precalculus

[slender river]

wym its not a cofunction

it's still a transform of some flavor

consider cot(x-h)

cot(x-pi/2) is just -tan(x)

[swift glacier]

Wait how?

[viscid thistle]

cot(pi/2 - pi/7) = tan(pi/7) by the co-identity.

And then oddness of cot completes it

[swift glacier]

Oooh I see

[rare zephyr]

Can someone help me with No.7. Second question

[atomic zodiac]

so you have the gradient at (2,-5)

you just need another point that has the same gradient

set y'= gradient at (2,-5), should be a quadratic

with one solution at x=2 and the other solution will be the other one

@rare zephyr

[rare zephyr]

Thx

[drifting imp]

can someone help me out with this logs question. Given log base 3 of 5=x and log base 3 of 2=y, write log base 3 of (12/5) in terms of x and y.

[rocky bison]

So what do you have @drifting imp

,tex \log_3\left(5\right)=x\ \ \log_3\left(2\right)=y

[obsidian monolith]

Pseudo:

[rocky bison]

This?

[drifting imp]

yes

[rocky bison]

Ok

So first we want log_3(12/5)

Now

If we do

,tex \log_3\left(2\right)=y\\log_3\left(2\right)+\log_3\left(6\right)=y+\log_3\left(6\right)\\log_3\left(12\right)=y+\log_3\left(6\right)

[obsidian monolith]

Pseudo:

[rocky bison]

Yes?

[drifting imp]

yea

[rocky bison]

Now let's look at

[drifting imp]

so solve it for y and then solve the other by x?

[rocky bison]

,tex \log_3\left(6\right)

[obsidian monolith]

Pseudo:

[rocky bison]

Hol up

Now we know this is equal to

,tex \log_3\left(6\right)=\log_3\left(3\right)+\log_3\left(2\right)

[obsidian monolith]

Pseudo:

[rocky bison]

Yes?

[drifting imp]

ok

[rocky bison]

Now log_3(3) is 1

log_3(2) is y

So

,tex \log_3\left(12\right)=y+y+1\\log_3\left(12\right)=2y+1

[obsidian monolith]

Pseudo:

[rocky bison]

Yes?

[drifting imp]

yea

[rocky bison]

ok

So now we have

,tex \log_3\left(12\right)=2y+1\ \ \log_3\left(5\right)=x

[obsidian monolith]

Pseudo:

[rocky bison]

Now if we subtract those two equations from one another

,tex \log_3\left(12\right)-\log_3\left(5\right)=2y+1-x

[obsidian monolith]

Pseudo:

[rocky bison]

Now by log rules

,tex \log_3\left(\frac{12}{5}\right)=2y+1-x

[obsidian monolith]

Pseudo:

[drifting imp]

kinda lost with the last part

[rocky bison]

Ok

So we've got

,tex \log_3\left(12\right)=2y+1\ \ \log_3\left(5\right)=x

[obsidian monolith]

Pseudo:

[rocky bison]

Now on this equation

,tex \log_3\left(12\right)=2y+1

[obsidian monolith]

Pseudo:

[rocky bison]

We're going to subtract x on both sides

,tex \log_3\left(12\right)-x=2y+1-x

[drifting imp]

ok

[obsidian monolith]

Pseudo:

[rocky bison]

Now we know that

,tex x=\log_3\left(5\right)

[obsidian monolith]

Pseudo:

[rocky bison]

So, on the left hand side of that equation we can say instead of x

log_3(5)

Giving us

,tex \log_3\left(12\right)-\log_3\left(5\right)=2y+1-x

[obsidian monolith]

Pseudo:

[rocky bison]

Yes?

[drifting imp]

ok

[rocky bison]

Now

One of the rules of logs is

,tex \log_c\left(a\right)-\log_c\left(b\right)\equiv\log_c\left(\frac{a}{b}\right)

[obsidian monolith]

Pseudo:

[drifting imp]

yea

[rocky bison]

So applying this to the left hand side of our equation

,tex \log_3\left(\frac{12}{5}\right)=2y+1-x

[obsidian monolith]

Pseudo:

[rocky bison]

Yes?

[drifting imp]

yea

[rocky bison]

and we're done

[drifting imp]

ok thanks for the help

[rocky bison]

that's log base 3 of 12/5

in terms of x and y

np

What?

@viscid thistle That's not much of a question

Proving what

It's not a question rn

Unless there's some weird notation of | I'm not aware of

[flat turret]

a|b means that a divides b

[rocky bison]

ah ok

=pup 3 mod 20^2-1

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@rocky bison

[rocky bison]

[spring thunder]

20^2 -1 mod 3 tho

[rocky bison]

oh lmao

=pup 20^2-1 mod 3

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@rocky bison

[slender river]

lol

[rocky bison]

lemme get some paper for this lmao

im trying it now

ok

I got it

wait no nvm

ok no

it's just wrong lmao

=pup \left(20^2 * 2 - 1\right) mod 3

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@rocky bison

[flat turret]

yo sry I was afk

[spring thunder]

wat mate

[flat turret]

I think it should be $20^{2n}-1$

[obsidian monolith]

Gonzo17:

[spring thunder]

well the thing he posted is ambiguous also ye

[rocky bison]

Oh

Lemme try again lmao

[flat turret]

then it's true

Try again? It's just $20^{2n}-1\equiv (-1)^{2n}-1=1-1=0$ mod 3

[obsidian monolith]

