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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

$\huge y= \frac{ e^{x} }{ 1+e^{x} }, x \epsilon [0,4]$

2. fozia Group Title

values maximum, minimum

3. burhan101 Group Title

yes, how tho?

4. burhan101 Group Title

i find the derivative first and then what?

5. fozia Group Title

first fint derivative that gives the result y'=ex/(1+ex)2 then eqaute it to zero gives ex=0

6. burhan101 Group Title

$\huge y'=\frac{ e^x }{ (1+e^x) }$

7. fozia Group Title

take 2nd derivatv now put the value of x =0 in 2nd derivativ if relt is postv then its min otherwise max

8. burhan101 Group Title

@fozia wait, you and I got different derivatives

9. tcarroll010 Group Title

@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 Group Title

11. Abhishek619 Group Title

check if the function is increasing or decreasing?

12. tcarroll010 Group Title

Here's a graph to show you that it is always increasing

13. burhan101 Group Title

how do i check if its inc/dec?

14. Abhishek619 Group Title

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 Group Title

Read my first post. That will tell you. That numerator of the first derivative is always positive. So is the denominator.

16. tcarroll010 Group Title

No extreme values for domain of all x. min and max though at the endpoints for the given problem domain.

17. tcarroll010 Group Title

That's because "e" is a positive base, so any exponent you put on it will result in a positive number.

18. tcarroll010 Group Title

Do you see that fozia's derivative is correct? That's where you should start.

19. tcarroll010 Group Title

Use the quotient rule to get the derivative. Do you need help with that?

20. tcarroll010 Group Title

@burhan101 I have 2 pending questions posted for you here.

21. burhan101 Group Title

@tcarroll010 im trying to figure out how that derivative is right

22. burhan101 Group Title

& yes i am using the quotient rule

23. burhan101 Group Title

$\huge \frac{ (1+e^x)(e^x)-(e^x)(e^x)}{ (1+e^x)^2 }$

24. tcarroll010 Group Title

$\frac{ (1 + e ^{x})e ^{x} - e ^{x}e ^{x} }{ (1 + e ^{x})^{2} } = \frac{ e ^{x} + e ^{2x} - e ^{2x} }{ (1 + e ^{x})^{2} }$

25. tcarroll010 Group Title

In the numerator, you are left with only e^x and the denominator is the original denominator squared.

26. burhan101 Group Title

ohhhh is see it

27. tcarroll010 Group Title

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 Group Title

but dont i plug in 0 and 4 in f(x) ?

29. tcarroll010 Group Title

So, min at: e^0 / (1 + e^0) max at: e^4 / (1 + e^4)

30. tcarroll010 Group Title

All good now, @burhan101 ?

31. burhan101 Group Title

yes thanks !!

32. tcarroll010 Group Title

uw! Good luck in all of your studies and thx for the recognition! @burhan101

33. burhan101 Group Title

thank you very much, you do explain things very clear !

34. tcarroll010 Group Title

I just like to help. Kind words on your part!

35. burhan101 Group Title

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 Group Title

YO BURHAANN MY BRO

37. dan815 Group Title

are u done with this equation

38. burhan101 Group Title

dannn:) ! no im not, do i plug the endpoints into the function ?

39. dan815 Group Title

explain the meaning of an extreme value to me

40. burhan101 Group Title

an absolute mx or an absolute min

41. dan815 Group Title

ok

42. dan815 Group Title

so that means the slope there is 0

43. burhan101 Group Title

yes

44. dan815 Group Title

but u must check the end points too incase its just a local max or local min there

45. dan815 Group Title

first solve f'(x) = 0

46. dan815 Group Title

|dw:1371749721969:dw|

47. burhan101 Group Title

so i solve for x

48. dan815 Group Title

yes

49. dan815 Group Title

okay lemme give u some simpler exmaples

50. dan815 Group Title

|dw:1371749857123:dw|

51. dan815 Group Title

what are the extreme values for a parabola

52. burhan101 Group Title

the vertex

53. dan815 Group Title

okay and the slope there is?

54. burhan101 Group Title

0

55. dan815 Group Title

so that means the derivative there is?

56. burhan101 Group Title

|dw:1371749935529:dw|

57. dan815 Group Title

what nooo

58. burhan101 Group Title

ohh:(

59. dan815 Group Title

0/8 = 0|dw:1371749995868:dw|

60. burhan101 Group Title

0=cos4x?

61. dan815 Group Title

@bahrom7893 http://multiplayerchess.com/#!/5gqIqmm

62. dan815 Group Title

hello bahrom jhanny wants us to play a game

63. burhan101 Group Title

what's the next step ..

64. dan815 Group Title

1what values of x wud make cos4x=0 think about cos pi/2=0 so what much it be

65. dan815 Group Title

must*

66. dan815 Group Title

ill give u one of the values that is in between 0 and pi cos(4(pi/8)) = 0 thats one

67. burhan101 Group Title

okay

68. burhan101 Group Title

would i keep my calculator in degrees or radians?

69. dan815 Group Title

heyy dont be silly u shud know this stuff!! we been doing trig of a while remember where cosx graph = 0

70. burhan101 Group Title

|dw:1371751462417:dw|

71. Loser66 Group Title

@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 Group Title

i told him to look at those points too :) but its not really needed in this case