precalculus

[viscid thistle]

haven't heard of that

[serene heath]

you can factor out (x-1)^2 from the numerator

thats what kang is sayin

[viscid thistle]

no I can;t

[clever inlet]

?

[viscid thistle]

what

[serene heath]

=pup factorise (3x^4-4x^3+1)

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@serene heath

[clever inlet]

you'll have (x-1)^2 * P(x)

where P(x) is some quadratic

yeah

[serene heath]

never say never

[viscid thistle]

oh

dang

[clever inlet]

you get the remaining factor generally by long division

[viscid thistle]

I understand now

[clever inlet]

or any division method

sometimes, there are nice shortcuts

but i don't see any here

[viscid thistle]

😃 ty

.

[serene heath]

.

[viscid thistle]

what shd I do next

the answer is a o with a slash

@serene heath @clever inlet

ø

[clever inlet]

limit does not exist

[viscid thistle]

?

[clever inlet]

wait

hmm

[viscid thistle]

Can you explain a bit

[clever inlet]

yeah

'0/0' is an indeterminate form

[viscid thistle]

yep

[clever inlet]

meaning you can do more work to possibly reach some answer

[viscid thistle]

yes

[clever inlet]

for the limit

but here

there's nothing more you can do

[viscid thistle]

oh

[serene heath]

it diverges

so limit DNE

[viscid thistle]

so once I have found out theres nothing I can do, so the limit doesn't exist

right?

[clever inlet]

kind of i guess

[serene heath]

yea it doesnt

[clever inlet]

you start to learn cases where limit dne

[viscid thistle]

my teacher just gave us this ws

[serene heath]

=pup plot (x+2)/(x-1)

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@serene heath

[viscid thistle]

and he just showed us how to do the factorization wihtout even telling us about the indeterminate form lol

[viscid thistle]

@serene heath @clever inlet Am I gonna long divide this?

[serene heath]

yea

[viscid thistle]

but

[serene heath]

top and bottom have a factor of -1

[viscid thistle]

oh

[serene heath]

so you can factor out (x+1)

[viscid thistle]

k lemme try

oh it's somehow related to the use of remainder theorem right? @serene heath

[serene heath]

t!yt polynomial long division

[brisk mica]

📽 | ** https://www.youtube.com/watch?v=8lT00iLntFc **

[serene heath]

this should help

[little hazel]

im fixin it

[viscid thistle]

Bot does it 4 u

[little hazel]

oh

[viscid thistle]

Ye, ye

[little hazel]

https://cdn.discordapp.com/attachments/363224154469826562/537626227926892554/JPEG_20190123_083414.jpg

[viscid thistle]

Post again

,rotate -90

[obsidian monolith]

Couldn't find an attached image in the last 10 messages

[viscid thistle]

:l

Heeeh

[little hazel]

try this

oh wow there we go

[viscid thistle]

Alright

[little hazel]

well i need help figuring out part B

idk what the H is there for

guess ill move it to one of the channels down below

[eternal lotus]

I guess once you've simplified the arc between x=2 and x=2+h

You can sub h=1

To get the ARC between x=2 and x=3

[little hazel]

but how do i simplify the arc between those two if i dont know what h is

[viscid thistle]

=pup graph 2x^2-x

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@languid wind

[eternal lotus]

The way you simplified the ARC between x=-1 and x=0

Once you get an expression that gives the ARC between x=h and x=h+1 in terms of h of course u sub h=1

x=2 and x=2+h * sorry

[little hazel]

so h is just 1

[eternal lotus]

Yeah becaue of h=1 we get the ARC between x=2 and x=2+1

Which is the arc between x=2 and x=3

Which is what we want

[little hazel]

oh ok i see

you just try to guess what h is

not really guess but yeah

[hexed ermine]

@ocean mauve ?

What do you need

[serene heath]

theyre just practicing their camera angles dont mind them

[ocean mauve]

^^^^

[fair solar]

what are the key differences between a hyperbola and ellipse

[hexed ermine]

Both are very similar ellipse has a +, hyperbola has a -

[kind pier]

is there a way to find the equation of a half of a curve like in a supply and demand graph so you can estimate the future if the trend continues?

[viscid thistle]

uhh

arc length maybe?

[slender river]

wat

what is "half of a curve"

[viscid thistle]

id assume that he means something like arc length

say you have a curve in the domain 0 < x < 2

take half of that

.. or something like that XD

[slender river]

[fallen cloud]

Sounds like differentiability and Taylor series

[deft flume]

how do i solve sin(B)=10(4)/sin(36deg.) ?

The given were A=36deg., a=10, and b=4

Law of Sine, to solve the triangle

any help would be nice

oh nvvm, i knew i had to do the inverse sine to both sides

for some reason, i cant do thr inverse sin in the calculator

its in degree mode

[atomic zodiac]

why not

[deft flume]

idk

im doing sin^-1(10(4)/sin(36)

just gives me domain error

[atomic zodiac]

that would be because the domain of sin^-1(x) is -1≤x≤1

[deft flume]

nvm, i got it

[cosmic furnace]

Very small question regarding transformation of quadratics
Would a horizontal shrink by a factor of 2/3rds be (1 ⅔X)^2?

