sparrow]

The x=y^2 one sounds easier in my brain I'm guessing

$$\frac{\sqrt{y^2}-4}{(y^2)-16}$$

[ocean seal]

Huntifer

[shadow sparrow]

if this is what you mean

[dusk smelt]

Dude

Read what I said

[shadow sparrow]

I did

Its confusing

to me anyways

[dusk smelt]

Write it out

[woeful pulsar]

can you simplify this now?

[shadow sparrow]

So you're saying sqrt the bottom

so its sqrt 16 which is 4

and sqrt x

[shadow sparrow]

I honestly don't really know, I know there is an operation to cancel sqrt like this but

[dusk smelt]

The denominator can be factored to (√x-4)(√x+4) then you can cancel out with the numerator

[woeful pulsar]

square root is the inverse of square

[shadow sparrow]

when I think of it I think of it like

[dusk smelt]

Lol I'm talking to myself, have fun

[shadow sparrow]

Well

You aren't explaining it

I don't get how the denominator can be factored to that

I said that

[dusk smelt]

Multiply it out, and see if it gives u x-16

[woeful pulsar]

can you simplify this?

[shadow sparrow]

Nein, I'm unaware how

Or at least I'm not confident in how

or do you mean

x=y^2

so you turn sqrt(y^2) into y

[woeful pulsar]

well yeah

you can do that

[shadow sparrow]

The way I think of it is

[woeful pulsar]

that gives $\frac{y-4}{y^2-16}$

[ocean seal]

Element118

[woeful pulsar]

and as x->16 y->4

[shadow sparrow]

$$\sqrt{x}^2 = 3^2$$

Or well, more in this form

[dusk smelt]

Wtf

[ocean seal]

Huntifer

[shadow sparrow]

Where the inverse square of the root, then you square both sides

[dusk smelt]

Good luck with whatever you're doing, but the answer to the limit is 1/8

[shadow sparrow]

I'm aware

I'm not here for just the answer

I'm here to know how, because this is a final review

[dusk smelt]

I told u what I need to do m8

[shadow sparrow]

And I told you I don't understand how thats done

I'm pretty visual in terms of learning

So I need to go through and see it in practice.

[shadow sparrow]

As x approaches 16, y approaches 4

what does that mean though in this case?

I guess more so how would it translate back?

[woeful pulsar]

since we did the substitution x=y^2

then y must approach 4

[shadow sparrow]

Err.. I sorta get it but I'm still kind of confused

I get how y must approach 4

So are you saying you substitute a x value with the trade for y?

[woeful pulsar]

since we only have y, we can write this as a limit as y goes to 4

[shadow sparrow]

so its y goes to 4

instead of x goes to 16

for the solving of it

?

So like you would substitute 4 instead of 16

[woeful pulsar]

can you substitute 4?

or does it result in 0/0?

[shadow sparrow]

Results in 0/0

[woeful pulsar]

probably time to do some factoring

[shadow sparrow]

I mean all I could see that you could do here is

$\frac{-4}{y-16}$

[ocean seal]

Huntifer

[shadow sparrow]

and I'm not sure if thats right either

[woeful pulsar]

erm

what did you do?

that doesn't look like factoring

[shadow sparrow]

If you factor it, would you be factoring a single value?

[woeful pulsar]

what do you mean?

[shadow sparrow]

Like

there is only x-4 on the top left

so would it be 1 and -4?

or just -4

[woeful pulsar]

you can't really factor x-4

it's linear

[shadow sparrow]

yeah

[woeful pulsar]

only the denominator can be factored

[shadow sparrow]

oh okay

well if there aren't two numbers

its just the x value for 1 and then the other number right?

so 1 and -16

is this what you meant by the other way?

[woeful pulsar]

yeah

that's another way

[shadow sparrow]

oh, well since the only number is -16

4

so (x+4)(x-4)?

so then it'd be

$\frac{x-4}{(x+4)(x-4)}$

[ocean seal]

Huntifer

[shadow sparrow]

or

$\frac{1}{x+4}$ @woeful pulsar

[silent path]

guys is this question correct?

The tangent line is the limiting position of the secant line as Q approaches P.

or the secant line is the limiting position?

