help-42

[hot prairie]

cosx

[jolly jacinth]

ok

so we have sin^3(x)

which means we can effectively use the power rule

do you know what that is?

[hot prairie]

yes

[jolly jacinth]

so

[hot prairie]

3sin^2(x)?

[jolly jacinth]

thats incorrect

you are integrating

which means the power increases by 1

[hot prairie]

oh mb mb

[jolly jacinth]

and you divide by the new power

all g mate

[hot prairie]

so sin^4(x)/4

[jolly jacinth]

yep

• c

does that make sense?

[hot prairie]

yes

ohh

I see now

[jolly jacinth]

if an integral doesnt make sense

sometimes you can differentiate your answer

[hot prairie]

yeah

[jolly jacinth]

and check that it comes back out

[hot prairie]

ok ok thanks

[jolly jacinth]

do ya have any other questions mate?

[hot prairie]

no, all goods

thanks

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[toxic haven]

when I factor out 1/13 from 27/13 I get 27/ 169, but not 27

i’m doing smth wrong idk what

[dull wagon]

you didn't factor out correctly

consider $\frac{27}{13} = 27 \times \frac{1}{13}$

[potent lotus]

ℝαμΩℕωⅤ

[toxic haven]

yeah

1/13 divided by 27/13

[dull wagon]

thus factoring 1/13 out of that leaves you with 27

[toxic haven]

but i did it by cross multiplication

i

[dull wagon]

you're dividing the wrong way

[toxic haven]

oh

[dull wagon]

to see what you're left with, you want to divide
what you're starting with: 27/13
by what you're factoring out: 1/13

(27/13) / (1/13) = 27/13 * 13 = 27

[toxic haven]

i u derstand but

how am i supposed to compute this?

thro cross multiplication?

[dull wagon]

as above

also cross multiplication isn't the correct terminology

[toxic haven]

oh

what is it

[dull wagon]

relation between division and multiplication

[toxic haven]

Is it possible to do it through cross multiplication because I just don’t really understand what you sent because I don’t understand how I would be able to compute that on paper

okie ty

[dull wagon]

cross multiplication refers to manipulation of equations doing:
$$\frac pq = \frac rs \implies ps = qr$$

[potent lotus]

ℝαμΩℕωⅤ

[toxic haven]

oh ok

my bad

okay

i got it now

tysm

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[lament sorrel]

how do i solve this equation for Z5 (all the whole numbers group between 0 and 4 including both)?

[lament sorrel]

do i do:
x^2 + 1 = 0 /+4
x^2 = 4
and then just plug in all the numbers 0<=x<=4 and see what works?

x*x mod 5 = 4, per the construction of Z5 as i know it (all possible reaminders of 5 group numbers)

[pseudo wedge]

yes that's correct

[storm spire]

sorry didnt mean to type haha

[lament sorrel]

but isnt that impractical? what if it was Z17, or anumber number group that's way bigger

[pseudo wedge]

I'm not sure if there's a better way tbvh

there probably is

[lament sorrel]

i think this is not the correct way to solve this problem

[pseudo wedge]

since 5 is a prime number I think you can factorise this

$(x+2)(x-2)\equiv 0 (\text{mod 5})$

[potent lotus]

kheerii

[pseudo wedge]

so $x+2\equiv 0$ or $x-2\equiv 0$

[potent lotus]

kheerii

[pseudo wedge]

note however this ONLY WORKS if your base (in this case, 5) is prime

[lament sorrel]

why?

i dont if it's not prime then it's not a field

but how does this matter? i mean what axiom

[pseudo wedge]

I'm not sure about the field theory aspect of it, so I can't help you there

I can tell you why it fails from a number theoretic standpoint

say you have $ab\equiv 0(\text{mod 10})$

[potent lotus]

kheerii

[pseudo wedge]

you can have $a\equiv 0$ or $b\equiv 0$

[potent lotus]

kheerii

[pseudo wedge]

but you can also have $a\equiv 2$ and $b\equiv 5$, this also gives you a valid solution

[potent lotus]

kheerii

[pseudo wedge]

that's why factorising doesn't work for non-prime bases

[lament sorrel]

what are a and b here?

[pseudo wedge]

the two factors

[pseudo wedge]

we started from this equation

[lament sorrel]

oh i see

but i dont understand how that's relevant

[pseudo wedge]

we can only conclude this because 5 is a prime number, that's what I was trying to explain to you

[lament sorrel]

since it's not a*b, but either a*a or b*b

[pseudo wedge]

$$(x+2)(x-2)\equiv 0 (\text{mod 5})\implies x+2\equiv 0 \text{ or } x-2\equiv 0(\text{mod 5})$$ but $$(x+2)(x-2)\equiv 0 (\text{mod 10})\text{ does not imply } x+2\equiv 0 \text{ or } x-2\equiv 0(\text{mod 10})$$

[potent lotus]

kheerii

[pseudo wedge]

apologies for the bad latex, I'm still learning

[lament sorrel]

how not? i mean it does mod 10 on the LHS too

[lament sorrel]

you're great at this

[pseudo wedge]

because, these are not the only possible cases

you can also have first factor = 2 and second factor = 5 for example

[lament sorrel]

i dont see how?

