## lgbasallote 3 years ago Find the domain: $h(x) = \sqrt{4-x} + \sqrt{x^2 - 1}$

1. lgbasallote

i smell a race....

2. ParthKohli

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

3. lgbasallote

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

4. lgbasallote

5. lgbasallote

^and then you did it again @Jonask

sorry,like you said its a race

7. lgbasallote

i hate races.....

8. swissgirl

Ok just try to see when each square root is negative

9. lgbasallote

anyway....

10. lgbasallote

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

11. swissgirl

yes x=0

12. lgbasallote

hmm oh yeah

13. lgbasallote

didn't think of that

14. lgbasallote

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

15. swissgirl

I would say greater than 4

16. sauravshakya

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

17. lgbasallote

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

18. Yahoo!

Take Seperately and Connect it by union (U)

19. lgbasallote

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

21. lgbasallote

^?

just a question that i wont interupt you can try

23. swissgirl

Domain is just 1 and 0

24. lgbasallote

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

25. lgbasallote

so the domain is 0<=x<=1

26. swissgirl

for f(g(x))

27. lgbasallote

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

28. Yahoo!

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

29. ParthKohli

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

30. Yahoo!

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

31. lgbasallote

"real" function?

32. ParthKohli

@Yahoo! I wasn't talking to you.

33. ParthKohli

Yeah, which has all ranges as real numbers.

34. Yahoo!

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

35. lgbasallote

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

36. ParthKohli

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

37. lgbasallote

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

38. ParthKohli

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

39. lgbasallote

i did

40. lgbasallote

when didn't i?

41. 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

42. ParthKohli

0 is included in all real numbers.

43. lgbasallote

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

44. ParthKohli

45. lgbasallote

....what exactly are you saying?

46. ParthKohli

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 ?

47. lgbasallote

48. 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.

49. 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

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

51. lgbasallote

^?

52. 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.

53. lgbasallote

54. ParthKohli

That 0 is not in the domain.

55. lgbasallote

and isn't that what we said?

56. ParthKohli

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

57. lgbasallote

when???

58. lgbasallote

59. 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?

60. ParthKohli

I was talking about the domain then.

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

62. lgbasallote

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

64. lgbasallote

it has been closed a long time ago @jonask