[shadow sparrow]

@silent path channel is in use

❤️

[silent path]

oh im sorry

[shadow sparrow]

Np 🙂

Rules for the math channels are #❓how-to-get-help

[silent path]

thank you

[ocean seal]

Huntifer

[shadow sparrow]

I think?

[woeful pulsar]

yeah

if you have y instead of x

[shadow sparrow]

yeah

And y = 4

[woeful pulsar]

therefore?

[shadow sparrow]

Actually, how did y become 4 there instead of 16? I'm guessing because 4^2 is 16?

1/8

I get it now, I'm just not exactly confident if I got a different problem I'd be able to do the same thing

I really just need to re-do algebra because they fucked me in high school so my understanding of algebra isn't great, but its sorta too late now that I'm doing calc stuff in college, don't have a go back button lmfao

[woeful pulsar]

yeah

[shadow sparrow]

so my options are to rationalize the equation

[woeful pulsar]

that's the idea of substitution

[shadow sparrow]

or to substitute a squared y value

[woeful pulsar]

well you can substitute anything

[shadow sparrow]

yeah

b^2 or whatever

[woeful pulsar]

or even x-10 if you wish

[shadow sparrow]

as long as it equals the approaching limit right

Anyways, I'll try to practice it but I should figure out that other one we kind of cut half way

So you said to take the x^3 and divide it by the x^2

[woeful pulsar]

divide numerator and denominator by x^3

[shadow sparrow]

ah

rationalization same as the sqrt thing?

[woeful pulsar]

because for x->infinity it's easier to evaluate 1/x

not really

divide by the fastest thing you see there

[shadow sparrow]

yeah

See I look at that

and the what I should do doesn't occur to me

Its just factoring or reducing or something

lmao

So divide the numerator and denom by x^3

[woeful pulsar]

so what do you get?

[muted hornet]

yes

bec all the other terms tend to 0

[shadow sparrow]

Something tells me I'm going to divide this wrong

[woeful pulsar]

just try

[muted hornet]

just divide everything you see by x^3

[shadow sparrow]

I think the premise here I need to clarify is

what to do when dividing with exponents

like, I honestly don't have any good grasp of how you'd do

[muted hornet]

x^5/x^3 = x^2

[shadow sparrow]

$\frac{8x}{x^3}$

[muted hornet]

so subtract the exponents

[ocean seal]

Huntifer

[shadow sparrow]

for instance

like would it just be 8x^2?

[muted hornet]

so thats just $\frac{8}{x^2}$

[ocean seal]

rokt enjr

[shadow sparrow]

yeah

see my brain sees that and goes

ah yes

8x^2

[woeful pulsar]

whoops

[muted hornet]

and when x tends to inf the entire term tends to 0

[woeful pulsar]

maybe you can try plugging in a number to check too

[shadow sparrow]

Like a negative number?

or wdym

[woeful pulsar]

like any positive number

8x/x^3 for x=10 is something<1

[shadow sparrow]

Is this even further reducible though?

[woeful pulsar]

but 8x^2 is 100

[shadow sparrow]

yeah

So

x^2/x^3

Will be x^5

8x will just be 8

?

like if you tell me to divide the numerator by x^3

I think this

$\frac{x^2-8x+7}{x^3}$

Or well no

that wouldn't even be right

[ocean seal]

Huntifer

[shadow sparrow]

Which then you would factor the top

[woeful pulsar]

factoring the top doesn't really help here

[shadow sparrow]

So, what confuses me is

Honestly idk, you'd divide x^2 by x^3 which is 1/x

then -8x would become -8/x^2

[woeful pulsar]

that's multiplying

[woeful pulsar]

???

[shadow sparrow]

Or sorry

Sorry I've been doing review all day so my brain is melting a bit

So

1/x

-8/x^2

and 7/x^3 isn't reducible

[shadow sparrow]

$\frac{1-8+7}{x^3}$

?

idk

[ocean seal]

Huntifer

[shadow sparrow]

Oh, okay now I see the fraction breakdown more

But the as approaches 0 thing confuses me

Like what I'm trying to understand is the base concept so I can apply it to different questions

as I'll have different stuff on the final I'm sure

So would this apply to basically any similar equation?

