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anonymous

  • 5 years ago

Find the relative rate of change fro each of the following functions. f(x)=x^3 -3x^2

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  1. anonymous
    • 5 years ago
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    I think you have to derive here, or am I wrong? Since it's about rate of change, right?

  2. anonymous
    • 5 years ago
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    no this is the problem

  3. anonymous
    • 5 years ago
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    the derivative would be 3x^2 -6x

  4. anonymous
    • 5 years ago
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    exactly

  5. anonymous
    • 5 years ago
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    that's how you compute the relative rate of change, by deriving ^_^

  6. anonymous
    • 5 years ago
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    im not understanding?

  7. anonymous
    • 5 years ago
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    so do i take f'(x) and divide it by f(x)

  8. anonymous
    • 5 years ago
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    lol, to find the relative rate of change, you have to find the derivative of the following function which is in this case :\[f(x) = x^3 - 3x^2\] so when you derive the function you'll get like what you've said : \[f'(x) = 3x^2 - 6x\] since there is no given for x, then that's the final equation I guess :)

  9. anonymous
    • 5 years ago
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    if they told you, for example, that x = 2 then substitute it in the derived function and you'll find the relative rate of change

  10. anonymous
    • 5 years ago
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    ^_^ clearer?

  11. anonymous
    • 5 years ago
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    or am I mixing it up with something?

  12. anonymous
    • 5 years ago
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    i got all that, but my teacher still counted 2 points off?

  13. anonymous
    • 5 years ago
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    hmm, I'm missing a point aren't I?

  14. anonymous
    • 5 years ago
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    sstarica, when you're done here, can you help dinaortega? I'm leaving soon.

  15. anonymous
    • 5 years ago
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    I'll try my best loki :)

  16. anonymous
    • 5 years ago
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    could you help me lokisan?

  17. anonymous
    • 5 years ago
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    Thanks.

  18. anonymous
    • 5 years ago
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    np

  19. anonymous
    • 5 years ago
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    relative rate of change is f'(x)/f(x)

  20. anonymous
    • 5 years ago
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    >_< thank you blexting

  21. anonymous
    • 5 years ago
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    blexting so how do i get that from f(x)= x^3 -3x^2 being the original problem and f'(x)=3x^2 -6x

  22. anonymous
    • 5 years ago
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    Sorry.. I don't know, but I knew the formula.. sstarica can you help?

  23. anonymous
    • 5 years ago
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    Post your assignment questions at www.aceyourcollegeclasses.com you can earn money answering questions there too!

  24. anonymous
    • 5 years ago
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    divide f'(x) with f(x)

  25. anonymous
    • 5 years ago
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    and simplify

  26. anonymous
    • 5 years ago
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    x^3-6x?

  27. anonymous
    • 5 years ago
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    6x/x^3

  28. anonymous
    • 5 years ago
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    \[\frac{f'(x)}{f(x)}=\frac{3x^2-6x}{x^3-3x^2}=\frac{3x-6}{x^2-3x}\]

  29. anonymous
    • 5 years ago
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    so you'll get :\[f'(x)/f(x) = (3x^2-6x)/(x^3-2x^2)\]\[= 3x(x-2)/x^2(x-2)\] \[= 3/x\]

  30. anonymous
    • 5 years ago
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    right?

  31. anonymous
    • 5 years ago
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    sstarica.. the original was 3x^2. I agree with lokisan

  32. anonymous
    • 5 years ago
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    oh , sorry my bad ^^"

  33. anonymous
    • 5 years ago
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    is it clearer now ballards?

  34. anonymous
    • 5 years ago
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    thank you guys that makes since, but what if you have a log problem like this... f(x)=5x-xlnx

  35. anonymous
    • 5 years ago
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    it is clear how to figure out the relative rate of change. Could someone help me figure out the f'(x) of that log then i maybe could get it from there

  36. anonymous
    • 5 years ago
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    if f(x)=ln(x) f'(x)=1/x

  37. anonymous
    • 5 years ago
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    same story, find the derivative then divide it by the original function ^_^ and you'll get : f'(x) = 5 - [lnx + 1] ( use u'v + uv' to find the derivatice of xlnx) then divide it by the original : \[f'(x)/f(x) = (5-\ln|x|-1)/ 5x - xlnx\] and try to simplify :) give it a try now.

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