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anonymous

  • one year ago

f(x) =(x/8)-3, g(x)=x^3 (f*g)^-1=

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  1. anonymous
    • one year ago
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    is that composite functions? \[(fog)\]

  2. anonymous
    • one year ago
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    yes

  3. anonymous
    • one year ago
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    pretty much substitute g(x) into the x of f(x)

  4. anonymous
    • one year ago
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    f(g(x))

  5. anonymous
    • one year ago
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    yeah but i got (x^(1/3)/8)-3

  6. anonymous
    • one year ago
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    is \[f(x)=\frac{ x }{ 8 }-3\]?

  7. anonymous
    • one year ago
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    yes

  8. anonymous
    • one year ago
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    \[(fog)=\frac{ x^3 }{ 8 }-3\]

  9. anonymous
    • one year ago
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    just plug x ==> x^3

  10. anonymous
    • one year ago
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    it said it was wrong

  11. anonymous
    • one year ago
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    @chris00

  12. anonymous
    • one year ago
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    pic me the question

  13. anonymous
    • one year ago
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  14. anonymous
    • one year ago
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  15. anonymous
    • one year ago
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    ohh inverse

  16. anonymous
    • one year ago
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    \[y=f(x)=\frac{ x }{ 8 }-3\] y+3=x/8 8y+24=x then let x=y 8x+24=y=\[f(x)^{-1}\]

  17. anonymous
    • one year ago
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    so \[f(x)^{-1}=8x+24\]

  18. anonymous
    • one year ago
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    do the same for y=x^3

  19. anonymous
    • one year ago
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    that would be x^1/3

  20. anonymous
    • one year ago
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    y^(1/3)=x therefore \[g(x)^{-1}=x ^{1/3}\]

  21. anonymous
    • one year ago
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    yep

  22. anonymous
    • one year ago
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    now\[(fog)^{-1}=8x ^{1/3}+24\]

  23. anonymous
    • one year ago
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    it says its wrong

  24. anonymous
    • one year ago
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    lol werid

  25. anonymous
    • one year ago
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    oh wait

  26. anonymous
    • one year ago
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    ( f o g)–1(x) = (g^–1 o f^ –1)(x)

  27. anonymous
    • one year ago
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    so we sub f(x) into the x function of g

  28. anonymous
    • one year ago
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    \[(fog)^{-1}=(8x+24)^{1/3}\]

  29. anonymous
    • one year ago
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    try that

  30. anonymous
    • one year ago
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    ah youre amazing!!!! thank you!!!!

  31. anonymous
    • one year ago
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    :D

  32. anonymous
    • one year ago
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    just try and know your properties. it a bit confusing http://www.purplemath.com/modules/fcncomp6.htm Thats a good link

  33. anonymous
    • one year ago
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    will do, thanks!

  34. anonymous
    • one year ago
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    : )

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