anonymous
  • anonymous
Determine the maximum and minimum. f(x)= 1+ (x+3)² ; -2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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TuringTest
  • TuringTest
how does one find maxima and minima of a function?
anonymous
  • anonymous
first find the derivative f'(x)= 2x+6
Jhannybean
  • Jhannybean
sure it isnt \(-2 \le x\le 6\) ?

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anonymous
  • anonymous
yes @Jhannybean
TuringTest
  • TuringTest
and what do we do with the derivative to find max and min (also known as critical points)?
anonymous
  • anonymous
x=-3
anonymous
  • anonymous
f(-3)=1 f(-2)=2 f(6)=82
TuringTest
  • TuringTest
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)
TuringTest
  • TuringTest
\(f'(x)\)=0
anonymous
  • anonymous
yes i calculated it that way x=-3
anonymous
  • anonymous
and sorry i forgot to put \[-2 \le x \le 6\]
TuringTest
  • TuringTest
and you still don't know the answer?
TuringTest
  • TuringTest
you want the max and min values of the function; when is f(x) largest and smallest according to your calculations?
anonymous
  • anonymous
yes i get 6,82 as my max and -3,1 as my min BUT my answer key says maximum:82, minimum:6
Jhannybean
  • Jhannybean
Use the points you tested to find your highest and lowest values.
TuringTest
  • TuringTest
-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
anonymous
  • anonymous
@TuringTest so there a problem with the ansswerkey?
TuringTest
  • TuringTest
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.
TuringTest
  • TuringTest
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
anonymous
  • anonymous
so what would the local minimum be?
TuringTest
  • TuringTest
the local mina and max are also the global min and max in this case
Jhannybean
  • Jhannybean
\[\large f(x) = 1+(x+3)^2 \ \ , \ -2 \le x \le 6\]\[\large {f'(x) = 2(x+3)(1) =0 \\ f'(x)= 2x+6=0 }\]\[\large x = -3\] 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
  • TuringTest
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
Jhannybean
  • Jhannybean
Sorry, Ab max = (6,82), Ab min = (-2,2) hehehe
TuringTest
  • TuringTest
well then we all agree :) nice job solving it though!
Jhannybean
  • Jhannybean
thanks! I typoed earlier :c
TuringTest
  • TuringTest
no worries, it happens to the best of us :P

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