[spice mountain]

For number 6) a) and b) my teacher graphed them like this. I don’t know how she did it. Can someone pls help explain?

And also I don’t know how to solve a quadratic inequality from looking at a graph. Can someone help explain?

[clever inlet]

do you have trouble with graphing quadratics in general

or just the regions for inequalities?

[spice mountain]

Regions for inequalities @clever inlet

Like I don’t know how she got there

[clever inlet]

well it's either inside or outside

[limber bone]

vegangod

what a terrible name

anyways

its too easy

if y is more

like y >3x+4

then u shade up

if y is less

y <3x+4

u shade down

=pup grpah y > 3x +4

[limber bone]

= pup graph y < 3x +4

[granite stirrup]

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Query made by

@limber bone

[limber bone]

more

above the line

u shade above the line

same with anything

imagine quadratics like two lines

if y > more

then shade above them

smae with absolute functions

and so on

got it?

[spice mountain]

For quadratics though I know how to do linear just not quadratic

[limber bone]

imagine quadratics as like

2 lines

shade above them

[clever inlet]

^

[limber bone]

shade above them quadratic it self

cuz thats y XD

[clever inlet]

when in doubt though, just test a point

[limber bone]

ye

but thats like

when ur really worried

like htink of it

if i want the values more than y

and i have y grpahed

then i shade higher thany XD

than y*

right?

=pup graph y > x^2

[limber bone]

=pup grpah y < x^2

[granite stirrup]

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Query made by

@limber bone

[spice mountain]

Like I know how to graph it I have to put it in the a(x-h)2+q

Too fastttt

[limber bone]

https://media.discordapp.net/attachments/363224154469826562/537936803554918410/result.png this is the graph of y > x^2

see how he shades above the lines

of the quadratics * assume they are lienes *

lines*

got it?

[spice mountain]

No sorry

[limber bone]

its ok

y > x^2 for example

to graph inequalities

all they differ from than normal equations

is that just the shading

and there are only two places where u can shade

above or below

how do u know? from the sign ofd the inequality

if y > then shade up

if y < then shade down

thats it

and note something small

if y > then do dotted lines at the end

if y>= then graph it normally and just do the shades

got it ?

[spice mountain]

y>x^2, why isn’t it at (0,0) for the origin of the parabola ??

Oooooo okay I understand the shading now

Also for this question: how do you find the inequality equation

??

The check mark is my teachers answer, I don’t know how she got it

[clever inlet]

well once you got the equation for the quadratic itself

you apply the same idea from above with the above or below

and the curve is dotted

which is used to represent strictly less than, or strictly greater than

strictly less than in this case

[spice mountain]

I don’t understand what’s your talking about

[limber bone]

he means

its the same thing as like wrinting an equation of a normal parabola

just differ the inequality sign

and if its weak or strong

here its dotted

and below the parabola

so y less

y <

go tit?

got it*

[spice mountain]

Ok so here we only have the x intercepts

Yes

[clever inlet]

was there working before this?

[spice mountain]

I understand that

[clever inlet]

cause she seemed to jump a bit ahead

you would generally need to find the leading coefficient

using a third point

usually the vertex

[spice mountain]

No she’s a bad teacher she just writes the answers and sits in for the rest of the time

[clever inlet]

here

[spice mountain]

If u ask her question she gets mad

[clever inlet]

she skipped an important step

i would think

ehh

[limber bone]

tell her u love her

[earnest nymph]

The y intercept maybe help?

[clever inlet]

what's the y int?

you can use that point

[earnest nymph]

-12

[clever inlet]

ah yeah

[spice mountain]

[clever inlet]

that point would be useful in finding the leading coefficient

your teacher wrote 2 for it

which i assume is correct

but with like no working out

y = a(x-3)(x+2)

plug in (0, -12)

solve for a

that's generally how you would do it

[spice mountain]

The drawing isn’t on zero so I got confused idk my friend said just say it’s zero cuz she drew the graphs up

[clever inlet]

what do you mean?

[spice mountain]

Why y=a(x-3)(x+2) where’s does this equation come from

[clever inlet]

It's the more general roots forn

[limber bone]

so u have problems with graphing tghe function as itself

not the inequalities

right vegan

?

[clever inlet]

Your two roots aren't enough to determine a unique equation for a quadratic

[spice mountain]

The vertex isn’t exactly on 0,-12 so I was confused with that

[limber bone]

^ ye but in most cases they are

but like

sometimes the parabola

is ab it stretched

from this a factort

so u need another poiunt

which kangaroux said

the y intercept

[clever inlet]

You can use any third point

[limber bone]

ye

but easiest

right?

[clever inlet]

(0, -12) isn't a vertex, but it is the y intercept

In this case

[spice mountain]

Oooo

Okay

Yes @limber bone

Ooo is that slope point form?

[clever inlet]

That's kind of specific to linear functions

[spice mountain]

The a(x-3)(x+2)

Ooo oof

[clever inlet]

a(x - r1)(x - r2)

Where r1 and r2 are roots

[spice mountain]

[clever inlet]

That's the general roots form for a quadratic

[spice mountain]

Yes

What’s the a then

[clever inlet]

You have your third point (0, -12)

Plug 0 into x, and -12 into y

And solve for a

[spice mountain]

Is a the slope?

