precalculus

[atomic urchin]

Yes

I do

N^1...2...3 etc

[wise kelp]

The degree is the highest exponent

[atomic urchin]

Right

[wise kelp]

So this is a third-degree polynomial

[atomic urchin]

Correct ok. I understand

[wise kelp]

So if this were a “reduced” polynomial as they’re defining it, there would be no term with x^2

If I understand the definition right

Is i. clear then?

[atomic urchin]

Could you go into a little further detail please.

Sorry I am asking so many questions. I just need to be sure.

[wise kelp]

No it’s fine

Further detail on reduced polynomials?

[atomic urchin]

Yes

Like how and when to know when to enact on a reduced polynomial. Can every polynomial be reduced? How can one reduce and know when to and when not to.

[wise kelp]

It seems like “reduced” is just a category

So it’s not like you can take a non-reduced polynomial and reduce it

[atomic urchin]

Ok

[wise kelp]

It’s just that some polynomials are reduced and some aren’t

Since whether a polynomial is “reduced” is defined by what terms it has

[atomic urchin]

Ok

[wise kelp]

And you can’t change a polynomial’s terms (without turning it into a different polynomial)

Is anything else unclear for i.?

I assume that “specific” is just the opposite of generic

[atomic urchin]

Nope. I now have a good reference of understanding

[wise kelp]

OK

So, on to ii.

[atomic urchin]

So the term is monic

[wise kelp]

The polynomial is monic

[atomic urchin]

Correct the polynomial is monic

The zeros would then be
X=8 (Multiplicity of 1)
x=2 + square root 3, 2 negative square root 3?

[wise kelp]

Yes

[atomic urchin]

Awesome that part I get

Now part three

I’m entirely lost

[willow bear]

can you repost the problem

[wise kelp]

g(x) = f(x+4) means that if you replace all of the x’s in f with x+4, you’ll get g

So (x+4)^3 - 12(x+4)^2 + 33(x+4) - 8

[atomic urchin]

Ahhh ok

I’ll be back I’ll do this by hand

[wise kelp]

Of course you’ll want to put that in standard form

OK

[viscid thistle]

someone plzz help id get this

[wise kelp]

@viscid thistle Do you know what the coordinates of those circles’ centers are?

[viscid thistle]

y3qh

-6, -8

-5 8

[wise kelp]

So the radius of the circle between them would be along a straight line crossing those two points, right?

[limber bone]

distance between the 2 centers

would the radius + the radius of the first circle and the secoond

[wise kelp]

Well

The diameter (of the middle circle) + the radius of the two circles

But yes

The answer I got isn’t one of the answers shown

Can anyone confirm?

[limber bone]

ye should be sqrt(273) - 2

?

right?

[wise kelp]

(sqrt(273)-2)/2 is what I got

Because it asks for radius, not diameter

[limber bone]

oh

i thought its asking for diameter

ye

[deft flume]

is anyone available to help me really quick?

im assuming that this is a geometric sequence unless it has a constant difference

[rocky bison]

What's the equation for the area of a trapezium?

[deft flume]

i dunno

[rocky bison]

,$ \frac{h}{2}\left(a+b\right)

[obsidian monolith]

Pseudo:

[deft flume]

but google did the job

[rocky bison]

So

What's a and b

They're the length of your top and bottom sides

and h is the height

First row has 14 bricks

18th row (last row) has 31 bricks

and there's 18 rows

So

a=14, b=31, h=18

plug in

You'll get an answer

[deft flume]

i was thinking that the a and b would be those values

ty so much tho

i tried solving for the magnitude then resulted in an incorrect answer for the direction angle

nvm i got it

[swift glacier]

would the slope of this just be the derivative?

[patent beacon]

The derivative evaluated at x = 3, yeah

[swift glacier]

to do that would I just plug in 3 as h?

[patent beacon]

Nop, that's x = 3 in the derivative

What is the derivative of x²?

[swift glacier]

2+h?

wait no 2

[patent beacon]

The derivative is 2x, you're missing the x somewhere

[grizzled orchid]

do you know the power rule?

[swift glacier]

Yeah I read my notes wron, it's 2x

the power rule?

[grizzled orchid]

yes

[swift glacier]

I don't think so?

[grizzled orchid]

the derivative of a function of the form x^c where c is a constant is cx^(c-1)

[patent beacon]

You may be expected to use the definition for now

[grizzled orchid]

ye

in other words

move the exponent to the coefficient, and subtract 1 from your exponent

bam. derivative. for very very simple functions

[swift glacier]

oooooh

I can see why it's called the power rule

[grizzled orchid]

you might want to learn the four basic rules (and maybe exponents that aren't constants?) before you start your calc class

[patent beacon]

The derivative of 2x³ is 6x², for example

[grizzled orchid]

will make it easier to follow along

yes

[swift glacier]

Thanks, I see what I did wrong the first time

[patent beacon]

It's possible you haven't learned these rules yet, but yeah they're going to be taught to you

[swift glacier]

oh ok

[grizzled orchid]

inevitably ur gonna need to

if you know it first and the proofs

will be easier in class

[swift glacier]

can you still list them just in case my teacher doesn't go over them?

