help-0

[unkempt robin]

So which point decides when the function stops increasing and subsequently makes it non-injective?

[subtle sun]

from when x = 0 ?

[unkempt robin]

Yes, but I am asking for a point

Also need y 🙂

[subtle sun]

ahhh alr sorry

[unkempt robin]

No problem!

[subtle sun]

(0, 2)

[unkempt robin]

Correct

So do you see how we can split the graph into 2 sub-graphs (ignoring the third part where it starts growing again) to make each sub graph injective again?

[subtle sun]

yes I can see

[unkempt robin]

So then what would you put instead of a

[subtle sun]

since the y=2, then the a=2 ?

[unkempt robin]

No, y is the vertical -axis

[subtle sun]

ah wait sorry I got confused with D1 for a second

[unkempt robin]

You are asking yourself from which x onward is the function non injective

[subtle sun]

In D2, a is the x coordinate so it would mean that a = 0?

[unkempt robin]

Yes, but in D1 as well

It is injective from (-∞, 0] right

And then when we split it it is also injective from [0, b]

And then what is b

[subtle sun]

b is 2?

[unkempt robin]

Why do you think that

We are looking for such an x which from the x onward it makes the function non-injective again

[subtle sun]

Sorry, I'm a bit confused. Tell me if I'm wrong, (-∞, 0] is the area of the graph which includes and is lower than Y = 0?

[unkempt robin]

We aren't talking about y here

We are finding x's

Earlier I just wanted you to tell me a point which needs a y coordinate as well

You can discard that now

You are looking at such intervals of (x1, x2) where the graph is injective

Domain of a function = x-values

[subtle sun]

hm okay I see, sorry this is taking me a while to process.

[unkempt robin]

If you take an interval x€(-2, -1) you see the graph during those x's is injective

No problem

Because no parallel line to the x-axis intersects the graph twice or more

But if we take an interval €(-1, 1)

We see it is no longer injective

Because you can draw a parallel line to the x -axis which intersects the graph twice

And so, still ignoring the third part of the graph where it starts growing again, you need to pick 2 intervals where the graph is injective in both

Can you think of an x which suffices both intervals (a)?

From -∞ to which x is the graph injective and from which x to ∞ is the graph injective (ignoring the third part). Let me give you a simpler example

Is this function injective?

[subtle sun]

No, it isn't

[unkempt robin]

Okay

So what would you do to make it injective

[subtle sun]

Split it at (2, -5)?

[unkempt robin]

Correct

And you'd get 2 intervals

From -∞ to that x and from that x to ∞

It is injective on the yellow interval and on the cyan

Correct?

[subtle sun]

Yes, I can understand

[unkempt robin]

Okay, so how would the two intervals look (D1 and D2)

D1 =

Think of it as the yellow line

[subtle sun]

Just to clarify again, D1 is (∞, 0] meaning that a = 0. Hence, D2 is [0, 4/3]?

And Domain only consists from values of x

[unkempt robin]

Yes that's correct

And then you are left with D3

[subtle sun]

Ahhh I see now, sorry I was thinking too much about x and y coordinates.

[unkempt robin]

Hehe

Also, in D1 it's -∞

Because you start on the left

The assignment forgot the minus

[subtle sun]

yeah that one is not on me though lol I'll cross the original out and replace it with -∞

so b= 4/3 hence D3 is [4/3, ∞)

[unkempt robin]

That's correct

You see that in all 3 intervals the graph is injective

[subtle sun]

Yes, I can see

[unkempt robin]

And that's it

[subtle sun]

okay, tysm

[unkempt robin]

No problem!

[subtle sun]

.close

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[finite knot]

prove that lnx<=x-1, for every x>0
with f(x)=x-1-lnx, x>0

[fickle sandal]

what are you struggling with here?

[finite knot]

i figured out how it works but cant really find a way to write it

[fickle sandal]

we're using differentiation here right?

[finite knot]

yes

[fickle sandal]

so can you explain in words first

[finite knot]

i differentiate f(x) and we get f'(x)=(x-1)/x, to find critical points we set f'(x)=0 and we get that x=1 is critical point, so we then know that f(x) is monotonically decreasing when x is between 0 and 1 and f(x) is monotonically decreasing when x is between 1 and +oo therefore knowing this we also know that lnx>=0 when x>=1 and all in all i can comprehand why this inequality is true but cant really mathematically prove it

[fickle sandal]

Looks good. First we can rearrange the inequality to get:

$f(x)\geq 0$

[ocean seal]

chlamydia

[fickle sandal]

which is much easier to look at and prove

then you've found the derivative that is always decreasing within the domain, but how do you know that f(x) won't cross 0?

[finite knot]

i dont

[fickle sandal]

we can show that that's the case

Sorry I've just missed something, you said f(x) is monotonically decreasing for x>1, but that's not true.

because f'(x)=1-1/x is positive for x>1

so from there, we can show that the global minimum of f(x) is at (1,0)

[finite knot]

but f'(x) can also be written as (x-1)/x and when we plug a number between 0 and 1 whe get a negative number

[fickle sandal]

yeah for 0<x<1, but that's not the case for x>1

[finite knot]

ok

[fickle sandal]

are you sure that's ok

[finite knot]

no

thats what i wrote

[fickle sandal]

yeah, so next is to figure out what happens at x=1, since it looks like there's a minimum there

right?

[finite knot]

perhaps

[fickle sandal]

i'll reword and say that there is a minimum there

[finite knot]

i know that there is a minimum there

f(1)=0

[fickle sandal]

so what are you stuck on now?

[finite knot]

proving lnx<=x-1

[fickle sandal]

so from what you've done above, we can see that f(x)>=0, right?