[clever inlet]

Ehh

In a sense, it affects steepness

This is the type of situation where playing on Desmos helps

[limber bone]

^

the more the a the narrower it gets

[spice mountain]

Sorry I don’t understand what this equation means the y=a(x-3)(x+2)

And why did the signs become opposite for 3 and -2 when you put it in

I just don’t understand what the equation means

[clever inlet]

Well, what happens when you plug in x = 3

y becomes 0 right?

This is what we want

[spice mountain]

Ooo yea

What’s does the a represent tho ?

[clever inlet]

essentially now narrow or wide it is

the cool thing is, no matter what a is in that form

it will always pass through the given roots

[spice mountain]

Ok thanks guys I understand now

[coral mulch]

hello

i have a question its very simple

but my brain isnt working atm so i need help

make (A/t )the subject of A=-A_0 *e^ -(lamda *t) where lamda = ln2/t_(1/2)

so this

make A/t the subject ^

[fickle moat]

how do i find my domain and range for this?

[viscid thistle]

When finding the domain and range of an inverse function we swap the domain and range of the parent function right? So the domain of tan is (-pi/2, pi/2) and the range is +/- infinity. This would mean the domain of arctan is +/- infinity and the range is (-pi/2, pi/2). Is that correct?

[viscid thistle]

<@&286206848099549185>

[serene heath]

@viscid thistle yes

but yea the domain is restricted so its one to one

[viscid thistle]

The domain of tan is restricted to (-pi/2, pi/2) or the domain of arctan is restricted?

@serene heath

Cause my understanding is that the domain of of arctan doesn't need to be restricted since it isn't periodic right? Wait is arctan a periodic function or no since we only map it from the restricted version of tan?

[serene heath]

sorry the range is restricted so that the function isnt many to one

not the domain, domain is all real

[viscid thistle]

Ahh that makes more sense. So can arctan be periodic? Is arctan able to be defined for other domains of tan? Like would we ever graph arctan from the tan of (-pi/2 + 2pi, pi/2 + 2pi)

What would it look like if we graphed arctan without the range restricted?

[serene heath]

no arctan is the inverse function of tan defined so its not many to one

if you change the defined range then its not arctan anymore

if you graphed the inverse of tan it'll just look like the tan graph but reflected across the line y=x

@viscid thistle

[viscid thistle]

"sorry the range is restricted so that the function isnt many to one" Thinking about this is a little confusing to me. I mean, I've never heard of range being restricted, except indicentally by restrictions to the domain. Does that make sense? Are you saying that the range of arctan is restricted by the restriction to the domain of tan?

@serene heath

[serene heath]

sorry im explaining this horribly

http://mathonweb.com/help_ebook/html/functions_2.htm

here this site explains it well

and sorry for the delays im watchin a footbal game lol

@viscid thistle

[viscid thistle]

No worries

Thanks for the link, i'll read it

[kind pier]

anyone have the ebook of this pre cal book? Author: Stewart
Edition: 7th
ISBN: 9781305701618
Copyright Year: 2016
Publisher: Cengage Learning

[viscid thistle]

I have actual books not books that you read online

[fresh crow]

@fickle moat should be all real numbers (-infinity, infinity) i believe

bc the domain goes from everything below zero to everythinf above and including zero

range should be same thing

bc it's a linear function

(were it a quadratic or something you'd have a min/max value or vertex for the range, even with a domain of all real numbers)

[spring thunder]

what about everything between 2 and 8 for the range ?

and that thing is 3 days old w/e

[viscid thistle]

@kind pier I highly recommend by AoPS' Precalculus book. A link to this is here: https://artofproblemsolving.com/store/item/precalculus

It is very comprehensive and even if you took the class, you will still benefit

[fresh crow]

oh yeah @spring thunder

piecewise

[scarlet basalt]

Alright, so on the unit circle, can someone tell me the pattern for radians? I identified the pattern for the degrees (like for example, the 1st and 4th quadrant, if you imagine the triangle that's 30 degrees upside down, it hits 330 degrees, and 330 + 30 = 360. This applies for quadrant 1 and 4, and quadrant 2 and 3.) and I know it with the coordinates (like Q1 would have (1/2, √3/2), and Q2 would have (-1/2, √3/2)), but what is the pattern for radians?

[rose pollen]

I'm not sure what you're asking here, but the pattern doesn't really change

It's just that the circle is expressed in terms of radians instead of degrees, so if we'd take your original 330 + 30 = 360 that'd be 11/6pi + 1/6pi = 2pi instead

[scarlet basalt]

Oh wait

I was trying to identify the pattern

of guessing what the radiant would be

but that helped me recognize it

thank you uwu

[rose pollen]

no problem uwu

[earnest finch]

My class just started polar graphing

How the fuck do you actually do that

I'm so confused

[odd yarrow]

graph the poles

https://www.dummies.com/education/math/calculus/how-to-plot-polar-coordinates/

dw you're not a dummy dad

[hollow plover]

How do find out if the graph is symmetrical to x axis y axis and or origin?
i know how to do it with equations but not graphs

[patent beacon]

If you flip the graph over the x-axis and it's the same, then it's symmetric about the x-axis

Similar idea with y

,ask graph cos(x)

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[patent beacon]

You can see from the higher graph that cos is even

The left is a mirror of the right

[hollow plover]

ok so basically if you want to find out if lets say a point on the graph is symmetrical to the y-axis, you have to take the x coordinate in (x,Y) and negate it?

sorry i just started my precalc class and we didnt really go into cos and stuff yet

although we learned that in geometry but i forgot most of that

this is from our first class

pls i beg

😦

[patent beacon]

Actually I derped. No function is symmetric about y because then it's not a function

[hollow plover]

so the graph is not symmetric to the x axis y axis or the origin?