Gonzo17:

[rocky bison]

ok I got it

So

,tex 3\ |\ \left(20^{2n}-1\right)

[obsidian monolith]

Pseudo:

[rocky bison]

Can be re written as

wait nvm spotted flaw in my work lmao

[flat turret]

There's at least 3 distinct proofs btw

[rocky bison]

i trying to think of one 😢

I think I got one

@flat turret Mind checking my process as i go?

or my thought process

[flat turret]

Ya sure

[rocky bison]

😄

If we consider

,tex 20^{2n}

[obsidian monolith]

Pseudo:

[rocky bison]

We can say that it's even for all natural n

And now consider the sequence

,tex 20^{2n}, 20^{2n}-1, 20^{2n}-2

[obsidian monolith]

Pseudo:

[rocky bison]

20^2n is going to be even

as will 20^2n-2

And given the width of the sequence is 3

One of them must be divisble by 3

Wait fuck

I'm retarded

nope

Nope nope nope

I assumed all odds were divisible by 3

ffs

I'm such a retard lmao

time to try again

OK

I fucking got it

So we prove for n=1

,tex 20^{2\left(1\right)}-1

[obsidian monolith]

Pseudo:

[rocky bison]

=pup 20^2-1 mod 3

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@rocky bison

[rocky bison]

So it's true

Now we assume that it's true for 20^2n

,tex 20^{2k}-1=3m

[obsidian monolith]

Pseudo:

[rocky bison]

Where m is some arbitrary integer

So we're assuming that this is true

Now, we prove it for n=k+1

,tex 20^{2\left(k+1\right)}-1\20^{2k+2}-1\20^{2k}\cdot20^2-1\20^{2k}=3m+1\\left(3m+1\right)20^2-1\3m20^2+20^2-1

[obsidian monolith]

Pseudo:

[rocky bison]

Now we've already proven that 20^2-1 is divisible by 3

And 3*m20^2 has to be divisible by 3

Therefore their sum is also divisible by 3

@flat turret Checks out?

[flat turret]

Yup

[rocky bison]

yey:D

on that note i go for dinner 😃

[flat turret]

Do you know modular arithmetics?

[rocky bison]

nop

don't belliev so

[flat turret]

Well if someone watches I can give a cool soln

[rocky bison]

👀

[flat turret]

$20^{2n}-1=(20^n-1)(20^n+1)$

[obsidian monolith]

Gonzo17:

[flat turret]

obviously $20^n$ isn't divisible by 3

[obsidian monolith]

Gonzo17:

[spring thunder]

i watch

[flat turret]

and one of $20^n-1, 20^n, 20^n+1$ is

[obsidian monolith]

Gonzo17:

[rocky bison]

Ohh ofc

[flat turret]

cause 3 consecutive integers

[rocky bison]

Yep

I tried to do some similar first but messed it up

[flat turret]

You can completely bash it with modular arithmetics though

$20^{2n}-1\equiv (-1)^{2n}-1=1-1=0$ mod 3

[obsidian monolith]

Gonzo17:

[spring thunder]

such precalc powa

[flat turret]

competition math >> precalc

[spring thunder]

or ze induction

3 mod 0 huh?

ahh

3 | 0

[flat turret]

tbh I still don't get howl's message

your last one

what 3 mod 0 means

There's a weird 400 in the middle

=400=

but if you just cancel that it's correct

btw this reminded me of a really cool problem

Find all n for which $14^n-9$ is prime

[obsidian monolith]

Gonzo17:

[flat turret]

How what got there?

There's no general way of proving things by induction

There's actually very much kinds of induction

Mainly you just try to see how you could use the stament for k

play around for a bit remembering your goal of proving the stament for k+1

and depending on the difficulty of the problem after some time you should be able to get it

Idk any recources except for problems from contests

but those are usually way harder than this

http://home.cc.umanitoba.ca/~thomas/Courses/InductionExamples-Solutions.pdf this seems like a decent resource

Proofs by induction can look very differently

It's always gonna be specific to the field you're using it in and so on

As I said, there's no general rule

You just have to try to prove the stament for k+1 knowing the stament for k holds

induction is a method of proving

recursion is a method of defining

np

[limber bone]

@rare arch

ask here

[rare arch]

the answer of this? I can't seem to get it

[limber bone]

if you can put them in the same base

then u can equate the exponents

so we can say 8 = 2^3 right?

[rare arch]

yea, I did this

but I just don't know how to remove the square

or it falls down

[limber bone]

2^root(2x+1) = 2^3^(root(x-3)

right?

[rare arch]

yes

[limber bone]

no

[rare arch]

3*

[limber bone]

not 2

3*

ye

[rare arch]

not 2*

ye

[limber bone]

you know a^b^c = a^bc

so now we got 2^root(2x+1) = 2^3root(x-3) right?

XD

right?

[rare arch]

i suppose so, yes

[limber bone]

you get it how we got here?

from this exponent law

a^b^c = a^bc

so 2^3^root(x-3) = 2^3root(x-3)

[rare arch]

I need to write this down, in my stupiud bug, there it isn't shown

[limber bone]

write it

[rare arch]

a^b^c should look like this?

[limber bone]

ye

like in our problem here

2^3^root(x-3)

= 2^3root(x-3)

right?

[rare arch]

yes,yes

[limber bone]

ok so now we got

2^root(2x+1) = 2^3root(x-3)

right?