[grizzled orchid]

sure lol

[swift glacier]

I just want to be able to search them up

[grizzled orchid]

the first four you will get is

power, product, sum, chain

they're all pretty simple

[patent beacon]

They will. A first year calc class is about using these rules

[grizzled orchid]

yeah

the teacher for sure will

but learning them early is always nice

[swift glacier]

^^^

fun

[grizzled orchid]

tbh calc I and II are not so bad if you are willing to review

often times concepts that you think are hard actually become quite simple when reexamined in these early classes

is why calc II can seem impossible first time through and them trivial when reviewing like 1 month later

[viscid thistle]

calc I is really easy, especially derivatives

[grizzled orchid]

yus

integrals are also simple here

because they are all closed

hardest topic in calc I is probably implicit differentiation

[patent beacon]

If you're interested in learning ahead, http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeIntro.aspx

Awesome site

[grizzled orchid]

yes

also 3blue1brown is excellent at introducing concepts

[swift glacier]

oh yes! Thanks so much!

[viscid thistle]

does everyone here just use paul's notes?

the heck

[swift glacier]

lmao

[patent beacon]

Yeah pretty much. They are easily the best text source on any calc

[grizzled orchid]

paul

is pretty much god

he owns us all

we sold our souls to him to pass calc II

or in my case, to learn calc II, which i will hopefully pass

[timber plinth]

@viscid thistle i used him for a quick introduction to laplace (and some other more pdf-y sources)

[echo plaza]

kaynex has unicode on lock

who needs latex

[patent beacon]

I have gboard, an app on my phone. I have a Greek keyboard downloaded. You can also make shortcuts. If I type rt I get √

[torn swift]

This is revolutionary

Greek keyboard is amazing lol

That is such a smart idea, lol

[patent beacon]

Most apps let you switch with a button press so γι'σιήλο δώρο

[dapper geyser]

imo for later calculus the lamar site has the crappiest explanations ever

[sturdy gyro]

Haha yeah this us good

Is*

Σδφ

[torn swift]

on laptop is nice as well

[viscid thistle]

$$f(x)=(x^x)/(x+1)$$ how do i find the reverse function?

[obsidian monolith]

exitmusik:

[lucid drift]

do you mean inverse?

,w inverse of f(x)=(x^x)/(x+1)

[obsidian monolith]

Display results online and refine query

Requested by killer_memestar#8571

[lucid drift]

[ruby otter]

,w inverse of f(x)=(x^x)

[obsidian monolith]

Display results online and refine query

Requested by Radical Ninja#7581

[copper lagoon]

what's W?

is it a weird function

[lucid drift]

ya

,w lambert w function

[obsidian monolith]

Display results online and refine query

Requested by killer_memestar#8571

[viscid thistle]

thanks but what is W?

lol nvm it glitched just saw the photo

[tropic crown]

would the limit as x apporaches pi of 5sin(x + sinx) = 0?

since sin(pi) = 0

[ruby otter]

ya

[honest igloo]

So I need help graphing asymptotes. I have a decent idea on how to find the x and y intercepts along with the horizontal and vertical asymptotes however graphing them is a bit foreign to me. Here's the problem: Draw the graph of f(x) = 3x/ 4 - x^2. Indicate the asymptotes.

So I know my horizontal asymptote is y=0 since the degree of the numerator is less than that of the denominator. My vertical asymptote is +/- 2 since those are the zeros of the denominator. I also know the y intercept is 0 since f(0) = 0. The x intercept is also zero because the zeros of the numerator are also zero.

[wise kelp]

You’ll want to plot some points to get an idea of the shape of the graph and see where it’s positive and negative

Then you can sketch it using the information you’ve already found

[honest igloo]

where do I get the points?

arbitrary points Im assuming?

[wise kelp]

Yeah

Doesn’t really matter, just pick points that you think will help

[honest igloo]

They dont have to be in a specific range?

[wise kelp]

You’ll want points near the vertical asymptotes

Try to get an idea of how it curves

And in this case you’ll want points on both sides of x=0 between the asymptotes

You can also get an idea of where it’s positive and negative just by looking at the equation

[honest igloo]

how many points would be adequate?

[wise kelp]

However many you need

I suppose two or three in each “region” is probably enough if you pick the right points

[honest igloo]

(-2, -11/2) (2, -5/2)

these are two points I got

Do I plot them?

[wise kelp]

Those points should be undefined

Since x = -2 and x= 2 are where the vertical asymptotes are

[honest igloo]

oh

ill get some others sorry

[wise kelp]

Two easy ones would be x = -1 and x = 1

[honest igloo]

alright I'll use those

(-1, -7/4) (1,- 1/4)

so these points would be better since they are within the range of the vertical asymptotes?

[wise kelp]

How did you get those y-values?

[honest igloo]

sorry are they incorrect?

ill double check

[wise kelp]

f(1) = 3(1)/(4-1^2) = 3/3 = 1

Gtg for a bit

[honest igloo]

alright Ill be back in a few

I accidently did the wrong problem

(1,1) (-1,-1)

[tropic crown]

as x apporaches infinity of (1-e^x)/(1+7e^x)

can i divide the numerator and denominator by e^x?

and plug in infinity, which would equal to -1/7?

[rigid beacon]

@tropic crown you can't divide by e^x because not every term has an e^x

[tropic crown]

how would u suggest i solve this?

[royal gull]

factor out e^x?

[tropic crown]

also what would tan^-1(infinity - infinity) be equal? i know tan^-1(infinity) = pi/2. would it simply be does not exist?

[royal gull]

???

