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

- lgbasallote

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

- schrodinger

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- lgbasallote

i smell a race....

- ParthKohli

Hint: you can't use positive numbers more than 4.

- lgbasallote

^i wish you would have let me try first....

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## More answers

- lgbasallote

- lgbasallote

^and then you did it again @Jonask

- anonymous

sorry,like you said its a race

- lgbasallote

i hate races.....

- swissgirl

Ok just try to see when each square root is negative

- lgbasallote

anyway....

- lgbasallote

i suppose \(\sqrt{x^2 - 1}\) will have no restrictions?

- swissgirl

yes x=0

- lgbasallote

hmm oh yeah

- lgbasallote

didn't think of that

- lgbasallote

then \(\sqrt{4-x}\) would be 4 above?

- swissgirl

I would say greater than 4

- anonymous

@lgbasallote was that the question of 100 level???

- lgbasallote

greater than 4...4 above...same shiz

- anonymous

Take Seperately and Connect it by union (U)

- lgbasallote

to be the best, one needs to master the basics @sauravshakya

- anonymous

\[f(x)=x^2,g(x)=\sqrt{1-x}\]
domain of\[gof,fog\]
domain

- lgbasallote

^?

- anonymous

just a question that i wont interupt you can try

- swissgirl

Domain is just 1 and 0

- lgbasallote

g o f would be sqrt(1 - x^2)
it would only be non-negative if 0 <= x <= 1

- lgbasallote

so the domain is 0<=x<=1

- swissgirl

for f(g(x))

- lgbasallote

f o g would be 1- x so all real numbers

- anonymous

\[Domain=(-\infty,0) U (0,4]\]

- ParthKohli

0 isn't included in the domain since \(\sqrt{0^2 - 1} = \sqrt{-1}\) and this is a real function.

- anonymous

Lol...@ParthKohli u see it is a open Bracket.....

- lgbasallote

"real" function?

- ParthKohli

@Yahoo! I wasn't talking to you.

- ParthKohli

Yeah, which has all ranges as real numbers.

- anonymous

oh.....Sorry...) @ParthKohli

- lgbasallote

now you confuse me...you're telling me they're wrong?

- ParthKohli

swissgirl Best Response 1
Domain is just 1 and 0
that ^^

- lgbasallote

hmm seems one of the answers here is wrong then...wonder which

- ParthKohli

Okay, let me ask you a question. Why didn't you include all real numbers in the domain? @lgbasallote

- lgbasallote

i did

- lgbasallote

when didn't i?

- 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

0 is included in all real numbers.

- lgbasallote

since when was \(\sqrt{-1}\) real @ParthKohli ?

- ParthKohli

You answered your own question @lgbasallote

- lgbasallote

....what exactly are you saying?

- 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

just state your point

- 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

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

\[D_g=(-\infty,1),D_f=\mathbb{R}\]
\[fog=\left| 1-x \right|\]\[gof=\sqrt{1-x^2}\]

- lgbasallote

^?

- 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

and your point is?

- ParthKohli

That 0 is not in the domain.

- lgbasallote

and isn't that what we said?

- ParthKohli

You said that you included all real numbers in the domain. -.-

- lgbasallote

when???

- lgbasallote

we were talking about RESTRICTIONS

- 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

I was talking about the domain then.

- anonymous

\[D_{gof}=[-1,1]\]
\[Dfog=x \in (-\infty,1]\]

- lgbasallote

you really tend to make things big when you misread, don;t you @ParthKohli

- anonymous

guys i think we can close the question i did not mean to create a big issue

- lgbasallote

it has been closed a long time ago @jonask

- anonymous

oh i dint notice,cos commentsn are still running

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