precalculus

[hoary valley]

Do they have different names?

[patent beacon]

They're both the definition of the derivative

[hoary valley]

@patent beacon Ok thanks a lot.

[hallow flicker]

the second one is better

its much more easier to solve for

[hoary valley]

"The slope of a curve at a certain point" is like saying "the slope of a tangent line to the curve"? @patent beacon

[willow bear]

the tangent line

if one exists

[hallow flicker]

Yea

[dusty wasp]

Hey

i need help calculating the sum of the thrill of the drops in this rollercoaster design

thrill of the drops is the product of the angle of steepest descent in the drop (in radians) and the total vertical distance in the drop.

but im confused on how to do it

[thorny bluff]

What's precalculus

(I live in a country that doesn't have precalc)

[wise bison]

I can't speak for everyone, but my text covers functions, trigonometry, matrices, analytic geometry, series and limits.

[thorny bluff]

I see

[languid crane]

The feeling when we don’t have limits before calc

[rigid beacon]

Basically it fills in the gaps between algebra 2 and calculus

It's really just Algebra 3

[warped willow]

^

[odd helm]

When I posted this yesterday someone said that it should be shifted 172 to the right because it should start on the first day

But the period is 365 right? Since 365 days in a year

And the summer solstice is in June

So to get to January you need to go to the right half a year right?

And 365/2=182.5 which isn’t right

So could someone help me figure out what I’m doing wrong thank you

[odd helm]

<@&286206848099549185>

[odd helm]

Bruh this is what I want to do

Too bad I’m pretty sure it’s required

Arch could you help me

11th

USA

Oh really

Alright I’ll look into it thank you

Nah that’s fine

Alright

And arch could you help me

With the problem I posted above

[odd helm]

<@&286206848099549185>

The problem is above I can repost it though

[odd helm]

<@&286206848099549185>

Ik it was 20

So uh I'll repost my question now

Basically I’m not sure why we go right 172

I think the key does it because it wants to start at the beginning of the year

But since we are starting in the summer and want to get to January we should be going right 365/2=182.5

Not 172 right?

[odd helm]

<@&286206848099549185>

[odd helm]

<@&286206848099549185>

...

[tardy crescent]

Hello Tenemos, I think I can help you with your problem. So the summer solstice is the 172nd day of the year,

Your amplitude and period are all correct

Look at your function without any horizontal shift, (So just 6.96*sin((2pi/365)(x)) + 12.41 ) and find where the first max occurs (which is when the argument is Pi/2, or (2pi / 365) * x = Pi/2). This is at 365/4 or 91.25.

That is what you want to shift from

Since it's your max, you want it to be at the summer solstice, which is the 172nd day of the year

So shift your whole function to the right by (172 - 91.25), or 80.75

So you get: 6.96*sin ((2pi / 365) (x - 80.75)) + 12.41

[odd helm]

I understand this makes sense

So I now know how to do it if I'm using sine

But what about with cosine? If I'm using cosine I'm already at the summer solcist and with cosine you start at the maximum point right?

So if I'm already at the maximum point, which is the summer solcist, why would I need to shift anything at all?

When using cosine

[uncut mulch]

x is being used to represent the day of the year

[odd helm]

I know so we want to get it so that it starts on the summer soloist I guess

But look

This is what it looks like when you draw a sign graph

Right?

But then we want to start at the summer Solstice

We are at spring right now when we start right? So then I would shift to the summer solicits by going forward 172-91.25=80.5 so the equation is 6.96 sin(2pi/365(x-80.5))+12.41

But when I do the cos graph I’m already starting at the max so idk why I need to shift anything at all

And in the key there is a shift to the right 172

I’m sorry if I can’t phrase my question right tbh the other problems I think I’ve been doing well on

So it doesn’t really matter ig

[uncut mulch]

what cos equation were you graphing?
(also 172-91.25 = 80.75)

when you did the shift for sin, you were shifting from the 172nd day

for the max of cos, you want the argument to be 0 on the 172nd day, i.e. when x = 172

cos( 2pi/365 ( x - C))
x - C = 0
172 - C = 0
C = 172

[hazy hull]

i need help with #2

[dapper prairie]

downloadable files are not safe, so i suggest you send a screenshot of the problem

@hazy hull

[trim fable]

question

how many solutions do u expect the equation 2x=0 to have?

[lilac pier]

One.

[trim fable]

oh what?

oops

i meant

cos2x

@lilac pier

[lilac pier]

cos2x=0?

[trim fable]

from a range of 0 to 2 pi

yeah

4 right?

also what about this

and how are the solutions of sin1/4x=1/2 related to sinx=1/2```

are there 16 solutions for sinx=1/2

[lilac pier]

Yeah there are 4 solutions.

[trim fable]

oh ok

[lilac pier]

for the cos one

[trim fable]

and for those questions^^

yeah ok

and sin 1/4 x will have double that so 32 solutions?

[lilac pier]

Yeah 4 solutions for sinx=1/2

[trim fable]

4??

with a range of

0 to 4pi?

[lilac pier]

0 to 4pi

[trim fable]

wait huh

doesn't it happen at every 1/4 pi then

o wait every half pi*

so

8 solutions

[lilac pier]

you want 1/2, don't include -1/2

[trim fable]

yeah

so 8?

[lilac pier]

so only 4 solutions

[trim fable]

how 4?

wont it occur a every 1/2 pi then?

are u talking about 4 for 2pi?

coz i agree with that but for 4pi?

isn't it 8?

[lilac pier]

we're talking about sinx or sin2x?

[trim fable]

and how are the solutions of sin1/4x=1/2 related to sinx=1/2

sinx

[trim fable]

oh..

UR RIGHT

LOL

i get what u mean

now

wont it be 5?

including the

origin so

[lilac pier]

Uh no, at origin, it's x=0

sin(0)=0

not 0.5

[trim fable]

ook

so its

OH ok

[lilac pier]

so now we have $sin(\frac{x}{4})$

That right?

