f(x)=-2/x^2+2x-3
Determine the domain of the function

- anonymous

f(x)=-2/x^2+2x-3
Determine the domain of the function

- jamiebookeater

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- anonymous

domain is negative infinity to infinity but there should be some vertical asymptotes in there so it is not continuously from negative infinity to infinity. gimme a sec

- dumbcow

x cannot be 0

- anonymous

no dumbcow, x cannot be -3 or 1

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## More answers

- anonymous

domain should be x not equal to -1 or -3 i.e R-{-1,-3}

- anonymous

oh whoa nevermind didnt see the negative sign in front of the 2

- anonymous

and R here is the set of real numbers:)

- anonymous

got it?

- anonymous

x cannot be 0 thats it so all real numbers except when x is 0

- anonymous

I'm trying. lol. Thank you for explaining; I sure do appreciate it. Got tons of this stuff to help someone else out.

- anonymous

it doesnt make sense even if x=0 as there is a '-3' in th e denominator ...lol

- anonymous

this is unbelievable i must be drunk final answer is all real numbers except when x equals to 1 or -3

- anonymous

All numbers except wen x=-1 or -3....

- anonymous

of course x can equal to 0!!!!!!

- anonymous

no its 1 or -3 not -1 or -3

- dumbcow

haha sorry you guys didnt read the problem correctly

- anonymous

yes it can be =0

- anonymous

yep

- anonymous

@dumbcow haha i hadnt the first time around either

- anonymous

now did u get it??

- anonymous

hello anybody there??

- anonymous

yes

- anonymous

so u got what i typed above?

- anonymous

Okay, so the domain is not (-inf, inf).....

- anonymous

no its not

- anonymous

So it's all real numbers except when x=0

- anonymous

no...its all real nos. except wen x=-1 or x=-3

- anonymous

now , is it alright?

- anonymous

Yes, thank you

- anonymous

you're welcome @JJ

- anonymous

So, if I had [f(x)=\sqrt{5-x}\] , would that mean, the Domain would be all real numbers? Just making sure I got that one right.I'm assuming this because I can plug anything in there.

- anonymous

no x cannot equal any negative number in that caase

- anonymous

u can plug anything there except the number/numbers that make the denominator go '0'

- anonymous

and yeah , sqrt can't be negative so u have to remeber that:) so domain wud be numbers<5

- anonymous

So, all real number <5

- anonymous

thinker i think youre a bit confused. x can equal indeed to 5, that would just mean y equals 0. The domain is R = (0,infinity)

- anonymous

I sure am confused. So sorry.

- anonymous

JJSONFIRE i am really off it today listen to thinker im wrong

- anonymous

Hey, trust me I understand.

- anonymous

yep youre right

- anonymous

hey ....if its(),infinity) that wud make the sqrt tern negative which is not Real!!

- anonymous

i know you do im admitting that im wrong. im tired but still willing to help

- anonymous

i just explained to you that i made a mistake

- anonymous

ive been working all day..... :/

- anonymous

yeah..thats ok...im tired too...nevermind

- anonymous

Do you think you guys could help me with one more..please

- anonymous

sure

- anonymous

This set of problems says I need to determine if "f" is defind by f(x)=2x^2+5. The problems are f(-2)............next is f(3x)............last is -f(x)............Please, explain the best you can ...i'm lost..the book is useless with explanations

- anonymous

just plug in -2 in the first one as it says "f(-2)"...in place of x & u will get the answer:) its simple:)

- anonymous

Wow, so i was right. Woohoo..

- anonymous

and similarly for the rest...if it says f(3x) then plug in '3x' in place of 'x'
got it?

- anonymous

ohh u were..?? good:)

- anonymous

now i gotta go...bye...hope i helped u:)

- anonymous

i've been at this all night...not sure whats been right and whats been wrong.

- anonymous

ohh..dont worry thats how u learn, right?

- anonymous

ACtually, the one with the negative f...not so sure actually

- anonymous

hmm..np...bye for now:)

- anonymous

-f(x)....how in the world...Is it as simple as just writing that out?

- anonymous

hmmm yupp:)

- anonymous

Thank you sooooooooooo much.

- anonymous

ohhhhh:):) youre welcome:) hope i help u next time too:)

- anonymous

Oh, I have 2 more sets to finish that are different...need to finish them today. I am helping someone who seriously needs the help that will never use this stuff, so i'lm trying my very best to help them. Its very important it gets done right you know.

- anonymous

hmm yeah...but i am really tired:(:( sry:(:( gotta go...next time..

- anonymous

Thank you. Nite.

- anonymous

gud nite:):) n byee

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