[finite knot]

yes

but f(x), x>0,

what about the values that are lower than 1 and higher than 0

[fickle sandal]

then we check x=0, or rather lim(x->0) to see that it's also greater than 0 there

[finite knot]

but the limit diverges to -oo which i presume is smaller than 0

[fickle sandal]

it doesn't, you have a -lnx term which goes to +inf

[finite knot]

o

that makes sense

[fickle sandal]

you sure?

[finite knot]

yes

because we know that the smallest a value of f(x) is 0 therefore any other defined value of f(x) is gonna be bigger therefore f(x)>=0 right?

[fickle sandal]

because the smallest value is 0, and the function only goes up from 0 on both sides

[finite knot]

fair enough

so how can i write this mathematically with set notation

[fickle sandal]

i would go with stating the minimum at x=1 from here, then showing the limit x->0

depends on how you're expected to set your working out

or are there other parts you want in set notation?

[finite knot]

i want to write as much as possible of the exercise with set notation, set notation is really nice

[fickle sandal]

I don't think there's anything else, frankly. The only intervals you need to deal with are x:(0,1) and x:(1,inf), right?

[finite knot]

yes

[fickle sandal]

or you can say that f(x):(0,inf) for x:(0,1)U(x=1)Ux:(1,inf)

i guess

[finite knot]

.close

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[hallow mesa]

Can anyone explain this question me

[lone heart]

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
5. I have a question about someone else's worked solution
6. None of the above

[hallow mesa]

1

[alpine sable]

how many triangles do you see

[hallow mesa]

4

[alpine sable]

so what’s the answer

[hallow mesa]

I don't know 😞😕

[alpine sable]

what are your options

[hallow mesa]

Wdym

2 , 5 , 3 , 8 , none

[alpine sable]

it’s a multiple choice question pick one answer

[alpine sable]

ok

and which of these is the correct one then

[hallow mesa]

None of these ?

[hallow mesa]

I don't have the solutins

[alpine sable]

ye that’s it

[hallow mesa]

🙏🏻🙏🏻

.close

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[daring pilot]

the first part, not sure what procedure is used here

[royal meadow]

if all else fails, try drawing a graph

a quick sketch

[daring pilot]

that would work but how would you do it using algebra

[royal meadow]

if you see from the graph that it's obviously not continuous

like, there's a big jump

you can say that the limit from the left converges to one thing but the limit from the right converges to another thing

so it can't be continuous

if they converge to the same thing then it is continuous

by definition

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[wild delta]

Helloooo, I’ve done this kind of question before but I genuinely forgot how to so could someone pls help me? 😭 tysm

[royal meadow]

sine rule will do it for b

a is just sorta obvious

triangle inequality

like, no one side can be longer than the sum of the other two sides

because then it just sorta

wouldn't work

[wild delta]

Wait sorry 😭😭 I’m so sorry I just kind of don’t get it

[royal meadow]

ok

so imagine you have two sides of fixed length

ok like

two sticks

joined at a hinge

[wild delta]

Yes

[royal meadow]

and you sorta just rotate them back and forth

[wild delta]

Yes

[royal meadow]

and then the third line is like idk an elastic band

between the two ends

ok

so imagine you want to stretch the elastic band to be as long as possible

then you'd want to sorta open it as wide as possible, right

[wild delta]

Yes

[royal meadow]

ok

[wild delta]

I think

[royal meadow]

so the maximum length is clearly kinda like, as long as the two sticks in a straight line

ok i'll draw a diagram

[wild delta]

Omg ty 😭😭

[royal meadow]

basically

triangle inequality says that the elastic band can't be longer than the other two sides added up

and then reverse triangle inequality says that the elastic band can't be shorter than the difference of the other two sides

and that's basically question a

[wild delta]

Ooooh

Oooh okay tysm this makes a lot more sense now!!

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[thick lynx]

The image shows an equilateral triangle with side lengths $1$, whose vertices are midpoints of circles with radii $r > 0$. These three circles touch another circle with radius $s$. \[5pt] How many possibilities are there for the circle with radius $s$ when $r = \frac 1 2$ and $r \neq \frac 1 2$?

[thick lynx]

Does anyone have a hint?

[subtle birch]

so the bottom two circles must touch the larger externally?

[thick lynx]

Hm, that's not mentioned in the problem statement, so I assume not?
But then, shouldn't we always have two possibilities (touching internally and externally)?

[subtle birch]

for certain radii of the smaller circle, there may exist a case where internal touching yields no equilateral triangle

lemme think more

[thick lynx]

If r = 1/2, then it should look like this

I think we can't have an internal circle when the three circles cover each other in the middle

So we need to find the r for which that happens

[subtle birch]

Oh well, that was neat

Yeah that is pretty much the only case with r=1/2

[thick lynx]

With r = 1/2, we can have both an internal and external circle

Only for some r >= p > 1/2, it will only be an external one because the triangle will completely be covered, and we need to find that p

[subtle birch]

yes

[thick lynx]

Hm, I'm thinking about finding the area of the internal circle in terms of r, and then setting that equal to 0, do you think that's a good idea?

[subtle birch]

for the case where all the three circles touch the larger circle externally, you have that centre of the larger circle and centre of the circumcircle of the triangle coincide

So you can easily get a relation between s and r

[subtle birch]

Do you mean larger circle by internal circle?

[thick lynx]

I mean the smaller circle, between the three circles with radius r

[subtle birch]

This?

[thick lynx]

Yes

[subtle birch]

why do you think that would help?

[thick lynx]

I thought that if the area of that becomes 0, then beginning from that r, the three circles with radius r overlap and completely cover the triangle, so there can only be the external circle

[subtle birch]

okay, that would be a specific case

but we are supposed to work for all radii >1/2

[thick lynx]

Well, say at r = p, the three circles completely cover the triangle.
Then we conclude that for 0 < r < p, there will be two possibilities for s (internal and external) and for p <= r, there will only be one possibility for s (external)

[subtle birch]

yes. you can conclude that

well, you need not introduce another variable p. The value of p is just 1/2

[thick lynx]

No?

at r = 1/2, the triangle is still not completely covered

p will be greater than 1/2

[subtle birch]

for any r>1/2, the three circles will overlap. So there would be no chance for an internal as wll as external touch

[thick lynx]

Do you mean an internal as well as external touch at the same time?