[patent beacon]

Well, there is ONE.

[hollow plover]

my teacher says its symmetric to the y axis

[rocky bison]

No function is symmetric about y?

[patent beacon]

OOF I'm mixing up my axis

Don't mind me just braindead

[hollow plover]

np

[patent beacon]

A function is even if the y-axis acts as a mirror for the function

We say it's symmetric about the y-axis

You can see cos(x) is even.

I'm not sure if your graph may be poorly drawn, but it doesn't look even. Almost though

[stuck palm]

Does anyone have any understanding on how to do this please? I need help, it’s confusing me :/

[patent beacon]

Polynomials are continuous everywhere, so no need to check anywhere but x = 1

A function is continuous at a point if:
-The limit exists there
-The function is equal to the limit at that point

[stuck palm]

So how would I check x=1 to see if it's continuous?

[spring thunder]

test the limit of your function at x=1

does it exist ?

[stuck palm]

Thats the issue, I don't even know to test the limit... I am so bad at bath

math*

I mean I would assume if the left = the right, then the limit exists, right?

[spring thunder]

yeah

check if the left lim = the right lim

[stuck palm]

I would say no? Because 1^3+6= 7 and 7 ≤ 1 does not make sense.

[spring thunder]

why 7 =< 1?

i mean, what does it have to do with our thing?

but yeah 7 is the left limit at x=1

and the lim to the right is ... ?

rip

[drifting imp]

can someone help with this question? its a logarithmic question based on exponential growth and decay

Dry cleaners use a cleaning fluid that is purified by evaporation and condensation after each cleaning cycle. Every time the fluid is purified, 2.1% of it is lost. The fluid has to be topped up when half of the original fluid remains. After how many cycles will the fluid need to be topped up?

Im not great with expo growth and decay in terms of finding an equation from it

[viscid thistle]

ok so you know you need to reach 50% of the original value

[drifting imp]

yea

[viscid thistle]

and you also know that your decay is 2.1% / purification

[drifting imp]

yea

[viscid thistle]

going by the formula of

,$ x_t = x_0(1+r)^t

[obsidian monolith]

yami:

[viscid thistle]

x_t is your 50%

x_0 is the original amount of fluid you had

t is your time

what is r?

[drifting imp]

wouldnt it be 2.1% or 0.21?

[viscid thistle]

0.21 = 21%

2.1% = 0.021

[serene heath]

theres no time dependencay tho

[drifting imp]

yea i know that

[viscid thistle]

then because its a decay you have to subtract that value

so you end up with

,$ 0.5 = x_0(1-0.021)^t

[obsidian monolith]

yami:

[drifting imp]

so x_0 would be 1? because of how much I initially started with

[serene heath]

[viscid thistle]

i guess yeah

[silk oriole]

wat is haeppning

[serene heath]

youre overcomplicating it

why is there a t

[drifting imp]

idek lol

thats what im getting confused on

[serene heath]

if 2.1% is lost each time then youre left with 97.9% yes?

if A is how much you had initially

then after i use youd have A*0.979

after 2 uses A*0.979^2

so on...

so in general after n uses

[drifting imp]

im lost tbh

[silk oriole]

this is complicated...

[serene heath]

ok lets restart

you have 100% to begin with

[drifting imp]

yes

[serene heath]

after you use it once yoy lose 2.1% of it

which means youre left with 97.9%

agree

?

[drifting imp]

so 1-2.1%?

[serene heath]

wdym

100-2.1 yes

[silk oriole]

yes

then

[serene heath]

or 97.9%

[drifting imp]

nvm i got that

[serene heath]

yea

[silk oriole]

then

[serene heath]

so after one use youre left with 97.9% of what u had originally

if originally you had A amount of fluid

then after one use youre left with 97.9% of A

or A*0.979

[drifting imp]

when you say "A" you mean A=100 right?

[silk oriole]

[serene heath]

yea

[drifting imp]

kk

[serene heath]

A is what we had

at the very start

[drifting imp]

so would the formula be y= A(2.1)^t ?

[viscid thistle]

t usually stands for time but it can be anything really

[serene heath]

in this case the formula is A*(0.979)^n

where n is the number of uses

that formula tells you much fluid you have left after n uses

[drifting imp]

I GET IT NOW LOL

thanks guys

[serene heath]

you sure?

you know how to answer it now?

