help-17

[strong grove]

this is ur question

which question are u solving ?

[native kiln]

what is u

[harsh canopy]

^

[native kiln]

sorry

i mean where do i sub in u

du/-3 = dx

in here

[obtuse sierra]

if u = 4-3x then 4-3x = u

so sub 4-3x with u in your problem

[native kiln]

i'm confuse

what law is this

[obtuse sierra]

no law

wdym

[native kiln]

bro i'

i'm brain dead

[harsh canopy]

Just the substitution method of integration

[native kiln]

$-\frac{1}{3}\int \left(4-3\left(4-3\right)\right)^{-3}$

[twin meteor]

svnset

[native kiln]

is this what you mean

[harsh canopy]

Wtf

[native kiln]

wait let me think

[harsh canopy]

Who are you pinging for? The guy with the question?

You successfully found dx in terms of du

From this, you have u = 4-3x, you have that exact thing in your integral, so substitute it out

[native kiln]

.close

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[woeful hull]

I have to prove this claim, however I am not quite sure where to start and how to progress, I know I have to do a direct proof by cases, which would be when x and y are the same sign or different sign.

[woeful hull]

I did something similar to this which was the triangle inequality I think, but I did that in class, and not sure how to do this.

[maiden iron]

hmm

well yea with the triangle inequality we have : |x+y| <= |x| + |y|

now the main problem is how to get max(|x|, |y|)

[woeful hull]

Yeah, i just don't how how to start or do anything, I started proofs not too long ago so its still mega confusing

[maiden iron]

can we say that |x| <= max(|x|, |y|)

I mean its kinda true because max(|x|,|y|) is either |x| or |y|

its true for |y| too

therefore |x| + |y| <= 2 max(|x|, |y|)

[woeful hull]

well yeah

[maiden iron]

in conclusion, |x+y| <= 2 max(|x|, |y|)

[woeful hull]

i dont get how this helps

[maiden iron]

?

[woeful hull]

i mean this is the claim

[maiden iron]

its the conclusion

Im already done with the proof

[woeful hull]

ok

so to start the proof itself, i said that x and y are of either the same or different signs

at least thats both of my cases

[maiden iron]

thats not what I said

but we can at least define max (|x|, |y|)

[woeful hull]

oh you are saying that either |x| or |y| is going to be returned from the max function?

so |x| = max{|x|,|y|} and same for |y| too?

[maiden iron]

$\operatorname{max}(|x|,|y|) = \begin{cases} |x| & |y| < |x| \ |y| & |y| > |x| \end{cases}$

[twin meteor]

Herels

[woeful hull]

yeah

i get that

[maiden iron]

thats why I said, |x| <= max(|x|, |y|)

same thing for |y| too

[woeful hull]

oh i see

[maiden iron]

since |x+y| <= |x| + |y|

and |x| + |y| <= 2 max(|x|,|y|)

we have :
|x+y| <= 2 max(|x|, |y|)

[woeful hull]

oh i see ok that makes sense

i have to follow a certain set of reasonings though let me send it

[maiden iron]

algebraic reasons

[woeful hull]

yea because a <= max{a,b} same for b

i have to do this in cases though so how would i do that

[maiden iron]

Sorry I don't know how to use your tables

here is how I would write it :

For $x,y \in \mathbb{R}$, we have, by the triangle inequality :
$$|x+y| \leq |x| + |y|$$.

We know that $|x| \leq \operatorname{max}(|x|,|y|)$ and $|y| \leq \operatorname{max}(|x|,|y|)$ with $$\operatorname{max}(|x|,|y|) = \begin{cases} |x| & |x| > |y| \ |y| & |y| > |x| \end{cases}$$.

Therefore : $$|x+y| \leq |x| + |y| \leq 2 \operatorname{max}(|x|,|y|)$$

In conclusion : $$|x+y| \leq 2 \operatorname{max}(|x|,|y|)$$

[twin meteor]

Herels

[woeful hull]

you can just state the triangle inequality?

but i cant really reason that

[maiden iron]

?

Its something that is true so we can use it

[woeful hull]

yeah its just i was never taught that idk

[maiden iron]

if you werent, you can just prove it

[woeful hull]

ok

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[upper willow]

In the household of family Huisman on a winter day a window with the surface of 3m² lets go 1,2MJ (1,200,000 Mega Joule) Of heat.

Make the formula for heat loss, Q in MJ and window surface in A m².

TRIED MY BEST TO TRANSLATE IT. Please help.

[upper willow]

A

<@&286206848099549185>

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[blissful sentinel]

(Please don't ping Moderators unless you're reporting something server-related like a spammer)

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[iron ingot]

how do you know if a linear system has infinitely many solutions vs just one

[full hatch]

Calculate them

[hidden niche]

You can also check the relations of the coeficients

[iron ingot]

very helpful

[split heart]

Does your system have a free variable?

[full hatch]

Thanks, it heavily depends on the system of equations.