[obsidian monolith]

Sup?:

[trim fable]

yeah

[lilac pier]

range is same?

0 to 4pi?

[trim fable]

yeah

and is = to 1/2

[lilac pier]

?

Ah yes

[trim fable]

so sinx=1/2 occurs at every pi right?

[lilac pier]

You'll see.

[trim fable]

oh ok

[lilac pier]

Draw the graph of sin(x/4) first, I'll do it for you.

From 0 to 4pi range.

[trim fable]

ill use desmos

i actually have it draw for me on the sheet we have to fill out

so ye

there

[lilac pier]

You need to understand yourself.

The coefficient of x in y=sin(x) tells you how many cycles will be there in a range of 0 to 2pi.

y=sin(x) means 1 complete cycle from 0 to 2pi.

y=sin(x/4) means only (1/4)th of the cycle will appear in the range of 0 to 2pi.

But since range is 0 to 4pi, I doubled it.

So there's only two solutions.

[trim fable]

ye

oh ok

so it occurs at every 2pi?

[lilac pier]

What does?

[rigid sun]

,w plot y=sin(x/4)

[obsidian monolith]

Display results online and refine query

Requested by maleb1964#6739

[hoary valley]

@frozen needle But how would I know if it's derivative exists.. I have to differentiate it first..

[frozen needle]

the existence of a derivative at $a$ is the existence of the limit
$$\lim_{h\to 0}\frac{f(a+h)-f(a)}h$$

[obsidian monolith]

Tuong:

[frozen needle]

$f$ is said to be differentiable at $a$ when
$$\lim_{h\to 0}\frac{f(a+h)-f(a)}h$$
exists, and in that case, the derivative of $f$ at $a$, $f'(a)$, is that limit

[obsidian monolith]

Tuong:

[hoary valley]

So basically this means, if I differentiate the function (f) at point (a) and it has a derivative at (a), I can call this function a "differentiable function at (a) " ..

Right?

[frozen needle]

you can't differentiate a function that's not differentiable

https://cdn.discordapp.com/attachments/363224154469826562/648265400412405780/94730921676115968.png

[hoary valley]

How do you know it's not differentiable without trying first?

[frozen needle]

everything's in there

https://cdn.discordapp.com/attachments/363224154469826562/648265400412405780/94730921676115968.png

the existence or non existence of $\lim_{h\to 0}\frac{f(a+h)-f(a)}h$ is what determines differentiability or non-differentiability at $a$

[obsidian monolith]

Tuong:

[hoary valley]

Ok what's a non-differentiability will look like in that derivative formula ? the outcome will be zero ?

[frozen needle]

if $\lim_{h\to 0}\frac{f(a+h)-f(a)}h$ does \textbf{NOT} exist, then $f$ is \textbf{NOT} differentiable at $a$, there isn't much more to say

[obsidian monolith]

Tuong:

[frozen needle]

$f$ is differentiable at $a$ if and only if
$$\lim_{h\to 0}\frac{f(a+h)-f(a)}h$$
exists, and when it does, the derivative of $f$ at $a$, denoted $f'(a)$, is that limit

[obsidian monolith]

Tuong:

[hoary valley]

But this function is not differentiable at 0

So how is that theorem right?

[frozen needle]

they only say "if", and not
"f is differentiable at a if and only if f is continuous at a"

the theorem is only a sufficient condition for continuity

[hoary valley]

Ok this function is continuous at 0, yet it's not differentiable at 0..

[frozen needle]

the theorem is only a sufficient condition for continuity, it's not a necessary condition for continuity

only a sufficient condition

do you understand the difference between "if ... then ..." and "... if and only if ..." ?

[hoary valley]

I mean yes..

[stuck lark]

f being d/dx-able at 0 guarantees f being continuous at 0, but f not being d/dx-able at 0 doesn't guarantee f isn't continuous at 0

[hoary valley]

Ok, I get it, but I don't like how it's written.

[frozen needle]

what don't you like about it? if you understand what "if... then..." means, then you shouldn't have any problem with it

[hoary valley]

Here : If (f) is differentiable at a, then (f) is continuous at a.
if (f) is continuous at a that does not guarantee its differentiability

It should be like that.

[frozen needle]

you just have issues with the "if... then..."

many theorems are just "if... then...", and there's no reason to specify every time they're only sufficient conditions because the if then already makes it clear

[prisma prairie]

How can i find the all the values?

[patent beacon]

You can factor (4 - x)² out

Thus 4 is a double root, and any others are solutions to
(4 - x) - 3x = 0

[prisma prairie]

when i check the answer the roots are x=1, x=4

i get the x=4 but not the x=1

@patent beacon

[wise kelp]

(4-x)^2 is one factor

What’s the other factor

[prisma prairie]

Idk

I forgot how to do this stuff

@wise kelp

[wise kelp]

Do you know how to factor out (4-x)^2?

[prisma prairie]

can i let a = (4-x) and do it like that?

its kinda hard to think with so much

[wise kelp]

Sure, if that helps

[prisma prairie]

Aright so i dont know if i did this right but yeha

[wise kelp]

Substituting a for (4-x)^2 should give you a^3-3(a^2)x

[candid onyx]

Can someone help me ? I don’t know where I’m going wrong

[patent beacon]

Cool whiteboard

,w partial fraction (6xxx - 7xx - 13x - 15)/(3xx - 5x - 2)

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[patent beacon]

,w remainder of (6xxx - 7xx - 13x - 15)/(3xx - 5x - 2)

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[patent beacon]

Your B equation has a mistake

Should read
B = -13 + 2A

@candid onyx

[candid onyx]

Thank you 🙏🏽

[candid onyx]

How can I use the bot you used to solve the problem without spamming the chat ?

