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TuringTestBest ResponseYou've already chosen the best response.1
how does one find maxima and minima of a function?
 10 months ago

burhan101Best ResponseYou've already chosen the best response.0
first find the derivative f'(x)= 2x+6
 10 months ago

JhannybeanBest ResponseYou've already chosen the best response.3
sure it isnt \(2 \le x\le 6\) ?
 10 months ago

TuringTestBest ResponseYou've already chosen the best response.1
and what do we do with the derivative to find max and min (also known as critical points)?
 10 months ago

burhan101Best ResponseYou've already chosen the best response.0
f(3)=1 f(2)=2 f(6)=82
 10 months ago

TuringTestBest ResponseYou've already chosen the best response.1
the critical points are found when the derivative is equal to zero (unless they are at the endpoints of the function, which you don;t need to worry about if you have no greater than or equal sign)
 10 months ago

burhan101Best ResponseYou've already chosen the best response.0
yes i calculated it that way x=3
 10 months ago

burhan101Best ResponseYou've already chosen the best response.0
and sorry i forgot to put \[2 \le x \le 6\]
 10 months ago

TuringTestBest ResponseYou've already chosen the best response.1
and you still don't know the answer?
 10 months ago

TuringTestBest ResponseYou've already chosen the best response.1
you want the max and min values of the function; when is f(x) largest and smallest according to your calculations?
 10 months ago

burhan101Best ResponseYou've already chosen the best response.0
yes i get 6,82 as my max and 3,1 as my min BUT my answer key says maximum:82, minimum:6
 10 months ago

JhannybeanBest ResponseYou've already chosen the best response.3
Use the points you tested to find your highest and lowest values.
 10 months ago

TuringTestBest ResponseYou've already chosen the best response.1
3 is not an option for x, but that leaves x=2, which still does not give 6... so I still do not see how that can be right
 10 months ago

burhan101Best ResponseYou've already chosen the best response.0
@TuringTest so there a problem with the ansswerkey?
 10 months ago

TuringTestBest ResponseYou've already chosen the best response.1
It's the only thing I see possible, and it certainly wouldn't be the first time I've seen the answer key be wrong.
 10 months ago

TuringTestBest ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=plot+f%28x%29%3D+1%2B+%28x%2B3%29%C2%B2+%3B+2%3Cx%3C6 how 6 is a minimum in this graph is beyond me
 10 months ago

burhan101Best ResponseYou've already chosen the best response.0
so what would the local minimum be?
 10 months ago

TuringTestBest ResponseYou've already chosen the best response.1
the local mina and max are also the global min and max in this case
 10 months ago

JhannybeanBest ResponseYou've already chosen the best response.3
\[\large f(x) = 1+(x+3)^2 \ \ , \ 2 \le x \le 6\]\[\large {f'(x) = 2(x+3)(1) =0 \\ f'(x)= 2x+6=0 }\]\[\large x = 3\] Test your end points and your critical points with the help of The Extreme Value Theorem. 1. your function is continuous since it's a polynomial 2. x=3 does not fall between the interval [2,6] so cannot be tested. 3.Test your end points \[\large {f(2) = 1+((2)+3)^2 = 2 }\]\[\large f(6) = 1+ ((6)+3)^2= 82\] Ab max = (2,2) Ab Min = (6,82)
 10 months ago

TuringTestBest ResponseYou've already chosen the best response.1
I agree, and think the answer key is wrong. I would guess it a typo where someone wrote the x=6 which corresponds to the max instead of the min. It really is not that uncommon; books can be mistaken :p
 10 months ago

JhannybeanBest ResponseYou've already chosen the best response.3
Sorry, Ab max = (6,82), Ab min = (2,2) hehehe
 10 months ago

TuringTestBest ResponseYou've already chosen the best response.1
well then we all agree :) nice job solving it though!
 10 months ago

JhannybeanBest ResponseYou've already chosen the best response.3
thanks! I typoed earlier :c
 10 months ago

TuringTestBest ResponseYou've already chosen the best response.1
no worries, it happens to the best of us :P
 10 months ago
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