Like for

[subtle birch]

yes

[thick lynx]

Oh

[thick lynx]

I didn't think about that case. So in total we have three cases actually

internal and external touch at the same time, only internal, only external

So there will be three cases for r < 1/2

[subtle birch]

and yeah, I understand you if the centre space left by the small circles is zero then there won't be any small circle who would be externally tangent to them

[thick lynx]

There will be two cases for some 1/2 <= r < p (only external, only internal)

and there will be only one case for p <= 1/2 (only external)

[subtle birch]

Okay, so here's my idea

Draw the circumcircle of the triangle. Label it centre as any point O. we also mark the centre of the larger circle as P
Now P and Q must lie on the same line and for some cases they may overlap.
Since we are given the length of side of equilateral triangle which is 1, you can conclude using a tip of trigonometry that the distance from centre of the circumcircle to the vertex of the triangle will always be 1/sqrt(3). Now, I'm a bit unsure how to proceed but I have a feeling that the cirumcircle and the larger circle have the same radii. It just something you may notice aswell after drawing a diagram (considering the case where the upper small circle is tangent internally while the bottom two are touching the larger circle externally)

I cam to this conclusion by observing behaviour of centre of circumcircle as the radius of smaller circle increases.
When the radius is zero, the larger circle is in fact the circumcircle of the triangle. As the radius increases, there is a down shift in the centre but it is equal to radius of the smaller circle

@thick lynx

by increasing the radii of the smaller circle, there is a shift in the circumcircle but the radius of the circumcircle remains constant

[lone heart]

@thick lynx Has your question been resolved?

[thick lynx]

Oh. I thought about this:

Now, R * (1 - sqrt(3)/2) = r, using trigonometry

Now, we should let r -> 0 and see what R we get

But that doesn't seem to work out nicely

Perhaps there is a mistake?

[subtle birch]

you should have something like R+r=1/sqrt(3)

[thick lynx]

By looking at one of the three smaller triangles formed by the larger triangle in this figure,

So cos(30°) = sqrt(3)/2 = R/(R + r)

So sqrt(3)/2 R + sqrt(3)/2 r = R

[subtle birch]

Okay, well but why are you solving for that case. Aren't you known that it is the case when R=1/2

[thick lynx]

Oh

[thick lynx]

No, I want to see what R we get when r -> 0

[subtle birch]

Okay, and I got it. The 'p' you were talking about is none other than 1/sqrt(3)

consider it the following way.
Construct circumcircle of the equilateral triangle. Label its centre as O.
Now that point would coincide with the point of intersection of all three circles. And the radius of the circumcircle (distance from centre to the vertex) would be equal to 1/sqrt(3) which the the required radius

[thick lynx]

Hm, [cos(30^\circ) = \frac{R}{R + r},] so [\frac{\sqrt 3 }{2}R + \frac{\sqrt 3}{2}r = R] and so [R\Bigl(1 - \frac{\sqrt 3}{2}\Bigr) = \frac{\sqrt 3}{2} r.]

Solving for r and letting r -> 0 makes R -> 0 for some reason

There must be a mistake

[thick lynx]

We should get R -> 1/sqrt(3) using this method too, no?

[subtle birch]

Does it help though?

Okay I get the mistake

This length is not necessarily 'R'

as the circles overlap, that distance becomes actually less than R

[thick lynx]

oh

And the circles will overlap, even when r is not 0 yet

[subtle birch]

yes

At this point all the circles have the same radii

and you know that radius of circumcircle of equilateral triangle with side length 1 is 1/sqrt(3)

[thick lynx]

I got 1/sqrt(3) using trigonometry too

Ok, so for r < 1/2, we have three circles, for 1/2 <= r < 1/sqrt(3), we have two circles and for 1/sqrt(3) <= r, we have one

[thick lynx]

Thank you!

[thick lynx]

2/3 * sqrt(3)/2, which simplifies to 1/sqrt(3), yeah

That's the other possibility

Thanks

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[dawn kettle]

hello may some of you help me

[dawn kettle]

i can send my math problem

[fickle sentinel]

go ahead

[chilly dove]

blud did not send the math problem

[near apex]

@dawn kettle Send your problem here.

[lone heart]

@dawn kettle Has your question been resolved?

[dawn kettle]

sorry i was struggling with some net problems

they said that we should demonstrate that first proposition implies the second

[near apex]

,rccw

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[minor totem]

How do you decompose this fractional expression

[vital jasper]

synthetic division to factor it?

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[unkempt vapor]

oh, Celeste player

is this trigonometry problem?

so physics?

sorry, can't help with this, i just know that it involves cos

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[thick lynx]

When solving an equation involving a root, e.g. [\sqrt{4 - x} = 8,] you need to check your solution because you squared, which is an irreversible operation. But when solving the same as an inequality, [\sqrt{4 - x} > 8,] then you don't need to check your solution, do you?

[tepid drum]

wdym by check your solution

[thick lynx]

Plug back in to the original equation

[tepid drum]

why would you have to plug

[thick lynx]

Extraneous solutions

[tepid drum]

by solving you get x=-60

no you wont

[tepid drum]

you just get this

square then subtract -4 and multiply by -1

[thick lynx]

[\sqrt{4 - x} = -8.]

If you decided to square here, you'd get the same, and that would be extraneous

[tepid drum]

no this isnt the case

its different when you are working from here and when you are working in reverse

[thick lynx]

What do you mean?