@muted hornet

[oak hound]

how to find volume again?

i wanna find this

[burnt grotto]

Try figuring out how to find the volume of each layer and then adding them together

[shadow sparrow]

volume would be L x W x H in this case

where the height is just 1 unit

per layer

[muted hornet]

this would apply to questions when the limit tends to infinity

bec imagine dividing anything by like 100000000

you get a very small number right

[shadow sparrow]

So basically

[muted hornet]

so we say that limits to 0

[shadow sparrow]

Lets say theoretically

this but its x -> inf

[muted hornet]

over here x tends to 16

not infinity

[shadow sparrow]

So it'll only be on equations similiar to the one?

[muted hornet]

you can use L'Hospital's Rule rule here

[shadow sparrow]

In terms of this

[muted hornet]

dividing num and denom by the largest power is a problem solving technique used in questions when x limits to inf

[shadow sparrow]

Okay

So I can do that in basically any of them if their limits approach inf

[muted hornet]

yes

most of the time

but math is math

[tardy scarab]

Hi guys

[shadow sparrow]

Yeah..

[muted hornet]

so there are questions when you have to use 1/h

[tardy scarab]

So um I tried all the ways ik from school but still can't get it

Can someone help?

[muted hornet]

1 litre = 7 km

[tardy scarab]

This is the whole

It's kinda confusing for me

[muted hornet]

3(2/7) how many km ?

its basic unitary method

[shadow sparrow]

7* 3 2/7

[muted hornet]

23/7*7

yes so 23 km

[shadow sparrow]

Yeah or that

[tardy scarab]

Ty

[muted hornet]

i hope this was not a test

[tardy scarab]

Imma revise this after I'm done with my homework

[shadow sparrow]

$\frac{23}{7} * 7$

[muted hornet]

okay.

[ocean seal]

Huntifer

[tardy scarab]

It's a homework my test is in thursday

Sundays andnthursdays

[muted hornet]

I see. Best of luck !!

[shadow sparrow]

my final is tomorrow

praying for you

[muted hornet]

oof, then you should get off discord

[shadow sparrow]

Nein, still studying for it

I'm on discord for this discord specifically rn

[muted hornet]

alright. i gtg, hope i was able to help you !!

[shadow sparrow]

Yep! That should work if I get any approaching inf questions

If it doesn't, oh well

[tardy scarab]

@shadow sparrow What grade?

[muted hornet]

practice more

[shadow sparrow]

Worst I can get is a C in calc, but I was hoping for a better one.

I want a 4.0

but I don't see it happening

[tardy scarab]

What grade r u ink @shadow sparrow

In*

[shadow sparrow]

College

[tardy scarab]

Damnnnn

[shadow sparrow]

Final semester of my associates.

I almost had a 4.0

[tardy scarab]

Me be in school terrified of homework and u here final semester

[shadow sparrow]

I'm prob gonna have a 3.92

because of this class

🙃

[tardy scarab]

Well gg

Good luck

Gl*

[shadow sparrow]

Good luck to you too, my tip for you is

pay attention to algebra and hunt it down in high school

[tardy scarab]

Well what we are learning is pretty new due to covid and all

[shadow sparrow]

if you go to college, it'll be pain if you don't understand algebra well

[tardy scarab]

Late schedule

[shadow sparrow]

I'm suffering because I didn't understand my algebra well because I had accelerated classes and my teachers didn't help much

but I should have went and seeked different teachers

[alpine sable]

getting an error when tan^-1(1.153), can anyone explain?

[shadow sparrow]

and so now I get to deal with somewhat scary stuff like this

🙃

[shadow sparrow]

pretty sure its because of the negative.

[alpine sable]

that looks like fun

[shadow sparrow]

Not fully sure though

[tardy scarab]

Guys if a question is like

-7 + 14 = 8 right?
So then what will be
14 - -22 = will it be a negative integer or what?

[alpine sable]

negative is needed when solving for an angle, it should be

[tardy scarab]

Top one

So will it become a negative integer?

[shadow sparrow]

Yeah, I think its maybe because tan is undefined in where you're calculating it?

[alpine sable]

I meant cosine mb lol

how did I get the two mixed

but still get an error

[shadow sparrow]

cosine, sine, tan, et cetera iirc aren't defined themselves iirc

they are defined when given at a value

[shadow sparrow]

Function of operations

[tardy scarab]

Operation pretty sure

[shadow sparrow]

It would be 14 + -22

then -7 + 15

[tardy scarab]

8

[shadow sparrow]

Everything inside the brackets first.