[iron ingot]

I ended like this

[split heart]

that's rref, good

That means you can plug in any x3 you want

[iron ingot]

is it infinite because x1 and x2 are equal to equations?

rather than a static number

[split heart]

That's one way to put it

[iron ingot]

fair enough

[split heart]

The key here, though, is that x3 is a free variable

you can have x1 and x2 be equations but x3 isn't

In which case, x1 and x2 will be fixed anyway

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[last umbra]

Hi

[last umbra]

This is a physics qn

But this uses a bit of match

Question 46

[cyan shadow]

!status

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6. None of the above

[last umbra]

2

So the initial velocity is 5.5 meters per second

How much should the runner accelerate at rate of 0.2 for the runner to complete 1100meters in exactly 3 minutes

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[vast shale]

I've posted the answer and question. But I'm not really sure about the details of the answer

why can't a point right of q2 be zero?

why does it have to be left of point q1?

is it because |q2| > |q1|?

yeah thats what it says

dam i silly

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[fleet salmon]

why did it shift to left

[fleet salmon]

i thought absolute takes negatives away

[halcyon wing]

-2 + 2 = 0

[fleet salmon]

wdym

[vast shale]

|-2 + 2| = 0

[fleet salmon]

how

[vast shale]

youre mistaken about how absolute values work

youre thinking of it as: |-2| + |2| = 4

[fleet salmon]

u do the math

then

u use absolute

ouu

[fleet salmon]

okay but

yes

but how

is it shifting

left

x can be positive

[vast shale]

your graph is y = |x + 2|. what is y when x = -2? 0.

we see that on your graph

y(-2) = |-2 + 2| = 0

(-2, 0)

[fleet salmon]

hows x -2

😭

[fleet salmon]

im supposed to graph it

[vast shale]

x = -2 here

[marble vine]

what plus 2 equals 0

[fleet salmon]

-2

[vast shale]

boom

[marble vine]

so x is -2

ur confusing the equation for |x| + |-2|

basically forget about the absolute value signs and complete the equation

because if you think x can be positive it will be |2+2|

which is 4

[fleet salmon]

so whrnevet plus is inside

u move left

if minus is inside

move right

[marble vine]

basically but dont go by this logic

just solve for x

[fleet salmon]

yes

[vast shale]

you should understand the logic behind it and not just memorize that idea

[fleet salmon]

yeayea so what makes it equal zero

[marble vine]

yes

[fleet salmon]

yesokok

i get it now

tyty

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[glass jewel]

So I have this function

[glass jewel]

I'm told to find f'(x) and show that it is positive for this interval

well

I get that f'(x) isn't defined in 3

meaning either I'm failing to understand something or the exercise is wrong and should've been ]3,infinity[ instead

am I missing something?

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@glass jewel Has your question been resolved?

[tawdry grove]

You did the exercise with a computer program?

[glass jewel]

it was allowed

so yes

[tawdry grove]

The program doesn't know the derivative at x = 3 because you told it the function is undefined at x < 3.

[glass jewel]

I defined it for x>=3, so it is defined at 3?

here

[urban jackal]

https://tenor.com/view/the-rock-the-rock-sus-the-rock-meme-tthe-rock-sus-meme-dwayne-johnson-gif-23805584

[tawdry grove]

f(x) is defined at x = 3, but derivatives are defined as a limit. Your program is unsure whether the limit exists at x = 3.

When is $f(x)$ defined? Is it only defined when $x \geq 3$?

[twin meteor]

chencking

[glass jewel]

yes

only when x>=3

[tawdry grove]

Why? Why can't you calculate f(2), for example?

[glass jewel]

I can, but the exercise says it should only be defined when x>=3

[glass jewel]

I mean, it defines f(x) like this

[tawdry grove]

Oh wait nevermind I missed the part of the question restricting the domain.

So one way to approach the problem is to just extend the domain of f(x) and using that plug it into your program.
The other way is to just calculate f(3) yourself and then use the program's derivative for the rest of the interval.

[glass jewel]

I guess

it's just kind of weird as they specifically require us to use this program and it seemingly has troubles with their own exercises

[tawdry grove]

Simply put though, your program could not find the derivative at x=3 due to the way they programmed it.

Referencing the screenshot in the comment I replied to, we know the limit is simply the limit when h approaches 0 from the positive direction. However, the program does not, so it made it undefined.

I am not confident, but my guess is that rather than restricting the domain to $[3, \infty)$, it seems the way you entered the function simply left it undefined on $(-\infty, 3)$.

If this is for a college course, you could try asking your professor or TA if there was a better way to enter it. Or perhaps a specific way they wanted you to use this output.

[twin meteor]

chencking

[glass jewel]

Yeah, I'll probably ask someone tomorrow, but thanks for the help, it was really useful :)

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[rose shard]

Is this correct?

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[simple mason]

Open a new channel

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[short river]

i ran this thru a calculator and it still says there are no removable discontinuities, i put DNE there and it said i was wrong

[pale perch]

is it not -3?