I dont understand

[echo plaza]

that's an indeterminate form

$\lim_{x\to \infty}\frac{1-e^x}{1+7e^x}$

[obsidian monolith]

CaptainLightning:

[echo plaza]

(this isn't the one I was talking about btw)

(u looked like u needed latex help)

[royal gull]

no I didnt

[echo plaza]

ic

[wise kelp]

I think what John said makes sense

Divide the top and bottom by e^x and you get

[royal gull]

yeah we've done that already

[wise kelp]

Oh

[royal gull]

the tan thing desotn make sense since infinity - infninity can be whatever you want

[tropic crown]

this is the original question

[echo plaza]

no

[tropic crown]

pretend DNE isnt there

i wrote that as a place holder

[willow bear]

you should first find the limit of x^2 - x^5 as x -> +∞.

[tropic crown]

until i get the right solution

[wise kelp]

You can’t do tan^-1(infinity)

[willow bear]

you sort of can if you know what you're doing

lmao

[wise kelp]

You can find the limit of tan^-1(x) as x approaches infinity

But you don’t plug in infinity directly

Maybe in higher level math you can, idk

But in precalc you don’t

But yeah, first find the limit of x^2 - x^5 as x approaches infinity

[tropic crown]

so it does not exist?

[willow bear]

what's it

[tropic crown]

so if x approaches infinity of x^2 - x^5, isnt it just infinity - infinity?

and since i cant conclude whether its positive or negative infinity

maybe it doesnt exist?

[charred hull]

[echo plaza]

try drawing a graph

if that helps

[tropic crown]

cuase i know arctan(positive infinity) = pi/2 and negative would be -pi/2

[echo plaza]

draw the graph of y=x²-x⁵

[wise kelp]

x^5 increases faster than x^2 does

So as x gets bigger the difference between x^5 and x^2 gets bigger

[tropic crown]

okay, that makes sense, so it would be -pi/2 in that case?

since -x^5 is increasing faster than x^2

[echo plaza]

yes

[tropic crown]

ty

[echo plaza]

np

[idle dust]

wah

isnt it ln(x)=1/x

this is ln(3x)??

shouldnt it be like 1/3x

[royal gull]

no

[charred hull]

chain rule

[idle dust]

mm

ok lemme try

[royal gull]

you also have to take derivative of 3x

whihc is 3

and it cancels out with the 3 in denominator

[idle dust]

1/3x times 3

OHH

[lucid drift]

you can also write it like ln(3x) = ln(x) + ln(3) and then the derivative of ln(3) is 0 since it's a constant

[idle dust]

INTRESTING

THANks a lot

❤

bruh

[royal gull]

rep me

[idle dust]

ok

[viscid thistle]

So ive done problem a)

I think

Ill post what I have

I don't get what question b) is asking me to find? I am confused.

And I dont know if I drew it right

[wise kelp]

I think b) is asking for the distance from he plane to the near end of the runway

@viscid thistle

[viscid thistle]

Someone told me it was the vertical distance of the plane to the ground

[wise kelp]

I don’t think so

It specifically says the near end of the runway

[viscid thistle]

You are right haha because, question c asks for the vertical distance lol

[viscid thistle]

someone help

they're all decreasing???

[echo plaza]

what makes you say that?

[raven crater]

You can use derivative in each interval

For the first one , slope is totally negative, so function is decreasing, and in second one, derivative is equal to (-4x) and it’s positive only in this interval [-1,0)

[cyan flame]

can someone help me quick plz

assingment due at 11:59

soo lost :/

[viscid thistle]

i can try

[cyan flame]

im supposed to graph it onto the dotted lines

no idea how to graph this

[viscid thistle]

alright

[cyan flame]

this is tricky

[viscid thistle]

so lets get the stretch factor

the first value

[cyan flame]

ok

[viscid thistle]

graph goes from -4 to 0

[cyan flame]

so -4

[viscid thistle]

the a value would be 2

[cyan flame]

nani

how

[viscid thistle]

a value is the distance from the center

to your maximum or minium

[cyan flame]

ohhh

[viscid thistle]

ye

y u put -4

[cyan flame]

so for instance this

would be

4?

[viscid thistle]

yes

[cyan flame]

oh crap

that made it soo much easier

thats all?

[viscid thistle]

yep

[cyan flame]

then the c is moving it on the y axis

what is the coefficent to the x do

[viscid thistle]

yes c is where the center is

the coefficient to x would be the period the graph is at

y= acos(bx)+c

all it does is stretches it horizontally

[cyan flame]

so a period of 3 would mean what?

and you made this sooo much easier thank u

would i look at how many times a full cicle happens

in 1 pi?

wow that worked

[viscid thistle]

i usually look how long 1 cycle is

in that case 1 cycle is 2π/3

the equation for period is PB=2π

p for period, b for the b in the parent eqn

to get b divide by P on both sides

B = 2π/P

2π divided by the period (in this case for that graph 2π/3)

and you end up with 3

[cyan flame]

im sorry im confused with that part

so how would this be 4 then?

at first i thought it was 2

because 2 cycles happen in 1 pi

[viscid thistle]

use the PB=2π equation

one cycle happens at pi/2

https://gyazo.com/ff1dc855600f0737806b4a8024432ad4

that is your P value

[cyan flame]

ahh

so if b is the b in the aprent equation

but i dont have b in the first place

[viscid thistle]

my bad

[cyan flame]

im saying that I dont start with b in the first place

[viscid thistle]

yes

you start with P (period)

P is given in the graph

[cyan flame]

oh

I think I understand that part

I was able to solve those

[viscid thistle]

what did you get for the b value for that graph

[cyan flame]

once I had solved it I clicked off

let me bring up another example questio

and can I type my thought process?