[tardy scarab]

Oh

[shadow sparrow]

The () are put in place usually with negative numbers

Just so its clear the operations being used is to add the numbers still

The brackets are a secondary way to separate equations

[tardy scarab]

The -7 + 15 = 8 I'm pretty sure I'm confused about 14 + -22

[shadow sparrow]

what is 22 minus 14?

[tardy scarab]

8

[shadow sparrow]

A negative number plus a positive number is always going to basically be subtraction

because if you're taking the -22 and moving it towards the positive.

[tardy scarab]

Ok

[shadow sparrow]

another way to think of it also is

14 minus 22

it goes past 0

with a remainder of 8 left over, so -8

[tardy scarab]

Ok

So 8x-8

That will be -64

[shadow sparrow]

wait

8x-8

where does that come from

https://cdn.discordapp.com/attachments/490557019623915520/841172966553157672/Screenshot_20210510-084046.png

left side is gonna be 8

Or actually let me try this maybe this will help explain it

$$[(-7)+15] * [14+(-22)]$$
$$[8] * [14+(-22)]$$
$$[8] * [-8]$$
$$8 * -8$$

Yes sorry

You said 8x-8 and I thought you meant the integer x, not multiply lol

[ocean seal]

Huntifer

[shadow sparrow]

So then yeah, -64

[tardy scarab]

@ocean seal Good bot

Whoever coded it be like I made math

[shadow sparrow]

Okay now help me solve these

jk

If you think it looks cool though, here are the derivative rules

[muted raft]

Apply power rule

[full wasp]

copiously

[shadow sparrow]

My brain isn't contemplating the sqrts on b again

I know to use the power rule, I'm just kind of blanking

[muted raft]

rewrite it in exponent form.

[shadow sparrow]

Or well, do I even need to do anything with the sqrt here?

Or just apply the nx^n-1

exponent form of that would be uhh

x^2/3

iirc

or 3/2

forgot which way it goes

[full wasp]

differentiating: now with malaise

[muted raft]

x^(2/3)

always add brackets as well.

[shadow sparrow]

Yeah

Well for example it says the answer is

So I'm confused a bit

[alpine sable]

Hello

[shadow sparrow]

Is it using the difference rule or something..?

[alpine sable]

I need some help with calculus. I have to submit my homework tomorrow evening.

[shadow sparrow]

@alpine sable Channel is currently in use I think(?), check out #❓how-to-get-help

[alpine sable]

Thanks for the reference

[shadow sparrow]

But yeah, idk how it managed to find b

aside from using the difference rule

[muted raft]

$\dv{x} \qty(x^{\frac{2}{3}} - 2x^{\frac{1}{2}})$

[ocean seal]

zslya

[shadow sparrow]

yeah

[muted raft]

power rule now

[shadow sparrow]

that makes sense to me

$\frac{2}{3}x(^\frac{2}{3}-1)$

[muted raft]

yeah I get your idea.

Do it for the other term as well.

[ocean seal]

Huntifer

[shadow sparrow]

kk

$-x^{\frac{1}{2}-1}$

Not sure how to do that for the whole exponent

w/ texit

[muted raft]

${}$

[ocean seal]

zslya

[muted raft]

$\frac{2}{3}x^{\frac{2}{3}-1}$

[shadow sparrow]

woo

So just more brackets

[ocean seal]

zslya

[muted raft]

pretty much

[shadow sparrow]

is that second one wrong?

[muted raft]

It should not be 1/2 - 2

What happens if you multiply 1/2 * 2

[shadow sparrow]

yeah thats what I was thinking

2/2

[muted raft]

which is

[shadow sparrow]

1

[muted raft]

yes

[shadow sparrow]

so x

or -x in this case

Okay so I was thinking right

[ocean seal]

Huntifer

[muted raft]

What is 1/2 - 1

[shadow sparrow]

1/2

lmao

[muted raft]

-1/2

[shadow sparrow]

or well -1/2 in this case

because the 1/2 is first

yeah

[muted raft]

Now what is 2/3-1

[shadow sparrow]

-1/3

[muted raft]

yes

So what is your final derivative?