[drifting imp]

yea I can solve it from here

[serene heath]

nice

[drifting imp]

all i needed was to understand how to find the formula for it

[slow wharf]

I am having problems with this problem

It is simple but I keep getting it incorrect

The sum of two natural numbers is 20

Find the numbers so that the product of the square first and cube second is maximum

And I am stupid

I keep writing product as + goddamnit

[uneven tendon]

Okay, so

Let's just say x+y = k for simplicity

x, y and k being natural numbers

Note that x = y-k

Thus $ x^2 + y^3 = (y-k)^2 + y^3 $

[obsidian monolith]

Nuke:

[slow wharf]

What

I think I got it

$a+b=20\ a^2b^3=max$

[obsidian monolith]

Autistic Hoodie:

[uneven tendon]

OH

[slow wharf]

$y=a^2b^3 \ y'=?$

[obsidian monolith]

Autistic Hoodie:

[uneven tendon]

Sorry man I'm fucking retarded

Misread it

[slow wharf]

I did too, lol

Wait this is confusing me

So I got

$b^2-32b+240$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

The result of that is 20 and 12

[uneven tendon]

Wait, no

[slow wharf]

But when I use the $x_{12} = \frac{-b=-\sqrt{b^2-4ac}}{2a}$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

I get it incorrectly

[uneven tendon]

You get $ a^2(20-a)^3 $

[obsidian monolith]

Nuke:

[uneven tendon]

Differentiate w.r.t a

[slow wharf]

$(20-b)^2b^3$

Work too

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Or should work too

[uneven tendon]

Yeah same thing

Wait fuck

[slow wharf]

$(400-40b+b^2)b^3$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

This is confusing me

Lets look at this polynomial

$x^2-32x+240=0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Wolfram says the answer is 12 and 20

Which is apparently correct

[uneven tendon]

I mean, you're not supposed to use wolfram

[slow wharf]

But when I use the formula

[uneven tendon]

Hold on a sec tho

[slow wharf]

$x_{12} = \frac{32+-\sqrt{(-32)^2-41240}}{2}$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

It gives incorrect answers

[uneven tendon]

Let $ f(b) = b^3(20-b)^2 $, thus $ f'(b) = 3b^2(20-b)^2 - b^3(20-b) $

[obsidian monolith]

Nuke:

[uneven tendon]

I mean

The answer are the roots of f'(b)

In fact there can only be

Yeah I'm not sure how many roots it has, tbh

But if you are allowed to use wolfram then you just need to get those roots

[slow wharf]

The polynomial is killing me

Why does the formula for the polynomial not work?

[uneven tendon]

Because honestly I have no idea where that polynomial of yours came from

[slow wharf]

Does it matter where is came from?

I made it up

Why is x_12 wrong Q_Q

[uneven tendon]

Ah

It's not wrong tho

The determinant is equal to 64

Thus x = 16 +- 4

Which is 20 and 12

[slow wharf]

Damnit

[uneven tendon]

[slow wharf]

I calculated 4*240 as 980

I think I need to go to slepe

I am too much deconcentrated

:/

[uneven tendon]

Yeah man that happens

Don't you wanna get done with this problem first tho?

[slow wharf]

I got it, it's correct

You wanna know how I got b^2-32b+240?

[uneven tendon]

Yeah sure

[slow wharf]

$y=(20-b)^2b^3 \ y = (400-40b+b^2)b^3 \ y = 400b^3-40b^4+b^5 \ y' = 1200b^2 - 160b^3 + 5b^4 \ y' = 5b^2(240-32b+b^2)$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

When is y' equal to 0

When 5b^2 = 0 or (240-32b+b^2) = 0

[uneven tendon]

Oh, that's true

[slow wharf]

Dunno if there was a simpler way

[uneven tendon]

I'd have needed to put this on paper to realize it

No no yours is perfectly fine

Well done!

[slow wharf]

Thank you ^_^

I need to practice math every day now...

I failed my college exam :/

[uneven tendon]

Oh man I'm so sorry to hear that :(

But hey you seem to be doing pretty fine now

If math was what pulled you down then if you keep practicing it'll stop being a bother in no time

[slow wharf]

Yeah

Motivation is the most important thing

[limber compass]

Can someone visually explain what we are getting when we are finding the gradient at a point ?

For example : http://bit.ly/2DFKkwM

I want someone to show me a visual presentation.

bcos i dont really understand. i am verry new

[eternal lotus]

Essentially we are finding the gradient of the tangent to the graph at x=2

https://www.desmos.com/calculator/xrvtve2rfs

[slender river]

get them the limit definition of the derivative

[limber compass]

Hmm. Sooo what is that blue line ?

Is it that tangent shiz i've been hearing about ?

@slender river

[slender river]

yes

[limber compass]

Can u explain a tangent to me 😃

[slender river]

okay!

first

u gotta tell me some things

just so u learn better

[limber compass]

Ok...

[slender river]

tell me the slope formula

[limber compass]

for what ?

[slender river]

slope of a line between two points

[limber compass]

to get m ?

[slender river]

yeah just generally

getting m

[limber compass]

Ohk. You get the derivative by differentiating

[slender river]

wat

[limber compass]

and then you plug in the number where x = __

[slender river]

u skipped a lot

lol

ok

[limber compass]

Ok I can give an example

[slender river]

uh okay

[limber compass]

Ok so find the slope of f(x) = x^3+4x-5 at x = 1

So first I have to differentiate that

which is f'(x) = 3x^2+4

and then i plug in X into that derivative

Is that it ...