[viscid thistle]

go ahead

[cyan flame]

a=3

c=2

p = pi/2

?

[viscid thistle]

p is pi

[cyan flame]

oh lol

so ther equation pb gives me

wait now im confused lol

[viscid thistle]

this is a full cycle for sin graph

1 cycle is pi

[cyan flame]

so what would I put as the B?

[viscid thistle]

use the eqn PB=2pi

solve for B

[cyan flame]

b= 2pi/pi

[viscid thistle]

yyes

[cyan flame]

so b = 2?

[viscid thistle]

yes

[cyan flame]

ohhh

ok

I think I understand

a= 3

c=2

oof

same question

[viscid thistle]

lol

[grizzled orchid]

lol

[cyan flame]

a=2

c=3

b=2pi/p

like nani

[viscid thistle]

one cycle completes at 2pi/3

*\

[cyan flame]

oh

so 1/3

wait no

3

[viscid thistle]

ye

[cyan flame]

OHHH

i got it

i understnad

thank u

lol

[viscid thistle]

np

[cyan flame]

the ultimate use of my new abilities

[viscid thistle]

given B now go backwards : D

[cyan flame]

this is so much harder

omg

[viscid thistle]

D :

lets first label y axis

the graph is not shifted at all

so the graph will go from [-8,8] on the y axis

im sure you got that part

b in this case is 6

so use P= 2pi/B

period is pi/3

1 cycle stops at pi/3

[cyan flame]

my time ended at 11:59 R.I.P

got it to 88%

and when I attempted to do it

I was sooo far off

wow

[viscid thistle]

F

[royal gull]

Maybe try to write f(X) as some ax+b

And integrate

[warm nebula]

Yeah so I have to integrate both of them? Then what?

When letting f x equals as plus b

[willow bear]

well, can you write $\int_0^1 (ax+b) \dd{x}$ and $\int_0^1 (ax+b)^2 \dd{x}$ in terms of $a$ and $b$?

[obsidian monolith]

Ann:

[mint widget]

g(f(x)); f(x) = x²; g(x) = $\sqrt{√x-7}$

what is the domain of g(f(x))

I thought it’d be [√7, ∞)?

[royal gull]

No

[mint widget]

Yeah, I got that

But that doesn’t really help me does it

[raven crater]

For what?

[willow bear]

is that $\sqrt{x} - 7$ or $\sqrt{x-7}$?

[obsidian monolith]

Ann:

[mint widget]

The latter

Ah man, i didn’t know about that command

[agile palm]

Please help me answer this question!! It’s multiple choice! Thanks!

If a function is not one-to-one on an interval I, it cannot have an inverse function on I.
a) True. If a function is not one-to-one then at least one of the range values of the function has more than one image when the inverse is defined.
b) False. The function can have an inverse as it is not one-to-one on the interval I.
c) True. If a function is not one-to-one then at least one of the domain values of the function has more than one image when the inverse is defined.

[raven crater]

C?🤔

I think the answer is C

[mint widget]

There we go

g(f(x)); f(x) = x²; g(x) = $\sqrt{x-7}$

what is the domain of g(f(x))

I thought it’d be [√7, ∞)?

[obsidian monolith]

Ryan_A:

[mint widget]

Oh

It can be negative too

[raven crater]

Yep

[mint widget]

Rip, alright thanks

[raven crater]

👍🏻✋🏻

[shell salmon]

While function is one to one, it would pass horizonal line test

[distant crypt]

@warm nebula maybe this can help

[leaden crow]

Your phi look a lot like a u, use clearer notation. No need to switch variable here, just keep it as x.

[willow bear]

phi should have a nice long descending tail

[viscid thistle]

how do you find domain?

[frozen needle]

by knowing domains of others functions and looking at where existence issues could be

[viscid thistle]

so its just something you have to remember?

[willow bear]

"it"?

[viscid thistle]

what?

[leaden crow]

it

[charred hull]

it

[finite helm]

I’m lost

[midnight summit]

can you rotate this

@finite helm

[spring thunder]

,rotate

[midnight summit]

ty

take it one by one

[limber bone]

ln(x+9) = (2ln(2)+2ln(5))/2

x+9 = e^(2ln(2)+2ln(5))/2)

x= e^((2ln(2)+2ln(5))/2)-9

[midnight summit]

i thought he asked about 35

[limber bone]

uh oh

[midnight summit]

@finite helm you there?

[limber bone]

28 - 3^(log_3(4)) - 4^(log_4(48))

28-4-48 ig

[midnight summit]

yes exactly

[finite helm]

Ye

It was 35

Thnx

[warped steeple]

Does the discriminant $\Delta$ of the quadratic formula $b^{2}-4ac$ have any geometrical significance on the graph of the quadratic.

[obsidian monolith]

boilhats:

[warped steeple]

It is the distance from the peak of the graph to the Ox axis?

