Need help.
Need to find the concavity and inflection point of
e^x/(8+e^x)

- anonymous

Need help.
Need to find the concavity and inflection point of
e^x/(8+e^x)

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- DanJS

have you started anythign yet?

- anonymous

I have the second derivative

- DanJS

what are the two derivatives you got

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## More answers

- anonymous

\[\frac{ 8e ^{x}(8+e ^{x})^{2}-16e ^{2x}(8+e ^{x}) }{(8+e ^{x})^{4} }\]
that's what I have for the second derivative

- lochana

your second derivative is correct. I just did it

- anonymous

so where do I go from there?

- lochana

inflections exist where \[f''(x) = 0\]

- lochana

and there are two types of concavity.
concave upward
concave downward
you need to find them too in your problem

- anonymous

I just don't know how to solve the second derivative=0 to get the inflection points

- lochana

okay. can you simplify numerator of second derivative.

- lochana

you need simplify numerator. there are some term that cancels each other.

- lochana

and numerator means upper part of a fraction

- anonymous

you can only cancel one of the (8+e^x) right?

- lochana

did you get
\[f''(x) = \frac{64e^x}{(8+e^e)^3}\]

- lochana

\[f''(x) = \frac{64e^x}{(8+e^x)^3}\]

- lochana

is it?

- anonymous

ya

- lochana

okay. To find inflection second derivative should be equal to 0
\[\frac{64e^x}{(8+e^x)^3} = 0\]\[64e^x = 0\]

- lochana

get it?

- anonymous

yes

- lochana

if you want a fraction to be equal to zero, you can't forget denominator(down part of fraction)

- lochana

so this means
\[e^x = 0\]
and this is the graph of e^x. I will draw it, so that you can understand easily.|dw:1447295142123:dw|

- lochana

give me a second

- anonymous

so there is no inflection point?

- lochana

mm..yes. that is what our solution says. but I want to be sure about that. please wait

- lochana

Our simplification is wrong

- DanJS

f ' (x) = (8e^x) / (e^x + 8)^2
f '' (x) = [ -8e^x*(e^x - 8)] / (e^x + 8)^3

- DanJS

the second can be zero to test for possible inflection points

- DanJS

the first derivative is never zero, the second derivative is zero at 3*ln(2)

- DanJS

test both sides of that x value and see if the concavity chanvges

- anonymous

I'm lost now

- DanJS

which part

- DanJS

used the quotient rule for the first derivative, then you have to use the chain rule for the quantities to powers to go to the second

- lochana

you should start from your second derivative.
and get
\[\frac{8x^x(8^2-e^{2x})}{(8+e^x)^4}\]

- lochana

tell me when you found it.

- lochana

@DanJS is correct.

- DanJS

yeah you can just factor and simplify if you want a bit more

- lochana

after that point, you can do further simplification
and get
\[f''(x) = \frac{8e^x(8+e^x)(8-e^x)}{(8+e^x)^4}\]\[f''(x) = \frac{8e^x(8-e^x)}{(8+e^x)^3}\]

- DanJS

(8+e^x) is common in the numerator

- DanJS

right

- lochana

yes. now you can find solutions to f''(x) = 0 @kdancer03

- DanJS

the non-simplified expression for the derivative doesn't stop you from solving for y '' = 0

- lochana

@kdancer03 are you following this?

- anonymous

yes

- lochana

now can you tell what value of e^x makes f''(x) = 0?

- lochana

well it is e^x = 8

- lochana

now find the value for x
\[log_{e}e^x = log_{e}8\]
\[x = ln8\]

- lochana

inflection point is x = ln8 and y = 0.5

- DanJS

pick values either side to confirm and determine the shape of the thing

- lochana

then you can use second derivative to find concavity,
|dw:1447297744106:dw|

- anonymous

I got it thanks

- lochana

good:)

- lochana

hope this helps. bye!

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