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anonymous

  • one year ago

HELP WILL FAN!!! l()()k<------------------------- Use the functions a(x) = 3x + 10 and b(x) = 2x − 8 to complete the function operations listed below. Part A: Find (a + b)(x). Show your work. (3 points) Part B: Find (a ⋅ b)(x). Show your work. (3 points) Part C: Find a[b(x)]. Show your work. (4 points)

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  1. anonymous
    • one year ago
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    @triciaal please help me

  2. anonymous
    • one year ago
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    Hello

  3. anonymous
    • one year ago
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    So this looks to be an excercise of dealing with functions. Functions can be added subtracted, multiplied, or divided much like numbers.

  4. anonymous
    • one year ago
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    yeah

  5. anonymous
    • one year ago
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    But crucially they can also be composed

  6. anonymous
    • one year ago
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    what do you mean

  7. anonymous
    • one year ago
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    For part (a) you need to add the two funtions a(x) + b(x)

  8. triciaal
    • one year ago
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    |dw:1444149834166:dw|

  9. triciaal
    • one year ago
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    group like terms and simplify

  10. anonymous
    • one year ago
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    for part a, it says to add (a+b)(x)

  11. anonymous
    • one year ago
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    This is the simplest example: a(x) + b(x) = (3x+10) + (2x-8) = 5x +2

  12. anonymous
    • one year ago
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    that is a shorthand notation meant to say you are adding the functions to one another regardless of what the x-value is

  13. anonymous
    • one year ago
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    I got that part but what about the x that needs to be multiplied to that solution of 5x + 2

  14. triciaal
    • one year ago
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    |dw:1444149917064:dw|

  15. anonymous
    • one year ago
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    When dealing with functional notation the (x) is meant to refer to the argument of the function

  16. anonymous
    • one year ago
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    triciall, for part a what about the x that needs to be multiplied

  17. anonymous
    • one year ago
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    so I can ignore the x for part a?

  18. anonymous
    • one year ago
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    It does not mean you are multiplying by the argument x... i.e. a(x) + b(x) can be writtedn (a+b)(x) or simply (a+b)

  19. anonymous
    • one year ago
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    Yes

  20. anonymous
    • one year ago
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    Now what do you think part (b) should be?

  21. anonymous
    • one year ago
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    Take a guess

  22. triciaal
    • one year ago
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    |dw:1444150022131:dw|

  23. anonymous
    • one year ago
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    part c is a bit confusing

  24. triciaal
    • one year ago
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    use b(x) as your x to find a(x)

  25. anonymous
    • one year ago
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    Which is called a composition

  26. triciaal
    • one year ago
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    it is not confusing if you actually read the response

  27. anonymous
    • one year ago
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    It wouldve been nice to try and have you work it out without knowing the answer first.... but the idea of a composition is that for a given x value you apply one function to it first.... then apply the second function to it

  28. anonymous
    • one year ago
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    A composition can be though of as just executing the functions in order 1 at a time... so given the value x=5 b(5)=2*5-8=10-8=2.... which means given x=5 b(5) gives 2

  29. triciaal
    • one year ago
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    |dw:1444150296553:dw|

  30. anonymous
    • one year ago
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    Then I have to do a(2)=3(2)+10=6+10=16

  31. anonymous
    • one year ago
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    Now to execute this process abstractly (without reference to any particular x) we write the composition as a(b(x)) as triciaal has shown

  32. anonymous
    • one year ago
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    If it helps you can think of it like this let y=b(x).... the composition of a and b (in this order) then means we are interested in a(y)=a(b(x))

  33. anonymous
    • one year ago
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    Does this make sense?

  34. anonymous
    • one year ago
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    3x+10 2x-8 So if y=b(x)=2x-8, Then a(y)=3y+10=3(2x-8)+10=6x-24+10=6x-14=a(b(x))

  35. anonymous
    • one year ago
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    Please let me know if this makes sense or if you are confused

  36. anonymous
    • one year ago
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    @ChanelAnderson203 Do you understand? Would you like another example?

  37. anonymous
    • one year ago
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    Well I have to go now hopefully this makes sense. Remember practice makes perfect and just like any language fluency in mathematics comes from learning all of the syntax and grammar as well as (unfortunately) all of the dialects that it contains. :D

  38. anonymous
    • one year ago
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    I'm still confused on C and sorry I was afk for a bit

  39. anonymous
    • one year ago
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    @triciaal I'm still confused on C

  40. anonymous
    • one year ago
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    for anyone here, I just need help with part c everything else I get please help its my last problem for my whole entire course pleasesss

  41. anonymous
    • one year ago
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    how are you confused on c

  42. anonymous
    • one year ago
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    I just don't understand what they mean if you could like start from the beginning of c that would be very helpful

  43. anonymous
    • one year ago
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    so can you help?

  44. anonymous
    • one year ago
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    yeah let me go get my book

  45. anonymous
    • one year ago
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    ok

  46. anonymous
    • one year ago
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    back

  47. anonymous
    • one year ago
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    kk

  48. triciaal
    • one year ago
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    |dw:1444152446510:dw|

  49. anonymous
    • one year ago
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    ok so far so good

  50. triciaal
    • one year ago
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    |dw:1444152528819:dw|

  51. anonymous
    • one year ago
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    3 (2x-8)+10

  52. anonymous
    • one year ago
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    6x-14??

  53. triciaal
    • one year ago
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    and then group like terms to simplify

  54. anonymous
    • one year ago
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    so is 6x-14 the answer for part c?

  55. triciaal
    • one year ago
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    |dw:1444152612850:dw|

  56. anonymous
    • one year ago
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    yes

  57. anonymous
    • one year ago
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    so the full answer would be a{b(c)}=6x-14

  58. triciaal
    • one year ago
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    not c x yes

  59. anonymous
    • one year ago
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    oh sorry ok thanks so much guys

  60. anonymous
    • one year ago
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    who does the best response go to? :/ idk who xd

  61. triciaal
    • one year ago
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    @StudyWithMe!

  62. anonymous
    • one year ago
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    do triciaal

  63. anonymous
    • one year ago
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    thanks so much for your help guys :))

  64. triciaal
    • one year ago
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    you are welcome

  65. anonymous
    • one year ago
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    no problem have a good day

  66. anonymous
    • one year ago
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    same goes for you @StudyWithMe!

  67. anonymous
    • one year ago
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    and of course @triciaal too

  68. triciaal
    • one year ago
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    thanks all the best to both of you

  69. anonymous
    • one year ago
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    see you some other time! :)

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