[tepid drum]

for example if you have $x^2=64$ then $x=\pm{8}$

[ocean seal]

calculus is fun

[tepid drum]

but you can start with x=8 and then squaring gives you x^2=64

[tepid drum]

you dont get into troubles of a new root here

bc you are squaring

you dont care if you are squaring

you care when you are taking sqrt of an even power

[thick lynx]

You can get extraneous roots when squaring

[half epoch]

It's not that you need to check your solution but that you are ignoring the domain constraints of sqrt when you do so

[thick lynx]

Take this as an example

[tepid drum]

i mean that you wont face this new root

[half epoch]

When squaring the left side and discarding the root, you are implicitly adding a constraint to your solution

Your squaring is correct if and only if 4 - x >= 0

[tepid drum]

bc you are just ignoring it

[subtle birch]

here you have the bound that 4-x>0. So you don't actually need to plug in the solution to check

[tepid drum]

if squaring will get you in troubles of new roots then why doesnt this root show up upon squaring $\sqrt{4-x}=8$

[ocean seal]

calculus is fun

[tepid drum]

square and you get 4-x=64 i dont see any root other than x=-60 here

[thick lynx]

Actually, even if you look at the domain, you still need to check if it satisfies it. Let me give you an example. \ [\sqrt{8 - 2x} = 1 + \sqrt{5 - x}.] The domain would be $x \leq 4 \wedge x \leq 5 = x \leq 4.$ Now, squaring twice, rearranging, you will end up wtih [x^2 = 16] and so $x_1 = 4$, $x_2 = -4$. \ Both satisfy the domain, so they are solutions, right? No.

Plug back in and you'll see x = 4 is extraneous.

[tepid drum]

this is different?

in the expression we initially had squaring led to a 1st deg pol

here you got into a 2nd deg thats why new false root

[subtle birch]

well, you still have another bound

squaring once, you get 2-x=2*sqrt(5-x)

here you must notice that 2-x must also be positive since the LHS is always positive

so 2>x. so 4 is rejected

[half epoch]

I guess that is what you mean by "checking", you also need to check the range

[tepid drum]

as long as squaring leads to 1st deg you wont face any new false roots

[subtle birch]

still you need to check the range

sqrt(-x)=-2

[tepid drum]

this needs to stop here bc sqrt(x)>=0

[subtle birch]

this is satisfied for x=-4 on squaring. So you must consider the range also

[thick lynx]

After squaring, isn't the solution set already changed, so you shouldn't make decisions based on that?

[tepid drum]

i am talking if it is defined from first

i mean if it is valid

[tepid drum]

you got an equation which cant be satisfied if the condition isnt

[subtle birch]

well, irrespective of the solution set, sqrt(anything)=y then both y and anything must be positive

so you can conclude to that

[thick lynx]

Yes, that's why we say the domain is x <= 4 and x <= 5, which means it's x <= 4.

Then the radicand is not negative

[subtle birch]

well on squaring, we have a new bound which we must also consider

since the conditions have changed on squaring, so has the bound

[thick lynx]

Well, the point is that squaring can turn something wrong (-2 = 2) into something correct (4 = 4), so we need to check our solution afterwards or sort the bad ones out some other way.
It's probably the fastest to just plug back quickly at the end

But we don't do that with inequalities that we squared, do we?

[tepid drum]

but you cant even say -2=2

you should stop here before squaring

bc this is wrong

sometimes false statements imply true statements

[thick lynx]

It doesn't have to be so obvious, it can be hidden that it's a false statement

Some equations don't have a root for example, and you set them equal to 0

That equation at the beginning would be wrong then aswell

I should move this to math discussion, I guess, maybe it fits better there

.close

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[dawn quail]

This is called implicit differentiation, right?

Why does the top answer not work?

It’s with respect to y, not r

But why am I allowed to just move that r over? Feels like I’m treating it like algebra for the second answer

[tepid drum]

bc you are differentiating wrt y not r

[coral flower]

u are differentiating the equaion
r = x^3 + t^3
d/dx (r) = d/dx (x^3 + t^3)

why is it wrt y tho , y isnt in there

[dawn quail]

The question asks for dr/dy

[tepid drum]

so d/dy=d/dx *dx/dy not (d/dx)(dx/dr)

[dawn quail]

Not d/dx(r)

[coral flower]

oh ok

[tepid drum]

be careful when using chain rule

[dawn quail]

I understand dr/dy is the same as d/dy(r) I’m just not sure why

Does anybody feel this notation is overly confusing? Or too late to change now we are stuck with it forever

[tepid drum]

so you are initially having d/dy (r) but to write it more nicely you can say dr/dy

it is same thing

just the way it looks

dr/dy is easier to write than d/dy (r)

now lets differentiate wrt y

$r=x^3+t^5\implies \frac{dr}{dy}=\frac{d}{dy}{(x^3+t^5)}$

[dawn quail]

d/dx(…) is a command. It says to find the derivative with respect to x
dy/dx is the same thing as d/dx(y)
So it does the same thing as d/dx(…)
But why do we have dy/dx if it’s just another way of writing d/dx(y)?

[ocean seal]

calculus is fun

[tepid drum]

faster and better to see

[dawn quail]

Lol…
Btw, your username is false 😂

[wheat isle]

But why would you want to write d/dx(y) over dy/dx

[dawn quail]

I dunno they mean the same thing

I think

[tepid drum]

now $\frac{d}{dy}{(x^3+t^5)}=\frac{d}{dy}{x^3}+\frac{d}{dy}{t^5}=\frac{d}{dx}{(x^3)}{\frac{dx}{dy}}+\frac{d}{dt}{(t^5)}{\frac{dt}{dy}}$

[ocean seal]

calculus is fun

[tepid drum]

derivative is distributive

and after that use chain rule

[dawn quail]

Ty

I think I understand

[tepid drum]

now $\frac{d}{dx}{x^3}=3x^2$ and $\frac{d}{dt}{t^5}=5t^4$

[ocean seal]

calculus is fun

[tepid drum]

then you are done

[dawn quail]

Ty

.close

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[wheat isle]

d*dx^-1

[tepid drum]

np have a nice day/night

[lone heart]

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[cerulean surge]

This was how far I got

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[blazing ruin]

can someone guide me through how to do both questions?