[stuck lark]

#bots @candid onyx

[wise bison]

So the question was "What type of function best models this set of data?". May I ask why this isn't a logistic function?

[rigid sun]

[wind cedar]

Trying to convert to polar... not really sure what I'm doing here. Answer is supposed to be x = 1+ 3cost and y = -2 + 3sint.......

[rigid sun]

Stop

now

[wind cedar]

I wish I could stop... but I don't know wtf I'm doing and I have a test tomorrow... teach me

I don't even understand how the answer got t's in it...

[rigid sun]

rcos is for circles with 0,0

rcos being the x component

get this, if all the x components move right by 1 unit, a circle with center 1,0

guess what

Rcos also goes up by 1

[wind cedar]

Number 6

[rigid sun]

Yeah

[wind cedar]

so

the -1 and +2 have something to do with something?

[rigid sun]

That’s the center

1,-2

[wind cedar]

so how do I find the parametric?

[rigid sun]

lol

if rcos describes the x for a circle at 0,0

Then

rcos+1 is a circle at 1,0

[wind cedar]

ok

[rigid sun]

so substitute

[wind cedar]

(rcos-1)^2+(rsin+2)^2=9?

[rigid sun]

Bruh no

[wind cedar]

I really need to understand this... 😦

[rigid sun]

that is the parametric

x=rcost +1

[wind cedar]

I haven't had any time to study this shit, because all my other professors have been slamming me with assignments

yes, I need it in terms of the parametic

[rigid sun]

y=rsint-2

Done

[wind cedar]

but that's not right

[rigid sun]

yes it is

[wind cedar]

like i said the answer is x=1+3cost

[rigid sun]

R=3

[wind cedar]

y = -2 + 3sint

[rigid sun]

that’s what I just said

[wind cedar]

so you're just plugging the actual r value in

in place of "r"

[rigid sun]

yes

r=r=radius

[wind cedar]

since its r^2 and sqrt 9 is 3

that makes sense

I get this shit, i'm a quick learner if something is explained to me. I just haven't had any time to study for this test now I'm stressing the night before

ok also

[rigid sun]

now ur done

[wind cedar]

the answer is t instead of theta, so am I just supposed to know that if it says parametric to replace theta with t

[rigid sun]

t is the variable people like to use

While using theta technically isn’t wrong

It just is kinda awkward

[wind cedar]

it's like the "interval" of the equation right? I know that's not exactly right

teacher said it was kind of like time, but not always

[rigid sun]

it would be like writing a quadratic y=q^2

Like

It’s right

But it’s not really how people like to write it

People use t most often because in physics, time is a parameter

[wind cedar]

I guess I just need to pay attention to the rectangular form tomorrow. if my x and y are squared like that then its a circle

I tried to expand the terms because I watched a youtube video where a guy did that. Like he had x^2 + (y-2)^2 = something and he expanded the y-2

to get an r^2 term

I dunno

[rigid sun]

That’s sped

Why wtf

there are technically many parametric equations to do a circle

you can even do x=t y=+-sqrt(r^2-t^2)

[wind cedar]

I dunno. I'll just stick to this way I guess lol. I went to the in school tutor the other day and the dude forgot how to do parametric/polar stuff... so he spent a lot of time guessing how to get the answer so I didn't learn shit

But I think I learned more in 5 minutes of reading what you said then I did in that whole hour I spent with the tutor

[rigid sun]

Yeah

The rest should be easy

[wind cedar]

[spark lance]

I don't think I'm approaching this problem right

How do I deal with the 2x squared?

[willow bear]

well for starters i'd pull a constant factor of 18 out of the denominator before doing the whole partial fractions thing just to make my own life easier

[spark lance]

uhh where?

[willow bear]

$2x^2(3x+2)^2 = 18 \cdot x^2(x + \tfrac23)^2$

[obsidian monolith]

Ann:

[spark lance]

woah

how does an 18 get out of that?

ooh

no now I see how

that's crazy, I never realized that

[rich kite]

how do I graph y=tan3x? How do I find the scale and the key points?

[willow bear]

do you know how to graph y = tan(x)

[spark lance]

after taking out the 18, what do I do with it?

[rich kite]

@willow bear yes

[willow bear]

compress

not stretch

stretching would give y = tan(x/3)

but yes, graph tan(x) then compress horizontally by 3

[spark lance]

@willow bear I still don't understand what I should do when it's a constant times x^2 instead of (constant + x)^2

[willow bear]

again
pull the constant out
do the partial fraction decomposition without it
then multiply it back in

[spark lance]

oh, you didn't mention the last two...

alright I'll give it a go

do I multiply everything by 18 or just denominators?

[willow bear]

uh

i mean

you're gonna get

1/18 * (fraction)

[spark lance]

uh oh

[willow bear]

once you pull the 18 out of the denominator

then once you are done with the pfd on that fraction

multiply it by 1/18 again

[spark lance]

so like this?

wait no this can't be right

[willow bear]

no???

god

no

no

[spark lance]

I'm really lost here

[willow bear]

i'm telling you to do partial fraction decomposition on $\frac{54x^3 + 127x^2 + 8x + 16}{x^2(x + \tfrac23)^2}$

[obsidian monolith]

Ann:

[willow bear]

and THEN multiply that back by 1/18

[spark lance]

so the equation I have at the bottom of my recent screenshot, I multiply each by 1/18?

[willow bear]

uh

yes?

