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burhan101 Group Title

Determine the extreme values:

  • one year ago
  • one year ago

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  1. burhan101 Group Title
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    \[\huge y= \frac{ e^{x} }{ 1+e^{x} }, x \epsilon [0,4]\]

    • one year ago
  2. fozia Group Title
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    values maximum, minimum

    • one year ago
  3. burhan101 Group Title
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    yes, how tho?

    • one year ago
  4. burhan101 Group Title
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    i find the derivative first and then what?

    • one year ago
  5. fozia Group Title
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    first fint derivative that gives the result y'=ex/(1+ex)2 then eqaute it to zero gives ex=0

    • one year ago
  6. burhan101 Group Title
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    \[\huge y'=\frac{ e^x }{ (1+e^x) }\]

    • one year ago
  7. fozia Group Title
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    take 2nd derivatv now put the value of x =0 in 2nd derivativ if relt is postv then its min otherwise max

    • one year ago
  8. burhan101 Group Title
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    @fozia wait, you and I got different derivatives

    • one year ago
  9. tcarroll010 Group Title
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    @fozia , you have correctly identified the first derivative, but it will never be equal to zero. Therefore, there is no need to take the second derivative. It is sufficient to just analyze the first derivative, realize that it is always positive, and that the original function is always increasing, so it has no extreme values.

    • one year ago
  10. burhan101 Group Title
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    • one year ago
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  11. Abhishek619 Group Title
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    check if the function is increasing or decreasing?

    • one year ago
  12. tcarroll010 Group Title
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    Here's a graph to show you that it is always increasing

    • one year ago
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  13. burhan101 Group Title
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    how do i check if its inc/dec?

    • one year ago
  14. Abhishek619 Group Title
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    its first derivative is always a positive, hence it is increasing for all [0.4]. the minimum value will be at x=0, maximu will be at x=4.

    • one year ago
  15. tcarroll010 Group Title
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    Read my first post. That will tell you. That numerator of the first derivative is always positive. So is the denominator.

    • one year ago
  16. tcarroll010 Group Title
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    No extreme values for domain of all x. min and max though at the endpoints for the given problem domain.

    • one year ago
  17. tcarroll010 Group Title
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    That's because "e" is a positive base, so any exponent you put on it will result in a positive number.

    • one year ago
  18. tcarroll010 Group Title
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    Do you see that fozia's derivative is correct? That's where you should start.

    • one year ago
  19. tcarroll010 Group Title
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    Use the quotient rule to get the derivative. Do you need help with that?

    • one year ago
  20. tcarroll010 Group Title
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    @burhan101 I have 2 pending questions posted for you here.

    • one year ago
  21. burhan101 Group Title
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    @tcarroll010 im trying to figure out how that derivative is right

    • one year ago
  22. burhan101 Group Title
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    & yes i am using the quotient rule

    • one year ago
  23. burhan101 Group Title
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    \[\huge \frac{ (1+e^x)(e^x)-(e^x)(e^x)}{ (1+e^x)^2 }\]

    • one year ago
  24. tcarroll010 Group Title
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    \[\frac{ (1 + e ^{x})e ^{x} - e ^{x}e ^{x} }{ (1 + e ^{x})^{2} } = \frac{ e ^{x} + e ^{2x} - e ^{2x} }{ (1 + e ^{x})^{2} }\]

    • one year ago
  25. tcarroll010 Group Title
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    In the numerator, you are left with only e^x and the denominator is the original denominator squared.

    • one year ago
  26. burhan101 Group Title
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    ohhhh is see it

    • one year ago
  27. tcarroll010 Group Title
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    So, there is no need to take the second derivative. It is sufficient to just analyze the first derivative, realize that it is always positive, and that the original function is always increasing, so it has no extreme values, if your domain is all x. But your domain is limited, so you have your min and max at the x-value endpoints.

    • one year ago
  28. burhan101 Group Title
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    but dont i plug in 0 and 4 in f(x) ?

    • one year ago
  29. tcarroll010 Group Title
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    So, min at: e^0 / (1 + e^0) max at: e^4 / (1 + e^4)

    • one year ago
  30. tcarroll010 Group Title
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    All good now, @burhan101 ?

