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

Question

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

lgbasallote · Answer

i smell a race....

parthkohli · Answer

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

lgbasallote · Answer

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

lgbasallote · Answer

^@Jonask

lgbasallote · Answer

^and then you did it again @Jonask

anonymous · Answer

sorry,like you said its a race

lgbasallote · Answer

i hate races.....

swissgirl · Answer

Ok just try to see when each square root is negative

lgbasallote · Answer

anyway....

lgbasallote · Answer

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

swissgirl · Answer

yes x=0

lgbasallote · Answer

hmm oh yeah

lgbasallote · Answer

didn't think of that

lgbasallote · Answer

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

swissgirl · Answer

I would say greater than 4

anonymous · Answer

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

lgbasallote · Answer

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

anonymous · Answer

Take Seperately and Connect it by union (U)

lgbasallote · Answer

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

anonymous · Answer

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

lgbasallote · Answer

^?

anonymous · Answer

just a question that i wont interupt you can try

swissgirl · Answer

Domain is just 1 and 0

lgbasallote · Answer

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

lgbasallote · Answer

so the domain is 0<=x<=1

swissgirl · Answer

for f(g(x))

lgbasallote · Answer

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

anonymous · Answer

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

parthkohli · Answer

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

anonymous · Answer

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

lgbasallote · Answer

"real" function?

parthkohli · Answer

@Yahoo! I wasn't talking to you.

parthkohli · Answer

Yeah, which has all ranges as real numbers.

anonymous · Answer

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

lgbasallote · Answer

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

parthkohli · Answer

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

lgbasallote · Answer

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

parthkohli · Answer

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

lgbasallote · Answer

i did

lgbasallote · Answer

when didn't i?

parthkohli · Answer

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 · Answer

0 is included in all real numbers.

lgbasallote · Answer

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

parthkohli · Answer

You answered your own question @lgbasallote

lgbasallote · Answer

....what exactly are you saying?

parthkohli · Answer

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 · Answer

just state your point

parthkohli · Answer

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 · Answer

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 · Answer

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

lgbasallote · Answer

^?

parthkohli · Answer

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 · Answer

and your point is?

parthkohli · Answer

That 0 is not in the domain.

lgbasallote · Answer

and isn't that what we said?

parthkohli · Answer

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

lgbasallote · Answer

when???

lgbasallote · Answer

we were talking about RESTRICTIONS

parthkohli · Answer

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 · Answer

I was talking about the domain then.

anonymous · Answer

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

lgbasallote · Answer

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

anonymous · Answer

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

lgbasallote · Answer

it has been closed a long time ago @jonask

anonymous · Answer

oh i dint notice,cos commentsn are still running