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lgbasallote
 4 years ago
Find the domain:
\[h(x) = \sqrt{4x} + \sqrt{x^2  1}\]
lgbasallote
 4 years ago
Find the domain: \[h(x) = \sqrt{4x} + \sqrt{x^2  1}\]

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ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Hint: you can't use positive numbers more than 4.

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0^i wish you would have let me try first....

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0^and then you did it again @Jonask

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry,like you said its a race

swissgirl
 4 years ago
Best ResponseYou've already chosen the best response.1Ok just try to see when each square root is negative

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0i suppose \(\sqrt{x^2  1}\) will have no restrictions?

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0didn't think of that

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0then \(\sqrt{4x}\) would be 4 above?

swissgirl
 4 years ago
Best ResponseYou've already chosen the best response.1I would say greater than 4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@lgbasallote was that the question of 100 level???

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0greater than 4...4 above...same shiz

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Take Seperately and Connect it by union (U)

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0to be the best, one needs to master the basics @sauravshakya

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=x^2,g(x)=\sqrt{1x}\] domain of\[gof,fog\] domain

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just a question that i wont interupt you can try

swissgirl
 4 years ago
Best ResponseYou've already chosen the best response.1Domain is just 1 and 0

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0g o f would be sqrt(1  x^2) it would only be nonnegative if 0 <= x <= 1

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0so the domain is 0<=x<=1

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0f o g would be 1 x so all real numbers

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[Domain=(\infty,0) U (0,4]\]

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.00 isn't included in the domain since \(\sqrt{0^2  1} = \sqrt{1}\) and this is a real function.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Lol...@ParthKohli u see it is a open Bracket.....

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0@Yahoo! I wasn't talking to you.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, which has all ranges as real numbers.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh.....Sorry...) @ParthKohli

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0now you confuse me...you're telling me they're wrong?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0swissgirl Best Response 1 Domain is just 1 and 0 that ^^

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0hmm seems one of the answers here is wrong then...wonder which

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, let me ask you a question. Why didn't you include all real numbers in the domain? @lgbasallote

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0lgbasallote Best Response 0 i suppose sqrt(x62−1) will have no restrictions? swissgirl Best Response 1 yes x=0 lgbasallote Best Response 0 hmm oh yeah

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.00 is included in all real numbers.

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0since when was \(\sqrt{1}\) real @ParthKohli ?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0You answered your own question @lgbasallote

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0....what exactly are you saying?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0ParthKohli 0 0 isn't included in the domain since sqrt(0^2  1) = sqrt(1) and this is a real function. <argument happens> ParthKohli 0 Okay, let me ask you a question. Why didn't you include all real numbers in the domain? @lgbasallote lgbasallote Best Response 0 i did lgbasallote Best Response 0 when didn't i? <some more convo> lgbasallote Best Response 0 since when was √−1 real @ParthKohli ?

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0just state your point

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0You are contradicting yourself. First, you say that you have included all real numbers in the domain. Then, by "since when was √−1 real @ParthKohli ?," you mean that 0 is not in the domain.

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0by the way 1) none of us said 0 is included in the domain (we were talking about restrictions) 2) sqrt 1 is not a real function

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[D_g=(\infty,1),D_f=\mathbb{R}\] \[fog=\left 1x \right\]\[gof=\sqrt{1x^2}\]

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0No, real function means that for all domains, the range will be real. We have to get a domain which keeps all ranges real.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0That 0 is not in the domain.

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0and isn't that what we said?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0You said that you included all real numbers in the domain. .

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0we were talking about RESTRICTIONS

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0ParthKohli 0 Okay, let me ask you a question. Why didn't you include all real numbers in the domain? @lgbasallote lgbasallote i did 10 minutes ago lgbasallote Best Response 0 when didn't i?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0I was talking about the domain then.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[D_{gof}=[1,1]\] \[Dfog=x \in (\infty,1]\]

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0you really tend to make things big when you misread, don;t you @ParthKohli

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0guys i think we can close the question i did not mean to create a big issue

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0it has been closed a long time ago @jonask

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh i dint notice,cos commentsn are still running
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