[rare arch]

Yes

[limber bone]

ok so if the bases are equal

then the exponents are equal

right?

so root(2x+1) = 3root(x-3)

right?

[rare arch]

they are, yes

and now, how do we get rid of the roots/

[limber bone]

square both sides

(root(2x+1))^2 = (3root(x-3)^2

2x+1=9(x-3)

[rare arch]

and we get, yes

[limber bone]

2x+1=9x-27

-7x+1=-27

-7x=-28

x = 4

got it?

[rare arch]

OH LORD

THANKS

haha, thanks you, much appreciated

[limber bone]

got it?

np man

[shrewd flame]

May someone help me with this equation? to prove this https://prnt.sc/m2vark

[slender river]

for proving ? ok

[shrewd flame]

yep!

[slender river]

wanna probably multiply by a conjugate kind of object

"1"

basically like a difference of squares or something

for the denominator

[shrewd flame]

O yes

[slender river]

so if the denominator is cosx+sinx

what binomial can you multiply it with to get something clean

?

[shrewd flame]

cosx-sinx

[slender river]

ye

but we want to work only in the Left hand side, and not want to change the value of the expression

so we do both numerator and denominator

make sure you foil correctly everywhere when doin this

frick now i gotta look up the double angle identities for secant and tangent lmao since i never learned them

[earnest nymph]

Lol

I can write it down

[slender river]

kek

[shrewd flame]

did i foil this correctly? https://prnt.sc/m2vho1

[earnest nymph]

@shrewd flame Yes

[shrewd flame]

😂

[slender river]

ye

[shrewd flame]

oo yay~

[earnest nymph]

Then u can do 3 things here

[slender river]

can we make that sin^2(x) in terms of cos^2x

wait no

wat am i doin

[earnest nymph]

That's extra work tho

[slender river]

yeah

big long way around

i see the double angle in the numerator

[earnest nymph]

And also his notation os wwirx

There are 2 double angles, 1 in the numerator and 1 in the denominator

And also we can combine cos^2 x and sin^2 x

[slender river]

ye

make it clean in the numerator

[dense zealot]

1- sin2x/coz2x

?

Simple..

[earnest nymph]

Yes

[dense zealot]

no

[shrewd flame]

so what part are we on

[slender river]

simplifying wat we got

cleaning it up and condensing terms

using some identities

[shrewd flame]

oki

so how do i clean up the numerator

[slender river]

what are the double angle identities you know?

well any and all trig identities tbh

[dense zealot]

There are infinite

[slender river]

(shh)

this is precal

lolol

[shrewd flame]

lmao

so the cos(2x)= cos^2(x) -sin^2(x) one?

[slender river]

uhhhhh

ye

you can use that one

had t oscroll up bc it's the one double angle identity that throws me off smh

[shrewd flame]

but theres a 2 in front of cosxsinx

[slender river]

ok that one in the numerator

that is a double angle identity for a differnt trig function

look at what Big Good/Bad Wolf posted earlier

[shrewd flame]

yeah i have that on my formula sheet

[slender river]

ok

so what you see, that 2cosxsinx...

[shrewd flame]

is it the same as 2sin(x)cos(x)

[slender river]

ye

multiplication is commutative

[shrewd flame]

oo yay

so do i also replace cosx^2 with 1-sin^2x?

[slender river]

in the numerator?

[shrewd flame]

yea

[slender river]

well sure if you want

remember sin^2(x) +cos^2(x) is just 1 anyway

so either way works

[shrewd flame]

so like this? https://prnt.sc/m2vmnz

[slender river]

oh resend it didnt embed

[shrewd flame]

one sec

[slender river]

:o owo

discord br0ke

[shrewd flame]

https://prnt.sc/m2vna8

:D

[slender river]

w0t

[shrewd flame]

lmao

[slender river]

that's one step yes

[shrewd flame]

o so im on the right track

[slender river]

now need to do more simplifications

yes

[shrewd flame]

oh ok good

so then i could cancel out the cosx^2 in both the numerator and denominator right?

[earnest nymph]

No

[slender river]

onoe

can use more identities that Big Good/Bad Wolf#7121 posted

https://media.discordapp.net/attachments/363224154469826562/530644566974529558/JPEG_20190104_151203.jpg?width=371&height=499

[obsidian monolith]

Pleb':

[slender river]

nice

[earnest nymph]

And also $ \cos^2{x} + \sin^2{x} = 1 $

[obsidian monolith]

Big Good/Bad Wolf:

[slender river]

those two and then you can get it pretty easily

[dense zealot]

no u

[slender river]

yes me

[shrewd flame]

like this? https://prnt.sc/m2vqfr

[slender river]

oh boy

[shrewd flame]

looks wrong

[slender river]

undo all of those moves

the move in the numerator is technically not wrong

so if you want, you dont have to undo it

but the ones in the denominator need to be undone bc it doesnt get you anywhere

[shrewd flame]

oh ok

[slender river]

you'll return back to (cosx)^2- (sinx)^2

[shrewd flame]

how come

oh

[slender river]

yeah

[shrewd flame]

wuts next

[slender river]

https://cdn.discordapp.com/attachments/363224154469826562/530650366870683649/167009558370320384.png

use this

and also simplify the top bc you can cancel some terms up there (notice the 2 sin^2(x)'s)

[shrewd flame]

for the formula u showed me, where do i use that one

nvm lmaoo

[slender river]

yea hah

it be like that

[shrewd flame]

also sin2x is the same as sinx^2 right?

