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anonymous

  • 5 years ago

Given the derivative of some inegrable function f(x) is f'(x)=e^x+(4/3)x^(-2/3). What is f(x) if f(1)=e?

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  1. anonymous
    • 5 years ago
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    and...what is (f^-1)'(3)? Soooooo confusing!!!!

  2. anonymous
    • 5 years ago
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    f(x) = (e^x) + (4x^1/3)

  3. anonymous
    • 5 years ago
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    How did you find that?

  4. anonymous
    • 5 years ago
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    integration and derivated from e^x is e^x integration from (4/3)x^(-2/3) is (-2/3 +1)=1/3 then (4/3)/(1/3)= 4x^1/3

  5. anonymous
    • 5 years ago
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    hmm...

  6. anonymous
    • 5 years ago
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    So, you just simply integrate the original equation?

  7. anonymous
    • 5 years ago
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    f'(x) means first derivative from f(x) to find f(x) you have to integrate it

  8. anonymous
    • 5 years ago
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    When you integrate don't you get e^x+4x^(1/3)?

  9. anonymous
    • 5 years ago
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    yup

  10. anonymous
    • 5 years ago
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    alright and then that is just simply f(x) right?

  11. anonymous
    • 5 years ago
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    f(1) means change x with 1

  12. anonymous
    • 5 years ago
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    Okay. How would I exactly find the inverse of f(x)?

  13. anonymous
    • 5 years ago
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    Follow this example. f(x) = (x+3)/5 To find its inverse, write y=f(x) y= (x+3)/5 Interchange x and y x = (y+3)/5 solve for y in terms of x 5x=y+3 y=5x-3 The inverse of f(x) is f^-1(x) = 5x-3

  14. anonymous
    • 5 years ago
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    I got this far...x=e^y+4y^(1/3) and now I'm stuck.

  15. anonymous
    • 5 years ago
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    log both side log x = y + 1/3 log 4y

  16. anonymous
    • 5 years ago
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    Then you get lnx=y+(1/3)ln4+(1/3)lny?

  17. anonymous
    • 5 years ago
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    isn't it ln x = y + 1/3 ln 4y, i dont understand this, confused too

  18. anonymous
    • 5 years ago
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    Well, I was thinking that you would add because usually when you have a product in a ln then that means you add. right?

  19. anonymous
    • 5 years ago
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    i made mistake, for example 2 ln 2y = ln 4y^2 so 4y^1/3 = 1/3 * 4 ln y

  20. anonymous
    • 5 years ago
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    Okay, thank you.

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