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