[blazing ruin]

for the first part but not sure if its correct

[tacit arch]

You should be using squeeze theorem

[blazing ruin]

i got stuck here

[tacit arch]

Divide everything by e^x

[blazing ruin]

e^x is 0 so both left and right sides are 0?

[tacit arch]

If that makes sense to you sure

[blazing ruin]

but is it really okay to divide by 0? i didnt know that was possible

[tacit arch]

Okay it doesn't make sense to you

e^x is never 0

It gets large as x goes to infinity

But it's in the denominator

[blazing ruin]

so approaching 0 instead right

[tacit arch]

Sure if that makes sense to you

[blazing ruin]

hmmm

[blazing ruin]

i somewhat have an idea but not quite sure what to write next on this

[tacit arch]

.

[blazing ruin]

can i just do this

[tacit arch]

That last line is wrong as written

The limit of both 3/e^x and -11/e^x becomes 0, but neither are actually equal to 0

You should read
https://tutorial.math.lamar.edu/classes/calci/computinglimits.aspx

[blazing ruin]

yea about that

idk how to present it

[lone heart]

@blazing ruin Has your question been resolved?

[blazing ruin]

<@&286206848099549185> can someone help me in this?

[coarse niche]

Do u need it for question a or b?

[blazing ruin]

b

[blazing ruin]

a im stuck halfway here

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[blazing ruin]

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[dusky kraken]

Have I differentiated this correctly?

[dusky kraken]

Here

[alpine sable]

yes it's good man

[uncut torrent]

id recomment using quotient rule tho

[alpine sable]

[uncut torrent]

a lot easier

[coral thorn]

why not do quotient rule, no negative powers

yeah

[dusky kraken]

That’s a great point, I forgot about that initially

[alpine sable]

i think the quotient rule is better

[dusky kraken]

.close

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[jaunty sierra]

is this correct?

[lone heart]

@jaunty sierra Has your question been resolved?

[magic arch]

problem with those is:

1. not all of us learn maths in English

!nosols

[lone heart]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[jaunty sierra]

how do i work through this?

[magic arch]

I'd love ot help but as I've already stated I don't know English Maths terminology that well

[jaunty sierra]

oh okay, np

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[worn fox]

"And"

[stable warren]

what is this sign

oh..

that makes sense

[worn fox]

$\wedge$

[ocean seal]

ΣΑCu

[stable warren]

so math made its own language?

💀

[worn fox]

It's a symbol from logic

[stable warren]

ahhhhh

thanks

♥️

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[manic jay]

What does it mean when 1 goes undee x ?

[fickle sandal]

it's a label for different x'es

[manic jay]

Different x.es ?

[fickle sandal]

i'm guessing you've got like a quadratic there or some equation

[manic jay]

Yeah

[fickle sandal]

so x_1 is being labelled as one particular solution

which is 3

maybe you've got x_2=-3

[manic jay]

Ur goddamn right

U gotta explain how you know

[fickle sandal]

so the numbers under x help identify them

because it says +-3 there

[manic jay]

I dont get it

[fickle sandal]

what about it?

[manic jay]

So

Here in the book

It says

The equation x^2 = 9 is x_1 = 3 and x_2 = -3

[fickle sandal]

so the equation has two solutions, and we can distinguish between the two solutions by labelling them?

imagine we're using a quadratic for some real purpose, where we only want the positive solutions

then i can say that i only care about x_1

and now i know where that 3 comes from because it's labelled

[manic jay]

So

All such equations has 2 answers

Both r correct

One is negative x negative = positive

Positive x positive = positive

[fickle sandal]

not all quadratics have real answers

[manic jay]

Why would we need negative x negative

It gives a positive product anyway

[fickle sandal]

what does the product of roots have to do with this

[manic jay]

Idk

U told me that

x_1 is 3

x_2 is -3

I think

@fickle sandal

[fickle sandal]

that is literally information from your book

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[little steeple]

I don’t know how to do this

[alpine sable]

B and D?

[little steeple]

Yh that’s it thank you

I’m in set 4 maths so I struggle a lot with the subject

[alpine sable]

Try to think of it as a paper cutout that you need to fold to get a cube

[little steeple]

I was trying but every time I did I would get confused and lost

[alpine sable]

Keep trying and you'll eventually understand it, ask you teacher for help as well

[little steeple]

Alright thanks

[lone heart]

@little steeple Has your question been resolved?

[little steeple]

Yes

[worn fox]

you are fine to open a new channel with a new question

just close this one first

[little steeple]

I closed the last one

Oh wait I didn’t sorry

[worn fox]

no i mean, opening a new channel is better for you as it puts your problem back on top of the list

so just close this one with .close

and open a new one with your new question

[little steeple]

K thanks

.close

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[fickle sandal]

try imagining building the net into the prism

when you roll the rectangles up, which length corresponds to the unknown

[little steeple]

Length 6?

[fickle sandal]

yeah

[little steeple]

Oh wow

Thanks

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[little steeple]

.reopen

[lone heart]

✅

[little steeple]

I got D B C but that was wrong

.reopen

[fickle sandal]

look at D carefully

it has two very narrow rectangles

but the sides you want to join it to aren't the same

[little steeple]

Oh so D is wrong

[fickle sandal]

also check E

[little steeple]

Therefore is must be A b c

E is right too

[fickle sandal]

A shouldn't be right

[little steeple]

Wait is that because two of the rectangles interfere with eachother

[fickle sandal]

yes

and it won't make a complete prism

[little steeple]

Oh k

Sooo what is it

[fickle sandal]

looks like it's BCE

[little steeple]

Ohhhhh I seee it now

[little steeple]

Wait it’s wrong tho

[fickle sandal]

oh sorry E's wrong, it's got an overlapping face

[fickle sandal]

it's got 7 faces

[little steeple]

Oh k

Now I have a new version of the question

[little steeple]

?