[echo plaza]

eh

$\frac{\sqrt{b^2-4ac}}{|a|}$ is the distance between the two roots

[obsidian monolith]

CaptainLightning:

[echo plaza]

if they're both real anyway

[west shore]

What are polar coordinates

[torn swift]

different coordinate system

[west shore]

Do you have an example

[torn swift]

positions are defined by a set distance from the origin, r, and an angle of rotation θ

like if you just have r, like r=1 then that's a circle of radius 1 centered on the origin

[west shore]

So is it on a circle

[torn swift]

polar is good for circular symmetry

[west shore]

So how would you write one

Like

2,pi/2

Assuming r=2

And you say set distance form origin, does that mean it can only lie of the circumstance of the circle

[torn swift]

you have these relations you can use to convert rectangular to polar and vice versa:\
\
$r=\sqrt{x^{2}+y^{2}}\
x=r\cos(\theta)\
y=r\sin(\theta)\
\tan(\theta)=\frac{y}{x}$

[west shore]

Rectangular?

[torn swift]

Cartesian

x.y

[west shore]

Oh okay

[serene heath]

r not r^2

[torn swift]

oh yea

lol

[obsidian monolith]

⚡Amphy⚡:

[west shore]

So given a point on the Cartesian you can use the relations

So how do we find theta

And is r arbitrary

[torn swift]

can use arctan

r is not arbitrary given x,y coordinates

[west shore]

How do we know what r is

[torn swift]

use the first relation listed

[west shore]

Oh nvm

I see now

So if x=r cos theta, how would we know What theta is

Can you just reverse the equation of something

[torn swift]

use arctan(y/x)

[west shore]

Okay

[torn swift]

that's the fourth relation I listed

[west shore]

Oh nice

So it comes together nicely

How would we write a polar coordinate

[torn swift]

the quadrant theta is in is based on the signs of y and x

recall unit circle relations to help you determine the quadrant the angle will be in

[west shore]

Okay

So is it just I. The form (a,b)

[torn swift]

to write something in polar, it's (r,θ)

[west shore]

Okay

That’s not too bad

I just wanted to know

So thank you

[serene heath]

np

[viscid thistle]

How would I get problem h

I looked at the solutions and the answer is pi/3

?

[short sorrel]

recall that tan is sin/cos; can you think of any values of x where sinx/cosx = sqrt(3)?

[viscid thistle]

So for calculating the inverse of tan I would have to look at the regular sine and cosine functions?

[short sorrel]

not always; if you can instantly recognize the tan value, you dont need it

but when you cant think of it

thinking in terms of sin and cos can be helpful

[viscid thistle]

I know that tan is sin over cos but I have no clue still...

Yup I still dont know what values of sin/cos would equal to sqrt3

[short sorrel]

well, we're looking for a sqrt(3) term

we know we can get that from sin(pi/3) or cos(pi/6)

both of those are sqrt(3)/2

in order to get rid of that division by 2

we need to multiply by 2

(which is the same as dividing by 1/2)

checking sin(pi/6) divided by cos(pi/6), we see it doesnt work; we get (1/2)/(sqrt(3)/2) = (1/2)(2/sqrt(3) = 1/sqrt(3)

but checking the other possibility we found, pi/3

sin(pi/3)/cos(pi/3) = (sqrt(3)/2)/(1/2) = 2*sqrt(3)/2 = sqrt(3)

so there's our answer: pi/3

we couldve got this without needing to recall the values

by looking at the unit circle

sin is a y value, cos is an x

so if we look at each point, and divide the y by the x

we'll eventually find the angle that works

pi/3, in this case

theres another solution, btw, in quadrant 3

but idk if the question wants both or only one

[viscid thistle]

Hold on I think I messed up with my drawing

[short sorrel]

the triangle you drew is not a conventional triangle

but you could make it such by dividing each side by 2

although it doesnt really matter, it's probably more familiar to you like that

either way, yeah, thats definitely another approach: draw a satisfying triangle

and figure out what angle it matches to

[viscid thistle]

Oh shit

I found m y way

but It takes too long

Actually it might not if he only asks for inverse tan

which most likely is too easy

So he wont only ask for inverse tan

oh nv

m

I can just remember that arcsin has the same input as arctan except its over 2

@short sorrel thanks for helping me understand it

[viscid thistle]

Is their any reason that in the solutions he left the denominator of the answer root of x^2 + 1 ?

Why not x + 1

This is my work btw

[torn swift]

why do you think that is is equivalent to x+1?

[viscid thistle]

Powers cancel out

[torn swift]

oh my goodness lol, that is not how that works

if it was just sqrt(x^2) then yes you can simplfy

but you got x^2 + 1 so with addition it's a different story

[viscid thistle]

Oh its not the same as sqrt(x^2) + sqrt(1)

[torn swift]

no no lol

[viscid thistle]

My algebra is bad

[torn swift]

that is not a valid thing to do with roots

[viscid thistle]

Ok mistake

and I understand

[torn swift]

addition and subtraction will not allow for that

[hard hornet]

every time you do that a kitty dies, that's what my middle school teacher told me, and i still believe it

traumatizing

[torn swift]

lol

[hard hornet]

$\frac{x+1}{x} \ne 1$

[obsidian monolith]

MemesPlease:

[torn swift]

was not the best use of algbra

[viscid thistle]

With fractions its sometimes easier for me to write it as x/x + 1/x so I made the mistake for square root

[torn swift]

have to be careful then lol

[silk sequoia]

when line L intersects plane P perpendicularly at point p, are there an infinite number of lines in plane P that pass through p and are perpendicular to L because you can make an infinite number of planes?

[willow bear]

are there an infinite number of lines in plane P that pass through p and are perpendicular to L
yes
because you can make an infinite number of planes?
no

[silk sequoia]

what is the reasoning for this then?

or should i say why are there an infinite number of lines in plane p perpendicular to L

ohhh wait

is it because you can rotate that perpendicular line 3 dimensionally

...but it says in plane P

hmm

[neat falcon]

Can anyone help me with the following equation?