[slender river]

ononono

[shrewd flame]

oh

[slender river]

sin(2x) is not sin^2(x)

[shrewd flame]

so i simply both of them right?

[slender river]

wym

[shrewd flame]

the -sin2x and the sinx^2

[slender river]

cant cancel??

leave the -sin(2x) alone for now

[shrewd flame]

oki

ok so this part

https://prnt.sc/m2vt9j

[slender river]

cool the bottom is correct

in the numerator tho

[shrewd flame]

what do i do there

[slender river]

dont want to convert both sin^2(x) and cos^2(x) bc it will return you to the same spot you were in before

[shrewd flame]

right

[slender river]

ty to rearrange it so you see the identity you want, which is sin^2(x) + cos^2(x) =1

[shrewd flame]

yeah that one

lol im so confused

[slender river]

try putting parens around each term in the numerator and making it a sum instead of there being a difference in the numerator

meaning

instead of something like a-b

you make it a + (-b)

and since addition is associative(?)

or something

whatever the word is to show that order doesnt matter with addition

(might be commutative, associative deals with parens or something)

you can easily rearrange it

[earnest nymph]

Communative, not associative

[slender river]

ye

[shrewd flame]

can u write it out for me

[slender river]

$\frac{\cos^2{x}-\sin{(2x)}+\sin^2{x}}{\cos{(2x)}} \text{becomes} \frac{(\cos^2{x})+(\sin^2{x})+(-\sin{(2x)})}{\cos{(2x)}}$

[obsidian monolith]

MetalNinja27:

[shrewd flame]

but what happened to the 1's?

[earnest nymph]

We are talking about this

[slender river]

yeah i meant to tell you to undo those steps in the numerator when you made them "1-sin^2(x)" and the other bc it doesnt actually help

[shrewd flame]

ohhhhhh

so then does it become https://prnt.sc/m2w28g

[slender river]

no

[shrewd flame]

wut

[slender river]

it's should be rearranged so that it looks like the step i did

and then that [ cos^2(x) + sin^2(x) ] can be replaced with "1" from the trig identity

[slow wharf]

What's the thing with minus zero in infinite limits?

[tiny verge]

Minus zero?

I think it probably means approaching 0 from the left hand side, or the negative side of the number line

[earnest nymph]

^

[slow wharf]

Yeah I get that

But does it in any way affect the result of the limit

Let me think of an example

$\lim_{x\to\infty^-}\frac{5}{\frac{1}{x}}$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

$\frac{5}{\frac{1}{-\infty}} = \frac{5}{-0}= -5$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

No?

FUck

That doesn't make sense

[tiny verge]

What you wrote was approaching infinity from the negative side..

and what you substituded was negative infinity

[slow wharf]

So

[frozen needle]

wtf ∞¯

[tiny verge]

I've never actually seen it written approaching infinity -

and btw

[frozen needle]

that's quite the esoteric notation

[tiny verge]

5/-0 isn't -5 -.-

[slow wharf]

$\lim_{x\to\infty^{-}} \ne \lim_{x\to-\infty}$

[tiny verge]

yeah

[obsidian monolith]

Autistic Hoodie:

[tiny verge]

but it's not very usual to write $\infty^{-}$

[frozen needle]

{}

[slow wharf]

\infty

[earnest nymph]

So we have a discovery now? 5/-0 = -5

[tiny verge]

phew

[obsidian monolith]

󠂪󠂪󠂪 󠂪󠂪󠂪:

[slow wharf]

That means infiinite from the left right?

[tiny verge]

Yeah, though

[slow wharf]

ANd it's not the same as minus infinte

[tiny verge]

Yup.

[earnest nymph]

So is it means it's approaching towards positive inf?

[tiny verge]

Yup

I don't even think you can approach positive infinity from the right side

that's the reason it's usually just written approaching infty without the indicator of approaching direction

[rocky bison]

You can't

If infinity is some point we can't reach then it's nonsensical to write approaching it from a direction beyond that point @tiny verge

[tiny verge]

yeah

I should have written, 'You can't approach' instead of 'I don't think you can approach'

[slow wharf]

How would I go about solving

$log_{10}\frac{5x+1}{2x-1} \geq 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

I think I get it

$10^{log_{10}\frac{5x+1}{2x-1}} \geq 10^0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

$\frac{5x+1}{2x-1} \ge 1$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Is it not?

[frozen needle]

Well it's a legal move

[slow wharf]

This is interesting

wolfram says it's an different number

but the domain are the same which I am trying to get

https://www.wolframalpha.com/input/?i=log{\frac{5x%2B1}{2x-1}}+\geq+0&assumption={"FunClash",+"log"}+->+{"Log10"}

https://www.wolframalpha.com/input/?i=\frac{5x%2B1}{2x-1}+\ge+1

[frozen needle]

Mathbot could display what you want in a nicer way tho

[slow wharf]

hmm

[slender river]

infinity ^ - is dumb

how else r u supposed to get to positive infinity smh

[slow wharf]

Wait

$\frac{5x+1}{2x-1} \ge 1$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

I'm trying to get the domain of that

$5x+1 \ge 1 \ 5x \ge 1 - 1 \ 5x \ge 0 \ x \ge 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Right?

Apparently it is supposed to be

$-\frac{2}{3}$

[obsidian monolith]

Autistic Hoodie:

[frozen needle]

where did the denominator go ?