[hexed ermine]

mhm

[limber compass]

Is that a yes? Or thinking like " Hmmmm "

[hexed ermine]

yeah

youre right

[limber compass]

😃

@slender river You there bud 😃?

[hexed ermine]

@∮c F ⃑(x(t),y(t)) ∙ dr ⃑(t) = 0 You there bud 😃 ?

[limber compass]

@hexed ermine Could ya explain it to me... Since he is afk

[hexed ermine]

Explain what

you did it correct

[limber compass]

Scroll up and read it...

[hexed ermine]

You want to know what a tangent is?

[slender river]

oh oof i was afk yes

sorry

[limber compass]

Its fine.. So i did it ^ the slope formula

[slender river]

yeah that's the derivative

and then at the point x it has the slope of what u evalueate the derivative at that x point

[limber compass]

😕 ( confused ) ^

[slender river]

uh

so u got like

[limber compass]

That blue line you showed me

[slender river]

3x^2 +4

[hexed ermine]

The derivative of a function calculates the slope of the tangent line at that given point

[slender river]

yes

all u need to do is get the position u want the tangent line is

[hexed ermine]

When you differentiate that function, it creates a function of the tangent line slopes to that original function

[slender river]

so u need the x and y value of the point on the curve u want

[hexed ermine]

so you can plug 1 into x for the derivative to get the slope of the tangent

[slender river]

when x=1 only

that particular slope is particular for x=1

[limber compass]

Ohk I think i get you guys.

So basically the tangent is the slope of the function at a give point.... rrriggght ?

http://bit.ly/2DFjpl2
So that blue line is called a tangent

[hexed ermine]

mhm

[limber compass]

Ohk.

[hexed ermine]

the instantenous change

to get an average of two points youd get a secant line

[limber compass]

isnt that f'(n)
where N is the number at where you want the slope for?

[fallen kite]

how much harder is calc 1 vs precalc2/trig

[hexed ermine]

but with a tangent its essentially the slope of the line but only 1 point is needed

Not very @fallen kite

[limber compass]

Ohk.

[fallen kite]

ok thats good to know, because im having enough trouble with this shit

u guys recommend any websites? ive just been looking up random youtube vids

[limber compass]

@fallen kite I am on the same page here

http://bit.ly/2DEIyfn

Yea. If u guys recommend any worksheet websites as well would be great.

[hexed ermine]

I self taught calc by watching Professor Leonard

he is really

good

[slender river]

good indeed

like

i love his work

it's a bit lengthy but it's fanstastic

like ur actually in the lecture hall and junk

super gud

[limber compass]

Hey @hexed ermine So i just showed you what level i am on in calculus. Should I be learning limits rn ?

[fallen kite]

hmm ill check that out how many hours a week were u going at it

[hexed ermine]

if you havent learnt them, yes

ehh, about 4

for about 2 weeks

@fallen kite

and now im here helping ppl with calc :v

[fallen kite]

2 weeks as in what, 80 hours?

[hexed ermine]

no no

about 8

I watched his vids at 2x speed

[fallen kite]

u learned calc in 8 hours starting fresh? i dont believe that lol

[hexed ermine]

His calc 1 series yes

I've did alot of practice on my own but ive spent 8 hrs of watching

[slender river]

yeh it's def possible

[hexed ermine]

his vid series

[slender river]

i am of the same boat sort of but i dont have enough practice bc i cannot find problem sets to save my life

[fallen kite]

its not possible to learn all in 8 hours.. sure maybe u can watch all the vids, but to understand and retain all of it you'd need to be a god

[hexed ermine]

Ive had a steady understanding

Helping others and doing problems on my own definitely helped

[fallen kite]

ok maybe in that case 😃

are u still in high school?

[hexed ermine]

Yes

[fallen kite]

thats a good idea to get ahead

[limber compass]

Did i do this right ?
f(x) = 3x^2+5 find the equation of the tangent line when x = 2
my answer was : y = 12x-17 ?

[viscid thistle]

derivative
6x
Slope at x=2, 12

y=mx +b

y=12x

Or

y-y1=m(x-x1)

X1=2 so y1 = 17

y-17=12(x-2)

@limber compass

[limber compass]

Thank you 😃

So i was right ! Yea

[viscid thistle]

,w graph (y-17=12(x-2)) and (y=3x^2+5)

Lets check

@limber compass

[limber compass]

Oh yea How do i check That is what i needed help on

[viscid thistle]

y-17=12x-24

y=12x-7

go to

[limber compass]

Can u teach me a way to check if the equation of the tangent line is correct?

[viscid thistle]

Demos.com/calculator

[limber compass]

I want to check without a calc

[viscid thistle]

Then type in the line and the original graph

Graphing is only visual way to confirm

[limber compass]

Ohk. I guess

Thank you man 😃

[viscid thistle]

I'll tell you though

Your- 17

Is wrong

Its -7

Via method I showed you

[limber compass]

Oh. what

Oh. Sorry that was a typo

[viscid thistle]

Oh you mean 7 turn

[limber compass]

sorry. I have it on paper.
C = -7

[viscid thistle]

Then*

Yeah

[limber compass]

17-24 = c

Yea.

