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

1. anonymous

i smell a race....

2. ParthKohli

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

3. anonymous

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

4. anonymous

5. anonymous

^and then you did it again @Jonask

6. anonymous

sorry,like you said its a race

7. anonymous

i hate races.....

8. anonymous

Ok just try to see when each square root is negative

9. anonymous

anyway....

10. anonymous

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

11. anonymous

yes x=0

12. anonymous

hmm oh yeah

13. anonymous

didn't think of that

14. anonymous

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

15. anonymous

I would say greater than 4

16. anonymous

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

17. anonymous

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

18. anonymous

Take Seperately and Connect it by union (U)

19. anonymous

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

20. anonymous

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

21. anonymous

^?

22. anonymous

just a question that i wont interupt you can try

23. anonymous

Domain is just 1 and 0

24. anonymous

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

25. anonymous

so the domain is 0<=x<=1

26. anonymous

for f(g(x))

27. anonymous

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

28. anonymous

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

29. ParthKohli

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

30. anonymous

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

31. anonymous

"real" function?

32. ParthKohli

@Yahoo! I wasn't talking to you.

33. ParthKohli

Yeah, which has all ranges as real numbers.

34. anonymous

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

35. anonymous

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. anonymous

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. anonymous

i did

40. anonymous

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. anonymous

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

44. ParthKohli

45. anonymous

....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. anonymous

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. anonymous

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

50. anonymous

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

51. anonymous

^?

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. anonymous

54. ParthKohli

That 0 is not in the domain.

55. anonymous

and isn't that what we said?

56. ParthKohli

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

57. anonymous

when???

58. anonymous

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.

61. anonymous

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

62. anonymous

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

63. anonymous

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

64. anonymous

it has been closed a long time ago @jonask

65. anonymous

oh i dint notice,cos commentsn are still running