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anonymous
 5 years ago
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anonymous
 5 years ago
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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sry, mistype, ill ask question now xP help with the attached file will be GREATLY appreciated, ty~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0take derivative of your function and equate it to 0. solve for x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0once you get x, substitute in f(x) and get the extrema

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hopefully 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0His 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, i got the derivative which is > 2/x^3 but that = 0 there is no solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that 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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