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burhan101
 one year ago
Determine the maximum and minimum.
f(x)= 1+ (x+3)² ; 2<x<6
burhan101
 one year ago
Determine the maximum and minimum. f(x)= 1+ (x+3)² ; 2<x<6

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TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1how does one find maxima and minima of a function?

burhan101
 one year ago
Best ResponseYou've already chosen the best response.0first find the derivative f'(x)= 2x+6

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3sure it isnt \(2 \le x\le 6\) ?

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

burhan101
 one year ago
Best ResponseYou've already chosen the best response.0f(3)=1 f(2)=2 f(6)=82

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1the 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)

burhan101
 one year ago
Best ResponseYou've already chosen the best response.0yes i calculated it that way x=3

burhan101
 one year ago
Best ResponseYou've already chosen the best response.0and sorry i forgot to put \[2 \le x \le 6\]

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1and you still don't know the answer?

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

burhan101
 one year ago
Best ResponseYou've already chosen the best response.0yes i get 6,82 as my max and 3,1 as my min BUT my answer key says maximum:82, minimum:6

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Use the points you tested to find your highest and lowest values.

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.13 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

burhan101
 one year ago
Best ResponseYou've already chosen the best response.0@TuringTest so there a problem with the ansswerkey?

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

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1http://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

burhan101
 one year ago
Best ResponseYou've already chosen the best response.0so what would the local minimum be?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1the local mina and max are also the global min and max in this case

Jhannybean
 one year ago
Best 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)

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1I 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

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Sorry, Ab max = (6,82), Ab min = (2,2) hehehe

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1well then we all agree :) nice job solving it though!

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3thanks! I typoed earlier :c

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1no worries, it happens to the best of us :P
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