[viscid thistle]

Ok I'm really stupid

[willow bear]

,rotate -90

[viscid thistle]

Can someone explain simply how to do this

[willow bear]

what have you tried

[viscid thistle]

Law of sines law of cosine etc

idk what I'm doing

[willow bear]

well there's a solution to this problem that involves both of those laws

it can also be done using the law of cosines only

either way it might come in handy if you know the length of CB

[viscid thistle]

Ok then

ok wut

[willow bear]

did you use a calculator

[viscid thistle]

Yes

[willow bear]

...how did you even get that number

[viscid thistle]

I have literally no idea

Maybe my thing is busted

[willow bear]

,calc 13^2 + 22^2 - 2 * 13 * 22 * cos(105 * pi/180)

[obsidian monolith]

Result:

801.04449379864

[willow bear]

can you show your calculator and your exact input

[viscid thistle]

I mean it's my phone one which I used through physics and never a had any problems with

[willow bear]

can you show your calculator and your exact input

[willow bear]

these parentheses should not have been there

it's not $(a^2 + b^2 - 2ab)\cos(\theta)$

[obsidian monolith]

Ann:

[willow bear]

it's $a^2 + b^2 - 2ab\cos(\theta)$

[obsidian monolith]

Ann:

[willow bear]

@viscid thistle

[viscid thistle]

Oohhh

Ok now I got 28.30

[willow bear]

no

it's just microsoft paint

lmfao

you can hold shift while using a line tool to make a line that's perfectly aligned to one of the 8 cardinal directions

well uh

horizontal, vertical or diagonal

shift + rectangle tool gets you perfect squares

shift + ellipse tool gets you perfect circles

really handy

[willow bear]

,rotate -90

[warm crescent]

I have searched internet. They use some sin(2x)=sinx*cosx2 identitiy whichbi havent learnt

[willow bear]

sin(2x) = 2sin(x)cos(x)?

there's no way around using that.

have you learned sin(x+y) = sin(x) cos(y) + cos(x) sin(y)?

[warm crescent]

No

[willow bear]

then you do not know enough to solve this question yet.

[warm crescent]

Ok

[green zenith]

Would someone help me with my solution and see whats wrong?

[wise bison]

I'm not very good at this stuff, but how did the third line become the fourth?

[green zenith]

I used the quadratic equation

[wise bison]

Seems very odd. The original equation is 3^(2x) = 7^x?

[green zenith]

Yes

[wise bison]

But I'm pretty sure 0 is the only solution to that. Let me take a look for a moment.

You can take the logarithm straight away, actually.

There's something not quite right with that quadratic equation. 7^x is not equal to 7/3 • y.

[green zenith]

What i did is i let 3^x = y

So 7/3×3^x = 7^x

[wise bison]

No.

7/3 × 3^x is 7 × 3^(x - 1), which is not 7^x.

Try to rewrite 3^(2x) as 9^x and take the logarithm of both sides of the original equation. You should be able to obtain the solution.

[green zenith]

Okay ill try

Wouldnt this cancel my x out?

Isnt x/x = 1?

[wise bison]

Well, let's see:
9^x = 7^x
log (9^x) = log (7^x)
x log 9 = x log 7
We didn't cancel out x here. You can still solve for x.

[green zenith]

And then x/x = log7/log9?

[wise bison]

Not quite. Continuing on,
x log 9 - x log 7 = 0
x (log 9 - log 7) = 0
Notice that log 9 - log 7 cannot be 0, so x has to be 0.

You cannot divide by x in this case because x is actually 0 and by doing that, you're dividing by 0.

[green zenith]

Oh i didmt know you could do that

[odd helm]

Alright ig I can post now?

Could someone explain to me what I did wrong because Ik it’s a cosine graph so the answer should be cosine right?

[wise bison]

What is the expanded numerator?

[odd helm]

Uhhh I don’t know what an expanded numerator is

[wise bison]

Like, expanding (sec x + tan x)(sec x - tan x) using FOIL gives?

[green zenith]

@wise bison thanks i appreaciate the help

[odd helm]

Yeah

I think the key did that but I didn’t do it

[wise bison]

sec² x - tan² x, right?

[odd helm]

I’ll try that

Yeah it is

[wise bison]

Which is?

[odd helm]

1?

[wise bison]

Yep.

And 1/sec x is?

[odd helm]

Alright

Cos

Oh

Ok thank you

[wise bison]

No worries.

[green zenith]

Hello again

I need help with this equation no 77

It 2^x - 8.2^-x - 7 = 0

I forgot how tp start it off

Do i multiply both sides by 1^x?

[lilac pier]

change 8 into 2

[willow bear]

by 1^x?

you do know that's just 1 right

[green zenith]

Oh

I was looking to remove the -x

[uncut mulch]

finding a way to remove the negative exponent is a good idea
you can multiply by something, but it shouldn't be 1^x=1

[green zenith]

So something other than 1?

[uncut mulch]

it might be more obvious if you represented
8 * 2^(-x) as a fraction

what can you multiply to get rid of the fraction

[green zenith]

Oh yeah!

So i multiply 2?

So i dont have to remove -x and do quadratics

[topaz junco]

Anyone gets this??

[uncut mulch]

i mean you could factorise the original expression as is,
but getting rid of the negative exponent completely, makes things easier to see

i.e. what happens when you multiply everything by 2^x

[spark lance]

A is supposed to be a different number

[willow bear]

uh

$8x^3 - 27 \neq 8(x-3)(x^2 + 3x + 9)$

[obsidian monolith]

Ann:

[spark lance]

oh

can you tell me why it's not? I checked it and it seems to be equal

[willow bear]

can you, on a separate piece of paper, expand out $8(x-3)(x^2 + 3x + 9)$

[obsidian monolith]

Ann:

[willow bear]

i want to see you do it

so i can point out where you fucked up

[willow bear]

alright but $8(x^3 - 27) \neq 8x^3 - 27$

[obsidian monolith]

Ann:

[spark lance]

oohhh

aahh tyvm

[scenic musk]

how do i find where (5/sqrt(2)), 5/sqrt(2)) is on the unit circle?

[spark lance]

are you looking for the quadrant?