[obsidian monolith]

Kildoes:

[midnight summit]

@neat falcon is this log to the base of x?

[neat falcon]

yeah

[obsidian monolith]

Kildoes:

[neat falcon]

Fixed it

[frozen needle]

What have you tried so far?

It seems like a good idea to rewrite those logarithms in such way there'll only be 1 base

[neat falcon]

I did

$log(2x-3)-log(x-1)-log(x+1)=log(x)$

[obsidian monolith]

Kildoes:

[frozen needle]

there's a little sign mistake

[neat falcon]

$log(2x-3)-log(x-1)+log(x+1)=log(x)?$

[obsidian monolith]

Kildoes:

[frozen needle]

Better

So now, how do you deal with this?

[neat falcon]

Idk..

maybe $log(\frac{(x+1)(2x-3)}{x-1})=log(x)$?

[obsidian monolith]

Kildoes:

[frozen needle]

Well that's a possibility

and then?

[neat falcon]

raise 10 to this so the log cancels?

[frozen needle]

Yeah

[neat falcon]

Thanks, it worked

[frozen needle]

You're welcome

[placid tusk]

How do you rearrange this equation such that $\theta$ is the subject? $x\sin(\theta) + y\cos(\theta) = k$

[obsidian monolith]

CabbageGuy:

[placid tusk]

Also, you aren't allowed to use the sin(a +b) identity

[serene heath]

y not

[placid tusk]

Cuz my syllabus doesn't have it and I'm afraid if I use it, I won't be awarded any points

Do you think it's possible?

[serene heath]

hmmm

prolly

[hybrid charm]

Let's see what we can do

$x^2\sin^2(y)+2xy\sin(y)\cos(y)+y^2\cos^2(y)=k^2$

[obsidian monolith]

♫ick (Far From Home):

[hybrid charm]

divide by cos^2(y) $x^2\tan^2(y)+2xy\tan(y)+y^2=k^2(\sec^2(y))$

[obsidian monolith]

♫ick (Far From Home):
Compile Error! Click the reaction for details. (You may edit your message)

[hybrid charm]

$x^2\tan^2(y)+2xy\tan(y)+y^2=k^2(1+\tan^2(y))$

[obsidian monolith]

♫ick (Far From Home):

[hybrid charm]

this is quadratic in tan(y)

[willow bear]

p sure you end up with some extraneous sols there

[hybrid charm]

then you can use quadratic formula for tan(y)

then y=atan(...)

and that y inside trig functions is $\theta$

[obsidian monolith]

♫ick (Far From Home):

[hybrid charm]

I should have used different symbol but okay

[ionic beacon]

Hi

How do u do this question

question 2b

wait i think i just did the calculations wrong nvm

Can somebody explain why theta must be in radians so the dimensions are correct

[short sorrel]

If theta is in degrees, we need to include an extra factor of theta pi/180 whenever we take the derivative of sin or cos

As $\dv{x} (\sin{x}) = \cos{x}$ only holds when x is in radians

[obsidian monolith]

Namington:

[short sorrel]

This is why radians are the most natural unit for calculus; they avoid pesky factors of pi/180

(since otherwise, chain rule says you'd need those)

[tawny nacelle]

wait

why does it only hold when x is in radians ?

[ionic beacon]

So for angles involved in derivatives is easier to use if we used radians

[short sorrel]

Well, it should be intuitively obvious that the rate of change of sinx is much lower than cosx when x is in degrees

Yes @ionic beacon

[ionic beacon]

what about integration

[short sorrel]

Same deal

[ionic beacon]

o

thanks

[short sorrel]

@tawny nacelle intuitive explanation: consider that sin(x°) varies from -1 to 1 over 360 degrees, and is never really that steep

If the derivative was cos(x°)

Which is frequently, like, 0.5ish

Then we'd expect sin(x°) to be much more steep and... Wavy - but it only goes from -1 to 1, which is bizarre for a measurement over 360 units if its slope were that high!

The rigorous reason is perhaps best seen through the limit of the difference quotient

But let's say you take it on faith that $\dv{r} \sin{r} = \cos{r}$ when r is in radians

[obsidian monolith]

Namington:

[short sorrel]

Then, if $x$ is in degrees, $r = \frac{x\pi}{180^{\circ}}$

[obsidian monolith]

Namington:

[short sorrel]

Substituting in this

$\dv{x} \sin{(\frac{x\pi}{180°})}$

[obsidian monolith]

Namington:

[short sorrel]

From here, it's a trivial application of chain rule

Since we know this is in radians

And we get cos(xpi/180) * the derivative of the inside, which is pi/180°

Which you'll notice is much more awkward than just using r

But yeah, if you want a fully rigorous explanatuon of why it works for radians

When it doesn't for degrees

Do the limit definition from first principles yourself

And it becomes fairly apparent

[tawny nacelle]

i c

[ionic beacon]

also

e doesnt change in both differentiation and integration right?

like if e^2

x

[tawny nacelle]

yeah

$\frac{d}{dx} e^x = e^x$

[obsidian monolith]

soap:

[tawny nacelle]

also $\int e^x dx = e^x + C$

[obsidian monolith]

soap:

[ionic beacon]

ty

[echo plaza]

note that $\dv{x}e=0$ and $\ \int e \dd{x}=ex+C$

[obsidian monolith]

CaptainLightning:

[echo plaza]

it's e^x which has that property, not e

[tawny nacelle]

E

[royal gull]

E

[short sorrel]

The real number with that property is 0

Tbh

$\dv{x} 0 = 0$ and $\int 0 \dd{x} = 0 + C$

....texit?

