Determine the maximum and minimum.
f(x)= 1+ (x+3)² ; -2

Mathematics
- anonymous

Determine the maximum and minimum.
f(x)= 1+ (x+3)² ; -2

Mathematics
- schrodinger

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- TuringTest

how does one find maxima and minima of a function?

- anonymous

first find the derivative
f'(x)= 2x+6

- Jhannybean

sure it isnt \(-2 \le x\le 6\) ?

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## More answers

- anonymous

yes @Jhannybean

- TuringTest

and what do we do with the derivative to find max and min (also known as critical points)?

- anonymous

x=-3

- anonymous

f(-3)=1
f(-2)=2
f(6)=82

- 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

\(f'(x)\)=0

- anonymous

yes i calculated it that way x=-3

- anonymous

and sorry i forgot to put \[-2 \le x \le 6\]

- TuringTest

and you still don't know the answer?

- TuringTest

you want the max and min values of the function; when is f(x) largest and smallest according to your calculations?

- 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

Use the points you tested to find your highest and lowest values.

- 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

@TuringTest so there a problem with the ansswerkey?

- 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

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

so what would the local minimum be?

- TuringTest

the local mina and max are also the global min and max in this case

- 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

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

Sorry, Ab max = (6,82), Ab min = (-2,2) hehehe

- TuringTest

well then we all agree :)
nice job solving it though!

- Jhannybean

thanks! I typoed earlier :c

- TuringTest

no worries, it happens to the best of us :P

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