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mathcalculus

  • one year ago

For the following function, h , determine the functions, f and g so that : h(x)= f(g(x)) If h(x)=(8x+8/11x-11)^7 then f(x) g(x)

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  1. mathcalculus
    • one year ago
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    helppppp :(

  2. kirbykirby
    • one year ago
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    Do you still need help even though your question is closed

  3. mathcalculus
    • one year ago
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    yes please

  4. kirbykirby
    • one year ago
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    Ok well for compositions, you have to imagine a function being INSIDE another function

  5. kirbykirby
    • one year ago
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    The basic functions are like: \(ln(x), cos(x), sin(x), x, e^x...\) ... so if you have like \(x^2|dw:1362456641846:dw|\)

  6. kirbykirby
    • one year ago
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    In your case we have |dw:1362456844212:dw|

  7. mathcalculus
    • one year ago
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    yeah

  8. mathcalculus
    • one year ago
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    then?

  9. mathcalculus
    • one year ago
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    ?????

  10. kirbykirby
    • one year ago
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    so, in math we'd say \(f(g(x))=h(x)\), so we let the inner function (the square box) be g(x)... since it's inside of f... \(g(x)=\frac{8x+8}{11x-11}\) Now you want to make all of g(x) to be raised to the power of 7. What we typically do is let \(u=g(x)\), now define \(f(u)=u^7\) |dw:1362457135131:dw|

  11. kirbykirby
    • one year ago
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    Ok no offense though but can you not see I'm still writing ? You know this openstudy site is voluntary for those to post answers, it's kind of rude to rush people with "?????????"

  12. mathcalculus
    • one year ago
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    don't waste my time if you're going to get emotional. & exactly, voluntarily. voluntarily leave.

  13. kirbykirby
    • one year ago
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    Well how am I wasting your time if I just posted the solution for you. And I am not wasting your time becuse it takes time to draw these pictures and code using LaTex

  14. mathcalculus
    • one year ago
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    lol just reading you reply is a waste. don't bother replying.

  15. mathcalculus
    • one year ago
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    thanks but no thanks! :)

  16. kirbykirby
    • one year ago
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    Wow I can't believe how ungrateful you are. You are even lucky there are people willing to voluntarily help you with your homework

  17. mathcalculus
    • one year ago
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    omg, enough. thanks kirby!

  18. mathcalculus
    • one year ago
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    i just saved you some trouble.

  19. mathcalculus
    • one year ago
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    I did not intend to be rude.. I just had a bad day, so please don't let it ruin yours, sincerely me.

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