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

Algebra
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i smell a race....
Hint: you can't use positive numbers more than 4.
^i wish you would have let me try first....

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Other answers:

^and then you did it again @Jonask
sorry,like you said its a race
i hate races.....
Ok just try to see when each square root is negative
anyway....
i suppose \(\sqrt{x^2 - 1}\) will have no restrictions?
yes x=0
hmm oh yeah
didn't think of that
then \(\sqrt{4-x}\) would be 4 above?
I would say greater than 4
@lgbasallote was that the question of 100 level???
greater than 4...4 above...same shiz
Take Seperately and Connect it by union (U)
to be the best, one needs to master the basics @sauravshakya
\[f(x)=x^2,g(x)=\sqrt{1-x}\] domain of\[gof,fog\] domain
^?
just a question that i wont interupt you can try
Domain is just 1 and 0
g o f would be sqrt(1 - x^2) it would only be non-negative if 0 <= x <= 1
so the domain is 0<=x<=1
for f(g(x))
f o g would be 1- x so all real numbers
\[Domain=(-\infty,0) U (0,4]\]
0 isn't included in the domain since \(\sqrt{0^2 - 1} = \sqrt{-1}\) and this is a real function.
Lol...@ParthKohli u see it is a open Bracket.....
"real" function?
@Yahoo! I wasn't talking to you.
Yeah, which has all ranges as real numbers.
oh.....Sorry...) @ParthKohli
now you confuse me...you're telling me they're wrong?
swissgirl Best Response 1 Domain is just 1 and 0 that ^^
hmm seems one of the answers here is wrong then...wonder which
Okay, let me ask you a question. Why didn't you include all real numbers in the domain? @lgbasallote
i did
when didn't i?
lgbasallote 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
0 is included in all real numbers.
since when was \(\sqrt{-1}\) real @ParthKohli ?
You answered your own question @lgbasallote
....what exactly are you saying?
ParthKohli 0 0 isn't included in the domain since sqrt(0^2 - 1) = sqrt(-1) and this is a real function. 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? lgbasallote Best Response 0 since when was √−1 real @ParthKohli ?
just state your point
You 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.
by 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
\[D_g=(-\infty,1),D_f=\mathbb{R}\] \[fog=\left| 1-x \right|\]\[gof=\sqrt{1-x^2}\]
^?
No, real function means that for all domains, the range will be real. We have to get a domain which keeps all ranges real.
and your point is?
That 0 is not in the domain.
and isn't that what we said?
You said that you included all real numbers in the domain. -.-
when???
we were talking about RESTRICTIONS
ParthKohli 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?
I was talking about the domain then.
\[D_{gof}=[-1,1]\] \[Dfog=x \in (-\infty,1]\]
you really tend to make things big when you misread, don;t you @ParthKohli
guys i think we can close the question i did not mean to create a big issue
it has been closed a long time ago @jonask
oh i dint notice,cos commentsn are still running

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