Determine the extreme values:

- anonymous

Determine the extreme values:

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- anonymous

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

- fozia

values maximum, minimum

- anonymous

yes, how tho?

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## More answers

- anonymous

i find the derivative first and then what?

- fozia

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

- anonymous

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

- fozia

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

- anonymous

@fozia wait, you and I got different derivatives

- 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

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- anonymous

check if the function is increasing or decreasing?

- anonymous

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

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- anonymous

how do i check if its inc/dec?

- 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

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

& yes i am using the quotient rule

- anonymous

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

- 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

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

- anonymous

ohhhh is see it

- 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

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

- anonymous

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

- anonymous

All good now, @burhan101 ?

- anonymous

yes thanks !!

- anonymous

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

- anonymous

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

- anonymous

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

- 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

YO BURHAANN MY BRO

- dan815

are u done with this equation

- anonymous

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

- dan815

explain the meaning of an extreme value to me

- anonymous

an absolute mx or an absolute min

- dan815

ok

- dan815

so that means the slope there is 0

- anonymous

yes

- dan815

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

- dan815

first solve
f'(x) = 0

- dan815

|dw:1371749721969:dw|

- anonymous

so i solve for x

- dan815

yes

- dan815

okay lemme give u some simpler exmaples

- dan815

|dw:1371749857123:dw|

- dan815

what are the extreme values for a parabola

- anonymous

the vertex

- dan815

okay and the slope there is?

- anonymous

0

- dan815

so that means the derivative there is?

- anonymous

|dw:1371749935529:dw|

- dan815

what nooo

- anonymous

ohh:(

- dan815

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

- anonymous

0=cos4x?

- dan815

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

- dan815

hello bahrom jhanny wants us to play a game

- anonymous

what's the next step ..

- dan815

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

- dan815

must*

- dan815

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

- anonymous

okay

- anonymous

would i keep my calculator in degrees or radians?

- dan815

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

- anonymous

|dw:1371751462417:dw|

- 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

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

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