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moongazer

  • 5 years ago

if f(x)= (x+1)/(x^2-1) and g(x) = (3x+7)/2x find f[g(x)]. Please answer clearly and explain the solution clearly and easy to understand. Thanks! (guaranteed medal for a good answer)

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  1. anonymous
    • 5 years ago
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    step on write \[f(g(x))\]

  2. anonymous
    • 5 years ago
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    step two replace \[g(x)\] by \[\frac{3x+7}{2x}\] to get \[f(\frac{3x+7}{2x})\]

  3. anonymous
    • 5 years ago
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    so far so good?

  4. moongazer
    • 5 years ago
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    yup

  5. anonymous
    • 5 years ago
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    then were we see an x in \[f(x)\] we replace it by \[\frac{3x+7}{2x}\] to get \[\frac{\frac{3x+7}{2x}+1}{(\frac{3x+7}{2x})^2-1}\]

  6. anonymous
    • 5 years ago
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    zat ok?

  7. anonymous
    • 5 years ago
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    is jsut an exercise in simplication

  8. anonymous
    • 5 years ago
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    simplification

  9. anonymous
    • 5 years ago
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    well actually why don't we simplify to begin with. \[f(x)=\frac{x+1}{x^2-1}=\frac{x+1}{(x+1)(x-1)}=\frac{1}{x-1}\]\]

  10. moongazer
    • 5 years ago
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    yup that is what my teacher told me

  11. anonymous
    • 5 years ago
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    so we could make this problem much easier if we use the second f i wrote. on the other hand this is not really valid

  12. anonymous
    • 5 years ago
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    those function are not identical. they are identical if x is not -1. if x is -1 the first function is undefined, whereas the second function is \[\frac{1}{-2}\]

  13. anonymous
    • 5 years ago
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    but if your teacher does not mind, neither do i. just write \[f(x)=\frac{1}{x-1}\] and so \[f(g(x))=f(\frac{3x+7}{2x})=\frac{1}{\frac{3x+7}{2x}-1}\]

  14. moongazer
    • 5 years ago
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    what??

  15. anonymous
    • 5 years ago
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    then multiply top and bottom by 2x to get \[f(g(x))=\frac{2x}{3x+7-2x}=\frac{2x}{x+7}\]\][

  16. moongazer
    • 5 years ago
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    what happened to the ( )^2?

  17. anonymous
    • 5 years ago
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    oops i lost you

  18. anonymous
    • 5 years ago
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    are we going to replace \[f(x)=\frac{x+1}{x^2-1}\] by \[f(x)=\frac{1}{x-1}\]?

  19. moongazer
    • 5 years ago
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    yes because i think it is easier

  20. anonymous
    • 5 years ago
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    ok in that case there is no square in the denominator

  21. moongazer
    • 5 years ago
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    ooooohhh now i know

  22. anonymous
    • 5 years ago
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    if we make that change in f, then \[f(g(x))=f(\frac{3x+7}{2x})\] yes?

  23. anonymous
    • 5 years ago
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    if so then everything i wrote above works yes?

  24. anonymous
    • 5 years ago
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    if not let me know

  25. moongazer
    • 5 years ago
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    isn't it it should first be 1/(3x+7)- 1?

  26. moongazer
    • 5 years ago
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    am i correct?

  27. anonymous
    • 5 years ago
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    no.

  28. moongazer
    • 5 years ago
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    why?

  29. anonymous
    • 5 years ago
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    \[g(x)=\frac{3x+7}{2x}\]

  30. anonymous
    • 5 years ago
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    not \[3x+7\]

  31. anonymous
    • 5 years ago
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    so when we replace x in f(x) by \[\frac{3x+7}{2x}\] we get \[\frac{1}{\frac{3x+7}{2x}-1}\]

  32. moongazer
    • 5 years ago
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    yup i forgot to type 2x sorry for that

  33. moongazer
    • 5 years ago
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    please continue

  34. anonymous
    • 5 years ago
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    ok good

  35. anonymous
    • 5 years ago
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    now you can simplify this compound fraction by multiplying top and bottom by 2x

  36. anonymous
    • 5 years ago
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    we have \[f(g(x))=\frac{1}{\frac{3x+7}{2x}-1}\] \[=\frac{1}{\frac{3x+7}{2x}-1}\times \frac{2x}{2x}\] \[=\frac{2x}{3x+7-2x}\] \[=\frac{2x}{x+7}\]

  37. anonymous
    • 5 years ago
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    k?

  38. moongazer
    • 5 years ago
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    what happened with the 2x below 3x+7?

  39. anonymous
    • 5 years ago
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    i multiplied top and bottom by 2x to get rid of it. canceled from the denominator

  40. anonymous
    • 5 years ago
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    \[\frac{3x+7}{2x}\times 2x=3x+7\]

  41. moongazer
    • 5 years ago
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    ohhh ok

  42. anonymous
    • 5 years ago
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    you cannot have a compound fraction as an answer, so i got rid of it by multiplying top and bottom by that annoying 2x in the denominator of the denominator

  43. anonymous
    • 5 years ago
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    clear or no?

  44. moongazer
    • 5 years ago
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    wait ill do it on a paper ^_^

  45. anonymous
    • 5 years ago
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    k

  46. moongazer
    • 5 years ago
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    how did it become 2x/2x+7?

  47. anonymous
    • 5 years ago
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    ok let us start with \[\frac{1}{\frac{3x+7}{2x}-1}\] ok?

  48. moongazer
    • 5 years ago
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    ok

  49. anonymous
    • 5 years ago
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    multiply top and bottom by 2x. what do you get?

  50. anonymous
    • 5 years ago
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    \[1\times 2x=2x\] that is your numerator

  51. moongazer
    • 5 years ago
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    ok

  52. anonymous
    • 5 years ago
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    \[\frac{3x+7}{2x}\times 2x = 3x+7\] that is why you did it to get rid of the pesky 2x in the bottom

  53. anonymous
    • 5 years ago
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    \[-1\times 2x=-2x\]

  54. moongazer
    • 5 years ago
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    ok

  55. anonymous
    • 5 years ago
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    of course you have to multiply the whole denominator by 2x, not just part of it.

  56. anonymous
    • 5 years ago
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    so you numerator is 2x

  57. anonymous
    • 5 years ago
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    your denominator is 3x+7-2x

  58. anonymous
    • 5 years ago
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    which give a denominator of x+7

  59. moongazer
    • 5 years ago
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    ok

  60. anonymous
    • 5 years ago
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    so our "final answer" is \[f(g(x))=\frac{2x}{x+7}\]

  61. moongazer
    • 5 years ago
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    i just got confused i forgot that 3x and -2x can be combined

  62. anonymous
    • 5 years ago
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    ah but clear now yes?

  63. moongazer
    • 5 years ago
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    yes! thanks^_^

  64. anonymous
    • 5 years ago
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    welcome

  65. anonymous
    • 5 years ago
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    ah but clear now yes?

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