## anonymous one year ago please help im so stuck Given the following functions f(x) and g(x), solve f over g (−5) and select the correct answer below: f(x) = 2x − 20 g(x) = x − 1

1. DanJS

f(-5) / g(-5) that right?

2. anonymous

-5 is one of the choices but im not sure

3. anonymous

could you explain to me how you got that answer?

4. anonymous

−5 5 one sixth 30 the coices

5. DanJS

$\frac{ f(-5) }{ g(-5) } = \frac{ 2(-5)-20 }{ -5 -1 }$

6. DanJS

right , yeah it is f(-5) / g(-5), just evaluate both and put them over each other

7. anonymous

ok thank you can you help with few more just trying out openstudy for the first time my friends said it helps

8. DanJS

yes, depends on who helps ya.. .. just ask a new thread

9. anonymous

you mean like post the question by itself?

10. DanJS

screw it, here is fine

11. anonymous

ok 1 sec

12. anonymous

given the functions f(x)and g(x), solve (f times g)(3) and select the correct answer below f(x)4x^2+12 g(x)=x-1

13. DanJS

f times g, or is it a small circle looking thing

14. anonymous

47 -47 96 -96 are choices

15. anonymous

its a small cirlce

16. DanJS

ok, then it is a compound function f[g(x)]

17. anonymous

so how would i set up the equation

18. DanJS

nope, it is gonna be multiplication

19. DanJS

f(3) * g(3) =

20. anonymous

so i would multiply the 2 equation they gave me?

21. anonymous

so (4x^2+12)3 and (x-1)3?

22. DanJS

you could do that first, or put in the value for x to start, [4(3^2) + 12] * (3-1)

23. anonymous

hows did you get the value of x

24. DanJS

THe problem says [f times g ](3) , x=3

25. DanJS

so you just evaluate both f and g at x=3 , then multiply the values together

26. anonymous

ooo ok

27. anonymous

let me try it

28. DanJS

I have time for one more if you want

29. DanJS

f(3) * g(3) = [4(3^2) + 12] * (3-1) = 96

30. anonymous

how did u get 96 i got 168

31. DanJS

[4*9 + 12]* 2

32. DanJS

48*2

33. anonymous

ok and wat a bout the part when you multiply it by 3-1

34. DanJS

that is what happened, the 2 at the end... f(3) * g(3) = [4(3^2) + 12] * (3-1) = 96

35. anonymous

ook

36. anonymous

i get it now i messed up omewhere but i see thay why its a 2 at the end cuz 3-1 is 2 ok can u hepl with one more?

37. DanJS

yep

38. anonymous

Given the following functions f(x) and g(x), solve f[g(6)] and select the correct answer below: f(x) = 6x + 12 g(x) = x − 8

39. anonymous

choices are −96 0 24 48

40. DanJS

These compound functions like that, you work from the inside most function first, then outwards carrying the values...

41. DanJS

figure out g(6), then use that value in f(x) for x

42. anonymous

so x is 6?

43. DanJS

yes, anytime you see g(#) or f(#) some number, that is what x is

44. anonymous

k so now i plug it in dont tell me ill tell u n u tell if its right

45. DanJS

So what is g(6)

46. anonymous

is it -2?

47. DanJS

no... First , what is g(6) ?

48. anonymous

i did x-8 and plugged 6 for x

49. DanJS

right, so g(6) = 6 - 8 = -2

50. anonymous

ok so now f

51. DanJS

now you use that value you just got in the f function for x

52. anonymous

f is 48?

53. DanJS

f(-2)

54. anonymous

6times 6 plus 12

55. DanJS

no, that is the thing with the compound functions, you have to use the value from g(6) = -2 for the value of x in f(x)

56. DanJS

Figure the inner function out, use that value in the next function

57. anonymous

o ok so i plug the -2 in instead of the 6?

58. DanJS

Yes.. f[g(6)] = f(-2)

59. anonymous

can you explain how you get that from f(x)=6x+12?

60. DanJS

ok, start over from the beginning...

61. anonymous

ok

62. DanJS

f [ g(6) ] What that means is instead of having the usual f(x), you have f( another function) Start with the inner function g(6) and evaluate. g(6) = -2 Now you can replace g(6) with -2 in the original problem

63. DanJS

Left to evaluate f(-2)

64. anonymous

oo ok

65. anonymous

so where do i go from here?

66. DanJS

just left to evaluate f(-2) = 6*(-2) + 12

67. anonymous

o ok so now u added the 6(-2)+12 wouldnt it be 0

68. DanJS

so it is zero i have to take off, you can fan me , ill prolly be on later

69. anonymous

ok how do i fan you?

70. anonymous

nvm n did it thanks