[short river]

why would it be -3

[pale perch]

the denominator factors into (t+3)(t+5) so there is a discontinuity at t=-3 but its removable since t+3 cancels with the numerator

[short river]

ahh thanks

i forgot about that because i had already factored it and simplified to 1/t+5

one more question

what do i do here

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[fleet salmon]

4a

[fleet salmon]

how r they both odd?

,rotate 270

[flat whale]

Uhh

did you write a formula for each f(-x) and g(-x)

And compare it to f(x) and g(x)

[fleet salmon]

no

but odd is if it has rotational symmtery

along origin

and the rational

x cannot equal 0

so

how is it odd

[flat whale]

Wot

Do you know what odd function means

[fleet salmon]

yes

[flat whale]

What does it mean

[fleet salmon]

it has rotational symmetry along the origin

[flat whale]

No

[fleet salmon]

(0,0)

istg

[flat whale]

f(x) is odd if f(-x) = -f(x)

For all x

[fleet salmon]

yes

[flat whale]

So do this

[fleet salmon]

1/-x

-x

[flat whale]

And

[flat whale]

Do they satisfy this

[fleet salmon]

uh

how is that

correct

[flat whale]

Does f(-x) = -f(x)?

1/(-x) = -(1/x)?

[fleet salmon]

yea

isnt that the same

f(-x)=-f(x)

it does work for both right

[flat whale]

Yes

[fleet salmon]

okok

ty

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[coarse ridge]

!status

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What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
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6. None of the above

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?

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[honest heath]

for problem 6, would the formula for the homothetic transformations also involve the step mentioned in the example from problem 5? like for 6a, would i need to shift the plane to be centered at i, then do the stretching, then shift back?

or would it just be a shift there and then stretch with no moving back?

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[vast shale]

i was asking ab this one yesterday im still completely lost ive been watching some videos and i think i have to rewrite the square root to get it to match the formula for arcsec however im not sure how to get there

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<@&286206848099549185>

[vast shale]

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[slate island]

The answer is supposed to be 120pisqrt26 but I can't seem to get it

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[flat whale]

Where even is the question

[slate island]

find the surface area of the volume generated when the following curves revolve around the y-axis.
If you cannot evaluate the integral exactly, use your calculator to approximate it.

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[slate island]

<@&286206848099549185>

[slate island]

posted the question

[slate island]

also its 197

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[mighty crater]

I am not sure how to go about this porblem

problem*

Could someone guide me in the right path please

[ancient knoll]

write an expression for F’(x)

[mighty crater]

alright

so basically in slope form just write it for f(x) ?

[ancient knoll]

F(x), and yes

[mighty crater]

so for f(x) we get f(x) = 1/2x + 2

g(x) = -2x+10

f(x) = 1/2x + 2

[ancient knoll]

and now you can evaluate F’(5)

[mighty crater]

Im not sure how to fully do that, this topic is still new to me. Would I plug in 5 to each of g(x) and f(x) ?

[ancient knoll]

the question asks for F’(5), i assume you were taught derivatives

note F’(5) isn’t F(5) which is what you were suggesting

[mighty crater]

right. Were still learning it. My professor has a really heavy accent so I am falling a bit behind. To be honest I am completly lost on what to do after to evaulate F'(5)

[ancient knoll]

okay, can you write an expression for F(x) without f(x) and g(x)

so not F(x)=f(x)+g(x)

[mighty crater]

ummm... possibly

I think it fould be F(x) = k * [f(x) + g(x)]

would*

[ancient knoll]

where’s k coming from?

[mighty crater]

would it not be the constant

of both functions

or am I completly speaking giberash and thinking of something else

[ancient knoll]

the constants are contained in f(x) and g(x) already

[mighty crater]

ahh okay

[ancient knoll]

so all you need to do is substitute f(x) and g(x) with the expressions you got above

[mighty crater]

okay so we would have F(x) = [1/2x + 2 + -2x +10]

[ancient knoll]

yes, and you can simplify that a bit

[mighty crater]

right we can combine like terms

so we would have F(x) = -1/2x + 12

[ancient knoll]

yes, now F’(x) refers to the derivative of F(x), or how the value of F(x) changes as you change x.

( i would start recommending reading up derivatives of a straight line somewhere online )

[mighty crater]

Got it,

so from here do I just find the derivative of the F(x)?