[viscid thistle]

Type those both into desmos

You'll see exactly why is right

[limber compass]

It will touch the point 2

right ?

[viscid thistle]

Yup

[limber compass]

Ohk let me try it

[viscid thistle]

Its a clear tangent

[limber compass]

should i write f(x)

or its fine if i dont

[viscid thistle]

Nah just put

y=12x-7

new line

y=3x^2+5

[limber compass]

😃 👍 http://bit.ly/2DFA9s6

[viscid thistle]

Yee

[limber compass]

One more _quick_question.

[viscid thistle]

Ok

[limber compass]

Do u know any websites that have worksheets realted to these type of questions ?

[viscid thistle]

Only one I could guess is Khan Academy

[limber compass]

what is this again? lol
equation of the tangent line at a point ?

[viscid thistle]

Right

Tangent lines

[limber compass]

right as in right or write ?

[viscid thistle]

You called it by the correct name

[limber compass]

Oh. : )

[coral cairn]

I've spent half an hour staring rock hard at this thing, browsed khan academy, etc; and I want to believe that the y-int is 1, what am I doing wrong?

[viscid thistle]

y-intercept?

[coral cairn]

f(0)

[viscid thistle]

well it is 1

[coral cairn]

okay

the solutions were given by my prof

she's known to make typos

[viscid thistle]

most definitely a typo

[coral cairn]

if anyone finds anything please tell, my prof sent an email correcting some typos but did not address this question

[limber compass]

@viscid thistle Hi bro 😃

https://www.desmos.com/calculator/jqdn5brojv

I am having problems finding the equation of the tangent.

[viscid thistle]

what point do you want the tangent at?

[limber compass]

x = 5

[viscid thistle]

what's the derivative of f at x=5?

[limber compass]

459

[viscid thistle]

good

what's the value of f at x=5?

[limber compass]

Woah. how did u get it that fast

[viscid thistle]

$f'(x) = 18x^2 + 9$

[obsidian monolith]

TendentiousTorturousTopics:

[limber compass]

what is that ^

OH that derivative lol ya soz i had a brain fart

[viscid thistle]

$f(5) = 800$

[obsidian monolith]

TendentiousTorturousTopics:

[viscid thistle]

soo y=459x-1495; you got it right

[limber compass]

Hmm. then why can't i see the line?

[viscid thistle]

because you need to zoom in

[limber compass]

tangent i mean

[viscid thistle]

the curve is very steep

[limber compass]

I did.

[viscid thistle]

derivative of 459 means the slope is 459 lol

the red thing is the curve, and the blue the line I think

[viscid thistle]

might be the other

yee Gamedolf to the rescue

[atomic zodiac]

you can adjust the axis with the gear icon on the right

[limber compass]

How ?

[atomic zodiac]

change the values

[limber compass]

OHH

what did u change too?

I dont know what to change it too lol

[atomic zodiac]

idk i'm on phone

[limber compass]

ohh

[atomic zodiac]

can just drag each axis individually

-10,000 ≤ x ≤ 10,000 or so

no

[limber compass]

Thank you 😃 @viscid thistle and @atomic zodiac

[atomic zodiac]

-10,000 ≤ y ≤ 10,000 or so

[limber compass]

love ya ( no homo )

Oh.

[atomic zodiac]

-10 ≤ x ≤ 10

it depends on your viewing frame

but ya you can just adjust

[limber compass]

thnx

[atomic zodiac]

np

[desert lantern]

How can I go about solving this inequality? Really confused about how to get x by itself when it's in the denominator

https://gyazo.com/f67ce13a3e8fe2e045fbf0d12475798d

[tawny nacelle]

add 5 and take the reciprocal

btw talking the reciprocal flips the direction of inequality

[atomic zodiac]

you could also consider each inequality separately to prove it to yourself

ie -1 < 120/x - 5
and 120/x - 5 < 1

[desert lantern]

@atomic zodiac Can you take each part and put it back together? I've never thought about doing that

@tawny nacelle That makes sense, thank you!

[atomic zodiac]

just do min < x < max

[tawny nacelle]

[desert lantern]

@atomic zodiac gotchya, thanks!

[queen lion]

Hey guys!!I want to ask if this answer is correct or not?

[limber compass]

Hmm. So what can u only get the derivative of ?

I know u can get it for a quadrative equation

but can it be ^4 ? ik that 3 is cubic

[atomic zodiac]

not sure what you mean @limber compass

that looks right @queen lion

[queen lion]

Logic is also fine,right?

[atomic zodiac]

not sure about getting rid of the sin(x)

but the answer is right

🤷

[queen lion]

Yup thats where I am in doubt too

Thanks anyways

[atomic zodiac]

why not keep it in?

[limber compass]

@atomic zodiac So..I tried this function
4x^4+3x^3 + 8
and seems like 4 as an exponential doesnt work

[atomic zodiac]

you're trying to differentiate that?

it does work

you get 16x^3 + 9x^2

[limber compass]

No i am trying to get the equation of the slope

[atomic zodiac]

yes the derivative

[limber compass]

I mean the equation of the tangent

[atomic zodiac]

yes

that is it

[limber compass]

wait I will give u the actual equation

7x^4+8x^3+9x p(0,4)

find the equation of the tangent when the coordinate is (0,4)

[atomic zodiac]

right

so find the derivative of the equation first

28x^3+24x^2+9 yes?

then plug in the x coordinate

28(0)^3+24(0)^2+9=9

which means your slope is 9

use the equation of a straight line with the given points to get the equation of the tangent

y=mx+c

4=9(0)+c so c=4

so the tangent line at p is y=9x+4

[limber compass]

Yes I got that.

