anonymous
  • anonymous
Determine the extreme values:
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\huge y= \frac{ e^{x} }{ 1+e^{x} }, x \epsilon [0,4]\]
fozia
  • fozia
values maximum, minimum
anonymous
  • anonymous
yes, how tho?

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anonymous
  • anonymous
i find the derivative first and then what?
fozia
  • fozia
first fint derivative that gives the result y'=ex/(1+ex)2 then eqaute it to zero gives ex=0
anonymous
  • anonymous
\[\huge y'=\frac{ e^x }{ (1+e^x) }\]
fozia
  • fozia
take 2nd derivatv now put the value of x =0 in 2nd derivativ if relt is postv then its min otherwise max
anonymous
  • anonymous
@fozia wait, you and I got different derivatives
anonymous
  • anonymous
@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.
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
check if the function is increasing or decreasing?
anonymous
  • anonymous
Here's a graph to show you that it is always increasing
1 Attachment
anonymous
  • anonymous
how do i check if its inc/dec?
anonymous
  • anonymous
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.
anonymous
  • anonymous
Read my first post. That will tell you. That numerator of the first derivative is always positive. So is the denominator.
anonymous
  • anonymous
No extreme values for domain of all x. min and max though at the endpoints for the given problem domain.
anonymous
  • anonymous
That's because "e" is a positive base, so any exponent you put on it will result in a positive number.
anonymous
  • anonymous
Do you see that fozia's derivative is correct? That's where you should start.
anonymous
  • anonymous
Use the quotient rule to get the derivative. Do you need help with that?
anonymous
  • anonymous
@burhan101 I have 2 pending questions posted for you here.
anonymous
  • anonymous
@tcarroll010 im trying to figure out how that derivative is right
anonymous
  • anonymous
& yes i am using the quotient rule
anonymous
  • anonymous
\[\huge \frac{ (1+e^x)(e^x)-(e^x)(e^x)}{ (1+e^x)^2 }\]
anonymous
  • anonymous
\[\frac{ (1 + e ^{x})e ^{x} - e ^{x}e ^{x} }{ (1 + e ^{x})^{2} } = \frac{ e ^{x} + e ^{2x} - e ^{2x} }{ (1 + e ^{x})^{2} }\]
anonymous
  • anonymous
In the numerator, you are left with only e^x and the denominator is the original denominator squared.
anonymous
  • anonymous
ohhhh is see it
anonymous
  • anonymous
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.
anonymous
  • anonymous
but dont i plug in 0 and 4 in f(x) ?
anonymous
  • anonymous
So, min at: e^0 / (1 + e^0) max at: e^4 / (1 + e^4)
anonymous
  • anonymous
All good now, @burhan101 ?
anonymous
  • anonymous
yes thanks !!
anonymous
  • anonymous
uw! Good luck in all of your studies and thx for the recognition! @burhan101
anonymous
  • anonymous
thank you very much, you do explain things very clear !
anonymous
  • anonymous
I just like to help. Kind words on your part!
anonymous
  • anonymous
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
dan815
  • dan815
YO BURHAANN MY BRO
dan815
  • dan815
are u done with this equation
anonymous
  • anonymous
dannn:) ! no im not, do i plug the endpoints into the function ?
dan815
  • dan815
explain the meaning of an extreme value to me
anonymous
  • anonymous
an absolute mx or an absolute min
dan815
  • dan815
ok
dan815
  • dan815
so that means the slope there is 0
anonymous
  • anonymous
yes
dan815
  • dan815
but u must check the end points too incase its just a local max or local min there
dan815
  • dan815
first solve f'(x) = 0
dan815
  • dan815
|dw:1371749721969:dw|
anonymous
  • anonymous
so i solve for x
dan815
  • dan815
yes
dan815
  • dan815
okay lemme give u some simpler exmaples
dan815
  • dan815
|dw:1371749857123:dw|
dan815
  • dan815
what are the extreme values for a parabola
anonymous
  • anonymous
the vertex
dan815
  • dan815
okay and the slope there is?
anonymous
  • anonymous
0
dan815
  • dan815
so that means the derivative there is?
anonymous
  • anonymous
|dw:1371749935529:dw|
dan815
  • dan815
what nooo
anonymous
  • anonymous
ohh:(
dan815
  • dan815
0/8 = 0|dw:1371749995868:dw|
anonymous
  • anonymous
0=cos4x?
dan815
  • dan815
@bahrom7893 http://multiplayerchess.com/#!/5gqIqmm
dan815
  • dan815
hello bahrom jhanny wants us to play a game
anonymous
  • anonymous
what's the next step ..
dan815
  • dan815
1what values of x wud make cos4x=0 think about cos pi/2=0 so what much it be
dan815
  • dan815
must*
dan815
  • dan815
ill give u one of the values that is in between 0 and pi cos(4(pi/8)) = 0 thats one
anonymous
  • anonymous
okay
anonymous
  • anonymous
would i keep my calculator in degrees or radians?
dan815
  • dan815
heyy dont be silly u shud know this stuff!! we been doing trig of a while remember where cosx graph = 0
anonymous
  • anonymous
|dw:1371751462417:dw|
Loser66
  • Loser66
@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.
dan815
  • dan815
i told him to look at those points too :) but its not really needed in this case

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