## anonymous 4 years ago Given that f(x)=sqrt{2 + x} and g(x)=sqrt{2 - x}, find formulas for the following functions, and their domains. In each case, enter the domain using interval notation. (a) f+g= and its domain is

1. anonymous

Have you searched the web? I havn't studied this yet /:

2. anonymous

I have.. I can't find anything similar :/

3. anonymous

@qpHalcy0n can you help me?

4. anonymous

5. jim_thompson5910

f(x) + g(x) literally means "add the functions f(x) and g(x)"

6. jim_thompson5910

but since f(x) = sqrt(2+x) and g(x) = sqrt(2-x), we know that f(x) + g(x) = sqrt(2+x) + sqrt(2-x)

7. anonymous

that's what I got.. But, I wasn't sure if f(x) + g(x) = sqrt(2+x) + sqrt(2-x) was correct or not.. how would I find the domain?

8. jim_thompson5910

yeah it seems too simple that it looks like a trick question, but it's not

9. jim_thompson5910

the domain is the set of allowable inputs, or x values

10. jim_thompson5910

remember that you can't take the square root of a negative number, so 2+x >= 0 and 2-x >= 0

11. jim_thompson5910

those two inequalities become x >= -2 and x <= 2 which combine to -2 <= x <= 2

12. anonymous

so, in interval notation.. it would be (-inf,-2)U(2,inf) ?

13. jim_thompson5910

no, [-2, 2]

14. jim_thompson5910

that represents $$\Large -2 \le x \le 2$$

15. anonymous

oh okay... and would this be correct? f(x) - g(x) = sqrt(2+x) - sqrt(2-x)

16. jim_thompson5910

yes that's 100% correct

17. jim_thompson5910

the domain is the same because the radicands (the stuff in the square roots) are the same

18. anonymous

f(x) * g(x) = sqrt(2+x) * sqrt(2-x) ?

19. jim_thompson5910

yes

20. jim_thompson5910

domains are the same (for the same reason explained above)

21. jim_thompson5910

domain is*

22. anonymous

f(x)/g(x) = sqrt(2+x)/sqrt(2-x) and same domain?

23. jim_thompson5910

the first part is correct, but the domain will be slightly different

24. jim_thompson5910

now we have to avoid dividing by zero, so 2-x can't be zero 2-x = 0 x = 2 so if x = 2, then 2-x is zero. So we toss out x = 2 from the domain So the domain is now [-2, 2)

25. anonymous

so now what is the difference between f+g and fog(x)?

26. jim_thompson5910

f+g means "add f and g" fog(x) means "start with f(x), and plug in g(x) as the input"

27. jim_thompson5910

fog(x) is the same as f(g(x))

28. jim_thompson5910

$\Large f(x) = \sqrt{2+x}$ $\Large f(g(x)) = \sqrt{2+g(x)}$ $\Large f(g(x)) = \sqrt{2+\sqrt{2-x}}$

29. anonymous

well it's with different numbers, I just wanted to know the difference.

30. jim_thompson5910

i gotcha, do you see how I'm getting that?

31. anonymous

f(g(x))=2+sqrt(2-x)?

32. jim_thompson5910

f(x) = sqrt(2+x) and g(x) = sqrt(2-x) correct?

33. anonymous

yes.

34. jim_thompson5910

Then $\Large f(g(x)) = \sqrt{2+\sqrt{2-x}}$ or $\Large f\circ g(x) = \sqrt{2+\sqrt{2-x}}$

35. anonymous

can you draw that? cause its not showing up correct.,. and im a lil confused

36. jim_thompson5910

|dw:1346027228896:dw|

37. jim_thompson5910

oh my bad, there should be a closing parenthesis after g(x)

38. jim_thompson5910

but everything else is correct

39. anonymous

okay so if f(x)=1/x and g(x)= 9x+1 fog(x) would be... 1/(9x+1)?

