Question

Determine the maximum and minimum.
f(x)= 1+ (x+3)² ; -2 12 years ago

Answer 1

how does one find maxima and minima of a function?

Answer 2

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

Answer 3

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

Answer 4

yes @Jhannybean

Answer 5

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

Answer 6

x=-3

Answer 7

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

Answer 8

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)

Answer 9

\(f'(x)\)=0

Answer 10

yes i calculated it that way x=-3

Answer 11

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

Answer 12

and you still don't know the answer?

Answer 13

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

Answer 14

yes i get 6,82 as my max and -3,1 as my min BUT my answer key says maximum:82, minimum:6

Answer 15

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

Answer 16

-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

Answer 17

@TuringTest so there a problem with the ansswerkey?

Answer 18

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.

Answer 19

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

Answer 20

so what would the local minimum be?

Answer 21

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

Answer 22

\[\large f(x) = 1+(x+3)^2 \ \ , \ -2 \le x \le 6\]\[\large {f'(x) = 2(x+3)(1) =0 \\ f'(x)= 2x+6=0 }\]\[\large x = -3\] 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)

Answer 23

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

Answer 24

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

Answer 25

well then we all agree :)

nice job solving it though!

Answer 26

thanks! I typoed earlier :c

Answer 27

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