[pseudo wedge]

okay maybe this specific case is a bad example

it probably isn't possible for this exact question

but I'm just telling you that you need to RULE OUT those cases separately

[pseudo wedge]

you cannot conclude this directly from that statement

[lament sorrel]

i dont follow

i mean i can

since it's a square euqals something in Z(n)

[ancient thistle]

heres an example

,, x^2 - 5x + 4 = (x - 1)(x - 4) \pmod {10}

[ancient thistle]

it seems that you have the solutions x = 1 and x = 4

but you also have
[ x^2 - 5x + 4 = x^2 - 5x - 6 = (x + 1)(x - 6) \pmod {10} ]

[ancient thistle]

so there is also x = -1 and x = 6

[pseudo wedge]

ahh yeah

I wasn't able to think of a good example

[lament sorrel]

why is it like that?

[ancient thistle]

because 10 has nontrivial factors

[lament sorrel]

I get it's because the group of reminders Z_p is not a field because p is not prime

but what axiom is not working?

[ancient thistle]

Z/10 is not a domain

Z/5 is a domain

a domain is where there are no zero divisors

in Z/10, there are zero divisors

2 divides 0, and 5 divides 0

[lament sorrel]

whats a factor?

[ancient thistle]

a factor is a factor...

2 and 5 are factors of 10

[lament sorrel]

not native

[lament sorrel]

alright so I understood

[lament sorrel]

wdym?

[ancient thistle]

2*5 = 0 in Z/10

in Z/5, there are no such elements

[lament sorrel]

10 | 2*x?

[lament sorrel]

oh

I understand, so the axiom of only 1 resetting multiplication number doesn't exist here

[ancient thistle]

i have no idea what that means

Z/10 satisfies all the axioms of a ring

it is in fact a perfectly fine commutative ring

it is not an integral domain however

[lament sorrel]

but how would you solve then for:
x^2+1=0 (for Z_10)

[ancient thistle]

you write it as x^2 - 9 = 0 and factor it into (x - 3)(x + 3)

so 10 divides (x - 3)(x + 3)

[lament sorrel]

0a=0 (only one such number can be "0" type)
so if ab=0 then it's not a domain

[ancient thistle]

then you have one of 10 | x - 3, 10 | x + 3, 2 | x - 3 and 5 | x + 3, or 2 | x + 3 and 5 | x - 3

[lament sorrel]

so the only solution is 3?

[ancient thistle]

thats the axiom of an integral domain

theres x = 3, x = -3 and no others

[lament sorrel]

integral domain is?

[ancient thistle]

a commutative domain

[lament sorrel]

-3 is not a member of Z_10

[ancient thistle]

yes it is

[lament sorrel]

no

[ancient thistle]

the equivalence class of -3 in Z/10 contains all the elements congurent to -3 mod 10

this includes elements like -3, -420693, 1337

[lament sorrel]

I see

my bad

[lament sorrel]

so these are the only solutions?

[ancient thistle]

you can check directly that the other two options dont work

[lament sorrel]

because we could then do 13,23,33, since those conguate to mod 10

to 3*

[ancient thistle]

those are all included in the x = 3 solution

because you're working mod 10

[lament sorrel]

but if you allow -3 to be a solution, so is -13,-23,-33

since you're working with mod 10

[ancient thistle]

those are all congruent to -3, and so is included in the x = -3 solution

but 3 is not congruent to -3, so x = 3 and x = -3 are distinct

[lament sorrel]

yes i didn't realize that sorry

[lament sorrel]

so then how would you solve this?

[ancient thistle]

you factor

and you say either

1. 10 | x - 1,
2. 10 | x - 4,
3. 5 | x - 1 and 2 | x - 4, or
4. 5 | x - 4 and 2 | x - 1

this gives you the 4 solutions that i listed earlier

[lament sorrel]

so it'll be 1,5,-1,6?

[ancient thistle]

the 5 should be a 4

i guess another way is by using the chinese remainder theorem

(x - 1)(x - 4) = 0 mod 10 means (x - 1)(x - 4) = 0 mod 2 and 5

you solve the equivalence in both moduli and combine it together

[lament sorrel]

I think I understand now

ty very much mate

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[ancient thistle]

👍

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[remote mural]

Is there way to find roots of cubic equation easily
Like there is [-b±√(b^2-4ac)]/2a for quadratic equation

[warm warren]

Yes

Not so easily but there’s a formula

[calm coral]

@remote mural Has your question been resolved?

[hollow totem]

,w x^3+x+1=0 exact solution

[potent lotus]

Results provided by WolframAlpha

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

Requested by a_mog

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[barren nymph]

is there an easy way I could approach this question

[barren nymph]

because I have no clue how I would find 2 factors so that the two vectors are equal to v1

[marsh agate]

Just write the linear dependance
You have ... = 0
Move the vector you want to the other side and divide by the appropriate constant (assuming its coefficient was nonzero, you can't always do that but here you can)

[barren nymph]

🤨

are you refering to c1v1 +c2v2+etc?

I do not understand sorry

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[remote mural]

Idk where to start

[low isle]

f(-1) = 0
f(0) = 1
f(1) = 2

[remote mural]

Yep

[low isle]

now some function g(x) = x + some constant b

which is always greater than f(x)

[remote mural]

Yep

[low isle]

g(1) = 1 + b
g(2) = 2 + b and so on

[remote mural]

Yes

[low isle]

is it a multicorrect question?

wait no

[remote mural]

Sorry wdym?