[royal aspen]

^

if you plug in x = 0

-1 >= 1

[slow wharf]

That isn't right?

[royal aspen]

$5x+1 \ge 2x - 1$

[obsidian monolith]

Trichloromethane:

[royal aspen]

this is only tru if

2x - 1 > 0

so consider both options

and now consider if 2x -1 < 0

[slow wharf]

What's the simplest way to get domain here :/

[royal aspen]

ok

$5x+1 \ge 2x - 1$ \ if x $>$ 1/2

damn

[slow wharf]

The procedure is that I put

Oh

[royal aspen]

it means 1/2

solve for x

[spring thunder]

a frac outside math mode don't really work

[royal aspen]

oh

[obsidian monolith]

Trichloromethane:

[slow wharf]

It's

$x \ge \frac{-2}{3}$

[obsidian monolith]

Autistic Hoodie:

[spring thunder]

he ain't finished mate

[royal aspen]

so we know that

$x \ge \frac{1}{2}$

[obsidian monolith]

Trichloromethane:

[royal aspen]

now consider the other case

where $2x-1 \ge 5x+1$ for x $<$ 1/2

[obsidian monolith]

Trichloromethane:

[slow wharf]

Um

$x \le \frac{-2}{3}$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

NVM

[royal aspen]

lol

so the domain of our inequality is

$(-\infty, \frac{-2}{3}] , (\frac{1}{2}, \infty)$

does the procedure make sense?

[slow wharf]

It's confusing to me

[royal aspen]

if we multiply both side of an inequality with a negative we have to change the direction of the greater than symbol

[slow wharf]

I do understand that

[royal aspen]

that is why in the first case i said that this inequality is true if x > 1/2

you solve for x and you see that this is true for

and you exclude any that do not fit the x > 1/2

[slow wharf]

$f(x) = \sqrt{\frac{2-x}{5+x}}$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

If it were to ask me the domain of this function

I know that a number under a root of 2, 4, 6 e.g. has to be greater or equal to 0

Which means

$\frac{2-x}{5+x} \ge 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

What procedure am I to use?

I am supposed to do

[royal aspen]

multiply both side by 5+x

[slow wharf]

I though I should calculate

$2-x \ge 0 and 5+x \ge 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Make the table with the plusses and minuses and get the domain from there?

[royal aspen]

that can work

[slow wharf]

Apparently not for all times since it didn't work for the previous one

You just did

$\frac{2-x}{5+x} \ge 0 \ 2-x \ge 0 * (5+x) \ 2-x \ge 0$

[obsidian monolith]

Autistic Hoodie:

[royal aspen]

i feel i got that answer wrong

=pup graph (5x+1)/(2x-1) >= 1

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@royal aspen

[royal aspen]

the intersection point looks like x = -2/3

[obsidian monolith]

Trichloromethane:

[slow wharf]

Yeah, that is correct

[royal aspen]

oh ok

so for this problem $\frac{2-x}{x+5} \ge 0$

[obsidian monolith]

Trichloromethane:

[slow wharf]

Yeah

[royal aspen]

if we multiply both side by x + 5

[slow wharf]

We just get $2-x \ge 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Right?

[royal aspen]

and solve for x

yeah

[slow wharf]

$x \le 2$

[obsidian monolith]

Autistic Hoodie:

[royal aspen]

yeah

that's it

=pup graph (2-x)/(x+5) >= 0

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@royal aspen

[royal aspen]

since the it is >= 0

we didn't need to worry about creating a negative number that would mess up the domain

[slow wharf]

Okay

So we also need to do

$x+5 \ge 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

To get the domain?

Or more precisely

$x+5 > 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Because it is not allowed to be 0

[royal aspen]

true

but we see that $x > -5$

[obsidian monolith]

Trichloromethane:

[serene heath]

what are you trying to do

[slow wharf]

So the domain is $Df = <-5, 2]$

[obsidian monolith]

Autistic Hoodie:

[royal aspen]

yes

[slow wharf]

Which is the correct answer

[royal aspen]

$(-5, 2]$

[obsidian monolith]

Trichloromethane:

[slow wharf]

When you said we see that x > -5 you mean from x+5>=0

[royal aspen]

yes

[slow wharf]

I finally know where I was failing in the procedure 😄

[slow wharf]

Thank you

[royal aspen]

good job for pointing out the x > -5

i missed that lol

[slow wharf]

Thank you 😃

[viscid thistle]

Hey, I have to calculate the derivative for this function:

$f ( x ) = \sqrt { 1 - 2 x } , x \leqslant 0.5 \text { for } x _ { 0 } = 0$

[obsidian monolith]

Ola Es:

[viscid thistle]

Can anyone tell me whether those calculations are correct?:

$f ^ { \prime } \left( x _ { 0 } \right) = \lim _ { h \rightarrow 0 } \frac { f \left( x _ { 0 } + h \right) - f \left( x _ { 0 } \right) } { h } = \lim _ { h \rightarrow 0 } \frac { \sqrt { 1 - 2 \left( x _ { 0 } + h \right) } - \sqrt { 1 - 2 x _ { 0 } } } { h }$
$f ^ { \prime } ( 0 ) = \lim _ { h \rightarrow 0 } \frac { \sqrt { 1 - 2 h } - \sqrt { 1 } } { h } = \lim _ { h \rightarrow 0 } \frac { ( \sqrt { 1 - 2 h } - 1 ) ( \sqrt { 1 - 2 h } + 1 ) } { h ( \sqrt { 1 - 2 h } + 1 ) } =$
$= \lim _ { h \rightarrow 0 } \frac { 1 - 2 h - 1 } { h ( \sqrt { 1 - 2 h } + 1 ) } = - 1$

[obsidian monolith]

Ola Es:

[viscid thistle]

Please ping me after you answer me.