[ancient knoll]

yes then find F’(5)

[mighty crater]

okay let me see real quick

ill post the derivative of F(x) if you can double check for me

wouldnt the derivative of our function F(x) just be - 1/2

[ancient knoll]

correct

actually you added the functions wrong :x

[mighty crater]

got is so if we have the derivative of F'(x) = -1/2 how would I evaluate F'(5) <-- ( If this statement is right )

oh

[ancient knoll]

but repeat the steps you just took will lead you to the right answer

[mighty crater]

: (

okay

just to confirm is this right F(x) = [1/2x + 2 + -2x +10]

i got -3/2x + 12

rooki mistake

[ancient knoll]

yes now it’s fine

[mighty crater]

gotcha

so how can I evaluate F'(5) from here if I know that F'(x) = -3/2

[ancient knoll]

you already have F’(x), so what do you do when you want to find F’(5)

it should be the same thing you do for any function when hon want to find it’s value at some point

[mighty crater]

oh so wouldnt F'(5) just be equal to -3/2

[ancient knoll]

yes

[mighty crater]

I see

Thank you so much for your help and time

I have one more practice I can do and repeat my steps

Thank you ❤️

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[vast shale]

Hi! First time ever doing stats, can someone PLEASE check over what I did, thanks

[vast shale]

Question 1 only

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@vast shale Has your question been resolved?

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@vast shale Has your question been resolved?

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[maiden iron]

x = 0 and x= 1 are parallels

[halcyon wing]

solve x=sin(4x) for 0<=x<=1

[maiden iron]

,w graph sin(4x) = x

[twin meteor]

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[dull maple]

Do you mean intersection points of y=x and y=sin(4x)?

I haven't seen term intercept being used for intersection points.

[maiden iron]

^

[dull maple]

Anyway, you can't solve conventionally, i think. Graphing tools seem like your only help or root estimation techniques.

[maiden iron]

or intermediate value theorem

[dull maple]

Of course, there is.

You just can't find it using your general isolation or such root finding methods.

,w solve x=sin(4x)

[twin meteor]

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[dull maple]

There you go.

What makes you think that number is in some surd form? Not every irrational number is in surd form.

On the top of that, we don't even know if x is irrational. Lol

[maiden iron]

intermediate value theorem is more of an existence theorem

[dull maple]

Yeah. That won't give you an exact value either.

[crude arrow]

If you made a small enough boundary tho

Idk prob doesn’t help

Sorry

[maiden iron]

hmm

I mean, its possible to find a value but I forgot how

[crude arrow]

It’s p useless tho

Cuz you would have to know the value

Also oops I hijacked

[maiden iron]

no but if you know the boundary of the x such that f(x) - x = 0

you can try dichotomy or smthing like that

[crude arrow]

That’s true

[maiden iron]

Welp it has been a while

Im too sure

im not *

[crude arrow]

Since calc?

[maiden iron]

maybe

[crude arrow]

No basically in a continuous function f across a domain (a,b)

$\forall y f(a) < y < f(b) \exists x$ s.t $f(x) = y$

[twin meteor]

NoSQLDB

[crude arrow]

Wait why is the latex spacing so shit

[maiden iron]

well, you can do spaces but i dont know how

[round plover]

you normally put quantifiers together at the beginning

[crimson jetty]

,,\forall y\in(f(a), f(b)):\exists x: f(x)= y

[crude arrow]

Doesn’t matter tho

[crude arrow]

Okay that’s wayyy too much

[crimson jetty]

🤨

[crude arrow]

It’s okay I love the effort

[crimson jetty]

if I haven't found a counter

its worth attempting to prove it

uh yh lol

this is what ur tryna prove

Yes thats how they differ

but it may turn out not to matter

im not what to say except youre going to need to expand the definition of max

in a symbolic proof

if im not mistaken it might not actually hold

but trying to prove it will hint you towards constructing a counter example

Doesnt matter - just try out the steps you would to prove it

Approach for a proof should be fairly systematic.

State what ur tryna prove

k(f + g) = kf + kg for any k, f, g

Then start with one side

k(f+g) =

and start expanding/manipulating

???

Let k, f, g.
k(f + g) = ???

No, you need to prove this algebraically

[crimson jetty]

the first step is to expand on definitions

yes geometry will give u the intuition needed

but this all needs to be symbolically manipulated

its all given to u in the question

k(f + g) only means one thing algebraically

it means

add f and g together

then

multiply by k

So first add f and g together - whats the definition of this

Let k, f, g.
k(f + g) = k max(f, g)

thats all i was looking for

you must expand definitions

Let k, f, g.
k(f + g) = k max(f, g)
= ...

So next you need to expand the definition of max

well you tell me what max means graphically

max f g

its a pointwise max right?

so ok we shall rewrite the proof pointwise

Let k, f, g, x.
k(f(x) + g(x)) = k max(f(x), g(x))
= ...

yes?

Then you can use the definition of max for real numbers

technically not but lets leave that aside

[crimson jetty]

to expand the definition of max, hint: consider cases

Let k be scalar; f, g functions; x in domain.
k(f(x) + g(x)) = k max(f(x), g(x))
= ...

