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ksandoval
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use analytic methods to find the extreme values of f(x)= (1/x) + lnx on the interval 0.5 ≤ x ≤ 4 and where they occur
 one year ago
 one year ago
ksandoval Group Title
use analytic methods to find the extreme values of f(x)= (1/x) + lnx on the interval 0.5 ≤ x ≤ 4 and where they occur
 one year ago
 one year ago

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ksandoval Group TitleBest ResponseYou've already chosen the best response.0
and i know that the derivative is f'(x) = 1/x^2 + 1/x but i dont know where to go from there... lol
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
Where do extreme values occur?
 one year ago

ksandoval Group TitleBest ResponseYou've already chosen the best response.0
i dont know thats what i need help finding.. lol. i just dont know how to find them.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

ksandoval Group TitleBest ResponseYou've already chosen the best response.0
okay sooo i set the derivative equal to zero...
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
and...?
 one year ago

ksandoval Group TitleBest ResponseYou've already chosen the best response.0
well i mean plugging in 1 for x would give you zero.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
okay, so 1 is one of our critical points
 one year ago

ksandoval Group TitleBest ResponseYou've already chosen the best response.0
now what do we do?
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
Where is f'(x) = 1/x^2 + 1/x undefined?
 one year ago

ksandoval Group TitleBest ResponseYou've already chosen the best response.0
when x = 0?
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
http://screencast.com/t/LJnPiplk
 one year ago

ksandoval Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
there is no local max... so i'm not sure where they got that.. might want to ask your teacher.
 one year ago

ksandoval Group TitleBest ResponseYou've already chosen the best response.0
oh.. ok. and how do they know that 1 is the min value? this is confusing for me ):
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
We already got that the minimum is at x = 1
 one year ago

ksandoval Group TitleBest ResponseYou've already chosen the best response.0
yea but how do you know its the minimum? :\
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.1
We know it is the minimum because at that point, the derivative = 0, this is the lowest point on the curve.
 one year ago

ksandoval Group TitleBest ResponseYou've already chosen the best response.0
thank you
 one year ago
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