precalculus

[viscid thistle]

is this correct

[uncut mulch]

that's not a graph of a parabola, and you completely ignored what the question is taking for

[lime bolt]

Lol interesting pfp

[viscid thistle]

lool

<@&268886789983436800>

[viscid thistle]

Also worth mentioning the 2 names above

[uncut mulch]

um what'd i miss

[hasty wolf]

@viscid thistle if you want a new nickname please ask.

[viscid thistle]

nsfw pfp, Ram

[hasty wolf]

Nicknames need to be pingable

@viscid thistle whose?

[viscid thistle]

somebody already deleted

[hasty wolf]

Oh good

[lament fiber]

I was just learning about composition of functions

so for two functions f(x) and g(x)
let g o f be represented by q(x)

It's mentioned that domain q= domain f
why's there no dependency on the domain of g
wouldn't the range of f need to be a subset of the domain of g, is there any condition that can be mentioned here

[lime bolt]

It is the codomain of f

[half star]

it's enough to assume that $\mathrm{Range}(f)\subset \mathrm{Dom} (g)$

[lime bolt]

Wait actually it is the range not coromain

[obsidian monolith]

bastian.uwu:

[lament fiber]

I meant g o f*

g(f(x))

[half star]

yep thought so

it's enough to assume that $\mathrm{Range}(f)\subset \mathrm{Dom} (g)$
@half star so to answer your question yes; it is very important to assume this if you want composition to be well defined

[lament fiber]

so you can't have the composition of any two functions?

just ones where range of the "inner" function is a subset of the domain of the "outer" function?
or is it just assumed to be the case for everything?

@half star (sorry for the ping)

[lucid oak]

ok

Is this the channel to ask questions

[lament fiber]

I think #help-1 to #help-0 for problems

or even these ones are fine

[lucid oak]

so can I ask my questions here?

[lament fiber]

depends on the topic

[lucid oak]

also rational functions?

[lament fiber]

perhaps that goes in #prealg-and-algebra

[lucid oak]

thanks

[lament fiber]

np

#precalculus message
@willow bear (I'm sorry for the ping, but can you please answer this q)

[misty ocean]

Consider an angle β with sec(β) = 3 and sin(β) < 0. Show that tan(β) = −2√2

i got a=2 b=√2 and c=√6

but that makes the sec(β)=√3

[lucid oak]

someone said that rational functions are not prealgebra

[lament fiber]

it's in algebra 1(or 2, I think) on openstax

I don't have this arrangement of topics in my country, so I just check the openstax books or Khan Academy to see what goes where

[lucid oak]

i m a junior

u r also from europe?

[lament fiber]

India

[gilded brook]

Isn't -20 D and if D<0 it's to the left |D| units?
https://gyazo.com/3199fded14a1d9366c3fcf6feb60b520

[half star]

just ones where range of the "inner" function is a subset of the domain of the "outer" function?
or is it just assumed to be the case for everything?
@lament fiber yes, that is the only case where you can properly define the composition. For instance, if you consider $f: \mathbb R \to \mathbb R$ given by $f(x)=3x$ and $g: (0,+\infty) \to \mathbb R$ given by the natural logarithm $g(x)=\log(x)$ then the composition $g\circ f$ is not well defined since the range of $f$ is not contained in the domain of $g$. For instance $f(0)=0$, and if $g\circ f$ was well defined on $0$ we'd have $$(g\circ f)(0)=g(f(0))=g(0)=\log(0).$$

[obsidian monolith]

bastian.uwu:

[lament fiber]

thank you for answering!

[flat maple]

Did I do this one right?

[gilded brook]

What did I do wrong here?
https://gyazo.com/233d20e50bf7b9fcc3c32ea34598bf0b

Oh it's sec not cos

oops

[viscid thistle]

When you find the second derivative of a function and get the answer you get the point of inflection right and if its positive its concave up and if its negative its concave down but if the answer is 0 does that mean its the inflection point? But what do we get when we set the second derivative to 0

[junior vine]

so im getting this stuff in my precalc homework right now it looks easy but im kinda braindead rn

am I supposed to make a function from the table or something

[drowsy sky]

Unless Im really dumb, how do you find the inverse of
y = (x^2)(e^x)

[patent beacon]

You can't without a W function

[half star]

it also doesn't exist in any interval containing -2 or 0, a fact you can prove using some calculus

[hearty vector]

can anybody help me with this?

im niot sure what to do

[quick mirage]

all you gotta do is invert the transformations

ill give you a hint

[hearty vector]

wait

i just swap x and y?

[quick mirage]

nope, thats not what i mean

[hearty vector]

hmmm

[quick mirage]

try to simplify the function so it reads just f(x)

but apply the inverse operations to the points while you simplify to f(x)

for example, if the point (2, 3) lies on the function f(x) + 1

then f(x) will have the point (2, 3-1) or (2, 2)

lets consider this function a f[k(x -d)] + c

from what i can tell ya, the "a" and "c" affect the y coordinate value, and the "k" and "d" affect the x coordinate value

[hearty vector]

my brain too small

im so confused

[quick mirage]

Ok, lets dumb it down a bit

[hearty vector]

yes pls

omegalul

[quick mirage]

lets assume f(x) = x

y = x basically

so if the point (2, 3) appears somewhere on the line of, lets say, y = x + 1, we know that the graph intersects the y intercept at 1

because from the linear equation y = mx+b, we know that b is the y intercept

right?

[hearty vector]

mhm

[quick mirage]

Okay, so if we know that

when we want to make y = x + 1 into just y = x

what would we have to do to (2, 3)?

[hearty vector]

OH

I GOT IT

lmaoooo

Ty for the help

❤️

[quick mirage]

glad you understand

if u need any help im online for a bit still

[hearty vector]

kk

tyty

[viscid thistle]

quick help?

[eager scroll]

how to do imaginary numbers?

[uncut mulch]

show your work

[viscid thistle]

wdym how do i do this, this worked for the last problem but it was on the denominator

[uncut mulch]

you did something wrong with your signs

[willow bear]

also consider that $\log_3(3)$ is a value you should know

[obsidian monolith]

Ann:

[viscid thistle]

oh yeah

[uncut mulch]

what's "it"?