There’s only one correct answer

[low isle]

the only value of b satisfying this is 2

[remote mural]

I don’t understand sorry

[low isle]

see g(1) = 1 + b right

and f(1) = 2

and it is in this pattern

[remote mural]

wait no isn’t g(1)=2+b?

[low isle]

g(0) = 0 + b
f(0) = 1

[low isle]

no, see g(x) = x + b

at x = 1

g(1) becomes 1 + b

did u understand that?

[remote mural]

Hmmm

Wait so

[low isle]

now look at this one too

the options for b are -1, 0 , 1 and 2

[remote mural]

Yes

[low isle]

at b = -1 or 0 g(x) is coming out to be less than f(x)

so they cant be right

[remote mural]

Right

[low isle]

at b = 1, g(x) is equal to f(x) but we are looking for g(x) greater than f(x)

the only remaining option is 2 whic is correct

[remote mural]

So it’s just only a matter of trial and error?

[low isle]

i suppose

or u can see that for b greater or equal to 2 satisfies it

i did so many steps just to make u udnerstand

[remote mural]

Right thanks

[low isle]

it just takes like 5 seconds of thinking

[remote mural]

Uhh yeah sorry for the inconvenience then

Bye

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[storm spire]

How do you know when to reslove forces and when to use f=ma (eg finding tension or reaction force)

[compact drum]

Could you give some context?

[blazing coyote]

show us the context

[storm spire]

ok one sec

this is my understanding - you can resolve forces if velocity is constanct (essentially f=m(0) so technically you use the f=ma for both), and use fma if there is acceleration

these questions have acceleration - but most questions state that the body is in equibilrium

[compact drum]

In equilibrium we still use fma
We find the forces that go in opposite direction and set them equal to find what we want

[storm spire]

they equal = as m(0) as constant velocity so no acceleration

[storm spire]

is this correct understanding

[compact drum]

You do have acceleration usually

Just that you have the same acceleration up and down

But the fundamental thing you set equal is usually f

[storm spire]

yh i mean in some question it will say its in equilbrium (so at rest or constant velocity)

so is it cirrect that techniwualy toy still use fma

like for q lets say there was no acceleration you can still resolve ?

sorry if confusing you hah

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@storm spire Has your question been resolved?

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[idle fractal]

i converted it into euler

e^(itheta)^4 upon e^(i(pi/2-theta))^5

[winter elbow]

Actually, you just need to acknowledge that $sin \theta$ = $cos( \pi/2$ - $\theta)$

[potent lotus]

Sufferrrrring rn D:

[idle fractal]

yes that is why i converted

[idle fractal]

.

[winter elbow]

Sorry, I’m not advance enough for Euler form XD

but it’s alright, at least we are referring to the same concept

De Moivre

Formula

[quaint sphinx]

use exponential property for dividing exponentials

and you will be done

[brittle isle]

numerator becomes e^(4i * theta) as done, denominator becomes (cos(pi/2 - theta) + isin(pi/2 - theta))^5 = e^(i * (pi/2 - theta) * 5)

so yeah

work all looks good, you can just do the division and convert back using euler's formula

so you have e^(4i * theta) / e^(5i(pi/2 - theta))

the denominator can be simplified actually

becomes e^(i * 5pi/2) * e^(i * -5theta)

and e^(i * 5pi/2) = i

so simplifies to e^(4i * theta) / (i * e^(i * -5theta))

= -i * e^(4i * theta + 5i * theta)

now you can just get your answer

and before a helper comes at me for providing sols, she dm'd me and was still confused so i explained it here in the help channel lol

[winter elbow]

You’re so nice

[idle fractal]

so the power comes in multiplication

Opps wait one term misread

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[jovial gyro]

If $\frac{z^2}{z-1}$ is always real, then z, can lie on a-
\(a) Real axis
\(b) A parabola
\(c) Imaginary axis
\(D) None of these

[potent lotus]

Pro_Hecker

[jovial gyro]

I tried using z = x+yi

and got 2x^2y-2xy-x^2y+y^3=0 by imaginary part =0

i don't know what to do next

[median oyster]

it doesn't look like that rearranges into any of the first three options, so just (d)

[jovial gyro]

How do you know?

[median oyster]

because of the cubic term

oh wait, that's just the imaginary part

[jovial gyro]

this simplifies to y(x^2 - 2x + y^2) = 0

so... x^2 -2x +y^2 =0

so it this a circle?

[median oyster]

the imaginary part is zero when y=0 or when it lies on that circle

i.e. if z is a real number, or if it's on that circle

[jovial gyro]

if y=0 then, can the answer be (a) real axis?

[median oyster]

yes

[jovial gyro]

ok thanks

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[trail hawk]

Pls help

[modern swallow]

Cross-multiply to eliminate the fractions

then distribute and simplify

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[formal shadow]

how to integrate this using partial fractions

[formal shadow]

$\frac{3x+2}{x^2+x+2}$

[potent lotus]

Pfannkuchen

[formal shadow]

I know i should split the 3x and the 2 but then idk what to do next cuz i forgot

[low isle]

it is a quadratic that cant be factorized

[formal shadow]

thats where im stuck

[low isle]

so what you do is 3x+2/(x^2 + x +2) = ax+b

[formal shadow]

i dont get it

isnt (3x+2)/(x^2+x+2) going to be a fraction

with x in the denominator

[low isle]

1 sec

[formal shadow]

kk

[low isle]

you do 3x + 2 = ax + b / (x^2 + x +2)

sorry did that by mistake

[formal shadow]

its fine

[low isle]

wasnt paying much attention, my bad :)

[formal shadow]

its ok

do i need to complete the square on the bottom?

and then use u sub

[low isle]

oopsie i did another mistake

wait no

[formal shadow]

?

