Calculus Help Needed for 5 Multiple Choice Questions.

[opaque salmon]

Could someone help me with 5 multiple choice calculus questions covering IVT, EVT, DERIVATIVES.

[cosmic shadow]

Is this a test again?

Do you know the answer on the question from yesterday?

[opaque salmon]

They will release it tomorrow.

[cosmic shadow]

Oh ok

What are your calculus questions?

[opaque salmon]

Assignment

Are you better in calculus than linear algebra?

[cosmic shadow]

Yes ;p

Lol

[opaque salmon]

Lol

This is first one:

[cosmic shadow]

So you need the IVT

Have you drawn a graph?

[opaque salmon]

I can draw a graph right now

One sec

[cosmic shadow]

Are all your assignment on some website...? Don't have to hand-in your reasoning?

[opaque salmon]

No reasoning for these lol. Just want answer.

[cosmic shadow]

Are they graded?

[opaque salmon]

Ye

[cosmic shadow]

Especially weird for that question yesterday, would be multiple points back when I was in school/college

[opaque salmon]

Yes it's very sad

Was 7 marks and I got 0 lol

But it happens

[cosmic shadow]

Wait what, that one question?

[opaque salmon]

Yes

[cosmic shadow]

Oh

How much did you get correct in the complete assignment?

[opaque salmon]

20/27

[cosmic shadow]

Also could you find something in the book about projections on a plane?

I think what we missed was the orthogonal basis vectors of the plane

[opaque salmon]

Yeah

I will find after this calculus lol

[cosmic shadow]

Ok

So can you explain the IVT?

[opaque salmon]

ye

If we assume that f is continous on a closed interval [a,b] and either f(a) < x < f(b) or f(a) > x > f(b).
Then there exists a c E (a,b) such that f(c) = x

[maiden matrix]

this picture might help

[cosmic shadow]

So what can you reason about the intervals in the answers?

[maiden matrix]

it's impossible for a continuous function to connect the two blue lines without crossing 0; or really, any other value between f(a) and f(b)

[opaque salmon]

So would it just be the third option

And the last option as well be that interval contains the third option interval?

nvm, it's only option 3 cuz f(-2) = -8 < 0 < f(1) = 1?

[cosmic shadow]

3, is correct

[maiden matrix]

Look closer at option 2

[cosmic shadow]

For the forth you don't know what f(9/2) is

Have you drawn it yet?

[opaque salmon]

Option 2 doesn't work? f(-1) = -9 < 0 > f(3) = -1

[cosmic shadow]

First look at every interval given

If you didn't know anything about f(1), that would be true

[opaque salmon]

Is option 2 correct because (-1,3) contains (-1,1) which works?

[cosmic shadow]

Yes

[maiden matrix]

Exactly

[opaque salmon]

Ok so only option 2 and 3 r correct?

[cosmic shadow]

But really draw it, or write down the intervals. That formatting in the question is really confusing.

[maiden matrix]

Yep. I agree with Lethe though, drawing the points and drawing function lines is really going to help you understand what's happening

[cosmic shadow]

Does (-2,-1) or (9/2, 6) contain any interval which contains f(x) = 0?

[opaque salmon]

No

So, then only option 2 and 3 r right?

[cosmic shadow]

Yes

[maiden matrix]

yes

[opaque salmon]

This one I'm not so sure about

[maiden matrix]

what kind of functions do these theorems apply to?

[opaque salmon]

Continuous

[cosmic shadow]

Little bit more?

[opaque salmon]

Continuous on a closed interval [a,b]

[cosmic shadow]

Right

So you need to check if they are continous in those intervals

[opaque salmon]

Yeah that's the part I'm not so sure how to do. What is the best approach?

[cosmic shadow]

And one answer you can eliminate even faster

What does a closed interval mean?

[opaque salmon]

Lol

Ok so option 4 won't work

[cosmic shadow]

Ok so the first question, is sin x continuous?

And what about cos(x) -1¿

Wait is that a x < pi, and not a x <= pi...?

[opaque salmon]

If a function is piece wise can it still be continuous?

[cosmic shadow]

When is a function locally continuous on an interval?

[opaque salmon]

If the limit of f(x) exists and the limit of f(x) = f(a)

This is the graph for the first one

[cosmic shadow]

Yes so can a piecewise function be continous?

[opaque salmon]

Yes

[cosmic shadow]

But i think this is a trick answer

[opaque salmon]

Based off the pic above, that first option is continuous?