[patent beacon]

@viscid thistle
Your steps are excellent. You have a few trivial simplifications you can make in the numerator

Oh wait, where did your x go?

[viscid thistle]

Thank you Kaynex. I calculate the derivative for the argument 0.

[patent beacon]

Oh I didn't see that, lol. Then yeah all looks good.

[viscid thistle]

Oh, I think it should go like this though:

[obsidian monolith]

Ola Es:

$f ^ { \prime } \left( x _ { 0 } \right) = \lim _ { h \rightarrow 0 } \frac { f \left( x _ { 0 } + h \right) - f \left( x _ { 0 } \right) } { h }$


$f ^ { \prime } \left( x _ { 0 } \right) = \lim _ { h \rightarrow 0 } \frac { \sqrt { 1 - 2 \left( x _ { 0 } + h \right) } - \sqrt { 1 - 2 x _ { 0 } } } { h } $


$f ^ { \prime } ( 0 ) = \lim _ { h \rightarrow 0 } \frac { \sqrt { 1 - 2 h } - \sqrt { 1 } } { h } $


$f ^ { \prime } ( 0 ) = \lim _ { h \rightarrow 0 } \frac { ( \sqrt { 1 - 2 h } - \sqrt{ 1 } ) ( \sqrt { 1 - 2 h } + \sqrt { 1 } ) } { h ( \sqrt { 1 - 2 h } + \sqrt { 1 } ) }$



$f ^ { \prime } ( 0 ) = \lim _ { h \rightarrow 0 } \frac { 1 - 2 h - 1 } { h ( \sqrt { 1 - 2 h } + \sqrt{ 1 } ) } = - 1$

[patent beacon]

1 - 2h - 1 = -2h

[viscid thistle]

I just go to nowhere with my calculations

Oh no, that's so sad...

I got it... 😃 thanks

[vernal lantern]

I want to test out of precalc this summer, but not sure on where to find a cirriculum. I took notes on all of the khan academy stuff, but I was looking something more official or somethin

doesnt anybody know where to find one

[viscid thistle]

why is a^logx = x^loga?

i cant work it out algebraically

[viscid thistle]

ah ok

i thought so, wasnt sure if i could flip the order

when i got to lnxlna

thank you so much

i guess lnx and ln a is just a number

[earnest nymph]

Multiplication is communative so u can

[viscid thistle]

yeah

thanks a lot =]

[native sequoia]

,rotate 90

[rare zephyr]

Can I get some help with no.11

[rare zephyr]

Nvm

[shrewd flame]

Is cot(x)^2 same as cot^2(x) ?

[spring thunder]

yus

[shrewd flame]

thank uu

[shrewd flame]

Halp

[spring thunder]

so waht have you tried ?

[slow wharf]

Is there a trick in factoring thing like these?

$20-13b-b^2$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

I know there's something with the multiple of two numbers being 20 and their sum being 13

or something like that

@shrewd flame You solved it?

[spring thunder]

idk i think he died

[slow wharf]

I got the answer as x = cos^-1(0)

That's

$x = \frac{\pi}{2} * 2n, n \in Z$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Right?

[spring thunder]

how'd get that?

(i mean the cos^-1(0))

[slow wharf]

Um, first of all

We can see that we can pull out the first ones

So it's

$cot^2(x) sin^2(x) = 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

Right?

Wait

[spring thunder]

those two are not equal tho

:/

[slow wharf]

I dun fukd up

I see where I failed

lemmie retry

[spring thunder]

there's no solution if we want to be very pedantic

otherwise it should be x=k*pi; k in Z (ie sin^-1(0))

[slow wharf]

What'd you get?

I'm getting

$cos^2(x)-cos(2x) = 0$

[obsidian monolith]

Autistic Hoodie:

[slow wharf]

That's only possible if they are equal

[spring thunder]

ie sin^2(x)=0

ffs cos^2(x) != cos(2x)

[slow wharf]

Apparently there is

[spring thunder]

cos(2x)= cos^2(x)-sin^2(x)

i even remember it and god knows i hate trig

so your equation is equivalent to sin^2(x) = 0

[slow wharf]

A solution exists

literally look at the graph

It's

[spring thunder]

ah yeah you're talking about that hehe

[slow wharf]

$x = \pi n, n \in Z$

[obsidian monolith]

Autistic Hoodie:

[spring thunder]

the cot^2(x)sin^(x) is the part where you can argue there's no sol

the original equation isn't even defined there

(cause of cot^2 sin^2)

[slow wharf]

Hm

Anyways um, I got distracted

$20-13b-b^2$

[obsidian monolith]

Autistic Hoodie:

[spring thunder]

so yeah for the quadratic shit

[slow wharf]

Any easy was to factorize it?

Is it even possible to do?

[spring thunder]

quadratic formula works lel

not possible every time tho in reals

[slow wharf]

If I remember correctly, I need to find two numbers who when multiplied give 20 and added give -13

That doesn't exist

[spring thunder]

ahh that formula

i see what you mean

but it only works when the coefficient in front of b^2 (in your case) is one

(this one is factorable)

[spice wadi]

Did you try completing the square?