40. jim_thompson5910

yes perfect

41. anonymous

and its domain would be all real numbers except ...?

42. jim_thompson5910

what makes 9x+1 equal to zero?

43. anonymous

anything less than zero?

44. jim_thompson5910

9x +1 = 0 x = ???

45. anonymous

-1/9

46. jim_thompson5910

good, so x can be any number but -1/9

47. anonymous

[-1/9,inf] ?

48. jim_thompson5910

(-inf, -1/9) U (-1/9, inf)

49. jim_thompson5910

start with (-inf, inf), which is the entire number line then poke a hole at -1/9 to get (-inf, -1/9) U (-1/9, inf)

50. anonymous

for the domain it it says.. its domain is all real numbers except __ so would I just put in -1/9?

51. jim_thompson5910

yes

52. jim_thompson5910

that's in a more understandable way (in my opinion)

53. jim_thompson5910

interval notation can be a little tricky to grasp

54. anonymous

haha ya! gof(x) would be 9(1/x)+1?

55. jim_thompson5910

yes, which becomes 9/x + 1

56. jim_thompson5910

the 1 isn't part of the fraction

57. anonymous

then to find the domain would I do x+1=0 x=-1?

58. jim_thompson5910

no, the denominator is just x

59. jim_thompson5910

$\Large f \circ g(x) = \frac{9}{x}+1$

60. anonymous

oh its (9/x)+1?

61. jim_thompson5910

yes

62. anonymous

so how would you do that domain then? would it be.. x cannot be zero?

63. jim_thompson5910

yes, x can be any number but 0

64. anonymous

would fof(x) be 1/(1/x).. so 1?

65. jim_thompson5910

1/(1/x) = (1/1)/(1/x) 1/(1/x) = (1/1)*(x/1) 1/(1/x) = x/1 1/(1/x) = x

66. jim_thompson5910

So 1/(1/x) simplifies to x

67. jim_thompson5910

We have to add the restriction that x can't be zero to make sure that the two expressions are completely equivalent

68. anonymous

fof(x)=x and its domain is all real numbers except 0?

69. jim_thompson5910

yes

70. jim_thompson5910

you nailed it

71. anonymous

gog(x)=9(9x+1)+1=81x+9+1=81x+10?

72. jim_thompson5910

perfect

73. jim_thompson5910

domain: all real numbers

74. jim_thompson5910

since the result is a polynomial

75. anonymous

it says... its domain is (__,__)

76. anonymous

-inf, inf?

77. jim_thompson5910

yes it is

78. anonymous

yyayyy! thank you!

79. jim_thompson5910

you're welcome

80. anonymous

one more thing :/ ?

81. jim_thompson5910

what's that

82. anonymous

How do I find the corresponding function values. 1. f(g(-2)) 2. f(g(0))

83. jim_thompson5910

1) what is g(-2)??

84. anonymous

that's the point at (2,-2)?

85. jim_thompson5910

look at g(x) and not f(x)

86. anonymous

(-2,2)?

87. jim_thompson5910

you are correct, the point (-2,2) is on g(x)

88. jim_thompson5910

so f(g(-2)) = f(2)

89. anonymous

so what would be the corresponding function value? how do I put that together?

90. anonymous

cause I tried f(2) and f(1) for the next one and they're both wrong.

91. jim_thompson5910

notice how (2,-2) is on f(x)

92. anonymous

yes

93. jim_thompson5910

so f(2) = -2

94. jim_thompson5910

which means overall f(g(-2)) = -2

95. jim_thompson5910

this example may be a bit confusing with all the 2's...

96. jim_thompson5910

but the idea is you start with g(x), plug in x = -2 then you get some result, which you plug into f(x) to get your answer

97. anonymous

so for the second one.. f(g(0)) (0,1)?

98. jim_thompson5910

close g(0) = 1 f(g(0)) = f(1) f(g(0)) = -1

99. anonymous

all right thank you!

100. jim_thompson5910

sure thing