[viscid thistle]

1

[eager scroll]

imaginary numbers, x^2-4x-5=0

[uncut mulch]

btw that wasn't the only thing wrong with it

[viscid thistle]

oh i mean the 21 @uncut mulch

[uncut mulch]

order of operations / distributive property and all that crap

[viscid thistle]

and i got the answer but idk why i have to use a + sign here but for the last problem i used a -

[uncut mulch]

understand what the answers are instead of blindly trying to apply "shortcuts"

apply the log rules 1 step at a time

log(ab) = log(a) + log(b)
log(a/b) = log(a) - log(b)
apply those properly and it should be clear

[brave frigate]

^^ I think i got it just want to confirm tho incase im wrong

[viral trail]

Where is my proof wrong?

Claim: if g is not injective, neither is g(f(x))

Proof: since g is not injective find a,b such that g(a)=g(b). Now find c, d such that f(c) = a and f(d)= b. Therefore g(f(c))= g(f(d)). Therefore gof(x) is not injective.

Where am i wrong?
Please tag me while answering

[willow bear]

Now find c, d such that f(c) = a and f(d)= b.
bold of you to assume those will exist @viral trail

[viral trail]

@willow bear suppose in some cases they exist.

So for those cases since g not injective => gof not injective, so by contrapositive can i say gof injective=> g injective?

[willow bear]

your proof works with the additional assumption that f is surjective

if you assume explicitly the existence of these specific c and d that's literally tantamount to assuming g o f is not injective

[lament fiber]

wouldn't this fail the vertical line test?

Wouldn't a vertical line drawn slightly to the right of the leftmost black vertical line would touch the graph at more than one point?

[somber yew]

Herb Gross

[lament fiber]

yes
the videos are quite nice

[somber yew]

Yes, it will fail the vertical line test for sure.

yes
the videos are quite nice
They are indeed!

[lament fiber]

then it's not a function, right?

[somber yew]

Nope, it shouldn't be, unless I'm missing some context here.

If it's an error, it must've been pointed out in the comments section on YouTube.

[lament fiber]

this was about inverse functions

[somber yew]

Oh

[lament fiber]

he mentions how the curve of an onto function won't have a break

and the curve of an into function won't double back

[somber yew]

If it's an inverse function, perhaps it takes in y as the input, which means it's a function of y but not x.

But he writes y=f(x)

Might have to take a look at the video to be sure; which one is it?

[lament fiber]

so the only way a function can have an inverse is if the curve has no breaks, and it's either rising or falling

Unit I: Lecture 3

lemme link this with the timestamp

[somber yew]

Cool!

[lament fiber]

https://youtu.be/elputTS7tAA?t=1063

[somber yew]

Hmm

He calls this function a "multi-valued function", which sounds like an alternative to the term "relation". Following this, he demonstrates that a multi-valued function of this sort can't have an inverse.

Ah

He clarifies everything. You should continue watching the video.

This guy knows his stuff ;)

[lament fiber]

thank you

I watched this part yesterday, then tried to see if I was making a mistake, then I watched it today and decided to ask here

[somber yew]

I see. He actually has the curve for a function which is not one-one(such as sine) below, and then he plots its "inverse" above which isn't a function at all. By restricting the function to some intervals, however, the multi-valued "inverse" can be reduced to a decent inverse function.

[lament fiber]

by single valued functions, does he mean one-one

because that's how he'd used it a min or so earlier

or does he mean functions
and refers to relations as multi-valued

the comments are all just complementing Herb Gross and him thanking them

I wish I could've complemented him on this tbh

can't anymore

[lament fiber]

it's going to take me some time to parse all he's said from that timestamp

[somber yew]

RIP to a great teacher.

By single valued functions, he just meant ordinary functions which produce a single image for every element in the domain.

Multivalued functions seem to be like relations in a sense

[lament fiber]

yeah looking them up it says they're a relation

[civic mason]

Can someone explain the question please?

[willow bear]

can you post a higher resolution picture

this is hard to read as-is

[civic mason]

sorry

[willow bear]

the whole thing

not just the bottom part

[regal kernel]

@civic mason For what equation can you input to f(x) to make |2-5x^2|

(f o g) means f(g(x)), or f composed of g

https://cdn.discordapp.com/attachments/690575246943715401/757617375624560711/AbsoluteValueFormula.PNG
I'm stuck on this question if anyone would want to help me a little

[willow bear]

you want a formula for f', not f itself.

read the question carefully.

[regal kernel]

Ahh alright thanks. My bad

[scenic slate]

can someone explain how the postive 1 goes into negative in the equation

[willow bear]

this is a geometric series

look up geometric series formula

[scenic slate]

ohhhhh

so there is no first term?

[willow bear]

??

if you insist... it's written backwards, but the first term is 1 and the ratio is 1.04

[scenic slate]

in the formula it has u1 as a apart of the numerator

ohhhh okay

thank you!

[viscid thistle]

How would I simplify this?

[scenic slate]

https://cdn.discordapp.com/attachments/755510384865050714/757667853515096284/unknown.png

can someone help me on this

and @viscid thistle im not sure

[blissful ridge]

$\sqrt[3]{x^3y} =\sqrt[3]{x^3}\times \sqrt
[3]{y}$

[obsidian monolith]

The Godfather:

[blissful ridge]

@viscid thistle

@scenic slate
Are you familiar with Binomial expansion?

[scenic slate]

yes

[blissful ridge]

Do you recognise (1+x)^n

[scenic slate]

i thought that (4x)^n-3 has to be (4x)^2

nah i don't recoignise that

[blissful ridge]

Write (4x+1)^n as (1+4x)^n
And write out first few terms

See what is the coefficient of x²

[scenic slate]

the coefficient of x^2 is 16

i know that 16 x "c" = 1248

but idk how to get c

[blissful ridge]

Huh?

[scenic slate]

like its 16 x nCr=1248

and nCr has to be equal to 78

[blissful ridge]

You have the value of r

[scenic slate]

what is the value of r? is it 3?