.close

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[little steeple]

.reopen

[lone heart]

✅

[fickle sandal]

what's the problem

[little steeple]

I don’t know how to do it

[fickle sandal]

you did it before

[lone heart]

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[tough lintel]

i ended up doing something along the lines..... of setting F_1 to be everything but with u_1 = -k, g

i then set F_2 to be everything but u_2 = -k, g

I subtract them and get zero, which shows the soluton is unique right?

and assumed that both u_1 and u_2 are both solutions to the BVP

[lone heart]

@tough lintel Has your question been resolved?

[tough lintel]

idk, am i supposed to use greens formula here???

[tough lintel]

.close

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[alpine sable]

How would i start off with this question?

I havent done this in class but ive done almsot all of the trignonometry section. I just have no clue where to start here and what this means.

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

[fickle sandal]

what don't you understand?

[alpine sable]

I dont get how to answer this question and i cant find any explanations online. I know what arcsin is and what a domain is and i know the graph of arcsin but what does xER mean and what does the function of x tending to pi/2 + arcsinx mean? How would i solve that?

[fickle sandal]

so the notation is foreign to you

the function isn't tending, it's mapping notation

it's the same as $f(x)=\frac{\pi}2+2sin^{-1}x$

[ocean seal]

chlamydia

[fickle sandal]

$x\in\mathbb{R}$

[ocean seal]

chlamydia

[fickle sandal]

means that x is (an element of) real numbers

all it is is setting up the question

[alpine sable]

oh okay i see

so do i rearrange this?

[fickle sandal]

rearrange what?

[alpine sable]

The top equation

[fickle sandal]

why would you need to rearrange it

[alpine sable]

Sorry i just read part a. just assumed it wanted me to rearrange it. I dont even know if rearranging is possible.

So part a is just to substitute x as 1/2 correct?

[fickle sandal]

but it's just asking for f(1/2)

yeah

[alpine sable]

Okay thanks so ive had a look at the other questions. Is part d asking me to rearrange then? as i was trying to before?

[fickle sandal]

yeah that'd be good

[alpine sable]

okay thanks so for part b, why isnt the range of f -Pi/2 to Pi/2? Because hasnt the arcsin graph got that range? im also assuming that by f they mean the y axis.

[fickle sandal]

arcsinx itself is -pi/2 to pi/2, but you've got some transformations on f(x) that make it different

[fickle sandal]

2arcsinx for example, has range -pi to pi

[alpine sable]

okay so then it would be -3/2Pi to 3/2Pi because 2arcsinx is -pi to pi so we add Pi/2 to both but the answer says -Pi/2 to 3/2Pi

[fickle sandal]

when you add pi/2 to the range, what happens to -pi

it doesn't turn into -3pi/2

[alpine sable]

Oh it doesnt affect it because were only adding a value

In that case, what if it was -2arcsinx + Pi/2

[fickle sandal]

range of -2arcsinx is still -pi to pi

so you get -pi/2 to 3pi/2

after adding pi/2

[alpine sable]

but thats just the same as the one above no?

[fickle sandal]

the range can be the same

what about sinx and cosx? their ranges are the same too

[alpine sable]

Oh right so what would change then?

[fickle sandal]

the arcsin shape flips

[alpine sable]

upside down or sideways?

[fickle sandal]

upside down because you use a minus

but they'll look the same either way

[alpine sable]

Okay thanks ill write this down in my notes

[alpine sable]

And for part c, am i correct to stretch the graph and move the centre point up to Pi/2?

Actually i just saw the graph but am i not allowed to sketch past like 180 degrees if you get what i mean because youre cuts off

[fickle sandal]

as in stretching arcsin?

[alpine sable]

yh by a factor of 2 right?

[fickle sandal]

yeah

[alpine sable]

Okay but am i not allowed to sketch past 180 degrees because youre one cuts off

[fickle sandal]

where is 180 degrees

[alpine sable]

I mean because it cuts off and the length of the curve respective to its bumps is usually 180 degrees. Im finding it hard to expalin

just do i need to have the graph cut off like here:

[alpine sable]

.

because my teacher also didnt finish the curves completely and i want to know if theres a way to draw these curves specifically

[fickle sandal]

yes, because otherwise it fails vertical line test

if it keeps going

[alpine sable]

Oh okay thanks for all your help

.close

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[lone heart]

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
5. I have a question about someone else's worked solution
6. None of the above

[analog robin]

4

[unkempt robin]

Huh

[gray isle]

you didn't facotrise the numerator correctly

what you have would expand out to
2x^2 + 4x

[unkempt robin]

Oh nevermind I see it now. That's an x

[analog robin]

i dont undertand

[unkempt robin]

If you expand the numerator back, you get this: $2x(x+2) = 2x^2 + 4x$

[ocean seal]

USS-Enterprise

[unkempt robin]

Which isn't equal to $4x^2 + 4x$

[ocean seal]

USS-Enterprise

[analog robin]

now do i simplify the 2x with the 2

to get x

?

[unkempt robin]

As we said, you didn't factor 2x out in the numerator correctly.

Fix that before doing anything else

[analog robin]

i do not understand do u mind showing me stepo by step

[gray isle]

identify common factors and factor them out properly, you could consider coefficients and variables separately if you want

[analog robin]

can u shbow me pkls

[gray isle]

looking at the coefficients of 4x^2 and 4x,
what's the hcf of 4 and 4

[lone heart]

@analog robin Has your question been resolved?

