Hey Veeky Forums trying to get back into maths and need some help:

Hey Veeky Forums trying to get back into maths and need some help:
What's the domain of this function?
Two possible solutions come to mind, but I have no idea wether any of them are right/useable.
Any help would be very much appreciated!

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wolframalpha.com/input/?i=Solve p=1-sqrt((2f+u)/(5f-v)) , f
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No. Go away.

Factor the polynomial and see for which values of x it becomes positive. -(x-a)(x-b)
+*+=+
-*-=+
+*-=-
-*+=-

Why no help? Did I break any rules?

read the sticky underage retard

Yes you faggot.
This is clearly homework. You can't ask for help for homework, except maybe in /sqt/, if it is clear that you are struggling and you are polite.

I'm 26 years old and working, I want to study but I'm missing tons of school maths.
This is what I've learned over the last week w/ khan academy.
The question is NOT homework b/c I'm not in school anymore.
I just want to understand how to do this, and the "Find a domain of a function"-chapter of khanacademy isn't much help.
"This is clearly math homework" my ass.

this is middle school shit, just calculate the roots of the polynom and since a is negative, the polynom is negative "outside" of the roots. you'll get x€[-1;3]

...

Can you elaborate? I know this may seem ridicoulusly easy for you, but I would appreciate it a lot.
I get that "-x^2 + 2x + 3" has to be bigger or equal to zero, and I can solve this equation for -1 and 3 (see the crossed part in my original picture), but how do I know, that the range from -1 to 3 is the only applicable one?

Technically, the domain could be the entire real line or even more, but you probably want y to be real, so you need to have the term inside the square root be positive. So

[math] x^2 -2x -3 \leq 0 [/math]
[math] x^2 -2x + 1 - 4 \leq 0 [/math]
[math] (x - 1)^2 \leq 4 [/math]
[math] -2 \leq x - 1 \leq 2 [/math]
[math] -1 \leq x \leq 3 [/math]

Because the function is only positive between -1 and 3 (see pic related). It's a rule to follow, the polynom is the sign of the coefficient a outside of the roots. Sorry if i'm not very clear, i'm not a native english speaker

Oh, great. That helps a lot, thank you!

Sooooo, that also really helps, but I'm confused. Shouldn't it be "x^2 - 2x - 3" bigger than/equal to 0 instead of smaller than/equal to 0?

Multiplying by -1 turns around the inequality, just like [math]3\geq 0[\math] is equivalent to [math]-3\leq 0[\math]

You have the negative of that term within the square root. -term >= 0 is the same as term

I mean:
Multiplying by -1 turns around the inequality, just like [math]3\geq 0[/math] is equivalent to [math]-3\leq 0[/math]

Cool! Thanks a lot, guys!

Yeah whatever.
Did you understand this: ?

Nah... I mean I get that +*+ = + etc., but I don't really understand how it's related to my question.
I got the other two answers though.
and

A polynomial [math] f(x) [/math] has root [math] r [/math] if and only if [math] (x-r) [/math] divides [math] f(x) [/math] , i.e. [math] f(x)=(x-r)g(x) [/math] for some other polynomial [math] g(x) [/math] .

Your polynomial has two roots: [math] -1 [/math] and [math] 3 [/math] .
Therefore [math] f(x)=-(x+1)(x-3) [/math] .
So, [math] f(x) \geq 0 [/math] if and only if [math] (x+1)(x-3) \leq 0 [/math] .
This happens if and only if " [math] x+1 \geq 0 \text{ and } x-3 \leq 0 [/math] or " [math] x+1 \leq 0 \text{ and } x-3 \geq 0 [/math] ". (+- , -+ )
which happens if and only if " [math] x \geq -1 \text{ and } x \leq 3 [/math] " or " [math] x \leq -1 \text{ and } x \geq 3 [/math] ".
The second part is impossible.
The first part says that [math] x [/math] must be between [math] -1 [/math] and [math] 3 [/math] .

This works in general.
[math] f(x)=c(x-a)(x-b) [/math] (with [math] c0 [/math] , then it is positive outside the roots.

Oh and c is always the coefficient of x^2.

Wow, thanks for the thorough answer, mate!

Guys I haven't done anything related to maths in over 7 years and can't remember how to do this.
Any help?

wolframalpha.com/input/?i=Solve p=1-sqrt((2f+u)/(5f-v)) , f

Move one to the left, square both sides, multiply by 5f-v, expand everything, condense the expression with the f's and take the f out, put that f*(something) expression on one side and the rest on the other side, divide by (something).

Yeah I know the steps it just keeps going awol around the middle for some reason, I'm probably jsut being lazy.

nvm I got it, I fucked up my BODMAS like a retard.