No.

what is max(a, b) in general

you need a definition for this otherwise ur stuck

and one way to do it is piecewise

yes

Let k be scalar; f, g functions; x in domain.
k(f(x) + g(x)) = k max(f(x), g(x))
= kf(x) if f(x) > g(x) else kg(x)

u can write that neater in braces

well yes? piecewise functions

so what im gonna say is that looks ugly and to split it into 2 cases

Let k be scalar; f, g functions; x in domain. Suppose f(x) > g(x).
k(f(x) + g(x)) = k max(f(x), g(x))
= kf(x)
= ...

ok now we're kinda stuck here, and what would be helpful in rough is to think about what we want to show

we want this
= ...
= max(kf(x), kg(x))
= kf(x) + kg(x)

yes?

so then u have to think for yourself when they are equal. And more importantly if they are always equal

well yeah, its just notation for the same thing

,,\max\cdots=\begin{cases}case A\case B\end{cases}

[crimson jetty]

I think u have all the pieces to finish the proof and either conclude distributivity holds or it doesnt with the above ^

so will leave it to you

it probably will read better on paper

If it doesnt, u need to come up with an explicit counterexample

but the point of writing everything above is to help u come up with it

(supposing there is a counter)

i probably wont be around

but someone should be if u just type

yh i did above, promise. reread all of it carefully

before i disappear f+g is defined pointwise.
but "f+g" itself is a legitimate new function. That is defined in that way

with regards to the earlier notation

technically i miswrote when i wrote
f+g = max(f, g)
its an abuse of notation

max is takes 2 real numbers as input

unless i define a new max which takes 2 functions as input

which does exactly what that oplus operation does

ie. max pointwise

so when ur tryna prove

kf+kg = k(f+g) this is functional equality

2 functions are equal iff they are equal everywhere pointwise

kf+kg = k(f+g)
hence ur tryna prove this for all x in the domain

note even kf is a function. Defined pointwise by (kf)(x) = k(f(x))

obvsly on paper stick to oplus, not +. super confusing in text

error on...

5th equals does not follow

thats all fine

u have good ideas

but unfortunately this does not hold in general

nope.

dont wanna give it away

but yeah think on this carefully

in particular note a proper proof states what k, x, f, g are. As in what kind of objects they are

and that's important

Whats k.

eh???

f, g are functions (of the type stated in the original question)
x is in the domain

k is ...

like what set

yes, now rethink this claim

well come up with some examples of them

and check a > b => ka > kb works or not

there are plenty which dont.

Right, but now u see this point. Its a key point with regards to the original question

well state me some k, a, b where it doesnt work

to be sure we re on the same page

well yh

just to be clear this claim doesnt generally hold

no.

ok shelving this aside (since i do have to disappear)

once uve found k, a, b. Great.

Now go back to your original proof

think about if it can be rescued

If it cant, that means uve probably come up with a counterexample

and then u should state it explicitly (let f = ..., etc)

oh yh and nitpick: one of your cases need to include the case f(x) = g(x).
for all the casing ur doing

well there u go lol

at the very least

k can be any real right

right ok i see your point

there is another k

half the k's dont work

show me what u get with negative k...

eh.

thats the point of expanding the proof

though

to break it into chunks

in this case yeah u wouldve missed it

but for someone marking or helping it helps pinpoint the issue

[crimson jetty]

like this

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[crimson jetty]

what didnt work

ur gonna have to pick a specific x as well

-1 . max(2, 4) =/= max(-1 . 2, -1 . 4)

thats correct. theres only one faulty line in your proof

now its a question if it can be fixed. or if what ur tryna show is false.

this is like correct, no problems

but u wanted to have it inside for... reasons.

huh

no?

no no no

ur either doing a proof to show theyre always equal

or ur stating a counterexample

to show theyre not

Theyre equal for all choices of k, f, g, x

is what u wanted to show

the negation of this statement is that there are some k, f, g, x where they not equal

yes but no. you need to give specific k, f, g, x

maybe but why not just come up with something concrete

to show 2x not equal to x^2

u should like

x = 1 disproves it

cus it does hold for some x maybe. like 0 and 2

well no, it defo works with positive k

u havent exactly shown it doesnt work for negative k

not with your working

its much much easier for a proof verifier

to check 1 specific case

rather than reading your proof of a certain class of cases breaking it

we only need 1

yes thatll do

x = something too

it distributes for x = 1

and x = 0

yes

k(a+b) = ka + kb

thats all it means

-1 max(f(0), g(0) = max(-1 f(0), -1 g(0)) is indeed true

np, this is a common thing to struggle on

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[fiery egret]

pls can one tell how to solve ?

y can be 81

[sudden compass]

substract 9^3x * y from both sides, take 9^3x common, and then solve for x and y

hmm actually nvm

just divide both sides by 9^3x

[fiery egret]

y eah and y=81

[sudden compass]

x could be any integer tho

so D imo

[vast shale]

Yeah so as I was saying since you know y is 81, then you effectively have 9^x = 9^x which means X can be anything

In the other channel

X could be smaller, could be bigger

[fiery egret]

righht i was away sorry

so replied late

[fiery egret]

yes

thanks

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[outer arrow]

Can anyone give me a quick real life example of a biased and an unbiased estimator?

