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anonymous

  • 5 years ago

I don't understand how to find the absolute extrema on an open interval.

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  1. anonymous
    • 5 years ago
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    on an open interval?

  2. anonymous
    • 5 years ago
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    For example, if the domain is on the interval negative infinity to positive infinity. Not a closed interval.

  3. anonymous
    • 5 years ago
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    well, that interval is not closed...

  4. anonymous
    • 5 years ago
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    i understand that intuition may tell you that it is

  5. anonymous
    • 5 years ago
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    f(x) = 4x^3- 3x^4 on the interval negative infintiy to positive infinity

  6. anonymous
    • 5 years ago
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    okay

  7. anonymous
    • 5 years ago
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    that is easier to approach

  8. anonymous
    • 5 years ago
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    first note that \[\lim_{x\rightarrow \infty}f(x)=\lim_{x\rightarrow- \infty}f(x)=-\infty\]

  9. anonymous
    • 5 years ago
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    do you see that? or should we discuss it?

  10. anonymous
    • 5 years ago
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    My prof. taught us to find the derivative and at the critical points, weshould look how thederivative behaves, but that doesn't always work..

  11. anonymous
    • 5 years ago
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    yes, but we are doing some preliminary investigation about the behavior of this function

  12. anonymous
    • 5 years ago
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    my point is, this function will have an absolute max because of the result of those limits

  13. anonymous
    • 5 years ago
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    oh, can you please explain that

  14. anonymous
    • 5 years ago
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    the give f(x) is a polynomial, yes?

  15. anonymous
    • 5 years ago
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    correct

  16. anonymous
    • 5 years ago
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    of degree 4

  17. anonymous
    • 5 years ago
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    so its end behavior will be the same at \[\pm\infty\]

  18. anonymous
    • 5 years ago
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    yes, I understand that

  19. anonymous
    • 5 years ago
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    in this case since the leading coefficient is (-1), both ends go to \[-\infty\]

  20. anonymous
    • 5 years ago
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    okay so whatever this function does, it can not have an "absolute" minimum right?

  21. anonymous
    • 5 years ago
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    no, it's suppose to only have an absolute max

  22. anonymous
    • 5 years ago
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    right

  23. anonymous
    • 5 years ago
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    so, find the critical points

  24. anonymous
    • 5 years ago
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    i.e. where \[\frac{df}{dx}=0\]

  25. anonymous
    • 5 years ago
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    its at x = 1 max = 1

  26. anonymous
    • 5 years ago
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    So, I understood we always look at the leading coefficient of the polynomial?

  27. anonymous
    • 5 years ago
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    yes, it as well as the degree will tell you a lot

  28. anonymous
    • 5 years ago
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    yes I mean that too

  29. anonymous
    • 5 years ago
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    Thank you so much! Just one last question..

  30. anonymous
    • 5 years ago
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    what level of calculus are you studying and what are your future plans in math?

  31. anonymous
    • 5 years ago
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    this is calc 1 and the onlysemester of calc i have to take.. thankfully! I'm a pharmacy major..

  32. anonymous
    • 5 years ago
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    oh ok

  33. anonymous
    • 5 years ago
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    good luck then, i have had many students that were pre-pharm

  34. anonymous
    • 5 years ago
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    Thank you. One last question, if the equation was f(x) = 2x^3 - 6x +2.. the polynomial is of degree 3 and leading coef is positive

  35. anonymous
    • 5 years ago
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    so the behavior of function: from negative infinty to positive infinity..

  36. anonymous
    • 5 years ago
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    and the critical points are -1 and 1

  37. anonymous
    • 5 years ago
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    but the solution says there is no extrema and I don'tunderstand why

  38. anonymous
    • 5 years ago
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    right, so it will have no "absolute" extrema

  39. anonymous
    • 5 years ago
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    do you see why?

  40. anonymous
    • 5 years ago
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    No I can't seem tosee it

  41. anonymous
    • 5 years ago
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    its all in that one word "absolute", absolute extreme , implies that the function achieves no values greater/less than, respectively, that value

  42. anonymous
    • 5 years ago
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    if your function runs off to \[+\infty\] one way and \[-\infty\] the other, then there is no absolute max/min

  43. anonymous
    • 5 years ago
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    There are however possible "local" max/min which is where many students get confused

  44. anonymous
    • 5 years ago
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    Oh okay, I will look into it more... Thank you so much foryour help! I really appreciate it!

  45. anonymous
    • 5 years ago
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    awesome, good luck!

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