[low isle]

i think tht shud work

[formal shadow]

ok

ill try it

it kinda got rly complicated

[low isle]

i think it's supposed to be 3x+2 = A(x^2 + x + 2) + bx + c

[formal shadow]

?

wait

i got it

thanks fo rthe help

i realized

that the question was asking for if the improper integral diverges or converges and it diverges

so ty

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[fossil forum]

How do i solve this

[fossil forum]

i tried to get (a) by using the formula 20-4x/4

but i did not get it

wait is 20-4x/4 == 20/4-4x/4 ??

[versed silo]

yes

[fossil forum]

how is that so?

[versed silo]

because fractions are divisions

i don't understand what you're asking

[fossil forum]

oh wait i forgot basic algebra

.closr

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[unreal sonnet]

[inverse trigonometric functions]

how do you work out the exact values of cos(arcsin(4/5)).

[verbal finch]

ok so

first draw a right angle triangle

[unreal sonnet]

thatw as fast

[verbal finch]

lol

[unreal sonnet]

btw 4/5 i emant

typo

not 3/5

[verbal finch]

yea okay

assume some x = arcsin(4/5)

then sinx is?

[unreal sonnet]

mhm

[verbal finch]

what is it?

[unreal sonnet]

so x = arcsin(4/5)?

[verbal finch]

yes

[unreal sonnet]

sin(arcsin(4/5))?

[verbal finch]

yes

which is?

[unreal sonnet]

4/5

[verbal finch]

yes

from here

find cosx

can u?

[unreal sonnet]

that'll just be cos(4/5)

but it'

[verbal finch]

nop

[unreal sonnet]

o

[verbal finch]

think again

sinx is 4/5

how do u get cosx

@unreal sonnet ?

[unreal sonnet]

im still thinking

[verbal finch]

sin^2x + cos^2x = 1

[unreal sonnet]

i dont think i can work it out

[remote mural]

make a right triangle with hypot 5 and a side 4 then ur gonna get the third side by using the pythagoras theorem

[verbal finch]

ever heard of this

[verbal finch]

works

[unreal sonnet]

yea but I'd only remember if mentioned

[verbal finch]

thats fine

ok so find cosx

[unreal sonnet]

is it suppose to be easy or smth and am i just overthinking

[fleet verge]

Write the equation here

[verbal finch]

what did u get?

it is pretty simple

[unreal sonnet]

cos(3/5)?

[verbal finch]

uh

only 3/5

cosx is 3/5

[unreal sonnet]

oh yea

BROO

[verbal finch]

now what is x from here?

[verbal finch]

ye?

[unreal sonnet]

u jus find out the adj

and ez

💀

[verbal finch]

yea

that also works

[unreal sonnet]

so like

sin(arctan(3/7))

is = 3/7.6

[verbal finch]

??

wait

[unreal sonnet]

huh

[verbal finch]

u asked cos(arcsin(4/5))

[unreal sonnet]

nono this is a dif question

[verbal finch]

u got that?

[unreal sonnet]

ye?

[verbal finch]

okay

[unreal sonnet]

is it wrong

[verbal finch]

mhm

[verbal finch]

nop 3/5 is right

[verbal finch]

assume x = arctan(3/7)
find sinx

[unreal sonnet]

7/3

oh awit

3/7

[verbal finch]

sry mb

[unreal sonnet]

o

[verbal finch]

sinx not tanx

[unreal sonnet]

7/7.6

[verbal finch]

?

what

[unreal sonnet]

no?

[verbal finch]

wait a sec

yes correct

[unreal sonnet]

phew alr

can we do one more

[fleet verge]

As an approximation

[unreal sonnet]

actually

ill do the question and u tell me if it's correct or not

arccos[sin(pi/4)]
= 4/2.48?

[fleet verge]

No

[unreal sonnet]

oh fuc btw we got this incorrect cause opp is 3

[fleet verge]

Notice that sin(pi/4)=cos(pi/4)

[unreal sonnet]

how

[fleet verge]

What is sin(pi/4)?

[unreal sonnet]

that's 45 degrees

[fleet verge]

What is cos(pi/4)

[unreal sonnet]

wait

[fleet verge]

Sin(pi/4) = 1/sqrt(2) = cos(pi/4)

[unreal sonnet]

i still

dont get it

i think you got it wrong

cause the if u draw a triangle

[fleet verge]

Think about it a bit more

[unreal sonnet]

cos = sqrt(16+pi^2)/4

huuh

alr

and not pi/4

how does cos = pi/4

so both adj and opp are equal and make 4

[verbal finch]

ooo yea mb

[unreal sonnet]

allgs

ok actually ima headoff

i gotta sleep early for tmr

yall have been a great help

wish yall a great day

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[supple vector]

It isn’t because it can derive the same string with multiple parse trees

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[narrow ridge]

quick question

[narrow ridge]

thats correct since it's already in standard form

right?

