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anonymous

  • one year ago

Find fog and gof and their domains for the pair of functions F(x)=2x^2+3 and g(x)=4x-5 (I have fog, I cannot figure out gof)

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  1. FibonacciChick666
    • one year ago
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    so first, what is g composed f?

  2. FibonacciChick666
    • one year ago
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    the function I mean,

  3. anonymous
    • one year ago
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    g should be \[2x^2+3\]

  4. FibonacciChick666
    • one year ago
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    that's what you have for f?

  5. anonymous
    • one year ago
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    g(x)=4x-5

  6. FibonacciChick666
    • one year ago
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    ok, so I am asking, when you plug f into g, what is the resulting function?

  7. FibonacciChick666
    • one year ago
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    ie. \(g(f(x) ~or~ g ~o~ f\)

  8. kc_kennylau
    • one year ago
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    Just making things tidier for everyone: \[\Large f(x)=2x^2+3\]\[\Large g(x)=4x-5\]\[\Large(f\circ g)(x) = 2(4x-5)^2+3\]\[\Large(g\circ f)(x) = 4(2x^2+3)-5\] (Please correct me if I'm wrong)

  9. anonymous
    • one year ago
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    \[32x^2-80x+53\]

  10. anonymous
    • one year ago
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    that was fog

  11. anonymous
    • one year ago
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    i dont know what gof is...I need help with that I only know fog, which I just gave an answer of

  12. FibonacciChick666
    • one year ago
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    how did you get fog, what did you do? (gof is the same process)

  13. FibonacciChick666
    • one year ago
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    or ok, let's say let x=2, what would you do?

  14. FibonacciChick666
    • one year ago
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    you'd put a 2 everywhere you see an x right?

  15. FibonacciChick666
    • one year ago
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    so in the equation g(x) replace every x with f.

  16. FibonacciChick666
    • one year ago
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    then replace every f with what f equals, it's just substitution

  17. anonymous
    • one year ago
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    I did \[fog=f(4x-5)\] \[fog=2(4x-5)^2+3\] which equals \[32x^2-80x+53\]

  18. anonymous
    • one year ago
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    that answer is for fog and now I don't know how to do gof

  19. UsukiDoll
    • one year ago
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    for g o f you place your f(x) function into the g(x) function

  20. FibonacciChick666
    • one year ago
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    same exact way as above

  21. FibonacciChick666
    • one year ago
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    you did it correctly

  22. FibonacciChick666
    • one year ago
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    its just g(f) instead of f(g)

  23. FibonacciChick666
    • one year ago
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    do you see what I mean?

  24. UsukiDoll
    • one year ago
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    some people use different terminology symbols... I don't use that but I do use g o f a lot... read this from right to left... your f(x) function should be placed inside the g(x) function

  25. anonymous
    • one year ago
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    I did and it keeps telling me I'm wrong. This is how I would do it... \[4(32x^2-80x+53)-5\] which would give me \[128x^2-320x+207\]

  26. FibonacciChick666
    • one year ago
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    ah, you put the wrong function in

  27. anonymous
    • one year ago
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    how do I put it in :(

  28. FibonacciChick666
    • one year ago
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    you put your answer for part a in instead of doing the new problem

  29. FibonacciChick666
    • one year ago
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    you plugged f composed with g in instead of just f

  30. UsukiDoll
    • one year ago
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    do you see the x in g(x) = 4x- 5 replace that x in g(x) with that entire f(x) function

  31. FibonacciChick666
    • one year ago
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    so what is f? Plug that in instead of your answer for part a, then I will double check it.

  32. FibonacciChick666
    • one year ago
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    you did set it up correctly, just misread your paper.

  33. anonymous
    • one year ago
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    ok so \[g(x)=4(2x^2+3)-5\]

  34. FibonacciChick666
    • one year ago
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    much better :)

  35. UsukiDoll
    • one year ago
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    yes so all you have to do is distribute that 4 for 2x^2+3

  36. FibonacciChick666
    • one year ago
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    go ahead, try and simplify

  37. anonymous
    • one year ago
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    final answer should be \[8x^2+7\] right?

  38. UsukiDoll
    • one year ago
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    we are multiplying

  39. FibonacciChick666
    • one year ago
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    well, how do you distribute the 4?

  40. UsukiDoll
    • one year ago
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    so what is 4 times 2x^2 and what is 4 times 3

  41. FibonacciChick666
    • one year ago
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    so just remember, if you have \[a(c+b)=ac+ab\] You have to send the a to b and c

  42. anonymous
    • one year ago
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    8x^2+12

  43. anonymous
    • one year ago
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    \[8x^2+12x-5\]

  44. FibonacciChick666
    • one year ago
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    ok now that is what happens when you distribute correctly, now, you still had that - five hanging out

  45. UsukiDoll
    • one year ago
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    we don't need the x for the 12

  46. UsukiDoll
    • one year ago
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    \[g(x)=8x^2+12-5\] what is 12 - 5 ?

  47. anonymous
    • one year ago
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    so its \[8x^2+7\]

  48. FibonacciChick666
    • one year ago
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    yea

  49. FibonacciChick666
    • one year ago
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    You were right to begin with, I apologize

  50. FibonacciChick666
    • one year ago
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    I got lost in usuki's comments

  51. anonymous
    • one year ago
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    Oh thanks...yeah I was confused that you guys said it was wrong haha

  52. FibonacciChick666
    • one year ago
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    yea, I'm sorry about that. I read the we are multip[lying and assumed you added, I didn't notice they come out the same. But anyways, does this make sense now?

  53. anonymous
    • one year ago
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    Yeah, is that the final answer?

  54. UsukiDoll
    • one year ago
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    yeah

  55. FibonacciChick666
    • one year ago
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    well, yes and no

  56. UsukiDoll
    • one year ago
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    how is this no?

  57. FibonacciChick666
    • one year ago
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    so that is the final answer for what is g composed f, but we are not just loooking for that

  58. FibonacciChick666
    • one year ago
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    now, we need to find their domain, for both bantaray

  59. anonymous
    • one year ago
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    we are looking for domains

  60. FibonacciChick666
    • one year ago
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    now, what does domain mean bant?

  61. UsukiDoll
    • one year ago
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    oh geez total miss. . .

  62. FibonacciChick666
    • one year ago
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    I got this usuki, you can go

  63. anonymous
    • one year ago
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    the domain for fog is \[(-\infty,\infty)\]

  64. FibonacciChick666
    • one year ago
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    h/o I need to find what that was again haha

  65. FibonacciChick666
    • one year ago
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    sounds right though

  66. FibonacciChick666
    • one year ago
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    yep, that's right

  67. FibonacciChick666
    • one year ago
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    now for g?

  68. FibonacciChick666
    • one year ago
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    side note, you can also say All real numbers

  69. anonymous
    • one year ago
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    Ok, I got the domains now...I figured them out really fast :) I think we are done

  70. FibonacciChick666
    • one year ago
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    yep, what did you get for g(f)?

  71. anonymous
    • one year ago
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    I have one more question.....Would you mind staying to help me?

  72. FibonacciChick666
    • one year ago
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    can you pop open another question and tag me?

  73. FibonacciChick666
    • one year ago
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    just close this one to make a new one. It's a bit of a long thread

  74. anonymous
    • one year ago
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    yeah sure!

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