...texit?

[royal gull]

Is gone 🦀

[short sorrel]

$\dv{x} 0 = 0$ and $\int 0 \dd{x} = 0 + C$

[obsidian monolith]

Namington:

[short sorrel]

!!!wow!!!

What a magical number!

[ionic beacon]

integration is the anti derivative

so what does the result mean

wait

if integration is to find the area under a curve

then what is differentiation for

[royal gull]

To find the growth of something

Kinda

[ionic beacon]

so indefinite integrals have C

[rigid beacon]

Integration: find net change

Differentiation: find rate of change

that's a very surface level look

[ionic beacon]

thanks

[rigid beacon]

you'll get more into the details in a calc class

which will be fun

[ionic beacon]

or the average age for it

[tawny nacelle]

fun

[royal gull]

Calc is good

[ionic beacon]

When do i get to be in a calc class

oh my message delayed

[tawny nacelle]

y r u so hyped about it

[ionic beacon]

because my current teacher is not very good

😃

and I dont think he teaches calc

[rigid beacon]

you're in HS right?

[ionic beacon]

yes

[rigid beacon]

what grade?

[ionic beacon]

2 more years until i graduate

[rigid beacon]

I mean your school doesn't offer a calc class like AP calc AB or BC?

[ionic beacon]

no

[rigid beacon]

ok

[ionic beacon]

I mean we have IB

[rigid beacon]

does your school offer any AP classes?

Oh yea IB is a thing

can you do IB math?

[ionic beacon]

I heard its harder

[rigid beacon]

idk how the system works really

[royal gull]

Where are you from minecraft?

[ionic beacon]

Im from hong kong

[royal gull]

Hmm idk then

[rigid beacon]

gotcha that makes sense

I mean I'm sure there is a math class for you next year

[lethal skiff]

In the uk can we do calc in yr 11 the gcse curriculum is boring

For anyone in the uk^

[echo plaza]

yeah there's ADmaths at the same time as GCSE

[lethal skiff]

What’s that

@echo plaza

[echo plaza]

uh

that's it's a thing with calculus in Y11

in the uk

[lethal skiff]

How do u get in

I’m in yr ten

[ionic beacon]

additional maths starts from year 10

its a 2 year course

[lethal skiff]

Fuck me

We have further maths next year if that covers it

I need to get a nine but it’ll be fine

[ionic beacon]

further maths sounds hard

[lethal skiff]

It’s all outside of school shit

[echo plaza]

it's just a normal exam not something you have to "get in" to

[ionic beacon]

Actually

there's further pure maths so maybe not as hard

[echo plaza]

oh yeah year 10 sorry

[lethal skiff]

So I can?

[echo plaza]

I mean you'd probably need a teacher to teach it to you

or you'd have to study in your own time

[lethal skiff]

I’d be fine with that

[echo plaza]

this is the one I did

https://www.ocr.org.uk/qualifications/fsmq/additional-mathematics-6993/

bear in mind it isn't all calculus

also it assumes FTC without proving it but A level is like that as well

[lethal skiff]

Is that just an ocr version of the aqa further maths

[echo plaza]

maybe

idk

[lethal skiff]

Like ima have to do 3 extra papers for a shot at an a star star

Is that it?

[echo plaza]

when I did gcse with edexcel you could get a 9 without any extra papers

[lethal skiff]

9 is only an a star

[echo plaza]

ADmaths was just another thing on the side

[lethal skiff]

I’m getting mines

Nines

Oh ok I’ll look into it

Ty man

[echo plaza]

AQA has always been a bit weird

np

I've seen what they do in A level biology from my friends and it's some proper dodgy stuff

[lethal skiff]

https://www.aqa.org.uk/subjects/mathematics/aqa-certificate/further-mathematics-8360

That’s it but wdym

@echo plaza

Sry if this is spam

[echo plaza]

like they'll ask the same question 2 different years in biology and the mark scheme will ask for different answers

[lethal skiff]

Yeah that sounds like aqa

They also ask broad questions like what helps respiration and expect very specific answers

[ionic beacon]

its me again

I don't get why x = 2

so the positive x-axis is 1?

nvm

just figured out ln1 is 0

[raven basin]

I need help plz. I dont understand this

[elfin night]

NO

@severe verge NO

[serene heath]

tf

[broken minnow]

What does it mean by restrict the domain to form a new function

[steel granite]

i think it just means to find the domain of g(x) such that f(g(x)) is defined

[broken minnow]

domain of g(x)

is all real numbers tho

[steel granite]

i mean if you restrict it

[broken minnow]

but what does restrict mean for this

[steel granite]

like if it were actuall [5,infinity)

for example

[broken minnow]

whats the point

of that

what does it do

[steel granite]

idk but if you restrict domain of g(x) to [1,infinity) then f(g(x)) is defined for all numbers on g's domain

the restricted domain for g

i mean

i think that the way they word it is stupid

[broken minnow]

I c

It rly is

Isn’t it just asking for the domain of the composite function ?

[swift wagon]

The only way it’s not defined is if x<1

It should say “state when” not “state why” unless I’m missing something

[main geyser]

Yo guys quick question

Do you think that precalc is possible to test out of?