[ocean seal]

Azaradichta

$$\dv{x} x^{\textcolor{yellow}{n}}=\textcolor{yellow}{n}x^{\textcolor{yellow}{n}-1}$$

[alpine sable]

I have no context, but it seems like you're doing some power rule problem, so I sent this

[muted raft]

cool colour

[shadow sparrow]

$\frac{2}{3} x^{-\frac{1}{3}} -x^{-\frac{1}{2}}$

oops

lmao

uhhh

oh

[muted raft]

$\frac{2}{3}$

[ocean seal]

zslya

Huntifer

[shadow sparrow]

I saw it lmao

[muted raft]

You can leave it like that or bring x down since the exponents are negative.

[shadow sparrow]

Yeah

So they would be sqrts?

or

Like you'd change them back into sqrt

so it'd be

[muted raft]

you can sure.

[shadow sparrow]

Oh how do you do roots in texit

of a sqrt

[muted raft]

$\sqrt[n]{x}$

[ocean seal]

zslya

[shadow sparrow]

$\sqrt[3]{x}$

[ocean seal]

Huntifer

[shadow sparrow]

Then mix it with 2/3

to make

$\frac{2}{3\sqrt[3]{x}}$

[ocean seal]

Huntifer

[shadow sparrow]

?

[muted raft]

yes

[shadow sparrow]

Okay, thats what it did

thats really funkopop

[muted raft]

you mean funky

[alpine sable]

He meant fun-kpop

I suppose

[shadow sparrow]

silly

kpop cant be fun

they are dreading their lives

anyways

[alpine sable]

idk I don't watch it

(or listen to it whatever it is)

[shadow sparrow]

Wait a fucking minute

So d is a trap

Because its

$\frac{5}{\sqrt[3]{x^2}}$

[ocean seal]

Huntifer

[shadow sparrow]

This is just the same thing

[alpine sable]

Send your original problem

And your answer

I will check it for you

[shadow sparrow]

I finished b

It makes sense

Here

[alpine sable]

[shadow sparrow]

These are the other two problems

[alpine sable]

Try doing yourself

[shadow sparrow]

Yeah

[alpine sable]

Just use power rule

[shadow sparrow]

I think you can just use power rule on the denominators

for c

[alpine sable]

No

[shadow sparrow]

mmm

[alpine sable]

You can't distribute derivative on fractions

[shadow sparrow]

Would you uh

I don't think you could revert them to sqrt

Combine?

[alpine sable]

$$\frac{2}{x^4}=2x^{-4}$$

[ocean seal]

Azaradichta

[shadow sparrow]

[alpine sable]

Why?

[shadow sparrow]

Brain isn't processing how that works for some reason

That a mathematical rule?

[alpine sable]

you can bring negative exponents to the denominator, right?

[shadow sparrow]

I honestly don't know, I'm assuming yes

lmao

[alpine sable]

You can leave it like that or bring x down since the exponents are negative.

[shadow sparrow]

I don't remember that

[alpine sable]

You did the same thing above

[ocean seal]

Azaradichta

[alpine sable]

You did this just above

[shadow sparrow]

with the 2/3 and the sqrt

yeah

?

[alpine sable]

Here we are just reversing it

[shadow sparrow]

Sorry, I've been doing calculus review all day

[alpine sable]

$$\frac{1}{x^4}=x^{-4}$$

[shadow sparrow]

My brain is kinda I think melting a bit

[ocean seal]

Azaradichta

[shadow sparrow]

yeah

[alpine sable]

You should relax, then.

[shadow sparrow]

Final tomorrow morning

The prof told us of the review less than a few days before said exam

🙃

Really bad professor this semester

[alpine sable]

:(

[shadow sparrow]

Book isn't very useful for explaining and we have had to teach ourselves mostly

That makes sense now that I see it

[muted raft]

Basically online learning.

[shadow sparrow]

Its just inverse

[alpine sable]

nice

[shadow sparrow]

I've used it before, brain just forgot it existed

okay so

In this case 2x^-4 and 3x^-3

[alpine sable]

Yea

[shadow sparrow]

$2x^{-4}-3x^{-3}-4x$

[ocean seal]

Huntifer

[shadow sparrow]

So power rule now

[alpine sable]

Yeah differentiate it now

[shadow sparrow]

Differentiate first?

or differentiate with power rule you mean