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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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sry, mistype, ill ask question now xP help with the attached file will be GREATLY appreciated, ty~
1 Attachment
take derivative of your function and equate it to 0. solve for x.

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once you get x, substitute in f(x) and get the extrema
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.
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.
ok, i got the derivative which is --> 2/x^3 but that = 0 there is no solution
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.
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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