[cosmic shadow]

Take a look at the interval, is it closed?

[opaque salmon]

Yes

[cosmic shadow]

The functions domain?

But maybe this is a typo...

@maiden matrix what do you think?

[opaque salmon]

Yeah it's [-pi,pi]

[cosmic shadow]

The function isn't defined for x = pi

[opaque salmon]

It's 0

When x = pi?

[opaque salmon]

Ah

[maiden matrix]

first option? it's defined at pi

oh! no it isn't!

[cosmic shadow]

Trick answer...

[maiden matrix]

yeah thats mean if its intended

[cosmic shadow]

And if it's a typo the question is wrong ;p

[opaque salmon]

Well, option 2 is correct right?

[cosmic shadow]

Have you drawn it?

[opaque salmon]

Yes

Here:

[cosmic shadow]

Does the limit exist?

[opaque salmon]

No

[cosmic shadow]

Is it continous?

[opaque salmon]

Oh wow

So, the only correct answer is option 3?

[cosmic shadow]

Well is it continous?

[opaque salmon]

Between [-1,1] yes

[cosmic shadow]

Good

[opaque salmon]

So, only option 3 is right here?

[maiden matrix]

correct

[cosmic shadow]

When is a function differentiable?

[opaque salmon]

If the derivative exists at each point in the domain

But how do I check that?

[cosmic shadow]

Well now you only need to check if it's differentiable at x = 0

[opaque salmon]

So if f'(0) exists?

[cosmic shadow]

Yes

Although

Well yes

[maiden matrix]

for the piecewise functions, f' has to be continuous at 0. so if you switch functions partway through, then their slopes need to match up

right?

[cosmic shadow]

Formally the limit needs to exists

[opaque salmon]

Hmmm

So

[cosmic shadow]

Not really sure about it, let me check my book ;p

[maiden matrix]

two-sided limit

[opaque salmon]

For the second one

Second option

f'(0) does not exist

So option 2 is wrong right?

[maiden matrix]

yes

[opaque salmon]

For the last option

f'(0)= 0

[cosmic shadow]

Yes two isn't even continous at x = 0

[opaque salmon]

So last option is correct?

[maiden matrix]

the absolute value breaks the derivative farther away. but it doesn't have any effect near 0; basically, as far as x=0 for this function and its derivatives is concerned, |cos| = cos

[opaque salmon]

Ok

So then option 4 is correctM

?*

[cosmic shadow]

Yes

[maiden matrix]

ye

[opaque salmon]

Why is option 1 incorrect?

[maiden matrix]

if you plug in 0 for both functions, then you get 0, so it's continuous; but the derivatives aren't the same, so it isn't smooth, so the derivative has a hole

[cosmic shadow]

What's the derivative of sqrt x at x=0?

The limit in this case, from the right side

[opaque salmon]

It's 0

[cosmic shadow]

What is the derivate of sqrt x?

[opaque salmon]

x^2

[cosmic shadow]

No

[opaque salmon]

No no

Sorry

[cosmic shadow]

Square root, not squared

[opaque salmon]

Ah ok

Ye ye

it ends up being undefined

Ok

So now last option to check is option 3

[cosmic shadow]

Ok

Show your reasoning

[opaque salmon]

Well

For option 3

Is x>=0 f'(0) is 1

And for x<0

f'(0) is 0

So option 3 is also wrong?

[cosmic shadow]

Can you show both derivatives?

[opaque salmon]

2x +1

2x

Those r the derivatives

[cosmic shadow]

Ok

Correct

So f'(0) = 1, while limit from x to 0- is 0

[opaque salmon]

So only option 4 is the correct option for this question?

[cosmic shadow]

Yes

[opaque salmon]

Hmm ok

I will go with option 4

Bruh

[cosmic shadow]

?

[opaque salmon]

I think we missed an option

It said it was wrong

Lol

Option 1, 2, 3 are all wrong like we said right?

[cosmic shadow]

Well for the first the limit does not exist

Second isn't even continous

Third the derivatives are not equal

[opaque salmon]

Wait

It's reloading

Ok we were right lol

It was glitching

[cosmic shadow]

Lol

Was looking at the fourth option, but that is continous and differentiable at x = 0

[opaque salmon]

yea

[cosmic shadow]

Got another question?

[opaque salmon]

Just 2 more

[cosmic shadow]

So what is the definition of continuity?