The way it is structured at my school is basically where first semester is a refresher of algebra 2 and second semester is trig

And I spoke with my math teacher and he said that it’s not recommended but if you think you can handle self-teaching it to yourself it’s okay

[torn swift]

I didn't do pre calc lol

I did self teach as well

[main geyser]

Really?

[torn swift]

yea, skipped that class and did calc

turned out fine, but this is all based on how much personal effort you put into it

[main geyser]

You think it’s possible to pull it off in a little over 2 months?

[torn swift]

sure

[main geyser]

Have you got any tips about self-teaching?

[torn swift]

Khan Academy is nice place to start

[timber plinth]

@main geyser precalc is ez you should have no problme

[rigid beacon]

Alot of stuff from precalc I don't use in calc at all but I'm sure it'll pop up in higher math

Haven't seen matricies or oblique asymptotes

Trig is really really important

So are limits

Having an intro to polar, vectors, and parametrics is important

[rocky bison]

[rigid beacon]

Precalc is just Algebra 2 2

[willow bear]

small brain: precalc
expanding brain: algebra 3
galaxy brain: algebra 2 2

[viscid thistle]

lol

[sturdy gyro]

Alpha brain Algebra 01100101001

[rare zephyr]

according to my syllabus there aint no calculus or precalculus in secondary school

the system set this "topic" to be a clump some of the topics known in A-level maths together

and it's like shallow asf

like when I read some questions about functions in this chat i get confused

cause the american-syllabus dive that sh*t deep

and i guess this goes to other subjects too

now i understand why american teenagers hate their high school academy

[devout raft]

[viscid thistle]

Someone please explain this

Would the answer be 1?, since f(1)=-2

[patent beacon]

Basically you're solving
x³ + x - 4 = -2

x³ + x - 2 = 0

You called that x = 1 is a solution, so can that get factored out?

[viscid thistle]

Isn’t that equal to 1

[patent beacon]

Which?

[viscid thistle]

The first equation you said

[patent beacon]

You're right that 1 is a solution to the problem. Is it the only solution?

[hard hornet]

:mindblown:

[viscid thistle]

There are imaginary solutions but idk if we put those

[patent beacon]

,w roots of x^3 + x - 2

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[patent beacon]

Fair, so yeah you could leave those out and 1 is the only solution

[viscid thistle]

Yeah

We aren’t allowed a calculator on this so idk if we put those

[patent beacon]

You can still find them by solving the quadratic

Up to your teacher, but likely not to include

[echo plaza]

it said f^-1 so I think you're ok for that

[viscid thistle]

K

[hollow horizon]

how do i do thissss

[viscid thistle]

What do you generally think about trying to do when tasked with simplifying an expression?

It's a bit of playing around, isn't it? Manipulating what you're given in ways you're not so certain will get you anywhere, but often reveal something about the expression you're looking at nonetheless

[hollow horizon]

simplifying to the simplest possible form

i subbed in numbers just to see and it turned out that the result was 0

i just dont know how to get there

i tried multiple different approaches then i just get stuck

[viscid thistle]

Give that a go, try what you know about square roots, exponents, logs, exponentiation, to play around with the expression

Experimenting with values is good! Nice, yeah it's zero and I'm sure you see that it might be but just can't justify it to yourself

What kind of approaches did you try?

[hollow horizon]

well i noticed when subbing in values

that when i simplify the b part

[viscid thistle]

Type things out in $\text{\LaTeX}$ here :)

[obsidian monolith]

Deconstructed:

[viscid thistle]

Or show scratch work screenshots

[hollow horizon]

alright sure

basically i noticed that

b = a^x

then the b part will always be b^1/sqrtx

i t ried that it didnt really work out

or

i figurd this might be of use but i didnt get anywhere

[viscid thistle]

Hmm I'm not sure how to push you in the right direction without giving you a hint

But first, what is this for?

The whiteboard said "tough", is this some optional question that interested you?

[hollow horizon]

we have a thinking task tomorrow which is basically we choose the amount of questions that add up to 13 marks

thats just one of the example questions of last year i believe

[viscid thistle]

Cool, I'm glad whoever is teaching you encourages individual study that you're interested in, and hopefully not boring trivial problems that you can't glean much from

Well, back to the problem:
Hint 1- put the problem into the form e^(something)-e^(something else)
When dealing with logarithms, things often show their true nature when you start working within the exponent
That's not a provable result really, just one of many problem solving methods

But once you do put the question in that form, try to manipulate things more

if e^(something)-e(something else)=0, then something= something else ye? Go from there

[hollow horizon]

hmmm

but i cant in the question assume that its equal to zero right

so

hmm

[viscid thistle]

Assuming the conclusion doesn't prove it, you're correct

However, since you've shown that the expression is 0 for some a and b, and your choices for those a and b seem arbitrary, you can make a guess and say:
"Okay, I think the expression might be 0 for every nonzero a and b. In that case, putting the equation into the form e^(beep)-e^(boop) will help me because beep should then equal boop so that the expression is 0."

Since you don't have any precise known method of solving a problem like this, using heuristic methods like making guesses can help

[hollow horizon]

yeah i understand that im just at a loss of what to do to get to the e^something - e^somethingelse

which is why im stuck 😢

[viscid thistle]

well you have

$a^(\sqrt(\log_a b))$

[hollow horizon]

haha i think

i understand

what ur trying to get at

dont worry

[viscid thistle]

I didn't finish my sentence even 😂