.close

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[coral dew]

Why is this false?

[pallid halo]

suppose x = -1

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[lucid knot]

are these correct?

[lucid knot]

A school committee needs 2 more students. How many ways can the committee be selected from a group of 10 students? i say 45 because 10C2

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[toxic haven]

For the question on top I’m kind of confused because I don’t understand how you get absolute minimum or maximum

[toxic haven]

i have the critical nimber

[potent smelt]

The critical numbers are local maxima, minima, or saddle points. The absolute minimum or absolute maximum is simply the smallest or largest value of the function on some domain.

It may be the case that a function does not have an absolute minimum or maximum

[toxic haven]

yeah but for this i used the cpt theorm which only has one critical number to test

[potent smelt]

Yes, so you test the critical number. Figure out if it's a maximum or minimum or neither, and then look at the limiting behavior of the function

[toxic haven]

how do I use a critical number to find the absolute maximum or minimum because I’ve searched up two different ways, and one is through the second derivative test or finding the limits but i want to know the most efficient way

[potent smelt]

If, for instance, the critical value is a minimum, but the limiting behavior as x->0 is to diverge to negative infinity, then your minimum cannot be an absolute minimum, for instance

@toxic haven the second derivative test is only to determine if a point is a minimum or maximum

[toxic haven]

this seems more straightforward

[potent smelt]

So the second derivative test can be inconclusive

[toxic haven]

how

[potent smelt]

Consider y = x^4

And y = x^3

We know that y = x^4 has a minimum at x = 0, but y = x^3 does not

What does the second derivative test say for these functions?

[toxic haven]

to use the first decorative test

derisive

but ur right it’s better to use linits

in this case how would i?

i don’t understand what function we’re talking about whenever we say as d approaches to infinity or zero

x*

[potent smelt]

So the stuff with limits is to determine if there is an absolute minimum or maximum at the endpoints of the function, (if the domain is finite) or if the function attains a larger (or smaller) value that any of our maximums (or minimums) found by the critical point tests

You need to do both

Additionally, you need to look at discontinuities in the function, if any

And the limiting behavior approaching those

[toxic haven]

ohh

okay

so how do i start

for the questions i sent

[potent smelt]

Here's how I approach problems like this

Generally

[kind prawn]

Using limits on an open set to determine extrema and ignoring the difference between maxima/minima and suprema/infima in their screenshot makes me want to roll my eyes

[toxic haven]

the screenshot i sent did it wrong kinda?

[kind prawn]

It doesn't really matter at this stage, it just makes me judge the lecturer

[potent smelt]

First, I find and classify all of the critical points, and find their values. Keep the largest local maximum and smallest local minimum. Note that any discontinuities in the function will be captured in the critical points, but sometimes the definition of the function can make you miss removable discontinuities, so

Next, I make extra careful I got all of the discontinuities, and took care of weird cases not covered by things like the second derivative test, for instance, locally infinite slopes for finite functions, cusps, and so on, I categorize these as well.

Third, I look at the end points of the domain, are these open or closed? What values are these approaching?

Finally, I've gathered all of the information I need to figure out which point is the absolute maximum or minimum. I just find the point which is the largest or smallest. Sometimes though, this is not possible, and the function doesn't have an absolute maximum or minimum.

[strange lichen]

This gotta get pinned in #calculus or something man

[toxic haven]

i’m still reading it 😭

[potent smelt]

Feel free to freeboot it.

[toxic haven]

okay i got it

[toxic haven]

so how would u approach that with the question i have cause we are usuing critical point theorem

[potent smelt]

Well, I told you how I would approach the problem, so how about this, using the above approach, give it a shot, and if you get stuck, I can try to help you

[toxic haven]

okie

usuing the critical number e^1/14

so we find local values?

[potent smelt]

You find the value of the function at that point, and also the value of the second derivative

[toxic haven]

okay the value of dunc if 1/14e

and the e^14 plugged in second dericirve is -4.46

[toxic haven]

what would i do now

[potent smelt]

Ok, so the second derivative is negative

[toxic haven]

yeah

[potent smelt]

This implies that the critical point is what, a local max or a min?

[toxic haven]

min

waits it’s

relative max

[potent smelt]

It's not a min, so it clearly must be a max, so maybe it will be helpful to reason through why it's a max

Why do you think this is a max?

[toxic haven]

vause its concave down

[potent smelt]

Right, but more specifically, the second derivative measures concavity of the original function

But what does it measure with respect to the first derivative?

[toxic haven]

i have no idea

just where it changes?

[potent smelt]

So the second derivative is the derivative of the first derivative

Seems silly to say it out loud

But this means everything that you know to be true about the derivative is true about the second derivative as it applies to the first derivative.

[toxic haven]

yeah

[potent smelt]

So at the critical point we know that the derivative of the function is 0

And so we know that if we consider the derivative as a function in its own right, it either goes from negative to positive or positive to negative, or it bounces off in some way.

If the second derivative is negative, this means the derivative is trending down at the 0.

[toxic haven]

yeah

[potent smelt]

Going from positive to negative.

[toxic haven]

yeah

[potent smelt]

So if we know the derivative is going from positive to negative, what can we say about the original function?