[scenic musk]

looking for the value at that point

trying to convert complex numbers

[willow bear]

wdym "convert complex numbers"

[scenic musk]

into polar form

z = 5/sqrt(2) +5/sqrt(2)i

[willow bear]

uh huh

and what's giving you trouble exactly

[scenic musk]

i need to express it as. z = re^(i*theta)

[willow bear]

uh huh

[scenic musk]

to find theta it says I need to find where 5/sqrt(2)

is on the unit circle

[willow bear]

uh

[scenic musk]

ive forgotten my trig

[willow bear]

have you found r already

[scenic musk]

yes, r = 5

[willow bear]

so

the number on the unit circle is 1/sqrt(2) + 1/sqrt(2)i

so what angle θ between 0 and 2pi has cos(θ) = 1/sqrt(2) and sin(θ) = 1/sqrt(2)

[scenic musk]

pi/4

[willow bear]

well there you have it

[scenic musk]

oh

wow tyvm

[clear glade]

How would I graph this?

What I did was cross multiply and say rsinx = 3 so y = 3 and graphed that

so it's like a diagonal line from bottom left to bottom right, is that how it should be done?

[dim jungle]

Should be except there should be holes in the graph

Cuz sin cannot be 0 or ur dividing by 0

[clear glade]

ok ty

[shrewd pebble]

@green ocean

[fallen cloud]

It's completely fine

Because the "holes" are actually not holes

They are the endpoints of the straight line that you see in polar and the asymptotes you see in y=sec(x)

[mellow scaffold]

How do I find the zeros of a function like this

3x^3+10x^2 ?

[uncut mulch]

$f(x) = 3x^3 + 10x^2 \ $ ?

[obsidian monolith]

ramonov:

[atomic bone]

in this case you can factor out an x^2 and get them like that

factoring by grouping will be necessary sometimes

[rare crow]

can someone get their cute ass in the voice chat and help me with pre calc:)

[prisma prairie]

f(x) = xlnx

how do we evaluate f(1/e) without a calculator?

[patent beacon]

ln(1/e) go

[prisma prairie]

no idea

[patent beacon]

Maybe try it with a calculator and see if you can explain the answer

[prisma prairie]

true

[patent beacon]

I don't have to be like this though I could just explain it

Lol which do you want?

[prisma prairie]

you could explain it i guess

i got -1

[patent beacon]

1/e = e^(-1)

ln(e^-1) = -ln(e) = -1

[prisma prairie]

oh i see

okay

thakns

thanks

[hoary valley]

It should be sqrt x^3 right?

[rigid sun]

??

sure i guess

that doesn't really matter

at all

[hoary valley]

@rigid sun Why it doesn't matter?

[rigid sun]

$x^{3/2}=x*x^{1/2}$

[obsidian monolith]

maleb1964:

[willow bear]

$1 + \frac12 = \frac32$

[obsidian monolith]

Ann:

[hoary valley]

@rigid sun Sure but then you have to remove the 2

[willow bear]

what do you mean remove the 2

why would you remove the 2

[rigid sun]

its still there

[hoary valley]

If you want this , then we have to remove the 2..

[rigid sun]

$x^{1/2}=\sqrt{x}$

[obsidian monolith]

maleb1964:

[rigid sun]

the two never left

you can think about the x as the dividend and the sqrt x as the remainder part that stays the same way

[willow bear]

@hoary valley what do you mean remove the two

are you refusing to accept that $2 \cdot x^{3/2} = 2 \cdot x \cdot x^{1/2}$?

[obsidian monolith]

Ann:

[hoary valley]

It should be exactly like that.

[rigid sun]

thats the same thing

just

less simplified

[hoary valley]

But then why are they not writting it like that?

[rigid sun]

its not as simplified

i doens't matter if you write it that way

it just makes problems easier to factor in the future

[hoary valley]

Hmm

[rigid sun]

literally the same thing

[willow bear]

are you refusing to accept that $2 \cdot x^{3/2} = 2 \cdot x \cdot x^{1/2}$?

[obsidian monolith]

Ann:

[hoary valley]

No

[willow bear]

you seem to be under the impression that each expression has one and only one "Simplified Form™" and any other forms one could put it into are Incorrect By Default™ and Should Not Be Used™

which just isn't true

[hoary valley]

Can you tell me What's the general idea behind this next step ?

[rigid sun]

thats kinda just

doing what it says

[willow bear]

...adding the the fractions together???

[green ocean]

@shrewd pebble

[hoary valley]

what's this equal to ?

[charred burrow]

As x approaches 0, that becomes 1

[hoary valley]

no limits

simply dividing

[uncut mulch]

don't know what you're looking for exactly

what is x?

[hoary valley]

Apparently you can factor out the GCF which is x... So I was wondering what's sin x/x

[uncut mulch]

can you post the original question / whole context?

[hoary valley]

It's literally " Factor out the GCF "

I think I know how to do it

[uncut mulch]

the gcf is just x
how are you getting sin(x)/x?

[hoary valley]

Idk why I thought 2x and sinx is 2 seperate things

.

^ is this equal to :

or

[uncut mulch]

sec^2 (x)

[hoary valley]

@uncut mulch Is it cuz it's a trig function ?

[uncut mulch]

not specific to just trig functions, but functions in general

the input is unaffected

[hoary valley]

Man you are wrong

[stuck lark]

how does that make ramonov wrong?

[serene heath]

[hoary valley]

Figure it out.

[uncut mulch]

as in
f(x) = x + 3
(f(x))^2 \neq f(x^2)
(f(x))^2 \neq x^2 + 3

[hoary valley]

My bad...

For some reason I thought sec and x are 2 functions..

@uncut mulch You are not wrong, I appreciate the help.

[pseudo sonnet]

can someone tell me if i did this correctly

particularly if 4pi/3 is a solution

how do i know if i can write a general solution for these terms

how do i know if pi/3 and 4pi/3 are equidistant

like sometimes you can combine them into 1 solution

my teacher said you can write a general solution for all the answers if theyre equidistant

how come they didnt write 4pi/3?

so i dont write 4pi/3 in my answer?

just the first one

[wise bison]

Yeah, it's already implied that π/3 + kπ includes 4π/3 and its corresponding solutions.

