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

1. TuringTest

how does one find maxima and minima of a function?

2. burhan101

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

3. Jhannybean

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

4. burhan101

yes @Jhannybean

5. TuringTest

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

6. burhan101

x=-3

7. burhan101

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

8. 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)

9. TuringTest

\(f'(x)\)=0

10. burhan101

yes i calculated it that way x=-3

11. burhan101

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

12. TuringTest

and you still don't know the answer?

13. TuringTest

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

14. burhan101

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

15. Jhannybean

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

16. 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

17. burhan101

@TuringTest so there a problem with the ansswerkey?

18. 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.

19. 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

20. burhan101

so what would the local minimum be?

21. TuringTest

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

22. 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)

23. 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

24. Jhannybean

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

25. TuringTest

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

26. Jhannybean

thanks! I typoed earlier :c

27. TuringTest

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