[toxic haven]

so as it goes to infinity it’s getting shorter

it’s decreasing?

[potent smelt]

Well, if a function's slope is positive then it's increasing

And if a function's slope is negative then it's decreasing

[toxic haven]

yeah

[potent smelt]

So what happens if you have a function whose slope changes from positive to negative?

[toxic haven]

decreasing

[potent smelt]

So imagine the function is like a ball traveling through the air

It goes up then it comes down

Did it reach a maximum or minimum height?

[toxic haven]

it hit a max height

[potent smelt]

Exactly

[toxic haven]

so it would be absolute max?

[potent smelt]

It is a local max right now

[toxic haven]

yeah

what now lol

[potent smelt]

So that's paragraph 1

We've handled all of the critical points, are there any discontinuities or anything like that?

[toxic haven]

no

[potent smelt]

Ok, now, what is the behavior at the edges of the domain?

First, what is the domain?

[toxic haven]

(0. infinity)

,

[potent smelt]

Cool, and what's the limiting behavior?

[toxic haven]

like

meaning what’s the limit?

idk what that means

[potent smelt]

Yes

What's the limit as x->0+ and x->inf?

[toxic haven]

when x approaches infinity it iincreases?

[potent smelt]

Can you show your work?

[toxic haven]

for this is where i’ve been getting stuck

because i did do a whole chapter on limits

i just don’t know how to do it for this mean i don’t really know what it means

like when u say x do u mean like plugging in smaller numbers to see if it increases or decreases as it reaches 0?

[potent smelt]

Ok, no worries, so in this case for the limit to infinity we can do a power comparison test

[toxic haven]

okie

[potent smelt]

ln(x), because it is logarithmic, has a smaller effective exponent than any positive n, x^n

On the other hand, the x^14 is just 14

So which term grows faster?

[toxic haven]

this one’s

?

[potent smelt]

Yup

So if the denominator dominates the numerator, what does this mean?

[toxic haven]

no idea 😭

[potent smelt]

Ok, so if we have some fraction, a/b

And I make b bigger, what happens to a/b?

[toxic haven]

it gets smaller

[potent smelt]

Yup

Let's say I make b massive, it's a million times bigger than a, how big is a/b?

[toxic haven]

very small?

[potent smelt]

Sure, but if it's exactly 1,000,000 times bigger than we know the value a/b

a/(1,000,000 a) = 1/1,000,000

So it becomes one millionth

[toxic haven]

yeah

[potent smelt]

So we can make b as big as we want, is there any limit to how small we can make a/b?

[toxic haven]

it can’t be 0?

[potent smelt]

It can't! But we can get as close as we want to it

So that's what we mean by limit

[toxic haven]

okok

[potent smelt]

If you give me a super small number, which we call ε, I can find some value of b which makes a/b smaller than ε

[toxic haven]

yeah

[potent smelt]

But the rule is ε must be > 0

So as a result the limit as x->inf is 0

Because x^14 grows much faster than ln(x)

So we can always find some x big enough to make the value smaller than whatever ε > 0 we choose

[toxic haven]

what would it be for 0?

[potent smelt]

Well, we can do a different dominated convergence test, this time for x->0 we have that ln(x) -> -Infinity faster than any x^n for finite n.

Or you can use L'hopitals rule

Actually, you could have used L'hopitals for the last one too

[toxic haven]

what is this math called

like limits?

[potent smelt]

Calculus

[toxic haven]

like more specifically

[potent smelt]

Or perhaps "Analysis" if you're seeking a math degree

[toxic haven]

how would i practice this

[potent smelt]

Well, what textbook are you using?

Stewart?

If so, one of the major strengths of Stewart is the sheer volume of excellent and interesting problems.

Just assign yourself homework from the chapter on limits

[toxic haven]

okie

okay again

why do we test the limit of inf of ln x and not x^14

but test the limit of x^14 for 0

[potent smelt]

We do both

It's just that x^14 dominated as x -> inf, but ln(x) dominates as x->0

Well, technically we have infinity/0 which isn't even really a question of where that's going

The numerator says infinity, the denominator says also infinity.

[toxic haven]

ik i was still abit confused on that

when we say x approaching 0 or inf we mean the x in the orginal function right?

[potent smelt]

Or just the x in the numerator or denominator, depending on context.

[toxic haven]

what about this context

[potent smelt]

Well when we're considering x^14 and ln(x) separately, then the context is whichever one we're talking about.

But when we put it all together, it's all together again and the context is the entire function

[toxic haven]

can we consider it all together?

[potent smelt]

Sure

[toxic haven]

and we are trying to test if it’s decreasing or increasing v

?

[potent smelt]

Just seeing where it is going

[toxic haven]

is this connected with it

like figuring it out if it’s abs max or min

[potent smelt]

Ah yes, sorry

You mentioned this a while back, but I didn't recognize the name

Yes, you can just use that

And make your life easier.

[toxic haven]

😭😭😭

i’m just confused because

they say if u find relative max it’s also the absolute max?

[potent smelt]

I'm trying to do a super general step by step process for the possibility of a really misbehaving function.

Whereas you have a rule for a really nice function

Which I was unaware you could use

And that makes this problem super easy

[potent smelt]

Only if you have a continuous function with a single critical point on the interval

[toxic haven]

because we did the second derivative test and we got that the function is a relative maximum so that also means that the thing is also absolute maximum?