[pseudo sonnet]

how do i know that though?

[wise bison]

Because 4π/3 = π + π/3, which is π/3 plus an integer multiple of π.

And that is the general solution proposed by π/3 + kπ.

[pseudo sonnet]

i see thanks

for 2pi/3 and 4pi/3

does 2pi/3 imply 4pi/3 as wel?

well*

[pseudo sonnet]

can i combine the general solutions for sin(x)=0 into anything?

[serene heath]

Not really

[willow bear]

yes you can

x = kπ

[pseudo sonnet]

yeah thats what I did ^

what can i do from here?

[willow bear]

uh

fix your parenthetical mismatches

[pseudo sonnet]

yeah i just noticed

doing it rn

[willow bear]

but also this just seems like (tan(2x)-1)(sec(2x)-1) = 0

[pseudo sonnet]

wait what lol

i factored out the sec(2x) in the beginning

how did u get to that @willow bear

[willow bear]

get to what

my factorization?

well the first step is not mistaking -a + b for -(a+b)

[pseudo sonnet]

ohhh

i see now

couldnt see at first because the problem was given with brackets for some reason

these are the final solutions i got

oh shit wait lol

cos one is wrong

[willow bear]

sec(2x) = 1.

[pseudo sonnet]

those are the solutions i got

[willow bear]

looks ok to me

[pseudo sonnet]

for this one

should i have divided by root 3

then squares both sides

like that

[uncut mulch]

you didn't divide the cos by sqrt(3) in that

try dividing everything by 2

[uncut mulch]

@pseudo sonnet you made an error solving
cos(x) = -1/2, ||( cos(pi/3) \neq -1/2)||

note that squaring can create extraneous solutions

1-cos(x) is non negative, so
sqrt(3)sin(x) must also be non-negative
which will restrict the domain of the principal angle to (0 <= x <= pi)

[pseudo sonnet]

you’re right i just checked

good catch

[uncut mulch]

what should the correct solution for the cos component be?

alternatively you can also try:

$\frac {\sqrt{3}}{2} \sin(x) + \frac12 \cos(x) = \frac12$

[obsidian monolith]

ramonov:

[uncut mulch]

and note that sqrt(3)/2 and 1/2 are special ratios

[viscid thistle]

I need help just understanding this part on distribution I may have forgotten, 3(12-x^2/x)^2 I get that the exponent goes in to 12 and -x^2 to get 432 and 3x^4. I just don't see where -72x^2 comes from to get an answer of 3x^4-72x^2+432

[uncut mulch]

$3\left( \frac{12-x^2}{x} \right)^2$?

[obsidian monolith]

ramonov:

[uncut mulch]

$(a - b)^2 \neq a^2 + b^2$

[obsidian monolith]

ramonov:

[viscid thistle]

1st one

oh no I just got it

thanks

[glossy rock]

Anyone know any good online calculators

[vernal anchor]

wolframalpha

[flint stirrup]

symbolab

[marble sand]

My teacher asked to prove it using just the graph (but not geometrically) and I'd love some help.

[hoary valley]

Can I factor out 2 from the numerator?

[willow bear]

can you?

[hoary valley]

It's written like that as the final answer in Chegg

but I just want to make sure

I think I can yes

But Idk, even symbolab doesn't factor out 2

[willow bear]

yes you can

but just because you can does not mean you have to

[marble sand]

Could anyone help me out?

[willow bear]

you haven't posted a problem yet

so until then, no

[serene heath]

he posted a while back

[heady jewel]

lol

[marble sand]

Yes i did

[serene heath]

what are u stuck with

[marble sand]

Well

[willow bear]

care to repost

[marble sand]

The "teacher's method" is the whole problem

[serene heath]

https://cdn.discordapp.com/attachments/363224154469826562/649223929126191104/20191127_131838-1.jpg

is the teacher method that one at the bottom

[marble sand]

Yeh

Start with

2<alpha<3

[serene heath]

have u come across the intermediate value theorem

[marble sand]

No

Well he made a reference to it

But since we havent studied continuity yet

[serene heath]

seems like its what he wants u to use here

[marble sand]

Well, i supposed so

But he clearly stated that we mustnt use it

I could maybe prove a specific case of the ivt

And he also mentioned something about f(x)-g(x)'s sign change may inmply an intersection

[serene heath]

yes that coupled with the fact that its continous

thats IVT

[marble sand]

Anyways, have you any idea on how to tackle this without ivt? Maybe use the expressions of f and g in an indirect way?

[hoary valley]

How is that denominator equal to the other one?

[vernal rapids]

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

[obsidian monolith]

DarK:

[marble sand]

Im really stuck here

Would really appreciate some help

[serene heath]

Why cant you use ivt

[hardy abyss]

I think IVT must be the intended method

if you try you can solve it explicitly, but in order to do it you need to find the roots of a cubic with three irrational roots

[tidal rain]

hey for b here is the only way to solve this quadratic formula

[willow bear]

you could divide out by 2 first

[viscid thistle]

@tidal rain do you want to know how to solve it with or without the quadratic formula?

you can solve it by factoring

[tidal rain]

Yeah but when i factor i got 2(x^2+10x-3)=0

[marble sand]

@serene heath we havent got there yet
@hardy abyss im pretty sure there must be some way to do it without cubic roots and ivt, maybe a kind of intuitive ripoff of ivt? Idk

[wise bison]

@tidal rain - I guess you could complete the square, but the quadratic formula seems to be easier.

[viscid thistle]

you should just use the quadratic formula

[wise bison]

The discriminant isn't a perfect square, so there's no easy way to factor this.