[blissful ridge]

No

[scenic slate]

then how do i know the value of r?

[blissful ridge]

Show me your work

How are you getting r as 3?

[scenic slate]

because the the first term would have the exponent as 0

the second would have the exponent as 1

[blissful ridge]

First term will have exponent as 0

[scenic slate]

so it would have to be the third term

because the third term would make x^2

[blissful ridge]

First term will have exponent as 0

[scenic slate]

yea i meant 0 not 1 for the first term

[blissful ridge]

Now what's the value of r?

[scenic slate]

r=3?

or would it be r=n-3?

[blissful ridge]

$(1+4x)={}^n C_0(4x)^0 + {}^nC_1(4x)¹+{}^nC_2(4x)²\ldots$

[obsidian monolith]

The Godfather:

[scenic slate]

ohhhh

so nC2 has to equal 78

[blissful ridge]

Huh?

[viscid thistle]

Oh godfather

[scenic slate]

nvm thats wrong

[viscid thistle]

Would the 3sqrtx^3 cancel itself?

[blissful ridge]

Yes

[viscid thistle]

So the answer would just be 3sqrty?

[scenic slate]

idk what to do from here

[blissful ridge]

So the answer would just be 3sqrty?
@viscid thistle

No, why is a 3 there

Write nCr in factorial form

[viscid thistle]

It doesn't distribute to both?

[scenic slate]

i got: n!/2(n-2)!

is equal to 78

[blissful ridge]

Okay I get what you mean by 3

Now,where did your x go

[viscid thistle]

It's by itself?

[blissful ridge]

Satan, try to simplify more

[viscid thistle]

So x 3sqrty?

[scenic slate]

simplified: n!/(n-2)!=156

[blissful ridge]

Yes

@viscid thistle

$n!=n(n-1)!$

[obsidian monolith]

The Godfather:

[scenic slate]

how did you get that?

[blissful ridge]

Do you know what factorial means?

[scenic slate]

yea, it means it is multiplied by the numbers lower than it so for like 5 it is 5x4x3x2x1

is that the formula for it?

[blissful ridge]

$5!=5.\underbrace{4.3.2.1}$

[scenic slate]

yea so whats the picture above

[obsidian monolith]

The Godfather:

[blissful ridge]

4.3.2.1=4!

Therefore
5!=5.4!

[scenic slate]

ohhh okay

[blissful ridge]

Use that to simplify your equation

[queen forum]

Hello, for # 36 I am not sure how to find instantaneous velocity

[scenic slate]

@blissful ridge I LOVE YOU!!!! thank youu i understand it now

i understand how the factorials work

@blissful ridge blessss

[blissful ridge]

@queen forum

I guess you are not introduced to Calculus yet

And how did you find instantaneous velocity for 35¿

[queen forum]

I was only assigned #36

I know the average velocity formula for part a anb

Sorry, should I post this under calculus?

[blissful ridge]

Do you know Calculus?

[queen forum]

Currently taking first year of calculus, we were reviewing precal and trig concepts

So barely any calculus

[blissful ridge]

Instantaneous velocity is
Derivative of distance

If you know Calculus it's pretty easy

v(x)=s'(x) at x=2

Edit: corrected t as x

[stuck lark]

don't mix x,t

or just say v(2)=s'(2), idk why the fixation with x

[terse ravine]

' means prime correct?

[stuck lark]

we say ' as prime

[terse ravine]

ok

[queen forum]

so s'(t) would be instant velocity?

[hard mulch]

I keep getting
The wrong answer

Can someone see why

[stuck lark]

since v=s' by definition, v(2) & s'(2) both denote velocity at time 2

[queen forum]

I found average velocity to be 2+h using the formula and the interval approaches 2

[viscid thistle]

Why isn't this correct

[uncut mulch]

where's the ln

[viscid thistle]

Oh oops lmao

I am big dumby

[viscid thistle]

Easy function composition question giving me trouble

Never mind pretty sure it's B

[hard mulch]

Okay, Im really stuck here.. why does when graphing 3x-6. Why is the slope 0,2?

Shouldn't it be 3/1

Why not at point 3? On the x axis

[boreal geyser]

The slope is 3

[viscid thistle]

h(x) = f(g(x)) where f(x) = sqrt(x) and g(x) = (sqrt(x))^3, what is h(x).

[blissful kayak]

√(√x^3)

It's exponent properties from there

h(x)=x^3/4

[viscid thistle]

Thank you. That is what I thought but the automatic homework kept saying the answer was wrong. I just had to email a teacher about it

[versed heron]

can soemone help me with #5

[dim jungle]

substitute y=sqrt(x) @versed heron

[viscid thistle]

@versed heron

[willow bear]

@versed heron (2x + 9sqrt(x))^2 is NOT 2x^2 + 9x

this is not only too much work but also fails to check for extraneous solutions

x=25 is not a solution to 2x + 9sqrt(x) = 5.

[viscid thistle]

o

[willow bear]

why not substitute u = sqrt(x) and get a quadratic right away? with smaller coefficients than yours too

2u^2 + 9u - 5 = 0

[viscid thistle]

rip

[willow bear]

u = (-9 ± 11)/2 => u = -10, u = 1/2
u=-10 is obviously extraneous so take u=1/2 to get sqrt(x)=1/2 => x=1/4

[viscid thistle]

thanks for correcting.

me

Since i am not good in latex i am actually tring to work out almost all problems in latex today.

[willow bear]

the secane line through the points (-1, f(-1)) and (1, f(1))

[terse ravine]

what points are y2-y1/x2-x1

[willow bear]

did you mean to ask which point is supposed to be (x1, y1) and which is (x2, y2)?

[terse ravine]

I see the secant line at -1

[willow bear]

if so, it does not matter. the formula will give the same result if the points switch places.

also, the formula for the slope is not $m = y_2 - \frac{y_1}{x_2} - x_1$.

[obsidian monolith]

Ann:

[terse ravine]

f(x)-f(a)/x-a?

