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Dodo1

  • 2 years ago

Calculus question ; photo attached

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

  3. Dodo1
    • 2 years ago
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    9+(2/x^2)=0?

  4. abb0t
    • 2 years ago
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    Great. Now that you have that, solve for "x".

  5. stamp
    • 2 years ago
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    @abb0t that derivative is incorrect.

  6. abb0t
    • 2 years ago
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    Hmm, is it now? Well, @Dodo1 you need to try again then.

  7. Dodo1
    • 2 years ago
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    how is it wrong?

  8. stamp
    • 2 years ago
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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

  9. Dodo1
    • 2 years ago
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    9+(2x^(-2))=0?

  10. stamp
    • 2 years ago
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    no

  11. stamp
    • 2 years ago
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    \[9x+2x^{-1}\]\[9+(-1)2x^{-1-1}\]\[9-2x^{-2}\]\[9-2/x^2\]

  12. stamp
    • 2 years ago
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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

  13. Dodo1
    • 2 years ago
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    oh not plus ok. I see

  14. Dodo1
    • 2 years ago
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    next step is mutiply x^2 both side?

  15. Dodo1
    • 2 years ago
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    @stamp

  16. stamp
    • 2 years ago
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    @Dodo1 It is algebra I at this point. Solve for x.

  17. Dodo1
    • 2 years ago
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    x=Sqrt(9/2) @stamp

  18. stamp
    • 2 years ago
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    \[9-2/x^2=0\]\[9=2/x^2\]\[x^2=2/9\]\[x=+/-(\sqrt2/3)\]

  19. Dodo1
    • 2 years ago
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    Ops sorry i did 9X^2+2=0

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

  21. stamp
    • 2 years 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.

  22. stamp
    • 2 years ago
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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.

  23. Dodo1
    • 2 years ago
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    Yea, I do need to prepare derivative problems. Are you going?

  24. stamp
    • 2 years 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.

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