Doesn't that result in factored answers?

I think you can also do this by doing some weird attempt at grouping

[spring thunder]

he wants to do it the vieta way ig

[spice wadi]

albeit it isn't gonna be whole in the last half

oh

[slow wharf]

-b +- \sqrt ... formula, you mean?

[spring thunder]

for quad formula yes

[spice wadi]

you can write that as factors end result

but not really factoring

[slow wharf]

20-13b-b^2 looks so stupid though and we are supposed to solve it without a calculator

I probably failed something earlier

[spring thunder]

🤷

[spice wadi]

You could do this through $20-12b-b^2 = 1b$

[obsidian monolith]

William:

[spice wadi]

and factor, divide and solve

That may work

[slow wharf]

Lolwat

[spring thunder]

that would give 20-14b-b^2=b tho

:/

[spice wadi]

yes I know

[spring thunder]

nono

it's getting kinda late lel pardon me

[spice wadi]

you end up with $\frac{b}{?}=20-12b-b^2$

[obsidian monolith]

William:

[spice wadi]

adding 1b??

-13b+b =/= -14b

[slow wharf]

Getting late? 😃

[spice wadi]

hmmm

[slow wharf]

It's 3:22 AM where I live and I'm learning math

fml

[spring thunder]

(it's also 3:22 am here)

[spice wadi]

Its only 6pm over here in American

[spring thunder]

[slow wharf]

hehe 😄

From?

[spring thunder]

france

[slow wharf]

croatia here

don't mention world champ pls

xD

[spring thunder]

🍻 ok

[spice wadi]

I watched that

[slow wharf]

hehe 😄

[viscid thistle]

Do I need to learn matrices before taking calculus 1?
Will they even be useful?

[fringe stream]

nope

[viscid thistle]

Huh

Would you mind telling me what else in the subjects of this course are not relvent?

https://www.khanacademy.org/math/precalculus

@fringe stream

[fringe stream]

a lot of those aren't necessary for learning calc

since they just branch out as their own extensive fields

you should have a good grasp of trig and algebraic manipulation and a good understanding of how functions work in general

[viscid thistle]

I feel like my grasp of those things is good, but I want to practice my weak points a bit more before diving into the calculus

maybe I'll just take swing at my calc textbook

Guess I'll figure out what I'm bad at along the way

[slender river]

i mean all u need for calculus 1 is just strong highschool level algebra foundation, quicc maffs maybe, and a really good grasp on like how functions behave (like u should be able to graph the parent functions and transform them in ur head if u want to have a really fun time for calc possibly)

stuff like understanding "change" will help i guess

[shrewd flame]

Yeah I got it my dudes

So like sin^2x same as sinx^2 right??

[limber bone]

sin^2(x) is the same as(sin(x))^2

dont square the argument

sin^2(x) = sin(x)*sin(x)

[shrewd flame]

So yes

[slow wharf]

Does L'Hospital's rule apply for limits that are not a fraction?

[limber bone]

you can

make them a fraction

lim x--> 0 x^2ln(x)

for example

thats not a fraction

but what you can do is write x^2ln(x)

as ln(x)/1/x^2

algebra

then do lhopital

got it?

brb

[slow wharf]

Hmm, interesting

[rocky girder]

Can someone please fully run me through how to solve 3^(x-1)=4^(2x+3)

[gritty blaze]

Hint: ln.

[slow wharf]

Yeah

[slender river]

change of base yuH

[slow wharf]

$log_3(4)^{2x+3} = (2x+3)log_3(4)$

[obsidian monolith]

Autistic Hoodie:

[royal gull]

and put x's on one side

[slow wharf]

Remember that

[fallen cloud]

Yo... you should make the 2x+3 inside the parenthesis

otherwise it looks like ln(4)^4 instead of ln(4^4) which are Very diffferent things

[plucky locust]

How come L'Hopital's rule can only be applied for indeterminate forms such as infinity/infinity

Why can't it be applied for all indeterminant forms?

[vagrant vapor]

Anyone on

[rocky bison]

No

Everyone on the server is offline right now

[viscid thistle]

I don’t understand this.... 1a) is <= 0, and 2a) is too.....
But the answer to them are different: 1a uses the interior region ( 1<= x <= 3 ) ans 2a is x = real number

Why wouldn’t it 2a) be x=2

(The answer to 2b)

It seems they’re inverted but don’t see why, and flipping signs doesn’t seem to work

[limber bone]

can u

rotate the picture? 😄

[fallen cloud]

,rotate

[fallen cloud]

,rotate

[limber bone]

fuck

my braain

[fallen cloud]

,rotate

[limber bone]

UHH

[fallen cloud]

,rotate

[fallen cloud]

yey

[limber bone]

from the graph

you want the solutions to x^2-4x+3 <=0

for example

u want the value of x that f(x) starts to become 0 and then less

note its an interval

cuz f(x) will not be <=0 forever

i can see that at x = 1

f(x) becomes 0

and after that untill x = 3

between x=1 and x= 3

its negative less than 0

right?

so the solution set is any x between 1 , 3

with closed intervals cuz they are with us they make f(x) = 0 aswell

so [1,3]

the same for number 2 just different function

for number 3 for a for example it askes if x =4 for x^2-3x-10 > 0

it askes if x = 4 makes x^2-3x-10 > 0

how do you know you just plug in x = 4

see if it makes it bigger than 0

if it does then its a solution

if it doesnt then it isnt

got them all?

