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Adi3

  • one year ago

will medal Please help. find the inverse of f(n) = 2(n-2)^3

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  1. Adi3
    • one year ago
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    @Jhannybean

  2. Adi3
    • one year ago
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    @AaronAndyson

  3. Adi3
    • one year ago
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    @ganeshie8

  4. IrishBoy123
    • one year ago
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    \(y = 2(n-2)^3 \) \((\dfrac{y}{2})^\frac{1}{3} = n-2\) etc

  5. Adi3
    • one year ago
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    how did you do that

  6. IrishBoy123
    • one year ago
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    \(f(n) = 2(n-2)^3\) \(y = 2(n-2)^3\) yes?

  7. IrishBoy123
    • one year ago
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    \(\frac{y}{2} = (n-2)^3\) ok?

  8. Adi3
    • one year ago
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    why would you do y/2, isn't it |dw:1443341707529:dw|

  9. IrishBoy123
    • one year ago
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    \[\dfrac{y}{2} = \dfrac{2(n-2)^3}{2}\]

  10. IrishBoy123
    • one year ago
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    you want the inverse, is that right, ie \(n = n(y)\)

  11. IrishBoy123
    • one year ago
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    \[\dfrac{y}{2} = \dfrac{2(n-2)^3}{2} \implies \dfrac{y}{2} = (n-2)^3\]

  12. Adi3
    • one year ago
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    f(n) = 2(n-2)^3

  13. Adi3
    • one year ago
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    we need to inverse this function

  14. IrishBoy123
    • one year ago
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    yes, indeed label the \(f(n)\) as y just because it makes it easier to write out. then we need to find and equation in form \(n = n(y) = f^{-1}(n)\) so \(y = 2(n-2)^3\) \(\frac{y}{2} = (n-2)^3\) like so.

  15. Adi3
    • one year ago
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    but isn't it like n= 2(y-2)^3

  16. Adi3
    • one year ago
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    i like this way

  17. Adi3
    • one year ago
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    @Miracrown

  18. Adi3
    • one year ago
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    @insa

  19. IrishBoy123
    • one year ago
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    ok let's go that way \(n= 2(y-2)^3\) \(\frac{n}{2} = (y-2)^3\)

  20. Adi3
    • one year ago
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    alright, next

  21. IrishBoy123
    • one year ago
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    \[(\frac{n}{2})^{\frac{1}{3}} = y-2\]

  22. IrishBoy123
    • one year ago
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    yes?

  23. Adi3
    • one year ago
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    wait wait, where did 1/3 come from

  24. IrishBoy123
    • one year ago
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    take cube root of each side

  25. Adi3
    • one year ago
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    why cant it be |dw:1443342930062:dw|

  26. IrishBoy123
    • one year ago
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    \(a^3 = b\) \(a = b^{\frac{1}{3}} = \sqrt[3]{b}\) same thing

  27. Adi3
    • one year ago
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    \[\frac{ \sqrt[3]{n} }{ 2}\]

  28. Adi3
    • one year ago
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    lets go this way if you dont mind

  29. IrishBoy123
    • one year ago
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    if that way, than \[\large (\frac{n}{2})^{\frac{1}{3}} = \sqrt[3]{\frac{n}{2}}= y-2\]

  30. Adi3
    • one year ago
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    yup, then \[\frac{ \sqrt[3]{n} }{ 2 } + 2\]

  31. IrishBoy123
    • one year ago
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    be careful the cube root must include the 1/2 its \[y=\sqrt[3]{\frac{ n} { 2 }} + 2\] NOT \[y=\frac{ \sqrt[3]{n} }{ 2 } + 2\]

  32. Adi3
    • one year ago
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    yeah

  33. Adi3
    • one year ago
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    so is that it

  34. IrishBoy123
    • one year ago
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    good!

  35. Adi3
    • one year ago
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    one more question can i ask please

  36. IrishBoy123
    • one year ago
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    shoot but if it's a new question, new thread is best

  37. Adi3
    • one year ago
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    find the inverse of g(x) = 1/x - 2

  38. Adi3
    • one year ago
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    i know that x = 1/y - 2

  39. Adi3
    • one year ago
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    @IrishBoy123

  40. IrishBoy123
    • one year ago
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    you like to do it this way \[x = \frac{1}{y} - 2 \] \[\frac{1}{y} = x + 2 \] \[y = \frac{1}{x+2}\]

  41. Adi3
    • one year ago
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    and the inverse of -4x + 1 is \[\frac{ x-1 }{ 4 }\] right

  42. IrishBoy123
    • one year ago
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    |dw:1443344105564:dw|

  43. Adi3
    • one year ago
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    alright thanks. no more questions

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