Basic Calculus isn't basic at all

[fervent marten]

I don't know how to start, anyone enlighten me?

[gleaming smelt]

1. Do not ping the Moderators, unless someone is breaking the rules.
2. Do not ping the Helper Moderators, unless there is a conflict between helpers.
3. Do not ping other members randomly for help.
4. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
5. Wait patiently for a helper to come along.
6. If the Helper has answered your question, remember to thank them with the Mathematics Ranks bot and close the thread with:

+close
Feel free to nominate the person for helper of the week in #helper-nominations
If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[obsidian violet]

It gives you the perimeter

And one of the lengths

Figure out how to express the other length in terms of x

[fervent marten]

So is 500 the perimeter and the other length is 250?

That's my guess

[obsidian violet]

How

[fervent marten]

Well, since it says the garden is 500m of fencing materials, I just assumed that it's the perimeter, and then since the perimeter is P=2l+2w, I just kinda assumed that the length and width are both 125, and since it's 2l and 2w, I just multiplied them to 2

It literally doesn't make any sense at all

Help😭😭😭😭

[obsidian violet]

You can't just assume that it's 125

You know that P = 2l + 2w

You can substitute X into it

[fervent marten]

So P=x+2w?

[obsidian violet]

2

2x

[fervent marten]

Aight, then?

[obsidian violet]

Well you know what P is?

You're trying to find w

In terms of x

[fervent marten]

So that's gonna be P-2x=2w?

So 500-2x=2w?

[obsidian violet]

Yes

[fervent marten]

Then divide both sides so it's gonna be 250-x=w?

[obsidian violet]

Yes

And now find the area

[fervent marten]

Area is l×w so since l is x, l=250-x

A=250-x(w)

What?

Doesn't seem to make sense

[obsidian violet]

What

250-x is w

????

[fervent marten]

Oh yeah

So A=x(x-250)

[obsidian violet]

Yes

[fervent marten]

A=x²-250x?

[obsidian violet]

Yes

[fervent marten]

Ahhhh

So how the b?

[obsidian violet]

Well the area can't be negative can it?

[fervent marten]

What does that mean

...

[obsidian violet]

It means what X values can it have

[fervent marten]

A=250x-x²?

What

[obsidian violet]

what

Just find when the function isn't negative

[fervent marten]

Any number greater that 0?

[obsidian violet]

Area greater than 0

You can think of it as a quadratic inequality

[fervent marten]

Wait

[fervent marten]

Isn't this supposed to be the correct one instead of A=x²-250x?

[obsidian violet]

yes

wait no

it's x(250-x)

so it is 250x - x^2

[fervent marten]

Yeah, that's what I'm saying

So when x is 250, the area would be 0

So x should be less than 250 ig?

[obsidian violet]

yes but there's another condition to satisfy

[fervent marten]

And thats?

For the area to be not negative nor 0, x should be -250<x<250

Correct?

[obsidian violet]

no

[fervent marten]

How?

[obsidian violet]

the other root is x = 0

[fervent marten]

0<x<250?

[obsidian violet]

yes

[fervent marten]

That makes more sense

So on the part "show that it is continuous on its domain", what am I supposed to say?

"the function is continuous on the interval (0, 250) because it doesn't equate to 0 nor a negative integer"?

[obsidian violet]

no

continuous means it doesn't just take on discrete values

you've basically already shown it's continuous

it can take on any real value between 0 and 250

[fervent marten]

So what am I supposed to say then?

[fervent marten]

If I leave it blank, my instructor may mark the answer as incorrect for not answering one of the questions

I mean if I didnt put any explanations why it is continuous

[unique brook]

Well, you can assert that a general polynomial is continuous on it's domain by stating that it does not have any asymptotes, jumps, or holes within it's domain in the reals, therefore this specific functions is continuous.

[fervent marten]

Got it, thanks

Another one...

Is the answer k?

[unique brook]

Show your work.

[fervent marten]

Is there supposed to be? I mean it says direct substitution so by looking at it, k will be the answer when x=-3

[unique brook]

But it says $\lim_{x\to c}$.

[thin sail]

Kocher

[fervent marten]

So what am I supposed to do then?

[unique brook]

Break it into cases; when x is not -3, and when x is -3.

[fervent marten]

What?

[fervent marten]

What if I just factor x²-9/x+3?

That'll give me -6 when I substitute x with -3

[unique brook]

Yes. Which is your k-value.

[fervent marten]

What?

[fervent marten]

So basically, it's -6?

I've been studying this subject for the past 7 hours

[fervent marten]

So the answer is -6?

Or is it k?

I'm going insane

Or maybe I'm just dumb

[unique brook]

Do you know how continuity works?

[fervent marten]

Well, yeah, ig

[fervent marten]

Wym?

[unique brook]

Define it.

[fervent marten]

There's 3 definitions

1. F(c) exists

2. Limit of the function exists

3. F(c) and the limit is equal

[fervent marten]

.

[unique brook]

Yes

[fervent marten]

So f(c) is k right?

[unique brook]

Yes

Well not necessarily, only at x=-3.

[fervent marten]

And I'm pretty sure the limit does not exist

[unique brook]

It does exist.

[fervent marten]

How

[unique brook]

If you set $k=\lim_{x\to -3}\frac{x^2-9}{x+3}$

[thin sail]

Kocher

[fervent marten]

There's no from the left or from the right limit or something

[unique brook]

Else the function isn't continouous

[fervent marten]

Still don't get it

[unique brook]

There is

$\lim_{x\to -3^-}\frac{x^2-9}{x+3}$ and $\lim_{x\to -3^+}\frac{x^2-9}{x+3}$

[thin sail]

Kocher

[fervent marten]

So the limit does exist

[fervent marten]

So just like you said earlier, -6 is my k, and the limit is also equal to -6 so basically the answer is -6?

Correct?

[unique brook]

As x->-3, yes.

[fervent marten]

Here, another one, same concept different situation,

f(c) is 8

and limit from the right is 2

So the answer would be DNE no?

[unique brook]

No

[fervent marten]

Howww

[unique brook]

It asks only for the limit from the right.

[fervent marten]

Yeah, and the limit from the right is 2

Which is not equal to 8

[fervent marten]

.

[unique brook]

The overall limit does not matter here.

[fervent marten]

What am I supposed to do then?

[unique brook]

Use only the right hand limit.

[fervent marten]

Yes, which is 2

Wait, is the answer 2?

[unique brook]

👍

[fervent marten]

Omg

Thanks