[lost talon]

Hey could someone please check that I did this correctly? I don't want to do all of the problems to find out I'm doing them wrong.

[tawny nacelle]

yeah it looks about right

[lost talon]

Thank you! One more question, how do I find the inverse of this function? Here are my attempts, but it just doesn't look right and I ended back up with the original equation.

[thick raptor]

You have to switch x and y

Everywhere that had x, you replace with y, and vice versa

[white lion]

What’s a log?

[thick raptor]

The inverse of an exponential function

[white lion]

So when they say Log A - 5 Log B

What the hell does that even mean?

What am I supposed to do?

I’m taking algebra and geometry so I haven’t started precal yet

[royal gull]

log (in base 10 lets say) A = logA means solving 10^x = A @white lion

[white lion]

I see

Thanks

[main ledge]

yo guys im taking a course that goes over logarithm functions, but i never really understood what "ln" is. My book briefly explains it as "a logarithm with the base
e". Now I have to solve "In2 - 0,5In16" but I have no idea how to solve it since there weren't any proper examples in the book. I could simply solve it with my calculator, but i would like to understand what ln is so that i could solve it without one

hopefully someone can explain it

[obsidian monolith]

Yue:

[solemn tiger]

Similar to how log(x) on its own implies base 10, ln(x) is in base e

[main ledge]

so how about that "In2 - 0,5In16" then? is it possible to solve it by hand or is the only possibility to use a calculator

[solemn tiger]

You could possibly condense it down to one natural log, but the solution won't look nice.

[main ledge]

ok, thank you

[spring thunder]

-ln(2) isn't nice enough for @solemn tiger ?

[solemn tiger]

I did say it could be condensed down to one natural log 🤔

Evaluating that natural log, however

Euler sucks

[spring thunder]

[thick raptor]

nani

u can compute ln(2) pretty fast tbh

[solemn tiger]

e^x = 2

So x should be < 1

Pretty sure it's greater than sqrt(e)

So 0.5 < x < 1

[thick raptor]

[solemn tiger]

Past that is a pain 🤔

[thick raptor]

No

Bad

Bisection is trash

trash

[solemn tiger]

[thick raptor]

🚮

$$-\ln(2)=\ln(1/2)=-\sum_{n=1}^\infty\frac1{n2^n}$$

[obsidian monolith]

Simple_Art:

[thick raptor]

Error of nth partial sum can be bounded by geometric series

So pretty gud convergence

[main ledge]

but my teacher wanted me to solve "In2 - 0,5In16" without a calculator which is weird

[thick raptor]

Ya you use log rules

[main ledge]

[patent beacon]

Where do you get that sum from?

[solemn tiger]

^

[main ledge]

from my homework

i can send a pic

[patent beacon]

@main ledge
These rules apply for any logarithm:
log(ab) = log(a) + log(b)
log(a/b) = log(a) - log(b)
log(aˣ) = xlog(a)

[thick raptor]

@patent beacon @solemn tiger it's just Taylor expansion of log

[main ledge]

ahh thank you, so i can use them even if its ln

it doesnt matter

[patent beacon]

Oh it is

[solemn tiger]

tfw completely forgot Taylor/Maclaurin series expansion

[patent beacon]

That's fast. I didn't realize simply flipping the log made convergence that much easier

[thick raptor]

Alternatively you can be lame and just expand ln(2) directly for alternating harmonic series, which converges hella slow

But since it's alternating, you can accelerate it with the Euler transform

Which interestingly returns the same as expanding ln(1/2)

[distant flume]

How do I solve this using Hopital's rule guys?

[main ledge]

ahh im kinda bad at this, can I move the 0,5 so that it becomes an exponent of 16?

[patent beacon]

Faster would be to write 16 as 2⁴

[main ledge]

ahh and then 4 * 0,5

awesome

[distant flume]

@patent beacon Can you give me a hand, please?

I've been stuck for some time

[thick raptor]

[patent beacon]

If you plug in directly, you get ∞*0. There's a trick for solving limits of that form

[thick raptor]

#calculus tho @distant flume

[distant flume]

@thick raptor Very sorry, I didn't know

[patent beacon]

Indeed, please keep the convo on the correct channel and avoid interrupting others

So, know the trick?

[distant flume]

I don't know actually

[thick raptor]

Rationalize the last part

[distant flume]

Do I just use L'Hopital's rule and get the derivative of each individual one?

But I tried I think

[spring thunder]

,lhban

[thick raptor]

[distant flume]

I'll try to rationalize again

[patent beacon]

You want to "flip a term twice".

You can do
lim (1/t + 1/√t) ÷ 1/(√[t + 1] - 1)

Notice that's ∞/∞

[distant flume]

I understand

Would that give me 1 or do I keep going from there

[thick raptor]

tbh L'H looks worse than just doing algebra

[patent beacon]

Yeah the L'h is ugly here

But if you want to L'h, you can do it to that

Try t = tan²(u) for some simplifications I think

[distant flume]

I'm not going to use tan's identity here though

Thanks anyway, I'll try both methods

[thick raptor]

wtf y would u use tan sub

Just rationalize

[main ledge]

sorry if im interrupting, but i need to ask one last time if this is correct