[vast shale]

Suppose you randomly sample only the tallest students from a school to estimate the average height of all the students. This would be a biased estimator

[outer arrow]

And unbiased would be randomly sampling in such a way that then mean of their height is very close to the actual mean?

[vast shale]

yes

[outer arrow]

So the estimator basically depends on our sampling?

[vast shale]

yes the performance of an estimator can depend on how you're sampling the population

[outer arrow]

Hmm okay

But the estimator itself would be a function that for example calculates the mean of our sample

And its value is our estimate

Is that correct?

Ty anyway, the example helps

.close

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[midnight marsh]

How to do this x²-11x+30=0
Quadratic equation

[midnight marsh]

Can I have explanation

[paper depot]

@midnight marsh do you know how to solve quadratic equations generally?

[midnight marsh]

Yeah

I'm using this form

I forgot what to do next

So? Mhn?

[midnight marsh]

U there?

@paper depot

[deft mortar]

thats right

now solve powers

solve the root

@midnight marsh

[midnight marsh]

This right?

[deft mortar]

yeah

continue

do that root

[paper depot]

sorry had to go do urgent shit irl

[midnight marsh]

Oh it's okeh

This right?

[paper depot]

yeah checks out

[midnight marsh]

How will ik who's the x1 and the x2?

Like this

Is 6 x1 and 5 the x2?

[deft mortar]

that doesnt matter i think

but you could do the x1 when it is a plus

or x1 the result of the biggest number

is the same i guess

[midnight marsh]

Oh okie thank you

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[hushed nexus]

hey i was wondering how i would find the derivative for this.

i was thinking i could bring the denomenator to the top

to get rid of the fract but honestly i dont really know how thats

[cyan shadow]

use the 'quotient rule'

[hushed nexus]

suppose to look like

[obtuse sierra]

use the 'chain rule'

[hushed nexus]

wait..... i thought quotient rules only for an x variable divided by an x variable?

[cyan shadow]

yeah this works too

[hushed nexus]

if i were using chain rule

would i just find the derivative of 1/x

then multiply

it by the derivative of 5x^2+6x?

[sudden gulch]

I uh think you should use quotient rule

[sudden gulch]

that would be a different answer

(-10x-6)/(5x^2+6x)^2

substitute 2 in

huh

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@hushed nexus Has your question been resolved?

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[vast shale]

The working is correct

I hope you meant this expression

$ax^{a-1}sin\left(\frac{1}{x^b}\right) + x^acos\left(\frac{1}{x^b}\right)\left(-\frac{b}{x^{b+1}}\right)$

And not $ax^xsin\left(\frac{1}{x^b}\right) + x^acos\left(\frac{1}{x^b}\right) -\frac{b}{x^{b+1}}$

[twin meteor]

- Miles12345

[obtuse sierra]

x^x?

[tawdry grove]

I believe he meant $x^{a-1}$

[twin meteor]

chencking

[vast shale]

Yes. Apologies for the typo

[twin meteor]

- Miles12345

[elfin moon]

Yes

So what should I do next?

[tawdry grove]

Do you understand what it would mean for $f'(x)$ to not be continuous at 0?

[twin meteor]

chencking

[elfin moon]

I don't understand

Please explain

[tawdry grove]

Assuming $b > 0$, $\frac{1}{x^b}$ blows up as $x$ approaches 0. So the limits of $\sin(\frac{1}{x^b})$ and $\cos(\frac{1}{x^b})$ are undefined. However, $\sin$ and $\cos$ are bounded, so the limits of $f\sin(\frac{1}{x^b})$ and $g\cos(\frac{1}{x^b})$ exist when the limits of $f$ and $g$ go to 0 as $x$ approaches $0$.

Can you see how that is similar to your problem's $f'(x)$? What would your $f$ and $g$ be? Under what conditions does their limit approach 0 as $x$ approaches 0?

[twin meteor]

chencking

chencking

[tawdry grove]

Yes

So you need to figure out under which values the limits of your $f$ and $g$ will be 0. That way, the limit of $f'(x)$ at 0 will exist.

[twin meteor]

chencking

[elfin moon]

You meant I have to zero f,g?

Is this option D?

[vast shale]

Yes

[elfin moon]

lim x tends to 0
Xsin(1/x)
So this function is continuous at any point right?

[vast shale]

Not at any point unless you specially define F(0) = 0 (where F(x) = xsin(1/x)) in a piecewise fashion

Since 1/0 is undefined

[elfin moon]

Sin(1/x) is bounded

Xsin(1/x) will be 0 at 0

So it's continuous no?

[vast shale]

Not at all

1/0 is undefined

So 0sin(1/0) is still undefined

But if you specifically define f(x) as xsin(1/x) for x≠0 and f(x) = 0 for x=0, then f(x) is indeed continuous over R

@elfin moon for continuity, you need to check the Right Hand Limit (as x->a^+) = Left Hand Limit (as x->a^-) = f(a). This boils down to saying that there should be the existence of the limit of f(x) as x->a and should be equal to f(a).

