burhan101
Determine the extreme values:
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burhan101
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\[\huge y= \frac{ e^{x} }{ 1+e^{x} }, x \epsilon [0,4]\]
fozia
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values maximum, minimum
burhan101
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yes, how tho?
burhan101
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i find the derivative first and then what?
fozia
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first fint derivative that gives the result y'=ex/(1+ex)2 then eqaute it to zero gives ex=0
burhan101
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\[\huge y'=\frac{ e^x }{ (1+e^x) }\]
fozia
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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
burhan101
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@fozia wait, you and I got different derivatives
tcarroll010
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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.
burhan101
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Abhishek619
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check if the function is increasing or decreasing?
tcarroll010
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Here's a graph to show you that it is always increasing
burhan101
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how do i check if its inc/dec?
Abhishek619
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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.
tcarroll010
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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.
tcarroll010
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No extreme values for domain of all x. min and max though at the endpoints for the given problem domain.
tcarroll010
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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.
tcarroll010
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Do you see that fozia's derivative is correct? That's where you should start.
tcarroll010
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Use the quotient rule to get the derivative. Do you need help with that?
tcarroll010
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@burhan101 I have 2 pending questions posted for you here.
burhan101
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@tcarroll010 im trying to figure out how that derivative is right
burhan101
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& yes i am using the quotient rule
burhan101
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\[\huge \frac{ (1+e^x)(e^x)-(e^x)(e^x)}{ (1+e^x)^2 }\]
tcarroll010
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\[\frac{ (1 + e ^{x})e ^{x} - e ^{x}e ^{x} }{ (1 + e ^{x})^{2} } = \frac{ e ^{x} + e ^{2x} - e ^{2x} }{ (1 + e ^{x})^{2} }\]
tcarroll010
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In the numerator, you are left with only e^x and the denominator is the original denominator squared.
burhan101
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ohhhh is see it
tcarroll010
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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.
burhan101
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but dont i plug in 0 and 4 in f(x) ?
tcarroll010
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So, min at:
e^0 / (1 + e^0)
max at:
e^4 / (1 + e^4)
tcarroll010
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All good now, @burhan101 ?
burhan101
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yes thanks !!
tcarroll010
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uw! Good luck in all of your studies and thx for the recognition! @burhan101
burhan101
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thank you very much, you do explain things very clear !
tcarroll010
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I just like to help. Kind words on your part!
burhan101
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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
dan815
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YO BURHAANN MY BRO
dan815
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are u done with this equation
burhan101
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dannn:) ! no im not, do i plug the endpoints into the function ?
dan815
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explain the meaning of an extreme value to me
burhan101
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an absolute mx or an absolute min
dan815
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ok
dan815
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so that means the slope there is 0
burhan101
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yes
dan815
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but u must check the end points too incase its just a local max or local min there
dan815
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first solve
f'(x) = 0
dan815
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|dw:1371749721969:dw|
burhan101
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so i solve for x
dan815
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yes
dan815
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okay lemme give u some simpler exmaples
dan815
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|dw:1371749857123:dw|
dan815
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what are the extreme values for a parabola
burhan101
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the vertex
dan815
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okay and the slope there is?
burhan101
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0
dan815
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so that means the derivative there is?
burhan101
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|dw:1371749935529:dw|
dan815
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what nooo
burhan101
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ohh:(
dan815
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0/8 = 0|dw:1371749995868:dw|
burhan101
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0=cos4x?
dan815
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hello bahrom jhanny wants us to play a game
burhan101
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what's the next step ..
dan815
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1what values of x wud make cos4x=0 think about cos pi/2=0 so what much it be
dan815
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must*
dan815
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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
burhan101
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okay
burhan101
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would i keep my calculator in degrees or radians?
dan815
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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
burhan101
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|dw:1371751462417:dw|
Loser66
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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.
dan815
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i told him to look at those points too :) but its not really needed in this case