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TrojanPoem

  • one year ago

Find the intervals of upwards concavity and downwards concavity and turning points if exists for the curve of function f as f(x) = x^2 +9 /x and find the absolute maximum and minimum for the function when x (- [1, 6]

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  1. anonymous
    • one year ago
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    start by taking the first and second derivatives

  2. TrojanPoem
    • one year ago
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    I know the steps, the problem lays in this example specifically. No turning points although there is upwards and downwards concavity.

  3. anonymous
    • one year ago
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    there is a vertical asymptote where denominator is equal to 0

  4. TrojanPoem
    • one year ago
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    Clarify a bit .

  5. TrojanPoem
    • one year ago
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    You mean something like this ? http://hotmath.com/hotmath_help/topics/rational-functions/image008.gif

  6. TrojanPoem
    • one year ago
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    So there is no turning points as the curve is separated into two parts ?

  7. anonymous
    • one year ago
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    \[f(x)=\frac{x^2+9}{x}=x+\frac{9}{x}\] right?

  8. TrojanPoem
    • one year ago
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    yep.

  9. TrojanPoem
    • one year ago
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    I will watch the video while answering

  10. anonymous
    • one year ago
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    \[f'(x)=1-\frac{9}{x^2}\]find the interval over what \(f'\) is positive, that will tell you where it \(f\)is increasing

  11. anonymous
    • one year ago
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    i always graph these problems first to get an idea of what im looking to identify

  12. anonymous
    • one year ago
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    as for the absolute max on \([1,6]\) check the critical point (where \(f'(x)=0\) and also check \(f(1)\) and \(f(6)\)

  13. TrojanPoem
    • one year ago
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    Good way of doing it, x = 0 -8, 0 > dec 0 , 8 increase 8 = infinite

  14. TrojanPoem
    • one year ago
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    From now , I will graph it too.

  15. TrojanPoem
    • one year ago
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    WHere is the turning point satellite ?

  16. anonymous
    • one year ago
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    in any case if you want a nice picture and pretty much everything else use this http://www.wolframalpha.com/input/?i=%28x^2%2B9%29%2Fx

  17. TrojanPoem
    • one year ago
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    turning point satellite ?

  18. anonymous
    • one year ago
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    if "turning point" means where it goes from increasing to decreasing and vice versa, those are the zeros of the derivative

  19. TrojanPoem
    • one year ago
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    Fine, so x= ?

  20. anonymous
    • one year ago
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    idk i didn't do it solve \[1-\frac{9}{x^2}=0\] and you will get them can probably do it in your head

  21. TrojanPoem
    • one year ago
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    @billj5 , Finding the asymptote takes decade !

  22. TrojanPoem
    • one year ago
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    This is the first derivative , we need the second one. and x = 0

  23. anonymous
    • one year ago
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    lol its not too bad once you get the hang of it

  24. TrojanPoem
    • one year ago
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    At least , I understood why x = 0 was not a turning point :D , Thanks !

  25. TrojanPoem
    • one year ago
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    Hey , may you help me with another question ? Will open new one.

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