Now in the question, it is specifically said that for x≠0, f(x) is defined as that expression, and for x=0, f(x) = 0 (since plugging in x=0 in f(x) which is defined for x≠0 gives an undefined value).

When you check the existence of the limit of f(x) as x->0, upon restrictions on values of a and b, you get the continuity at x=0.

So you need to find those restrictions

[vocal sleet]

@elfin moon Has your question been resolved?

[elfin moon]

I didn't understand your last restriction lines

Why specially x>0

[vast shale]

Apologies for the confusion. I meant -> symbol as the arrow symbol used in limits

[elfin moon]

@obtuse sierra please help in this question

[elfin moon]

I am confused because one user said b>0 and f,g should be zero

I am confused what should be zero and what should be postive

[obtuse sierra]

oh hey

i dont want to read this entire discussion (lazy), also is the answer D?

@elfin moon

ping me when you're back

[elfin moon]

Ha answer is D

@obtuse sierra

[obtuse sierra]

ok

[elfin moon]

Limit, DIFFERENTIATION, continuity sab mix ho gya🤦

[obtuse sierra]

we have this

[elfin moon]

Yes

[obtuse sierra]

now we want this to go to 0 when x = 0

[elfin moon]

Acha

[obtuse sierra]

whatever is the argument of sin and cos, they always stay between -1 and 1

wait a sec, brb

[elfin moon]

Yes

[obtuse sierra]

so you need to focus on the x that is in multiplication with those two

those two should make f'(0) = 0

that means there must be a term of x in the multiplication

but not in denominator

otherwise infinity

[elfin moon]

So that continuity exists

At x=0

[obtuse sierra]

understand what I just said?

[elfin moon]

Ek minute

Ha hona chahiye jo dono ko 0 bana de

Second wala toh samaj aa gya

[obtuse sierra]

second?

[elfin moon]

I meant when power of x (b+1-a)>0 then it will be 0

[obtuse sierra]

yes

similarily for the x with sin

[elfin moon]

But first wale ka malum ni

[obtuse sierra]

a - 1 > 0

[elfin moon]

a=0?

[obtuse sierra]

that term should also go to zero when x = 0

[elfin moon]

a-1>0 then how it can be 0?

It will be infinity no?

[obtuse sierra]

then there will be a x with sin too

just like cos

then both terms will go to 0

[elfin moon]

Not clear

First term is not clear

[elfin moon]

From here

[obtuse sierra]

lets take the first term

put x = 0 in the first term if a = 1

[elfin moon]

If x=0 then whole term 0

Doesn't matter a=1,2,-1,-5

Right na?

[obtuse sierra]

put a = 1 in the first term

what do you get?

[elfin moon]

x^(a-1) sin (1/x^b)

x^0 so 1
sin(1/x^b)

Which is 0

[obtuse sierra]

ohh?

you only have sin(1/x^b)

put x = 0, do you get x = 0?

,w plot sin(1/x) from - 1 to 1

[elfin moon]

Infinity

[twin meteor]

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[elfin moon]

Yes like that oscillation

[obtuse sierra]

so we dont want that

we want 0

so we want x

[elfin moon]

Like what?

so what will you do?

Any medicine for it 💊

we have -a+b+1 >0
a-1>b
By second thing

[obtuse sierra]

do you understand that we need x multiplied with sin(1/x^b)?

[elfin moon]

Yes

And cofficient of sinx should be 0

Or tends to 0

[obtuse sierra]

yeah after we sub in the limit, yes

so the exponent of x should be?

[elfin moon]

It should be 0

[elfin moon]

What limit? 0?

[obtuse sierra]

no

bro

[elfin moon]

🤦🤦🫡

Hindiiiiii kar lo yar

[obtuse sierra]

if the exponent of x is 0, then its just 1

lo

[elfin moon]

Ha 1

So?

[obtuse sierra]

agar sin ke saath x nai hua toh oscillating limit ho jayegi

[elfin moon]

a=1, x=1

[obtuse sierra]

agar x ki jagah 1/x ya 1/x^2 ya 1/x^3 hua toh infinity bhaag jaayegi

agar x hua ya x^2 x^3 x^4... toh 0 jaayegi.

[elfin moon]

Bahar wale x ki bat kar rhe na abhi?

[obtuse sierra]

yes

[elfin moon]

Na ki andr wale right?

Okay wait

[obtuse sierra]

nope

[elfin moon]

,w draw 1/x sin (1/x)

[twin meteor]

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[obtuse sierra]

[elfin moon]

,w draw x^5 sin(1/x)

[twin meteor]

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[obtuse sierra]

[elfin moon]

Feel ni aa rha h ye aakhir ho kaise rha h

[obtuse sierra]

x ko 0 daalo

aur kaise

[elfin moon]

1/x me x increasing karenge toh ye sin(1/x) ko jyada motive karega so infinity

And x× hoga jab usko niche layega which is 0

Yooooo

[obtuse sierra]

[elfin moon]

Ha x=0 se Jayda samaj aaya😂

Let's go back to point

So we want x to be positive so that it makes whole 0

a-1 should be greater than 0

Yooo both term connected now

[elfin moon]

a-1>b>0 yooo damn

[obtuse sierra]

[elfin moon]

f'(0) exist

Karwa hi diya tum

[obtuse sierra]

exist nai continuous

[elfin moon]

Thank youu so muchhhhhh

.close

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[gritty spade]

Is their a different method to factor x^3+4x^2-15x-18. I want to use something thats different than the rational root theorm.

