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anonymous

  • 5 years ago

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  1. anonymous
    • 5 years ago
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    h

  2. anonymous
    • 5 years ago
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    sry, mistype, ill ask question now xP help with the attached file will be GREATLY appreciated, ty~

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  3. anonymous
    • 5 years ago
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    take derivative of your function and equate it to 0. solve for x.

  4. anonymous
    • 5 years ago
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    once you get x, substitute in f(x) and get the extrema

  5. anonymous
    • 5 years ago
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    Hopefully you can recognize the graph of the function. (-1/x^2) is increasing as x increases and decreases as x decreases. So the maxima and minima are at the extremes of the intervals.

  6. anonymous
    • 5 years ago
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    His in the right pace. Since this is in a closed interval we know that the absolute min and max must occur in a critical point ( when f'(x)=0) or in the end points. Once you have all your x values just plug it in to the original function f(x). The largest values is your absolute max and your smallest is the absolute min.

  7. anonymous
    • 5 years ago
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    ok, i got the derivative which is --> 2/x^3 but that = 0 there is no solution

  8. anonymous
    • 5 years ago
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    its alright, i got the answer.... using the derivative i found that within the domain there is no value which 'does not exist' or gives 'zero' so i plugged in x=0.25 and x=5 into the original formula and found the extrema.

  9. anonymous
    • 5 years ago
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    that is absolutely right, as Romero said, Since this is in a closed interval we know that the absolute min and max must occur in a critical point ( when f'(x)=0) or in the end points.

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