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burhan101

  • 2 years ago

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

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  1. TuringTest
    • 2 years ago
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    how does one find maxima and minima of a function?

  2. burhan101
    • 2 years ago
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    first find the derivative f'(x)= 2x+6

  3. Jhannybean
    • 2 years ago
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    sure it isnt \(-2 \le x\le 6\) ?

  4. burhan101
    • 2 years ago
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    yes @Jhannybean

  5. TuringTest
    • 2 years ago
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    and what do we do with the derivative to find max and min (also known as critical points)?

  6. burhan101
    • 2 years ago
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    x=-3

  7. burhan101
    • 2 years ago
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    f(-3)=1 f(-2)=2 f(6)=82

  8. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    \(f'(x)\)=0

  10. burhan101
    • 2 years ago
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    yes i calculated it that way x=-3

  11. burhan101
    • 2 years ago
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    and sorry i forgot to put \[-2 \le x \le 6\]

  12. TuringTest
    • 2 years ago
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    and you still don't know the answer?

  13. TuringTest
    • 2 years ago
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    you want the max and min values of the function; when is f(x) largest and smallest according to your calculations?

  14. burhan101
    • 2 years ago
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    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
    • 2 years ago
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    Use the points you tested to find your highest and lowest values.

  16. TuringTest
    • 2 years ago
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    -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
    • 2 years ago
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    @TuringTest so there a problem with the ansswerkey?

  18. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    so what would the local minimum be?

  21. TuringTest
    • 2 years ago
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    the local mina and max are also the global min and max in this case

  22. Jhannybean
    • 2 years ago
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    \[\large f(x) = 1+(x+3)^2 \ \ , \ -2 \le x \le 6\]\[\large {f'(x) = 2(x+3)(1) =0 \\ f'(x)= 2x+6=0 }\]\[\large x = -3\] 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
    • 2 years ago
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    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
    • 2 years ago
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    Sorry, Ab max = (6,82), Ab min = (-2,2) hehehe

  25. TuringTest
    • 2 years ago
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    well then we all agree :) nice job solving it though!

  26. Jhannybean
    • 2 years ago
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    thanks! I typoed earlier :c

  27. TuringTest
    • 2 years ago
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    no worries, it happens to the best of us :P

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