[hard atlas]

there is the cubic formula but it's painful to use

rational root theorem is by far the best option

[gritty spade]

ok

[elfin moon]

I can tell you

[gritty spade]

sure

[elfin moon]

Try putting 1,-1,2,-2

If one of them satisfied then x-a will be one root then simply divide it and solve quadratic

[gritty spade]

ok

[hard atlas]

that is basically the rational root theorem

[gritty spade]

except your doing polynomial divison after finding out one number that satisfy's it but yeah technically same thing

[elfin moon]

Yes

[gritty spade]

thank you for your time .close

.close

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[hard atlas]

well the point is that the rational root theorem tells you which numbers to test

instead of always blindly testing 1,-1,2,-2,...

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[copper pebble]

hi

[copper pebble]

I need to find out whether this function is injcetive

I know that injective means that each element in the domain maps to only 1 in the codomain

But I don't know if I should approach this by doing a 'counter-argument' where I need to prove that it's not injective or what

it goes from R -> R

[vocal sleet]

@copper pebble Has your question been resolved?

[solemn cape]

Try to find the roots of your polynomial

[solemn cape]

Find the roots of your polynomial and see if it means that it’s injective or not

[copper pebble]

Okay, thanks

[copper pebble]

after finding roots, how can i determine if its injective or not

[solemn cape]

Like you said, a function is said injective if for all $a, a’\in A$ such that $f(a)=f(a’)$ we have $a=a’$

[twin meteor]

Joseph.P

[solemn cape]

Are you still here ?

[copper pebble]

Yes sorry

Still trying to figure it out

[solemn cape]

What did you find as a root ?

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@copper pebble Has your question been resolved?

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[warped chasm]

I am attempting no.2

and I have no idea how to do it

what does a(underscore) b(underscore) even mean

any help

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@warped chasm Has your question been resolved?

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[astral pebble]

I am stuck here

I think I need to use the lb and sin/cos + the angle

I just dk what combo is used for what

like is P1x 190sin(40)?

or is it 190cos(40)

or am I completely off base

[regal bane]

sin = opp/hyp

And the hypotenuse is always the actual length P1

Know how to draw the triangle, such that P1 is the hypotenuse? One side will be the x-axis, another side will drop down

[astral pebble]

not really tbh

ik I learned all this I am just burnt out rn

[regal bane]

Make sure you can do it the "correct" way by drawing a triangle and finding the length of its sides

[astral pebble]

for P3 I can use the fact its a 30 60 90 triangle right?

Could I draw a right triangle from it

[regal bane]

But I do have a bit of a hack. If you instead find the angle going ccw from the positive x-axis, then cos is always the x-component, and sin is always the y-component

So for example, you can get to P3 by going -30 degrees from the x-axis.

x-component is 120cos(-30)
y-component is 120sin(-30)

[astral pebble]

-30?

[regal bane]

Should be pretty normal if you've been working with the unit circle

[astral pebble]

oh shit

that just clicked

I should not be doing any form of math rn lmaooooo

[regal bane]

That's probably not the way you've been taught, but imo it's a cheaty way that allows you to ignore drawing triangles

[astral pebble]

wait do you have to take the absolute values of the answers?

bc 120sin(-30) is -60

can you have a negative force component?

[regal bane]

Sign is needed. That means "downward".

[astral pebble]

oh true

[regal bane]

Or "leftward" if you're working with the x-component

[astral pebble]

broooo I am so off my game rn

lol

hmm

@regal bane any ideas?

[regal bane]

You are disagreeing with your own answer

Oh nvm mb

[astral pebble]

oh I think ik why my last are wrong

[regal bane]

So I was talking about P3 above, not P1. What angle, starting with the positive x-axis, do you need to get to P1?

[astral pebble]

ohhhhh

I am dumb

one sec

I am all mixed up

for that I used 190cos(-140) and 190sin(-140)

where did I mess up?

[regal bane]

,calc 190cos(-140 degrees)

[twin meteor]

Result:

-145.54844419261

[regal bane]

Hmm. I like your answer. That's odd.

[astral pebble]

wait

do you think its the space between the - and 145

and - and 122

[regal bane]

?

[astral pebble]

theres a space between the number and the negative

neither of the other ones that are marked right have that

[regal bane]

Oh yeah, your negative does look very different. Type something wrong?

[astral pebble]

space accidentally

is it just 340/cos(55)