But then i tried using desmos to check

but i dont understan

[atomic zodiac]

well

your point p isn't on the curve

unless your equation was supposed to be 7x^4+8x^3+9x+4

[limber compass]

Ok so what should i be learning now ...

I have learned finding the equation of the tangent line .. now what ? @atomic zodiac 😃

[atomic zodiac]

what do you mean

idk lol i just do maths for fun

[limber compass]

fun.

Doesn't need help has left the chat ...

[atomic zodiac]

o/

[valid vector]

So my professor gave us this function to graph: f(x) = -2 (x - 4)^2 + 3. According to this site, it should end up looking like this: https://www.desmos.com/calculator/iettzueevt

However, if I wanted to do the table of values, she told us that we couldn't use any random numbers for the domain, instead we need to use specific values like -3, -4, 5 and 6, but why specifically those and not something like -1 through 2?

[serene heath]

you can use any numbers

just make sure you have a wide enough range

so you get the shape right

[valid vector]

OK, thank you!

[limber compass]

So my sister was telling me to learn something like maximum and minimum

any idea on what ? she was talking about

and any yt tutorials would be great if u could link me

[viscid thistle]

like extrema of a function?

[atomic zodiac]

probably

[viscid thistle]

https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-analytical-applications-new/ab-5-4/v/testing-critical-points-for-local-extrema

[limber compass]

Uh Ok 😃

Can someone explain me why we differentiate to find the rate of change ?

http://bit.ly/2UrZRGd

[atomic zodiac]

because slope is rate of change?

[limber compass]

Doesn't differentiating something just give us the gradient function of the original graph.

[viscid thistle]

wow they've really dumbed things down

made a cookbook to solve problems

excellent teaching; might as well just teach the computer to do that for you

wow there let's not talk about gradients

We don't need to get into the definition of gradient, which coincides with derivative for functions of 1 variable

differentiating something gives you the instantaneous rate of change

[limber compass]

http://bit.ly/2UukRMF ?

what is the instantaneous rate of change

[viscid thistle]

yes

the rate at which the height is increasing

[steady cove]

is that a better answer or is logx^2

or it doesnt matter

[atomic zodiac]

2logx

[steady cove]

is there a reason?

[atomic zodiac]

yes

you have 3logx on the left so you can simplify

[steady cove]

mmmk

but if they're the same thing why does it matter?

[atomic zodiac]

one is simpler than the other

[steady cove]

mk

[atomic zodiac]

same reason you don't write 10 as 100000/10000 every time

[steady cove]

ah ok

thxs for explaining

[atomic zodiac]

np

[limber compass]

How does this work ?

http://bit.ly/2DKvS6T

How did they get the vertex of the parabola at 2 ?

[clever inlet]

This is kind of calc

It's a point of inflexion

[limber compass]

?

[clever inlet]

Stationary points on the derivative are points of inflexions

Or possible points of inflexions

[limber compass]

ohk.?

[clever inlet]

It's where concavity changes

[limber compass]

Ohk. I don't understand these big words

[clever inlet]

Are you studying calculus?

I guess alternatively

You could think of x = 0 as where f(x) is most steep positively

With a slope of 2

Actually I can't read

Y coordinates are difficult

For derivatives

[atomic zodiac]

think about what the 2 means

it's the highest point on the parabola

meaning the highest slope will be at the same x coordinate

[clever inlet]

^

The actual value of 2 can be difficult to get when sketching derivatives

[atomic zodiac]

greatest positive slope rather

[clever inlet]

Like, it's hard to tell the slope is 2 exactly

But what's important is that's the greatest value of the derivative and it's positive

[slow wharf]

I've realized that I have big problems combining fractions with minuses

For example

$\frac{-3x^2-5x-2}{2\sqrt{1-x}} + -x\sqrt{1-x}$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

How would I interpret this?

$\frac{-3x^2-5x-2+2\sqrt{1-x}(-x\sqrt{1-x})}{2\sqrt{1-x}}$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Apparently it's incorrect?

[serene heath]

no thats correct

[slow wharf]

I failed somewhere else then...

damnit

[earnest nymph]

Probably need to expand the bracket and continue to do?

[viscid thistle]

What are you trying to do? Simplify?

[viscid thistle]

how am I supposed to do this

[tawny nacelle]

i thonk i know dis

[viscid thistle]

yeah?

I know it is 0

[tawny nacelle]

let the r + 1 th term be the what u want

[viscid thistle]

right?

the coefficient of x cube

[tawny nacelle]

and r + 1 th term is ${100 \choose r}(\frac{2}{x^2})^{100}(3x)^{100-r}$

[obsidian monolith]

soap:

[viscid thistle]

yes

so?

it doesn't prove anything

[tawny nacelle]

lemme finish bruh

[spring thunder]

(2/x^2)^r

[viscid thistle]

ok 😃