/sci/ - Science & Math » Thread #12239949

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Anonymous Fri 16 Oct 21:56:22 2020 No.12239949 View Reply Original Report

Quoted By: >>12240291 >>12240467 >>12240473 >>12240479 >>12240590 >>12240598 >>12241278 >>12241444 >>12241541 >>12241744

let's see how smart you guys are.

Is f(x) = pic related defined in the interval x = -5 -> 5?

Motivate your answer.

Anonymous

Anonymous Fri 16 Oct 2020 22:54:21 No.12240291 Report

Quoted By: >>12240415

>>12239949
In the Reals? No. Unless I'm being a midwit rn which wouldn't surprise me. I'm not the best at math.

Anonymous

Anonymous Fri 16 Oct 2020 23:23:57 No.12240403 Report

Quoted By:

homework thread

Anonymous

Anonymous Fri 16 Oct 2020 23:24:04 No.12240404 Report

Quoted By:

This your grade 6 homework, OP?

Anonymous

Anonymous Fri 16 Oct 2020 23:27:07 No.12240415 Report

Quoted By:

>>12240291
Yes. Why wouldn't it be?

Anonymous

Anonymous Fri 16 Oct 2020 23:34:24 No.12240435 Report

Quoted By: >>12241188 >>12241411

is the square root function treated to be C —> C?
if so, even if f is R —> R, the function is defined for all real values.
if the square root function is R —> R, then the function isn’t defined for x<1

Anonymous

Anonymous Fri 16 Oct 2020 23:47:40 No.12240467 Report

Quoted By: >>12240475 >>12240478

>>12239949
Undefined for x=1, otherwise defined

Anonymous

Anonymous Fri 16 Oct 2020 23:54:52 No.12240473 Report

Quoted By:

>>12239949
yes because i define it that way.
what are you gonna do about it now bitch?

Anonymous

Anonymous Fri 16 Oct 2020 23:55:58 No.12240475 Report

Quoted By: >>12240588

>>12240467
How the fuck is it undefined for x = 1?

Anonymous

Anonymous Fri 16 Oct 2020 23:57:20 No.12240478 Report

Quoted By:

>>12240467
retard

Anonymous

Anonymous Fri 16 Oct 2020 23:57:25 No.12240479 Report

Quoted By:

>>12239949
Do you not know interval notation?

Anonymous

Anonymous Sat 17 Oct 2020 00:45:07 No.12240588 Report

Quoted By: >>12240596

>>12240475
sqrt 0 is undefined

Anonymous

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Anonymous Sat 17 Oct 2020 00:46:41 No.12240590 Report

Quoted By:

>>12239949
no

Anonymous

Anonymous Sat 17 Oct 2020 00:48:56 No.12240596 Report

Quoted By: >>12240599

>>12240588
*multiplies 0 by 0*

heh, nothing personal kiddo

Anonymous

Anonymous Sat 17 Oct 2020 00:51:04 No.12240598 Report

Quoted By:

>>12239949
Yes.
if the argument of the square root is negative, it will become complex. however, the absolute value will return the magnitude which is real

Anonymous

Anonymous Sat 17 Oct 2020 00:51:27 No.12240599 Report

Quoted By: >>12240600 >>12240609 >>12241210 >>12241293 >>12241651

>>12240596
sqrt has two solutions for all x=0, therefore it's undefined

Anonymous

Anonymous Sat 17 Oct 2020 00:52:29 No.12240600 Report

Quoted By:

>>12240599
I meant x =/= 0

Anonymous

Anonymous Sat 17 Oct 2020 00:55:04 No.12240609 Report

Quoted By:

>>12240599
Lmao

Anonymous

Anonymous Sat 17 Oct 2020 05:07:11 No.12241184 Report

Quoted By:

you mean in the intervall [-5..5] I see negative 5 greater than or equal to (-5 -> 5) in the question, but yea its defined.

I assume this is a proof of course, so prove that f(x) = ABS[sqrt(x-1)+5]

is the absolute value of sqrt(-1) also known as 'i', defined? what is the absolute value of a negative square? that is a fun thought

Anonymous

Anonymous Sat 17 Oct 2020 05:10:43 No.12241188 Report

Quoted By:

>>12240435

Anonymous

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Anonymous Sat 17 Oct 2020 05:21:31 No.12241210 Report

Quoted By:

>>12240599
>for all x=0

Anonymous

Anonymous Sat 17 Oct 2020 06:05:42 No.12241278 Report

Quoted By:

>>12239949
fuck off with your Zoom midterm

Anonymous

Anonymous Sat 17 Oct 2020 06:12:09 No.12241293 Report

Quoted By:

>>12240599
>function has two values
lolno
$\displaystyle<br />\sqrt {x^2} \ne \pm x, \quad \sqrt {x^2} = \left | x \right |<br />$

Anonymous

Anonymous Sat 17 Oct 2020 07:26:58 No.12241411 Report

Quoted By:

>>12240435
why? the function will return reals even if its inner workings use complex numbers, but inner workings should not matter if the function returns values as it should in the given interval. x + i^2 = x - 1 but that does not mean that f(x) = x - 1 is complex

Anonymous

Anonymous Sat 17 Oct 2020 07:55:15 No.12241444 Report

Quoted By:

>>12239949
yes by construction

Anonymous

Anonymous Sat 17 Oct 2020 09:04:24 No.12241541 Report

Quoted By:

>>12239949
What is your question?

Anonymous

Anonymous Sat 17 Oct 2020 09:21:20 No.12241571 Report

Quoted By: >>12241704

Define $a: \mathbb{R} \to \mathbb{C}$ and $b: \mathbb{C} \to \mathbb{R}$ such that $a(x)= \sqrt{x-1} + 5$ and $b(x)=|x|$ . Then $f = b \circ a$ and $f: \mathbb{R} \to \mathbb{R}$ . So yes.

Anonymous

Anonymous Sat 17 Oct 2020 10:20:20 No.12241651 Report

Quoted By:

>>12240599
>undefined
*Indeterminate. Perfectly well-defined

Anonymous

Anonymous Sat 17 Oct 2020 11:06:22 No.12241704 Report

Quoted By:

>>12241571
This.

Anonymous

Anonymous Sat 17 Oct 2020 11:38:08 No.12241732 Report

Quoted By:

Correct me if i am wrong .
Say integral(f'(x))dx was defined from -5 to 5; after taking integral function became |f(x)|mapping from -5 to 5 and assume the C to be 0.
I got f(x) to be 2 + sqrt(6)

Anonymous

Anonymous Sat 17 Oct 2020 11:45:42 No.12241744 Report

Quoted By:

>>12239949
>Motivate your answer
Go on, answer, you can do it!

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