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burhan101

  • one year ago

Determine the extreme values:

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  1. burhan101
    • one year ago
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    \[\huge y= \frac{ e^{x} }{ 1+e^{x} }, x \epsilon [0,4]\]

  2. fozia
    • one year ago
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    values maximum, minimum

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

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

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

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

  7. fozia
    • one year ago
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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

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

  9. tcarroll010
    • one year ago
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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.

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

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

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

  14. Abhishek619
    • one year ago
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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.

  15. tcarroll010
    • one year ago
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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.

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

  17. tcarroll010
    • one year ago
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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.

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

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

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

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

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

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

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

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

  26. burhan101
    • one year ago
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    ohhhh is see it

  27. tcarroll010
    • one year ago
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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.

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

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

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

  31. burhan101
    • one year ago
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    yes thanks !!

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

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

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

  35. burhan101
    • one year ago
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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

  36. dan815
    • one year ago
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    YO BURHAANN MY BRO

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

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

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

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

  41. dan815
    • one year ago
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    ok

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

  43. burhan101
    • one year ago
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    yes

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

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

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

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

  48. dan815
    • one year ago
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    yes

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

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

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

  52. burhan101
    • one year ago
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    the vertex

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

  54. burhan101
    • one year ago
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    0

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

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

  57. dan815
    • one year ago
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    what nooo

  58. burhan101
    • one year ago
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    ohh:(

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

  60. burhan101
    • one year ago
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    0=cos4x?

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

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

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

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

  65. dan815
    • one year ago
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    must*

  66. dan815
    • one year ago
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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

  67. burhan101
    • one year ago
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    okay

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

  69. dan815
    • one year ago
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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

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

  71. Loser66
    • one year ago
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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.

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

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