Question

Determine the extreme values:

Answer 1

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

Answer 2

values maximum, minimum

Answer 3

yes, how tho?

Answer 4

i find the derivative first and then what?

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

@fozia wait, you and I got different derivatives

Answer 9

@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.

Answer 10

check if the function is increasing or decreasing?

Answer 11

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

Answer 12

how do i check if its inc/dec?

Answer 13

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.

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

& yes i am using the quotient rule

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

ohhhh is see it

Answer 26

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.

Answer 27

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

Answer 28

So, min at:

e^0 / (1 + e^0)

max at:

e^4 / (1 + e^4)

Answer 29

All good now, @burhan101 ?

Answer 30

yes thanks !!

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

YO BURHAANN MY BRO

Answer 36

are u done with this equation

Answer 37

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

Answer 38

explain the meaning of an extreme value to me

Answer 39

an absolute mx or an absolute min

Answer 40

ok

Answer 41

so that means the slope there is 0

Answer 42

yes

Answer 43

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

Answer 44

first solve
f'(x) = 0