lgbasallote
  • lgbasallote
Find the domain: \[h(x) = \sqrt{4-x} + \sqrt{x^2 - 1}\]
Algebra
schrodinger
  • schrodinger
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lgbasallote
  • lgbasallote
i smell a race....
ParthKohli
  • ParthKohli
Hint: you can't use positive numbers more than 4.
lgbasallote
  • lgbasallote
^i wish you would have let me try first....

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

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