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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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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.