[opaque salmon]

Ok well

Option 3 is incorrect right?

[cosmic shadow]

Yes

[opaque salmon]

Because continuity does not imply differentiability

[cosmic shadow]

Correct

[opaque salmon]

I'm trying to find definition of continuity

Can't find it

[cosmic shadow]

I think you have said it earlier

[opaque salmon]

Ok so option 1 is incorrect?

All of the options are false??

[cosmic shadow]

Have you found the definition?

Can you find a counter example to all?

[opaque salmon]

Well the definition says

Limit as x -> a f(x) = f(a)

Here it just says f(a) exists

Now that it's equal to f(x)

[cosmic shadow]

Correct

[opaque salmon]

So option 1 is wrong?

[cosmic shadow]

Yes

[opaque salmon]

Well

For option 2

A piecewise can be continuous but on a certain interval

So I'm confused

Cuz question doesn't say anything about intervals

[cosmic shadow]

Well you can take the interval the whole domain

Let's say f(x) = { 0, for x < 0 ; x , for x >= 0 }

Is that continuous?

[opaque salmon]

Yes

[cosmic shadow]

So there you have an example

[opaque salmon]

So option 2 is false

[cosmic shadow]

Yws

[opaque salmon]

And option 4 is also false

Piecewise can be continous

But not differtiable

[cosmic shadow]

It can be differentiable

[opaque salmon]

Rlly?

What's the example?

[cosmic shadow]

Like the |cos(x)|

[opaque salmon]

That is not pieswise tho?

[cosmic shadow]

Ho, well it's not continous on the whole domain

But just at x =0

[opaque salmon]

So

[cosmic shadow]

But yes you can have differentiable piecewise functions

[opaque salmon]

So

The answer is

None of the above?

[cosmic shadow]

Yes

[opaque salmon]

Ok last question

This one I will need to do quickly lol

[cosmic shadow]

Ok, try it

[opaque salmon]

Ok last option

Is false

Right?

[cosmic shadow]

Why is that?

[opaque salmon]

Approaching from right side of 2 gives 1/t

Left side gives 4t-6

Can't be same

[cosmic shadow]

Ok

[opaque salmon]

Am I right?

[cosmic shadow]

Yes

[opaque salmon]

How. Do I check option 1?

I have 10 min to do this question lol

[cosmic shadow]

So what can you say about option 1 and 2 with your answer for 4?

[opaque salmon]

It can be continuous

But how would we check?

[cosmic shadow]

Can it?

Whats the definition of continuity?

f(2) = lim t to 2 f(t) = lim t to 2- f(t) = Lim t to 2+ f(t)

Well is it?

[opaque salmon]

Oh

So it's not continuous on entire domain?

[cosmic shadow]

Correct

And not at t=2

[opaque salmon]

How do I check option 1?

[cosmic shadow]

Can a function that is not continuous be differentiable? (At some x)

[opaque salmon]

No

So option 1 also wrong?

[cosmic shadow]

Yes

[opaque salmon]

Ok so option 1, 2 and 4 r wrong

Now we check 3 quickly

Idk how to check 3 tbh

[cosmic shadow]

Well what do you know about the limit at t = 2? Does it exist?

See 4th option

[opaque salmon]

It doesn't exist

[cosmic shadow]

Correct

[opaque salmon]

So

Answer is

None of the above?

[cosmic shadow]

Yes

[opaque salmon]

Ya ya makes sense

Ok thank you so much!

I appreciate all the help!

You are amazing! 😘

[cosmic shadow]

Was easier than that plane thing...

[opaque salmon]

That plane question, made me 🥵🫠😵

[cosmic shadow]

So you got linear algebra and calculus at the same time?

[opaque salmon]

Yes I am in math program in university 😅

Always 3 math courses at same time

[cosmic shadow]

Right

[opaque salmon]

Yeah, thanks again.

[cosmic shadow]

Hm my linear algebra book saus something about orthogonality and projections...

Totally forgot that

[cosmic shadow]

No problem

[opaque salmon]

Ah I see

[cosmic shadow]

But it's like chapter 5...

Of 7

[opaque salmon]

Oh ok

[cosmic shadow]

Something about least square problem and projections...

Can't remember anything about this

[opaque salmon]

Yeah it's ok. I will let u know when I get the answer to that one.

[cosmic shadow]

Ok

I should sleep, can't read anymore

[opaque salmon]

Good night! 😴

[cosmic shadow]

Have a good night