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and i know that the derivative is f'(x) = -1/x^2 + 1/x but i dont know where to go from there... lol
Where do extreme values occur?
i dont know thats what i need help finding.. lol. i just dont know how to find them.
okay, so in theory, extreme values will occur where the derivative of the function is equal to zero (i.e. a horizontal slope where there is maxima or minima), and they also occur where the derivative is undefined.
okay sooo i set the derivative equal to zero...
well i mean plugging in 1 for x would give you zero.
okay, so 1 is one of our critical points
now what do we do?
Where is f'(x) = -1/x^2 + 1/x undefined?
when x = 0?
exactly, but if you would notice in the question it gave us restrictions of 0.5 ≤ x ≤ 4, so we dont take 0 into account... So our critical point is 1
oooh ok. so for the answer it says: max value is 1/4 + ln4 at x = 4 min value is 1 at x = 1 local max at (1/2, 2 - ln2) how did they get the max value and local max?
Well for the max value, they just put the biggest number they could, 4, into the function, they chose 4 because of the restrictions 0.5 ≤ x ≤ 4... 4 is the biggest number, i.e. giving the biggest value.
there is no local max... so i'm not sure where they got that.. might want to ask your teacher.
oh.. ok. and how do they know that 1 is the min value? this is confusing for me ):
We already got that the minimum is at x = 1
yea but how do you know its the minimum? :\
We know it is the minimum because at that point, the derivative = 0, this is the lowest point on the curve.