    • one year ago
  31. burhan101 Group Title
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    yes thanks !!

    • one year ago
  32. tcarroll010 Group Title
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    uw! Good luck in all of your studies and thx for the recognition! @burhan101

    • one year ago
  33. burhan101 Group Title
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    thank you very much, you do explain things very clear !

    • one year ago
  34. tcarroll010 Group Title
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    I just like to help. Kind words on your part!

    • one year ago
  35. burhan101 Group Title
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    For this function \[\huge f(x)= 2\sin4x+3 ; [0, \pi]\] i get the derivative as \[\huge f'(x)=8\cos4x \] To find the extreme values, do i plug in the endpoints into f(x) ? @tcarroll010

    • one year ago
  36. dan815 Group Title
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    YO BURHAANN MY BRO

    • one year ago
  37. dan815 Group Title
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    are u done with this equation

    • one year ago
  38. burhan101 Group Title
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    dannn:) ! no im not, do i plug the endpoints into the function ?

    • one year ago
  39. dan815 Group Title
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    explain the meaning of an extreme value to me

    • one year ago
  40. burhan101 Group Title
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    an absolute mx or an absolute min

    • one year ago
  41. dan815 Group Title
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    ok

    • one year ago
  42. dan815 Group Title
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    so that means the slope there is 0

    • one year ago
  43. burhan101 Group Title
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    yes

    • one year ago
  44. dan815 Group Title
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    but u must check the end points too incase its just a local max or local min there

    • one year ago
  45. dan815 Group Title
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    first solve f'(x) = 0

    • one year ago
  46. dan815 Group Title
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    |dw:1371749721969:dw|

    • one year ago
  47. burhan101 Group Title
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    so i solve for x

    • one year ago
  48. dan815 Group Title
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    yes

    • one year ago
  49. dan815 Group Title
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    okay lemme give u some simpler exmaples

    • one year ago
  50. dan815 Group Title
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    |dw:1371749857123:dw|

    • one year ago
  51. dan815 Group Title
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    what are the extreme values for a parabola

    • one year ago
  52. burhan101 Group Title
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    the vertex

    • one year ago
  53. dan815 Group Title
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    okay and the slope there is?

    • one year ago
  54. burhan101 Group Title
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    0

    • one year ago
  55. dan815 Group Title
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    so that means the derivative there is?

    • one year ago
  56. burhan101 Group Title
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    |dw:1371749935529:dw|

    • one year ago
  57. dan815 Group Title
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    what nooo

    • one year ago
  58. burhan101 Group Title
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    ohh:(

    • one year ago
  59. dan815 Group Title
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    0/8 = 0|dw:1371749995868:dw|

    • one year ago
  60. burhan101 Group Title
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    0=cos4x?

    • one year ago
  61. dan815 Group Title
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    @bahrom7893 http://multiplayerchess.com/#!/5gqIqmm

    • one year ago
  62. dan815 Group Title
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    hello bahrom jhanny wants us to play a game

    • one year ago
  63. burhan101 Group Title
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    what's the next step ..

    • one year ago
  64. dan815 Group Title
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    1what values of x wud make cos4x=0 think about cos pi/2=0 so what much it be

    • one year ago
  65. dan815 Group Title
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    must*

    • one year ago
  66. dan815 Group Title
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    ill give u one of the values that is in between 0 and pi cos(4(pi/8)) = 0 thats one

    • one year ago
  67. burhan101 Group Title
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    okay

    • one year ago
  68. burhan101 Group Title
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    would i keep my calculator in degrees or radians?

    • one year ago
  69. dan815 Group Title
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    heyy dont be silly u shud know this stuff!! we been doing trig of a while remember where cosx graph = 0

    • one year ago
  70. burhan101 Group Title
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    |dw:1371751462417:dw|

    • one year ago
  71. Loser66 Group Title
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    @dan815 I don't know why everybody here ignores the ends of interval when they find out the max, min. My prof always asked us to consider those points, too.

    • one year ago
  72. dan815 Group Title
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    i told him to look at those points too :) but its not really needed in this case

    • one year ago
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