[tidal rain]

Cool i was wondering if that was really the easiest way

Ty guys

[viscid thistle]

i get the roots to x_1 = -10.29, x_2 = 0.29

Factored: f(x) = 2(x + 10.29)(x - 0.29)

[tidal rain]

Yeah i have never factored with decimals lol

[tidal rain]

For an absolute value equation |a|=b

Which is more correct for the two cases

a=-b and a=b

Or -a=b and a=b

[willow bear]

the two are the exact same ¯_(ツ)_/¯

[dense oar]

How do I use half angle formulas to find cos(292.5°)

$-\sqrt \frac{1 + cos 225°}{2}$

I think it’s that but I’m not sure how to simplify

[obsidian monolith]

AutisticAri:

[dim jungle]

so

do it instead from the

cos(292.5) as the half angle

[elder junco]

Hello, may someone please help me what this dot means?

[dim jungle]

composition @elder junco

[elder junco]

Hmm... Okay thank you.

[dim jungle]

so you would have

[obsidian monolith]

Jbao:

[dim jungle]

@dense oar

thought ur way works fine as well

wait

i misread

ur question

[dense oar]

Wouldn’t it be other way around

[dim jungle]

mb

yeah

[dense oar]

But one I simplify I got this

[dim jungle]

$\cos{292.5}= \pm \sqrt{\frac{1+\cos{585}}2}= \pm \sqrt{\frac{1+\cos{225}}2}= \pm \frac{\sqrt{2-\sqrt{2}}}2$

[obsidian monolith]

Jbao:

[dim jungle]

$a^2+b^2=2$ and $2ab=-\sqrt{2}$

[bold isle]

I tried proving this limit but i need to assume that delta is equal to 1 but I don't know how to get rid of the abs(1/x)

[obsidian monolith]

Jbao:

[dim jungle]

which doesnt rly simplify nicely sad

[valid violet]

@bold isle what's delta, where does abs(1/x) come from

[elder junco]

So composition is basically imputing a function in a function’s x variables?

I think...

[valid violet]

$f\circ g(x) := f(g(x))$

[obsidian monolith]

gfauxpas:

[elder junco]

So yeah basically.

Thanks.

🙂

[valid violet]

👍

[bold isle]

i'm not sure about the 4th point @valid violet

the function is e^2x -1/x = 2 when x => 0

[valid violet]

do you have to use epsilon-delta to prove it?

[bold isle]

yes

[valid violet]

why are you setting delta=1

delta will generally depend on epsilon

for e-d proofs

[bold isle]

I've seen proofs that set delta to 1, so delta will have to be the minimum between 1 and the epsilon found

[valid violet]

thinking

where's the 3x come from

[bold isle]

I put -3x in each side of the inequality

[valid violet]

which inequality

[bold isle]

the ones in the first and third point

[serene heath]

How did your class define e^x?

[bold isle]

An exponential function(?

[valid violet]

in precalc usually they define it as a limit of a sequence

(1+x/n)^n

[serene heath]

Not sure how to do without using derivative of e^x

[bold isle]

oh i found that exercise in a book of calculus, we haven't defined that limit yet

[valid violet]

hold on I made a list

https://proofwiki.org/wiki/Definition:Exponential_Function

I made most of these uwu

but usually the sequence definition is the ones used in precalc because it doesnt use derivatives

and because it looks like kthe compount interest formula

why are you octagoning me bro

[viscid thistle]

no uwu-ing

[valid violet]

;_;

[tidal rain]

Did i verify the solutions correctly here

Or do i have to use the original equation

[wise bison]

It's better to use the original equation, in case the following steps are wrong. This helps you to spot mistakes, if there are any.

[wraith hawk]

how do you find roots of i?

so like the x^2=i, x^3=i

ive seen the formula but i want some reasoning behind it

[hoary valley]

Which one is the quotient rule ?
I tried solving a problem using these 2 formulas, I got 2 different answers...

[willow bear]

i mean yeah these differ by sign

it's the one on the right.

[hoary valley]

Thanks...

[willow bear]

also this is calculus

[hollow pelican]

hey can someone please help me with a question

the topic is logarithms

[willow bear]

ok

which of these two do you need help with

[hollow pelican]

both

pls

a and b

[willow bear]

okay let's start with a

[hollow pelican]

ok

[willow bear]

what is giving you trouble here

[hollow pelican]

so I think so far I have switched the x and y

But i dont know what to do after

[willow bear]

$x = -2^{\frac13y - 1} + 4$

[obsidian monolith]

Ann:

[willow bear]

so you have this?

[hollow pelican]

Yes

[willow bear]

well your end goal is to isolate y

[hollow pelican]

Ok

[willow bear]

there's one simple thing you can do to get you closer to that

[hollow pelican]

x-4=...?

I can move the 4 right

[willow bear]

there's no such thing as "moving" anything

you can subtract 4 from both sides; what will this give you?

[hollow pelican]

x-4

is equal to the RHS

and then 4-4=0

so those cancel out

What next?

[willow bear]

$x-4$ is not equal to $-2^{\frac{1}{3}y - 1} + 4$.

[obsidian monolith]

Ann:

[hollow pelican]

i was thinking x-4=-2^(1/3y-1)

[willow bear]

then why did you not write that

i can't read your mind so idk what you are thinking

[hollow pelican]

Mb

im confused as to what to do next though

[willow bear]

well now you have x - 4 = -1 * (something with y)

what can you do

[hollow pelican]

-1? isnt it -2?

can i divide by -2

[willow bear]

$-2^t$ is $-(2^t)$, not $(-2)^t$.

[obsidian monolith]

Ann:

[hollow pelican]

ok

[willow bear]

i mean, you can divide by anything you want.v just be mindful of what actually happens when you do that.

[hollow pelican]

so how would i find the inverse of the equation

[willow bear]

$x-4 = -1 \cdot 2^{\frac{1}{3}y-1}$

[obsidian monolith]

Ann:

[willow bear]

keep on isolating y, you're not done yet.

