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Dodo1 Group Title

Calculus question ; photo attached

  • one year ago
  • one year ago

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  1. Dodo1 Group Title
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    • one year ago
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  2. abb0t Group Title
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    Start by taking the derivative of your function. That will be your first step. Can you do that?

    • one year ago
  3. Dodo1 Group Title
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    9+(2/x^2)=0?

    • one year ago
  4. abb0t Group Title
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    Great. Now that you have that, solve for "x".

    • one year ago
  5. stamp Group Title
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    @abb0t that derivative is incorrect.

    • one year ago
  6. abb0t Group Title
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    Hmm, is it now? Well, @Dodo1 you need to try again then.

    • one year ago
  7. Dodo1 Group Title
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    how is it wrong?

    • one year ago
  8. stamp Group Title
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    @abb0t Good luck evaluation that term = 0 without using imaginary numbers. @Dodo1 Forgot his - sign when taking the derivative using the power rule

    • one year ago
  9. Dodo1 Group Title
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    9+(2x^(-2))=0?

    • one year ago
  10. stamp Group Title
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    no

    • one year ago
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    \[9x+2x^{-1}\]\[9+(-1)2x^{-1-1}\]\[9-2x^{-2}\]\[9-2/x^2\]

    • one year ago
  12. stamp Group Title
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    @Dodo1 How did you plan on solving 9 + 2/x^2 = 0? It is not possible using real numbers because any real number squared is positive, this was the first clue that something was wrong with the derivative

    • one year ago
  13. Dodo1 Group Title
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    oh not plus ok. I see

    • one year ago
  14. Dodo1 Group Title
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    next step is mutiply x^2 both side?

    • one year ago
  15. Dodo1 Group Title
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    @stamp

    • one year ago
  16. stamp Group Title
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    @Dodo1 It is algebra I at this point. Solve for x.

    • one year ago
  17. Dodo1 Group Title
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    x=Sqrt(9/2) @stamp

    • one year ago
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    \[9-2/x^2=0\]\[9=2/x^2\]\[x^2=2/9\]\[x=+/-(\sqrt2/3)\]

    • one year ago
  19. Dodo1 Group Title
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    Ops sorry i did 9X^2+2=0

    • one year ago
  20. Dodo1 Group Title
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    whats the next step? max at _sart(2)/3? @stamp

    • one year ago
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    You do not need to tag me in every response, Dodo, I am here. These values, x = +/- c, are critical points. They are max or mins, we do not know until we consider each point with respect to the second derivative. You have been familiarized with the second derivative test in lecture, now we are using it to see which of these critical points are max, mins, etc.

    • one year ago
  22. stamp Group Title
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    I have to go, but this should help you enough to get started. Somebody else should be able to help you.

    • one year ago
  23. Dodo1 Group Title
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    Yea, I do need to prepare derivative problems. Are you going?

    • one year ago
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    f(x) = http://www.wolframalpha.com/input/?i=9x%2B2%2Fx f'(x) = http://www.wolframalpha.com/input/?i=derivative+of+9x%2B2%2Fx f''(x) = http://www.wolframalpha.com/input/?i=second+derivative+of+9x%2B2%2Fx critical points where f'(x) is 0 = http://www.wolframalpha.com/input/?i=solve+derivative+of+9x%2B2%2Fx graph of f(x) http://www.wolframalpha.com/input/?i=plot+9x%2B2%2Fx Visually one can observe that the negative critical point is a maximum, while the positive critical point is a minimum. Analytically, you can observe the change of f''(x) to the left of the critical point and to the right of the critical point. Since f'( - c ) goes from + to - , - c is a maximum. Since f'( + c ) goes from - to +, + c is a minimum.

    • one year ago
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