[potent smelt]

And that critical point is a local max or min

[toxic haven]

yeah we have that

[potent smelt]

Yes exactly

[potent smelt]

We do, but I wanted to emphasize this because you don't always have this

You do need to check the conditions in order to use this fact

[toxic haven]

okay so we got that figured it out

[potent smelt]

Yup

[toxic haven]

yeah

[potent smelt]

🥳

[toxic haven]

okay finally

tysm✊🏼✊🏼

.close

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[keen grove]

why is this not a subspace?

[alpine stone]

Well, do you think it should be?

If so, try verifying the conditions for a subspace

[keen grove]

so would i use the form p = ax^3+bx^2+cx+d

[blazing coyote]

what's $\P$? The field of polynomials ?

[alpine stone]

You could

[potent lotus]

ƒ(Why am. I here)=I don't know

[keen grove]

ye degree less than or equal to 3

[frank heart]

could someone explain what happened here?

[blazing coyote]

!occupied

[calm coral]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[keen grove]

.close

[calm coral]

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[opaque ledge]

AB = CD
AD = BC

It is indeed slanted

Look at part B and it's distance to bottom, it's 5

From C to bottom is 1.5

It is slanted

Ah

I thought AD is also slanted

But yeah you're right ig it's not slanted

I would say it is not slanted, but if you have the opportunity to ask your teacher, then sure why not

Okay yea ask your teacher

Because if AD is slanted, then question is much easier solveable than when it's not

You can send it here and if I can't, someone else might

What do we know about the circular part

Alright

How do you think we can approach this problem?

It's fine

What is your initial idea

Imagine you know the curved part

Let's call it X (Imagine we know it)

How do you find the answer?

Hmm

Let's look at it in another way

If it was normal triangle, without the curved part, it would've been easier to solve this, do you agree?

So let's do something

Let's solve the question not considering the curved part

And then deduct the curved part from it

yes yes

I tried to replicate the question

Let's find the purple part, then deduct the blue part

Excuse my drawing skills

Yea but we'll get to that

So big triangle is 20

Now as for the blue part

It seems like the curve, is part of a circle

In other words like it is a quarter of a circle

Yes

[calm coral]

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[raven crypt]

Could someone help me here? I am stuck in this derivative

[hollow totem]

The mistake is from lines 3 to 4, you can't combine exponents like that

$a^b * c^d \neq (ab)^{c+d}$

[potent lotus]

thewizardofOU

[raven crypt]

how could I resolve it?

[upper flare]

okay so

key line you messed up

was after

$f'(x) = 4x\sqrt{2-x} + 2x^2 \left( \frac{1}{2} \cdot (2-x)^{-\frac{1}{2}} \cdot (-1) \right)$

[potent lotus]

poudel

[upper flare]

instead of turning 1/2 into 2^(-1), try multiplying all the numbers together as is

i.e. try simplifying by doing 2 x 1/2 x (-1) instead

[raven crypt]

I think i got it

thank you very much to both!

[hollow totem]

Np

[raven crypt]

how to close the channel?

.close

[calm coral]

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[tall moon]

a,b,x,y are real numbers such that
$$ax+by=3$$
$$ax^2+by^2=7$$
$$ax^3+by^3=16$$
$$ax^4+by^4=42$$
find $ax^5+by^5$

[potent lotus]

Skill_Issue

[cobalt basalt]

a

there was a trick

[blazing coyote]

binomial theorm perhaps?

or a variation thereof

[cobalt basalt]

(ax+by)whole^5=ax^5+ax^4by+ax^3by^2+ax^2by^3+axby^4+by^5

dam never mind

[blazing coyote]

that's wrong I think

[cobalt basalt]

idk i was think we take x and y common or somthin

[blazing coyote]

,w (ax+by)^5

[chilly lodge]

hecker

[potent lotus]

Results provided by WolframAlpha

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

Requested by physicsrocks

[chilly lodge]

I saw a solution to this on youtube

the trick is something like

(ax+by)(x+y)=ax^2+by^2+axy+bxy
3(x+y)=7+xy(a+b)

and forming another such equation

[tall moon]

i dont understand what you said

@chilly lodge

[chilly lodge]

$ax+by=3$\$\left(ax+by\right)\left(x+y\right)=3\left(x+y\right)$\$ax^{2}+by^{2}+xy\left(a+b\right)=3\left(x+y\right)$\$7+xy\left(a+b\right)=3\left(x+y\right)$\\$ax^{2}+by^{2}=7$\$\left(ax^{2}+by^{2}\right)\left(x+y\right)=7\left(x+y\right)$\$16+xy\left(ax+by\right)=7\left(x+y\right)$\$16+xy\left(3\right)=7\left(x+y\right)$

[potent lotus]

B-eard

[chilly lodge]

do similar with ax^3+by^3 and you'll have two linear equations in xy and (x+y)

[calm coral]

@tall moon Has your question been resolved?

[tall moon]

$$ax^3+bx^3=16$$
$$(ax^3+bx^3)(x+y)=16(x+y)$$
$$42+xy(ax^2+by^2)=16(x+y)$$
$$42+xy(7)=16(x+y)$$

what do i do now

[potent lotus]

Skill_Issue

[tall moon]

ok, i got xy=-38 and x+y=-14

[calm coral]

@tall moon Has your question been resolved?