[hollow pelican]

why cant 2 be negative?

[willow bear]

$-2^t$ is $-(2^t)$, not $(-2)^t$.

[obsidian monolith]

Ann:

[hollow pelican]

-x+4 = log base 2 *1/3y-1

[willow bear]

no

[hollow pelican]

ok

[willow bear]

$2^t$ is not the same thing as $\log_2(t)$.

[obsidian monolith]

Ann:

[hollow pelican]

oh

[willow bear]

...

[hollow pelican]

uwu

[willow bear]

...

[hollow pelican]

im really confused as to what to do now

[willow bear]

$4-x = 2^{\frac13y - 1}$ \
now take $\log_2$ of both sides, and get \
$\log_2(4-x) = \frac13y - 1$

[obsidian monolith]

Ann:

[hollow pelican]

y={log_2(4-x)+1}(3)

[willow bear]

that's a really obtuse way of writing it

but sure, yes, $y = 3(\log_2(4-x)+1)$

[obsidian monolith]

Ann:

[hollow pelican]

how did u know to take log 2 of both sides?

[willow bear]

,,, do you even know what log is

[hollow pelican]

damn ok

[willow bear]

if you do it's obvious, if you don't you need to learn what log is in the first place

[hollow pelican]

i guess so

my teacher has gone really fast in the past few days and im trying to understand what has been taught

its all pretty new to me

hence why i asked for help

thanks for the answer tho

[viscid thistle]

Log = logarithm

The only stance I know how to use it is with exponents.

[willow bear]

...you're late.

[viscid thistle]

Perhaps

Not only am I late, but it’s late. To the point that it’s early, thus Ima hit the pillow.

C’y’all later

[wise bison]

Quick question: The subscript w is in boldface, so am I required to put an arrow above it in writing?

[willow bear]

if you insist on putting arrows above all vectors in writing, then yes.

[wise bison]

Alright. Thanks.

[valid violet]

But it's usually apparent what's a vector even if you don't put arrows

Just by writing | | | | and •

[wise bison]

Ah. I wasn't sure because my text says all vectors should come with an arrow.

[willow bear]

once you've done enough linear algebra, you can usually distinguish with relative ease what is what

[wise bison]

Ah. I'm nowhere near proficient in that. These are just 2 sections in the trigonometry section of my precalculus text. 😰

[valid violet]

if it helps you not be confused, then use arrows or boldface for sure

nothing wrong with it

when I took my calc 2 exams I actually brought a fine point marker for vectors

when I was a little mathematician

@wise bison

[wise bison]

Ah. I guess I'll keep doing them for now since they don't really get in my way.

[tiny dove]

Ellipse: standard form satisfying
Center at ( 5,1 ) , V1( 5,4 ) ; End of Minor axis ( 3,1 )

since were just gonna subtract out the Vertices and the Minor axis what happens if it ends in the Major axis?

do we add???

or still subtract?

[distant basin]

quick question: if I want to find the gradient function of 3x^2, is it f'(x) = 2x or f'(x) = 2x + 3.
So do I only use f(x) = x^n -> f'(x) = nx^n-1 or do I have to use f(x) = mx+p -> f'(x) = m

[serene heath]

Its neither

[distant basin]

Oh wait I think I might understand now. 6x?

[stuck lark]

👍 and parentheses, nx^(n-1)

[distant basin]

So 3x^2 - 4x + 3 is 6x - 4

yes, ty 😄

[green zenith]

I was looking at my book and i didnt find any equations for solving the area for ssa triangles. So if i were given a ssa triangle to solve, i determine first if it has 1, 2 or no triangle. 1 and no triangles are simple. But if i had two triangles, i have to look for the area of the two triangles.

Is this correct?

[faint sapphire]

1/2 ab sin (C) where C is a co-interior angle to a and b is a formula for the Area of a triangle when given two sides.
Hope that helps?
@green zenith

[green zenith]

That is an triangle SAS, right?@faint sapphire

[uncut mulch]

with SSA, you have ambiguous cases and you'll have to check for all of them

[faint sapphire]

Yes. With two sides of known length and one angle in between them. (SSA or SAS).

[uncut mulch]

SSA is different from SAS

when represented as SSA, the angle between the two given sides is ambiguous

[green zenith]

Ahhh that is what i thought

Thanks

[obsidian monolith]

JY1853:

[sturdy haven]

^

Infinity is the idea of never ending growth

It keeps getting bigger

[obsidian monolith]

JY1853:

[sturdy haven]

How do you do notation on here

[valid violet]

@frail nest you familiar with M-N proofs? like epsilon delta but with infinite limits?

how about with regular epsilon delta proofs

so it works the same way but in the opposite direction

instead of saying "for |x-c| sufficiently small" and |f(x)-f(c)| sufficiently small"

you say "for x sufficiently large", and "for f(x) sufficiently large"

so

the same way you would tell me in an epsilon delta proof

"I want |f(x)-f(c)| to be epsilon close"

you say

"I want f(x) to be larger than M"

follow so far?

so

I want sqrt(x) > M

I claim x > N works if I choose N = M^2

can you finish the proof?

make it N=M^2 + 1 if you'd like

then sqrt(N) = sqrt(M^2+1)>sqrt(M^2)=M

so sqrt(N) > M for N = M^2 + 1, no matter how big M is

$\blacksquare$

[obsidian monolith]

gfauxpas:

[valid violet]

called an M-N proof instead of epsilon-delta

same idea

[sturdy haven]

So what would that mathematical proof actually look like

[valid violet]

Theorem:

$\lim_{x \to +\infty} \sqrt x = +\infty$

[obsidian monolith]

gfauxpas:

[valid violet]

Proof:

this limit statement is the assertion:

for every M > 0, there exists an N > 0, such that whenever x>N, sqrt(x) > M