[obsidian monolith]

Ann:

[willow bear]

vanish, i am trying to correct your incorrect rendering of formulas in plaintext.

fractions sometimes require parentheses to be properly rendered into plaintext.

but you seem very unreceptive to my attempts to do this.

[terse ravine]

use parenthesis in plaintext

(y_2-y_1) / (x_2-x_1)

hopefully that is correct..

[willow bear]

ok that's better, yes that's the slope formula now typed correctly

your points are (-1, f(-1)) and (1, f(1))

f(-1) and f(1) are to be read off the graph.

[terse ravine]

f(-1) is x and f(1) is y correct?

[willow bear]

no, f(-1) and f(1) are numbers

[terse ravine]

Ok I read the numbers on the graph what is next?

[willow bear]

can you please tell me what you think f(-1) and f(1) are

so that i am 100% sure you didn't mess up on that

[terse ravine]

secant line is f(-1) and f(1)

f(-1) and f(1) are numbers

[willow bear]

....

what is happening

ok let me be more clear in what i'm trying to get out of you

f(-1) = ???

[terse ravine]

0

[willow bear]

and f(1) = ???

[terse ravine]

1

[willow bear]

no

f(1) is not 1

no, that's even more wrong. what the fuck

f(1) = ???

[terse ravine]

idk

[willow bear]

no, f(1) is not 0 either!

look at the graph!

[terse ravine]

1

[willow bear]

are you just guessing numbers at random now?

[terse ravine]

f(1) = 1

[willow bear]

no, f(1) is not 1

[terse ravine]

0.1?

idk

[willow bear]

are you just guessing numbers at random now?

[terse ravine]

Idk what f(1) is

[willow bear]

how did you know f(-1) = 0? was that a guess too?

how did you know f(-1) = 0? was that a guess too?

[terse ravine]

Now you're just being mean

[willow bear]

i'm trying to find where the hole in your knowledge is

[terse ravine]

oh

[willow bear]

it's weird to me that someone would be capable of evaluating a function based on its graph at one point but not at another

[terse ravine]

I see the secant line at -1 to be at 0

[willow bear]

BRUH FORGET ABOUT THE SECANT LINE

[terse ravine]

😮

[willow bear]

THIS ISNT ABOUT THE SECANT LINE THIS IS ABOUT THE FUNCTION NAMED F

forget about the blue line, all you can see is the graph of y = f(x) in BLACK!

how is it not obvious from a SINGLE look on the graph that f(1) = 3???

[terse ravine]

f(1) is 3

[willow bear]

yeah now you can just parrot that back to me

[sand harbor]

[terse ravine]

assuming assumptions I was about to type that

[willow bear]

curious how i didn't get a single answer to any of my leading questions from you

[terse ravine]

you just are a mean person

[willow bear]

what assumptions. WHAT FUCKING ASSUMPTIONS.

[terse ravine]

yeah now you can just parrot that back to me

doesn't that sound mean to you?

Well we found the hole

[sand harbor]

Sounds funny to me

[willow bear]

i gave you an answer

i still dont know whether or not you understand why f(1) = 3

for all i know, maybe your eyes will fall out from looking at the graph with enough focus

[terse ravine]

You just told me to ignore the blue line..

[willow bear]

yes ignore the blue line

[terse ravine]

Ok pls cut out your attitude

[willow bear]

cause you were having A LOT OF TROUBLE with the SIMPLE TASK of evaluating f(1)

[terse ravine]

You are going off on a tangent again..

[willow bear]

there would be no attitude had you only been a little bit more cooperative and responsive from the start

anyway ok whatever

f(-1) = 0
f(1) = 3

[terse ravine]

You just dont know when to stop

just calm down

[willow bear]

$\frac{f(1) - f(-1)}{1 - (-1)} = ??????$

[obsidian monolith]

Ann:

[terse ravine]

I'm done with you.

[willow bear]

lol look how long that took you to type

[terse ravine]

You don't stop.

[sand harbor]

😦

[willow bear]

certainly longer than it'd take to calculate the slope of your secant line since you already know f(1) = 3 and f(-1) = 0 so like

what gives

[sand harbor]

-3/5

[terse ravine]

certainly longer than it'd take to calculate the slope of your secant line since you already know f(1) = 3 and f(-1) = 0 so like
@willow bear

What are you talking about?

What is this attitude?

I said I'm done with you.

[willow bear]

then block me and don't respond to me lmao

[sand harbor]

Gosh Ann

[willow bear]

you refuse to answer the simplest of questions and then complain about my "aTtiTUde" when i get miffed about that as if you, a random person on the internet, hold any authority over me

[terse ravine]

You just keep going...

I'm not mad at you at all. I just said I'm done with you.

Thank you for your help.

[lime bolt]

bruh

[sand harbor]

[willow bear]

bruh

[lime bolt]

you could just admit u made a silly mistake and continue

[willow bear]

if you're unsatisfied then don't thank me lmao

[terse ravine]

Again you just keep on going and laughing.

[sand harbor]

It's not his/her fault he doesn't understand, if she/he understood better she/he wouldn't be asking for help 😩

[willow bear]

i clearly provided zero help to you so this thankyou reads as if you value politeness over honesty

[terse ravine]

Again I said thank you for help.

You can leave me alone.

[terse ravine]

@sand harbor Do you know how to do these problems?

[sand harbor]

I wish I did

[terse ravine]

xD

[sand harbor]

:(

[left fable]

The graph is the pointing at the answer

[terse ravine]

hmm

[left fable]

The evaluation is where the curves intersect that was shown in your graph

those are your numerator values for the slope formula

Then the denominator is the x values where those intersections happen

[terse ravine]

Yeah Ann helped me get to the f(-1)=0 and f(1)=3

So those are our numerators?

[left fable]

Yes

it’s generally easier if you set the larger number first and then subtract the smaller number so it’s positive. The order doesn’t matter which you subtract because you’ll still get the same thing

just make sure the order you subtract up top is respective of the order of the subtraction on the bottom

[terse ravine]

(3/1)-(0/1) = 3

Is this correct?