[analog robin]

no

[lone heart]

@analog robin Has your question been resolved?

[lone heart]

@analog robin Has your question been resolved?

[lone heart]

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[strange stratus]

i understand how to get the answer (-20)

but i dont know if i should place greater than or lower than

[wary stream]

The sign of a

If it's positive, it opens up (looks like a U)

Negative, opens down (looks like a unpside U)

[strange stratus]

so if its positive its greater than

like >

[wary stream]

So because x^2 + 6x - 11, a is positive, it opens upwards meaning it has a minimum value

[strange stratus]

OHHHH

okkkkk

thank you so much

[mild fractal]

you could also think of it this way

[wary stream]

Btw this is what I mean

[strange stratus]

yeah yeah

[mild fractal]

since its a k^2-20 type of expression where k=x-3; the minimum value of a squared term is zero, or greater than that...That means from observation you can tell that at x=-3, youre getting a minimum when you convert it to the required form (ie a(x+b)^2+c); but as you increse or decrease the value of x, regardless of sign, you find that the expression is always greater than -20

so your expression will either be -20, or greater than that

[strange stratus]

😭 i think ill stick with this method

[mild fractal]

and yes i definitely recommend understanding the geometric essence of these expressions, because when you know how the graph behaves for a general case, you can quickly substitute the variables and catch upto what the question wants you to answer

[strange stratus]

no offence but i wont be able to understand this

[mild fractal]

dont worry, keep at it and you will

[strange stratus]

thxx

thanks for both of yall

[mild fractal]

that's how i did too :>

[strange stratus]

.close

[lone heart]

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[twin tapir]

im confused on what to do here, i got x=-1.47 and x=1.13 but its wrong

[gray isle]

show work

[twin tapir]

i done a=3 b = 1 c=-5 so x=-1+- square root thingy 1 squared - 4 x 3 x-5 over 2x3

then

x=-1+- square root thingy 1+60 over 6

so x either = -1+ square root thingy 61 over 6

or same with - instead of +

then put it in calculator

[gray isle]

you didn't round properly

[acoustic basin]

'square root thingy'

[twin tapir]

how else am i meant to do the square root thingy

[gray isle]

sqrt(stuff)

[twin tapir]

oh

[acoustic basin]

Just say 'root over 61' or this

what ramonov said

Not the square root thingy' fr

[gray isle]

root over is just as bad

,calc (sqrt(61)-1)/6

[acoustic basin]

Better than the 'square root thingy' tho fr

[twin tapir]

ok so where did i go wrong then

[ocean seal]

Result:

1.1350416126511

[gray isle]

you didn't round properly

[gray isle]

both are bad enough that neither should be used. doesn't matter if one might be slightly better than the other

[twin tapir]

wait so is it not -1.47 and 1.14

ugh it still says wrong

[acoustic basin]

True

[hollow vapor]

Round the other one again

Not just the 1.14

[gray isle]

show how you're entering your answer into the system

[acoustic basin]

Pretty sure that it was rounded up correctly

[gray isle]

other value seems fine

[twin tapir]

x=-1.47, x=1.14 cant take a pic cause it only shows one of them in the pic

[hollow vapor]

Oh lol i cant read

[twin tapir]

oh wait i accidentally added x=1.14 twice thats why

it worked now

[lone heart]

@twin tapir Has your question been resolved?

[lone heart]

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[blazing ruin]

how to do?

[blazing ruin]

all ive done for part a

this is the part a ans btw

[lone heart]

@blazing ruin Has your question been resolved?

[blazing ruin]

<@&286206848099549185> help..

[zealous trench]

What exactly are you struggling with?

[blazing ruin]

thats my working for part a then idk what to do next

[zealous trench]

What course is this? How much rigour is expected?

I'd say the left limit and right limits are both 1

and x^2 - 3 is cont. everwhere

and 5/(3+x) is cont when x is not -3

hence f(x) is cont. on [-3,\infty)

[blazing ruin]

i guess you can consider this calc 1

[blazing ruin]

where do you get left and right limit?

[zealous trench]

i'm not american idk what that means

[blazing ruin]

the course is just called engineering mathematics 1

[zealous trench]

plug 2 into each part of f and you get 1

[blazing ruin]

oh like what i did?

[zealous trench]

yes

[blazing ruin]

so the only discontinuity is x=-3?

[zealous trench]

well, no

because at x = -3 you are using the x^2 - 3 part

the points is that each part of f is cont. on it's interval

and at the potential problem point (the point where you chaneg how f is defined) has the same left and right limit

so it is cont. on [-3,infinity)

[blazing ruin]

can you rephrase that cause i dont quite get it @.@

[zealous trench]

so the graph looks something like that

okay

[blazing ruin]

yes

ohhhhhhhh

[zealous trench]

do you get it?

[blazing ruin]

yes

[zealous trench]

cool, sometimes it just helps to draw these things

[blazing ruin]

what about the other two parts? if you dont mind explaining

[zealous trench]

here is a graph

there is a "hole" at x = -2

[blazing ruin]

is it cause x is supposed to be 0?

[zealous trench]

that is where (x^2-4)/(x+2) is undefined

[blazing ruin]

oh

[zealous trench]

plug in -2 there and you get a zero in the denominatoe

[blazing ruin]

is it cause you can cancel out

[zealous trench]

it should actually look like that

where the black circle is the f(0) = 2

[zealous trench]

The fraction simplifies yes but you need to take the question as it is given

it simplifies to x-2 but that is not the same as (x^2-4)/(x+2) in this context

[blazing ruin]

but aside from all that, how are you able to sketch out the graph just from looking at the functions

[blazing ruin]

this is from the middle 2, x=0 right

[zealous trench]

yes

[zealous trench]

what do you mean? It is just a rough sketch. Do you know what the general shape of these functiosns are?