[hollow granite]

2 answers for xy given these 2

y=-38/x
y=-14-x

-14-x=-38/x

14x+x^2=38

x^2+14x-38=0

(x+7)^2-88=0

use quadratic fomrula

[tall moon]

wait a minure

$$ax^4+bx^4=42$$
$$(ax^4+bx^4)(x+y)=42(x+y)$$
$$ax^5+by^5+xy(ax^3+by^3)=42(x+y)$$
$$ax^5+by^5-38(16)=42(-14)$$

[potent lotus]

Skill_Issue

[tall moon]

wtf its 20?

ok then thanks yaal

.close

[calm coral]

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[calm coral]

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[blazing coyote]

well, what have you tried

[remote mural]

Options are
a)All necessarily real
b)non real except one psitive root
c)non real except 3 positive roots
d)non real for three positive roots of which exactly one is positive

[blazing coyote]

do you need the exact roots

or number of roots

okay

find the points of inflection

[remote mural]

okay

41,49,2009

3 points of inflection

oh wait

oops

there are none ?

[cerulean cradle]

Hey

[remote mural]

hello!

[cerulean cradle]

You see the function is always increasing ?

Since f'(x) > 0

So i think

[remote mural]

yes

[cerulean cradle]

It must cut the x axis at one single point

Can we use wolfram to plot the graph ?

[remote mural]

dang

[remote mural]

nope

[cerulean cradle]

Coz i am not that sure tbh

[ionic moon]

i believe it is

[remote mural]

answer is B

understood

[quaint sphinx]

it is definitely right though that's just the solution

[remote mural]

is there any other alternative method ?

[cerulean cradle]

I dont think so

[ionic moon]

yeah foil it out

[remote mural]

okay

[ionic moon]

lol dont

[quaint sphinx]

exponent of 2009 dominates the others

[jovial gyro]

wolfram alpha can't compute lmao

[remote mural]

yes

[lament ledge]

,w (x-41)^49+(x-49)^41+(x-2009)^2009=0

[potent lotus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by flappie.

[lament ledge]

rip

[ionic moon]

,w zeroes of (x-41)^(49)+(x-49)^(41)+(x-2009)^(2009)=0

[blazing coyote]

,w (x-41)^49+(x-49)^41+(x-2009)^2009

[potent lotus]

Results provided by WolframAlpha

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

Requested by physicsrocks

[ionic moon]

woah

[blazing coyote]

,w (x-41)^49+(x-49)^41+(x-2009)^2009=0

[potent lotus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by physicsrocks

[ionic moon]

oh =0 lol

didnt even see that

,w zeroes of (x-41)^(49)+(x-49)^(41)+(x-2009)^(2009)

[jovial gyro]

lmao, just expression

[lyric snow]

Can we do it by numerical method?

[quaint sphinx]

w/e

the root is around 207.384

but this is completely non-instructive

[jovial gyro]

how did you calculate?

[remote mural]

I will close it
Thank you guys!

.close

[calm coral]

Channel closed

Closed by @zinc olive

Use .reopen if this was a mistake.

[calm coral]

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[remote mural]

If its given that Det(A)=0, can we say anything about Det(Adj(A)) ?

[cerulean cradle]

Det(adj A) = det(a)²

[remote mural]

Ohhhhhhhh

dumb me

thanks!

.close

[calm coral]

Channel closed

Closed by @zinc olive

Use .reopen if this was a mistake.

[cerulean cradle]

Bro its

.reopen

[calm coral]

✅

[cerulean cradle]

For 3*3 matrix

For n th order matrix

Det adj (A) = det (a)^(n-1) generalised

@remote mural Note this bro

[remote mural]

i was talking about a 3*3 matrix anyways

thats what we get for almost all the questions

[remote mural]

I forgot this property

[cerulean cradle]

Ah !

[remote mural]

.close

[calm coral]

Channel closed

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[calm coral]

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[blazing coyote]

could someone explain this proof pleas

[blazing coyote]

*please

[glass heart]

you need to say which steps you dont understand

[blazing coyote]

step k

[glass heart]

what do you not understand about it

this isnt the first time we've had this discussion. you need to be clearer about what you dont understand

[blazing coyote]

like what's happening in step k

[glass heart]

look I dont know how to answer that. axler wrote quite precisely what is happening. you need to say which of what he wrote you dont get

I can repeat the same thing but that wont help

[blazing coyote]

ok, let me put it this way. I don't get what he's doing

so we first construt a subspace with the elmennt {0}

then {0,v_1}

(0,v_1,v_2,v_3..}

and so on?

[glass heart]

no

at the start, we have two cases, either U = {0} or not

in the first case we are just done

the space {0} is finite dim

[blazing coyote]

yes

[glass heart]

otherwise, we start by picking some vector u_1

now, we could either have U=span(u_1) or not

in the first case we are done

otherwise, we are picking some u_2 which is in U but not in span(u_1)

and again we have two cases, either U=span(u_1,u_2) or not

in the first case we are done

[blazing coyote]

got it

[glass heart]

otherwise, we are picking some u_3 which is in U but not in span(u_1,u_2)

and so on

[blazing coyote]

got it

thanks!

.close

[calm coral]

Channel closed

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[calm coral]

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[hearty heart]

this is the set