@left fable

The slope of the secant line is 3

[left fable]

The denominator is exactly like Ann put : 1 - (-1)

What is 1 - (-1)?

[terse ravine]

2

[left fable]

Exactly

[terse ravine]

-3/2

[left fable]

Where did the negative come from?

[terse ravine]

(3/1)-(0/-1)=3

no negative

It's not 3

What am I doing wrong

@left fable It's not 3

So is it 3/2?

[sand harbor]

So you had $\frac{3-0}{1+1}$

[obsidian monolith]

raelo:

[sand harbor]

Right

[terse ravine]

yeah

x-a

I forgot to add the minus

[sand harbor]

1-(-1)?

That becomes 1+1

[terse ravine]

2

yes

[sand harbor]

3-0

[terse ravine]

the formula is m_sec=(f(x)-(f(a)) / (x)-(a)

I shouldn't be doing this at 2:34 AM

xD

[sand harbor]

Oh yeh

You should rest

[terse ravine]

I love Math too much.

My apologies for earlier.

Good night everyone.

thank you

[sand harbor]

Gn

[umbral cloud]

Hey guys

I need help to solve this

[willow bear]

what's that graph there

[fading token]

what the graph is that

[umbral cloud]

There is no equation for the graph

[trim hemlock]

Thats weird

How could we find the limit if we dont know what f(x) is, or what it approaches to as x goes to -inf

[willow bear]

@umbral cloud that's not what i asked

is the graph y=f(x) or something else?

[umbral cloud]

Yeah y=f(x)

I didn't understand you bruh sorry

Note: x-> - inf =-1
x-> 0= +inf
x-> +inf=x-1

[willow bear]

okay well f(x) seems to approach -1 as x goes to minus infinity

so f(x)+1 -> 0

more precisely f(x)+1 -> 0+

[umbral cloud]

So x/0+ while x-> - infinty =?

Does it iqual to - inf?

[willow bear]

yes

[umbral cloud]

Thx

[mossy escarp]

Can someone help me

[violet granite]

can someone help me

[lament fiber]

Can someone explain to me how the number of mappings from a set A which has m distinct elements and set B which has n distinct elements in n^m
I don't know combinatorics or anything related to counting, I'll be taught that in a couple of months
I can't wrap my head around this for some reason
(Please ping me when you answer)
Here's what I get:
Each element in A has n choices in B
so it's n*m
but I know that this is wrong
I'm confused now
If I think about it another way:
we will already have n different functions if we consider functions where each element in A maps to a single element in B

I saw this on stackoverflow, but I still can't get it

why do we multiply, and not add

[lunar citrus]

id actually love to see this answer because it is starting to make my head hurt to think about

@lament fiber im gonna write it down and see what i get

[somber yew]

@lament fiber Every element of A can be mapped to any one of the n elements of B

[lament fiber]

I mean I can do it for a small set and see but that kind of approach won't work the higher I go in terms of the level of the topic

yeah
n choices for each element

there are m elements

[somber yew]

Thus total number of such mappings is n^m

[lament fiber]

yeah I don't get how it's n^m

[somber yew]

Taking into account all possible choices

[crisp heron]

n choices for the first, n choices for the 2nd ... n choices for the mth

[lament fiber]

isn't that n+n+n+n...m times?

how is it n *n *n *n

[somber yew]

No. Consider how many mappings you have from {1, 2} to {1,2,3}

[lament fiber]

I can see from this example

but still not wrap my head

[crisp heron]

say 1 goes to 2, you have multiple situations from there

so it cant be summing every situations

uh idk how this chapter is called in english

[lament fiber]

Combinatorics?

Counting?

[crisp heron]

yea basically

yea counting

[lament fiber]

Yeah they're going to teach us that, and also told us to skip this stuff in the book till then

[crisp heron]

draw the different situations with arrows linking both sets

[lament fiber]

but I wanted to try nonetheless

[crisp heron]

it should help you understand

[lament fiber]

yeah I did

say 1 goes to 2, you have multiple situations from there
this helped me understand it fully

thanks!

[crisp heron]

np

[lament fiber]

so if 1 maps to 2, for that one choice of mapping for 1, I can have different mappings for the other elements, so it's not summing, but multiplying

is saying this correct?

[crisp heron]

yea

[lament fiber]

ok
Edit: This got me thinking for a while, to the point that I got somewhat bored, so thanks for helping me out!

[patent flicker]

Can someone explain the process?

[proud raven]

do you have a given theta?

[sand harbor]

Uhhhh

[proud raven]

it's just a double angle identity

[patent flicker]

No the math. I did

[proud raven]

oh the combination of all those factors into a single fraction?

[patent flicker]

ye

[proud raven]

they're just multiplying through

all numerators multiplied by eachother, all denominators multiplied by each other

you get 64 on the bottom but theres a factor of 2 that drops out

[patent flicker]

OOOHHHH

got it

thnx

[proud raven]

[patent flicker]

cat

[edgy blaze]

is theta always the angle made with the x axis?

Like im helping someone with this and it doesnt say

[uncut mulch]

no, theta is whatever you define it to be

[edgy blaze]

i appreciate your technicality answer now how about the real one if you don't mind

[uncut mulch]

that is the real answer

also

angle made with the x axis
is ambiguous

theta is just a variable

since there were variables like alpha and beta in the question,
i initially assumed that your question were not directly about questions D,E and F

trigfunction(thing)
thing is the measure from the positive x-axis

[viscid thistle]

I tried change of base on the first term, but it got really messy from there. How do I do this? (Review for test)

[edgy blaze]

i don't understand what you mean

[viscid thistle]

I did a bad job of explaining that. I'm solving for X. I attempted to use the change of base formula to change the first term into base 3. I got lost from there.

[viscid thistle]

should i put everything in base 10?

[hushed sorrel]

what was your expression

also with this question

ive done it

except the first part 'prove that...'

doesnt come into the second part

at all (i did substitution for part 2)

which is rlly weird

can you guys see how the first part comes into the second?

[flint elbow]

f(x)=x^2-8x+16 I need to change out of standard form but I'm not sure how

[rich saddle]

do you need it in vertex form?