[blazing ruin]

um

[zealous trench]

the first fraction simplifies to x -2 which is a straight line so it is easy to sketch

[blazing ruin]

polynomials without dividing anything are just curves and if its a polynomial over something its an asymptote

thats all i know

[zealous trench]

the asymptote is clearly at x = 4 because that is where you get a zero in the denominator

[blazing ruin]

oh

[zealous trench]

maybe you need some practice in what graphs of the form $\frac{a}{x-b}$ generally look like?

[blazing ruin]

no

[ocean seal]

Scerball

[zealous trench]

?

[blazing ruin]

i cant picture that

[zealous trench]

They are called reciprocal graphs

[zealous trench]

maybe practice them?

[blazing ruin]

okay i will

but for now is there a way to do these questions without draw them out

cause my prof just works around with the inequalities

[zealous trench]

I mean you are searching for discontinuities

so for part c) find the places where the denominator of the fraction is 0

[blazing ruin]

can i do that for all questions?

finding the values where denominator becomes 0

i got 0, 2 and -1

from factorising the denominator

[zealous trench]

yeah

so it is cont. everywhere else

other than those points

idk why you'd try to do this without knowing what the fucntions look like

[blazing ruin]

ill try to brush up on my graph imagining skills

[blazing ruin]

but for now its okay to just do this and write the set notation that avoids those values right

[zealous trench]

yeah

[blazing ruin]

nice

thank you so much im enlightened now

ill be closing this then, cya around :)

.close

[lone heart]

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[gray jetty]

can someone explain this to me, i didnt get this question correct but would still like to understand the solution

[pale kestrel]

explain which parts didnt make sense.

or what of the solution

[gray jetty]

i dont know how to approach solving this correctly

what i did was like substitute in the compositions if this makes sense

[gray jetty]

so g composed of f should be R -> R

because f inputs r and G outputs R

[pale kestrel]

ok now i look at this, no way that is correct wtf

[gray jetty]

huh?

[pale kestrel]

this isnt right

is that the solutions they told u

[gray jetty]

no

thats what i put

[pale kestrel]

right ok

so

f o g means

Do g

then

Do f

[gray jetty]

im trying to figure out the solution

so

[pale kestrel]

(f o g)(x) = f(g(x))

so f o g would be

Do g then Do f
[0, inf) -> R -> [0, inf)

[gray jetty]

yeah i see why that is incorrect now

[pale kestrel]

Right, then check the rest of the options to see which is valid

[gray jetty]

i dont think the second one is true either

[pale kestrel]

you cant guess, you have to compute.

[lone heart]

@gray jetty Has your question been resolved?

[lone heart]

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[tough dawn]

Just want to make sure I do these right. Like for #11 would I just say shifted 43 units to the left?

[glacial crest]

Yep, sounds like that's what the question is asking for

[tough dawn]

Ok thought so thank you

.close

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[kind patio]

i understand most of this

[kind patio]

I’m just confused for when solving for B how they figured x=2

Already claimed

[azure turret]

Settle down we sent it at the same time…

[kind patio]

just letting yk my bad

[wary stream]

My guess, because of the (x - 1) part, when you make x = 2, you just have (2 - 1) which makes life easier

[kind patio]

i’m confused how they or where they got x=2 though in general

[wary stream]

I told you how

because of the (x - 1) part, when you make x = 2, you just have (2 - 1)

If you made x = 1232, you have 1232 - 1, or 1231 and you don't want to deal with big numbers

You can make x anything you want, because you already know what A and C are

So the easiest value of x, would be x = 2

[kind patio]

The way they found A and C was setting it equal to the x-1 (so x=1)or in C’s case since there was no 3rd factor setting X=0 but

Bx-1=x^2+9 doesn’t give u 8

or in this case 2*

[wary stream]

And as mentioned, you already found A and C, so you easily plug that into the equation, and all you have left is B and x, but you're solving for B, so you can just plug in any x value you want

They picked x = 2 because that's the easiest

[kind patio]

you are allowed to just do that’s

?

how come with the A, and C there was a reason they had to plug in x=1 or 0 but for this one there is no reason other than whatever number you like

[wary stream]

Because you wanted to eliminate the other letters so you just have one unknown

$A(x^2 + 9) + (Bx + C)(x - 1) = 10$

[ocean seal]

dldh06

[wary stream]

That's the equation

You have 3 variables to solve for, A, B, C and one equation

One of the ways to eliminate all the other variables, is to utilize what x can be

[kind patio]

wait wait

oh nvm sorry continue

[wary stream]

To solve for A, you want B and C to be eliminated, so to eliminate B and C, if you do x = 1, then you have $A(1^2 + 9) + (B * 1 + C)(1 - 1) = 10 \to A(1^2 + 9) + (B * 1 + C) * 0 = 10$ then anything times by 0 is 0 so B and C are eliminated so you are left with $A(1^2 + 9) = 10$

[ocean seal]

dldh06

[kind patio]

so they simply made x=2 because it was easier to compute

[wary stream]

Yes

[wary stream]

Do you understand up to here?

When you do all that, you get A = 1

Since A = 1, you can just plug that into the equation and leave it be $A(x^2 + 9) + (B * x + C)(x - 1) = 10 \to 1(x^2 + 9) + (B * x+ C)(x - 1) = 10$

Now all you have left is B and C to deal with, and you want to eliminate one of the two, the easiest is eliminating B, by making x = 0

[ocean seal]

dldh06

[wary stream]

So if x = 0 then $1(x^2 + 9) + (B * x + C)(x - 1) = 10 \to 1(0^2 + 9) + (B * 0 + C)(0 - 1) = 10$

B is now gone

[ocean seal]

dldh06

[wary stream]

B is now gone so you have $1(0^2 + 9) + (C)(0 - 1) = 10$, you can solve for C

[ocean seal]

dldh06

[wary stream]

And C = -1

[kind patio]

i got those 2