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TuringTest Group Title

Suppose that \(f\) satisfies the equation \[f(x+y)=f(x)+f(y)+x^2y+xy^2\]for all real numbers \(x\) and \(y\). Suppose further that\[~~~~~~~~~~~~~\lim_{x\to0}\frac{f(x)}x=1\]a) Find \(f(0)\) b) Find \(f'(0)\) c) Find \(f'(x)\)

  • 2 years ago
  • 2 years ago

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  1. TuringTest Group Title
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    I found \(f(0)=0\) but now I am not seeing how to proceed. Just a hint would be nice.

    • 2 years ago
  2. TuringTest Group Title
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    I further think I proved that the function is odd, though I think that doesn't really help me :/

    • 2 years ago
  3. Zarkon Group Title
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    use the definition of a derivative

    • 2 years ago
  4. Zarkon Group Title
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    the fact that it is odd does seem to help with the last part

    • 2 years ago
  5. TuringTest Group Title
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    I was thinking of using the definition of the derivative somehow but couldn't figure out how to get it into the right form....

    • 2 years ago
  6. Zarkon Group Title
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    \[f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}\]

    • 2 years ago
  7. TuringTest Group Title
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    limit f(x)-f(0) something...

    • 2 years ago
  8. TuringTest Group Title
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    ah dang but... oh, okay I guess I was trying to get things algebraically manipulated there, but I see now that since x=f(x) at x=0 the limit \(is\) the definition got it :)

    • 2 years ago
  9. TuringTest Group Title
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    okay so that is f'(0)=1, on to the last part you say the oddness could help?

    • 2 years ago
  10. Zarkon Group Title
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    yes

    • 2 years ago
  11. TuringTest Group Title
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    can I use the derivative of an odd function is even?

    • 2 years ago
  12. Zarkon Group Title
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    don't need it

    • 2 years ago
  13. Zarkon Group Title
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    use the definition of a derivative again

    • 2 years ago
  14. TuringTest Group Title
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    that's what I'm doing, let me toy for a sec...

    • 2 years ago
  15. Zarkon Group Title
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    I found f(x) ;)

    • 2 years ago
  16. TuringTest Group Title
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    haha, dang okay then go ahead and give me a little more help on finding f'(x) they don't even ask for f(x), jeez

    • 2 years ago
  17. TuringTest Group Title
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    ...if you would be so kind

    • 2 years ago
  18. Zarkon Group Title
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    if you know f'(x) and that f(0)=0 then finding f(x) is really easy :)

    • 2 years ago
  19. Zarkon Group Title
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    \[f'(x)=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}\]

    • 2 years ago
  20. Zarkon Group Title
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    \[f(y)-f(x)=f(y)+f(-x)\]

    • 2 years ago
  21. Zarkon Group Title
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    does that help?

    • 2 years ago
  22. Zarkon Group Title
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    ok

    • 2 years ago
  23. KingGeorge Group Title
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    If it helps, it's easier for me to see it when I write the derivative as \[f'(x)=\lim_{y\to0} \frac{f(x+y)-f(x)}{y}\]

    • 2 years ago
  24. TuringTest Group Title
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    that would be \[\lim_{y\to0}{f(y)+x^2y+xy^2\over y}=1+\lim...\]ah... it's \(f'(x)=x^2+1\), right ? :)

    • 2 years ago
  25. Zarkon Group Title
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    yes

    • 2 years ago
  26. TuringTest Group Title
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    sweet, thanks y'all I still want to see how your way was gonna go @Zarkon , but I'll work on that myself for a while

    • 2 years ago
  27. Zarkon Group Title
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    gotta go...fun problem...I'll post it later if you like

    • 2 years ago
  28. TuringTest Group Title
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    always a pleasure :)

    • 2 years ago
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