[flint elbow]

I think that's what it's called I need to get it into y=a(x+h)^2 +k

[rich saddle]

yeah thats vertex form

complete the square

[flint elbow]

explain that please

(sorry if that sounds rude)

[rich saddle]

oh actually you don't need to complete the square because it's already a square

What binomial $(ax + b)^2 = x^2 - 8x + 16$?

[obsidian monolith]

MarkusG:

[flint elbow]

oh so put it in that form '

[rich saddle]

yeah

[flint elbow]

so (1x+8)^2?

[rich saddle]

is 8^2 = 16?

remember $b^2$ has to equal $c$ in the original polynomial

[obsidian monolith]

MarkusG:

[rich saddle]

and also $2b$ has to equal $b$ in the original polynomial

[obsidian monolith]

MarkusG:

[rich saddle]

so you're looking for a number that doubles to equal -8 and squares to equal 16

[flint elbow]

so b is -4

[rich saddle]

badda bing badda boom

[flint elbow]

okay thank you

[rich saddle]

definitely look into completing the square though because that's going to come up

if you had $x^2 - 8x + 15$ the approach we just used wouldn't work

[obsidian monolith]

MarkusG:

[flint elbow]

I'm sure it will come up later on this homework so I'll come back

[rich saddle]

👍

[flint elbow]

back again, y= 4x^2-4x+21

[rich saddle]

https://www.mathsisfun.com/algebra/completing-square.html

one thing in addition to this

because your x^2 term has a coefficient of 4, you need to factor 4 out of the entire expression first

[flint elbow]

okay x^2-x+5.25

[rich saddle]

yep now follow the article for completing the square

and as a side note, i'd keep things in fractions for as long as you can

decimals are a waste of effort, especially in intermediate work

[flint elbow]

okay

[heady field]

what does this mean lol

[fiery wren]

is this right?

[viscid thistle]

Awkward punctuation on the first option

What’s wrong with option 2?

[valid estuary]

Hey could anyone help me understand how to find the second solution for sin(theta) = 0.4?

[patent beacon]

The two solutions are θ and 180 - θ

Using degrees

[valid estuary]

Right how do you get the 180/pi though

[patent beacon]

I imagine a unit circle haha

The first solution has the terminal arm somewhere around 2PM. I don't care exactly where it is, it's a picture in my mind!

[valid estuary]

Right

[patent beacon]

The second must have the same y-value. The only way that's possible is if the terminal arm is somewhere around 10PM

Man, the clock doesn't really work here. Seeing as they don't even go the same direction

Anyway, the relationship between these two solutions is their distance from the x-axis

That is, distance from 180

[valid estuary]

Hold on let me make sure I get it

[patent beacon]

Here's a good picture of the two solutions to sinθ = 0.71

The first is 45°. The second is 45° away from the 180° mark

[valid estuary]

Oh yeah yeah yeah I see that

[patent beacon]

That's generally true. If the first solution was 13°, the second would be 180°-13°

[valid estuary]

Ok so for something like sec(theta) = -4

You could turn that into cos(theta) = -1/4 and you can get the first solution

[patent beacon]

Aight so I'm lying a little bit. Everything I've explained works perfectly for sin.

Good news is that it's a slight adjustment to get cos working

[valid estuary]

It's just these style of questions that trip me up with no exact unit circle values

[patent beacon]

You can get one! Your calculator will give you the first solution

[valid estuary]

Right inverse function

[patent beacon]

,w arccos(-1/4)

[valid estuary]

Yeah I'm good with that it's just the second value

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Kaynex#7520

[patent beacon]

You seem to be cool with radians. Let's use that. So this terminal arm is somewhere near 4:00

Since cos only cares about the x-value, the second solution will be somewhere near 2:00. That make sense?

[valid estuary]

Nah not really hold on

[patent beacon]

Oh wait no I'm messing up

[valid estuary]

Hold on hold on so we got cos(theta) = -1/4

[patent beacon]

Forget I just said literally any of that haha. The first solution is around 11:00

As that has a slightly negative x-value

[valid estuary]

That puts it in the second quadrant less than (-1/2,radical3/2), so yeah 11:00

[patent beacon]

You want somewhere else on the unit circle with that slightly negative x-value

[valid estuary]

You also have it in the third quadrant just across the x-axis

[patent beacon]

Exactly

[valid estuary]

Like 7:00 or so

[patent beacon]

Right! Is there any relationship between these two solutions?

[valid estuary]

Well I know it's got something to do with adding 2pi or pi or subtracting

[patent beacon]

This one is actually very easy. One solution is the negative of the other

Just rotate in the opposite direction from the x-axis, and you get the other solution

[valid estuary]

What the key does is that it subtracts the ~1.82 from 2pi

2pi is a full 360

[patent beacon]

Oh I see. They don't want a negative answer so they're adding 2π to it

-1.82 is the same as -1.82 + 2π on the unit circle

[valid estuary]

Nah nah I mean that it's 2pi - 1.82

Equals roughly 4.46

And that is the second solution

[patent beacon]

Yeah -1.82 + 2π

[valid estuary]

Oh yeah I'm dumb you're right

[patent beacon]

I'm trolling a little haha

Sorry I could have wrote that correct I derped when typing it

[valid estuary]

But I don't understand where we're getting the 2pi and everything from

2pi is a full circle right

[patent beacon]

2π is a full rotation. You can always add/subtract 2π to an angle without changing what the angle "is"

[valid estuary]

Ok yeah I get that but couldn't you just get an infinite amount of answers then

[patent beacon]

-1.82 and 2π - 1.82 are the same angle. Your solution book went with the second solution because it is positive

[valid estuary]

Just keep on adding 2pi it's the same value

[patent beacon]

You can get infinitely many of the "same" angle, but there are exactly two "different" angles

[valid estuary]

Ok yeah no I get what you're saying

So let me throw another one in the mix

cot(theta) = 2

Turn that into tan(theta) = 1/2

Get the exact value from the calc

[patent beacon]

Haha okay, let's keep this going. tan is best visualized as the slope of the terminal arm

So you care about where the slope is 1/2

I would say 2:00 and 8:00

[valid estuary]

Ok I follow you now

[patent beacon]

I'm using these "clock" measures only to convey unit circle positions. Please don't actually think of the unit circle this way it's backwards af lol

[valid estuary]

Yeah yeah I get it

It would be like between (pi/6) and (pi/4)

[flat maple]

Is this correct?

[patent beacon]

Your calculator will hand you the 2:00 position. The 8:00 position will be π + that

[flat maple]

Also sorry if I interrupted

[patent beacon]

Just a half circle rotation between the solutions

[valid estuary]

Yeah but what's up with the pi now why not a 2pi?

I think it's cause tangents have a different period right

Repeats every pi

[patent beacon]

That's very related!

One might say that what we're doing can prove the period is π

[valid estuary]

So for all sorts of tangent problems I can find the second solution by adding/subtracting a pi

[patent beacon]

But by imagining the unit circle, I can realize that the solutions are a half-circle rotation apart. Adding π onto one gives me the other

And yes this will generally work. Up to you if you want to memorize it, I have just imagined a lot of unit circles

[valid estuary]

I think what's tripping me up is the sec(theta) = -4 one

Cause for sin(theta) = 0.4 we worked with pi

What's the difference with sec(theta) = -4? We go from quadrant 2 to quadrant 3

[patent beacon]

The solutions are related in that they have the same x-value

[valid estuary]

Yeah yeah I think I'm getting it now

[patent beacon]

Since that x value is slightly negative, -1/4, they'll be 11:00 and 7:00

Your calculator will give you 11:00

[valid estuary]

Yeah now I think I have a better grappling of it now

Just need to do some practice problems

[patent beacon]

7:00 is just the opposite rotation, if that makes any sense. So a teacher should accept -1.82

Okay, cool cool

[valid estuary]

Thanks much for your help man have a great night

[patent beacon]

Have a good one! Feel free to ask if you need anything else

[mighty onyx]

Shouldn't this be right?

[past meadow]

either a rounding error or it wants exact values

because that should round to -19.8619

[viscid thistle]

what is this question asking exactly?

oh nvm

i get it

what is f(t)g(t) in this instance?

sqrt(t+4) * t^2 ?

[patent beacon]

That's f(t)g(t), yes

That's not fog(t), in case you happen to want their composition

[viscid thistle]

nah

thanks

for 15

z^2-1-z^2?

I feel like I got that wrong

Like if we know z^2 = m(z), how do I evaluate m(z+1)?

(z+1)^2?

yup

Which is z^2 + 2z + 1

2z+1

well no

H e c k I cannot type

it's 2z+1 right?

Uhh

because once you foil and subtract z^2 that removes the z^2 from (z+1)^2

Yea seems like it...

gotcha thanks

mmk

Np

(-x)^3 can just be called

-x^3 right

[trim hemlock]

yes

they are the same thing

[viscid thistle]

and then the cube roots of -x^3

are

(-x)(-x)(-x)

the only cube root of -x^3 is -x?

[trim hemlock]

well yeah

[viscid thistle]

so like for any

root

like root n

there are a corresponding amount of n roots, correct?

to like construct that number or variable

[trim hemlock]

wdym?

[viscid thistle]

like

sqrt is 2

so there are 2 roots

cube root is 3

so there are 3 roots

so whatever the word is for like

the 4th root

there are gonna 4 roots right?

x^(1/4)

maybe that makes no sense

[trim hemlock]

well not exactly

square root of 1 would give you 2 roots, which are 1 and -1

but cube root of 1 would only give you 1

the number infront of a root doesnt always say how many roots there will be

[viscid thistle]

Oh ok, my apologies

f(x) = e^x-x

is this odd, even, or neither?

I got neither for both

[trim hemlock]

lets try pluggin in -x

we would get 1/(e^x) + x

doesnt seem like it equals to f(x) or -f(x)

thats the answer

[viscid thistle]

5e^(ln(A^2))

this evaluates to 5a^2, correct?

[willow bear]

no, it evaluates to 5A^2.

[viscid thistle]

oh well, yeah

So I start by taking the ln of both sides, right?

ln(7^x+2) = 17x

(x+2) * ln(7) = 17x

get the exponents out using that log rule

x * ln(7) + 2 * ln(7) = 17x

I get stuck here

[blazing parrot]

Move all the x to one side

[viscid thistle]

xln(7) -17x = -2ln(7)

uh maybe I did it wrong lmao

if anyone could point out where I went wrong, I'd appreciate it

ah

i see

I wasn't trying to factor

it looks correct so far

yeah once I did x(ln7-17) = -2ln(7) it made more sense

solve for t

I wanted to say

what have you tried

trying to think about how to write it on discord

since I don't know latex

$Q_{0}a^{nt}$ not $(Q_{0}a)^{nt}$

[obsidian monolith]

HoboSas:

[viscid thistle]

Ah

base is a not Q0a

(logQ-logQsubscript0)/n = t

where base is a

I should really learn latex after this exam

Thank you @viscid thistle

np

So when you have exponential equations

like

P= P0*a^t

pretend that's a subscript

P=P0*e^kt

k is basically a constant to balance it out in order to include e as the base right

But if the latter equation is used for continuous compounding, why does that end up being equivalent to a non-continuous exponential equation such as P=P0*a^t

Or am I missing something

yes correct

But if the one with e is continuous, then it compounds at a different rate

$a^x=e^{\ln(a)x}$

[obsidian monolith]

HoboSas:

[viscid thistle]

because it's not being updated on fixed intervals

k could also depend on other factors

so

e^x is commonly used because of its derivative

Yes

However my question is

what differentiates continuous rate interest, from non-continuous?

Because k=0.02 is not the same as a=1.02

maybe the explanation above is just confusing

Because k=0.02 is not the same as a=1.02
ln(1.02)=0.02

ah

so a faster way to find

or convert from there

[patent beacon]

Nothing. For any exponential "compounded annually" or something like that, there's some way to find it's continuous rate

[viscid thistle]

Oh, so they end up being the same for all values of x?

That's a bit confusing

I guess I'll just treat them as interchangeable with the continuous rate as easier to manipulate

I just felt like if one is at a continuous rate it wouldn't line up in the long term with a non-continuous one

maybe once I get into the chpaters that have more to do with continuity this will make sense

$a^x=e^{\ln(a^x)}=e^{\ln(a)x}$

[obsidian monolith]

HoboSas:

[viscid thistle]

oh yeah, that is a simpler way to look at it

it's basically just a constant

it's the same as saying that all log functions differ by some constant

Ok, thank you

so for an exponential equation

say p(t) = 1.04^t

the inverse can said to be f(p) = log(t)

base 1.04

[patent beacon]

Yes

[viscid thistle]

There's probably a better way to say, that, huh

[patent beacon]

log1.04(t) if you're terrible like me

[viscid thistle]

That does make sense

Sorry for the rapid fire questions, I had a job upon enrolling in university and didn't do much studying for calculus

the exam is in about like 4 days

so I'm just reviewing

1. f(t) = 1 + ln t

what is the inverse of this

f(g) = e^t

I'm not sure what to do with this 1 here

[patent beacon]

Inverse is always "swap x and y"

Basically, you've got:
x = 1 + ln(y)
What's y?

[viscid thistle]

Oh, yeah

gotcha

e^(y-1) = t

What's it called when you turn both sides into exponents for a base?

[viscid thistle]

How would I shift vertically by 3?

Do I just multiply the y value of each point by 3?

You need to add the y value of each point by 3

Do I multiply it or add it?

Add

Are you sure?

Yes

But that doesn’t make sense?

That would be the y value?

It does, you are "shifting" vertically

Means along the y axis

No no no

Say I have that graph

And I need to get

Y=3f(x+2)-1

The 3 is a vertical shift

The -1 means you subtract from the y value

3 is multiply right?

In the equation you gave above, you are multiplying (x+2) by 3f then subtracting 1, so yes, in the equation 3 is multiply. But shifting a coordinate vertically means adding 3 to its y value.

[heady field]

what is the parent function for this?

[fleet knoll]

would this be nCr (12, 6) x 2?

[wispy aurora]

seems like it yes

[sick seal]

why is b not 8.7 dont they have to travel at the same speed to get the same distance in the same time??????

[wispy aurora]

not true

if you notice, the formula for d involves a quad equation

[sick seal]

yea

[wispy aurora]

whereas constant speed is always linear

[sick seal]

im aware, but it is asking for the contant speed

which is confusing me to hell

[wispy aurora]

ye ok. So plug in 8.3 into the equation

and then subtract the value of t = 8

and divide by 0.3

that's all

[sick seal]

okie dokie.

[wispy aurora]

once you have the answer, and if you still have a q, I'll explain 🙂

[sick seal]

i just wrote everything down on paper and i got a lightshow of lightbulbs. thanks man

[wispy aurora]

Cool! Feel free to ping me whenever!

[sick seal]

got it. i really hate words im math. i get so confused easily.

[wispy aurora]

I get that

[sick seal]

thanks again

[fallow carbon]

Solving Radical Equations. Solve. Write your solutions in set notation.
√1 − 4x = x + 5

[wispy aurora]

Is the -4x part of the root?

[fallow carbon]

yes

[wispy aurora]

have you tried squaring on both sides?

[fallow carbon]

no

[wispy aurora]

Give that a go then

[strange arch]

Idk if this can be considered as prevalent but...

Can someone help me on this

[buoyant osprey]

Ok

[strange arch]

idk if it 4 or 8

[viscid thistle]

is there any way I can solve for x without graphing it

[wispy aurora]

@viscid thistle Are calculators allowed?

[viscid thistle]

its a practice question

but i dont really feel comfortable using the calculator

to graph because our teacher hasnt shown us how to do it yet

[wispy aurora]

idt there's a way to do it without graphing/calc

but ask others

They might know

[sick seal]

hello again, this came up in the homework and my notes dose not have this type of problem.

[strange arch]

Guys plz help me with my question too

[split fossil]

@sick seal Draw a triangle with base 1 and hypotenuse 5. Solve for height

[sick seal]

thank you

[split fossil]

You're welcome

[terse ravine]

Is this correct?

I get different answers on mathway and on my calculator. I'm probably not typing it correctly in my calculator.

I thought it was approaching 6 but then I see it approaching some DNE value at the end.

[split fossil]

@terse ravine You are typing it wrong. It should be approaching 5.5

You can see this from the values in the table, and later on in Calculus you can use L'Hopital's Rule

[terse ravine]

So, I just ignore the bottom 3 values?

[split fossil]

Yes pretty much, either your calculator was failing you, or you were typing it wrong

[terse ravine]

You can see for yourself when you plug in 0.001 for x

I get these 0.001675 ans

[split fossil]

I got 5.4999720417

[terse ravine]

What am I doing wrong here

I'll have to try another one.

[split fossil]

I tried it on Mathway. It looks like 0.001^2 might be too small for it to compute.

In any case, it isn't doing math correctly

[terse ravine]

My calculator has on Math/Deg/Norm1/d/c /Real

[split fossil]

The problem is degrees and radians

I think that Mathway is in radians

[terse ravine]

I'm in degrees and getting -8.681188 and Mathway gives me -24.311059

cymath also gives me -24.311059

[split fossil]

Take a picture of calculator

[terse ravine]

I typed in (cos(11x0.1)-cos(8x0.1)) / 0.1^2 straight across on my Casio fx-CG10

[split fossil]

This is a different problem?

[terse ravine]

yes I failed the first one.

[split fossil]

Can you show me the problem?

So i can compare it to what you typed in

[split fossil]

Weird. Mathway is right this tme

Actually, it appears that Mathway switches between radians and degrees depending on the question.

[terse ravine]

my fx-CG10 is in degree mode and gave me-15.018427

[split fossil]

Switch to radians

[terse ravine]

trying to figure that out

ok re entering it now

Ok, I got -13.98253987

rounded off to 6 decimal places is -13.982540

[split fossil]

That is not correct

It should be -24.311059