anonymous
  • anonymous
Need help. Need to find the concavity and inflection point of e^x/(8+e^x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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DanJS
  • DanJS
have you started anythign yet?
anonymous
  • anonymous
I have the second derivative
DanJS
  • DanJS
what are the two derivatives you got

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anonymous
  • 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
  • lochana
your second derivative is correct. I just did it
anonymous
  • anonymous
so where do I go from there?
lochana
  • lochana
inflections exist where \[f''(x) = 0\]
lochana
  • lochana
and there are two types of concavity. concave upward concave downward you need to find them too in your problem
anonymous
  • anonymous
I just don't know how to solve the second derivative=0 to get the inflection points
lochana
  • lochana
okay. can you simplify numerator of second derivative.
lochana
  • lochana
you need simplify numerator. there are some term that cancels each other.
lochana
  • lochana
and numerator means upper part of a fraction
anonymous
  • anonymous
you can only cancel one of the (8+e^x) right?
lochana
  • lochana
did you get \[f''(x) = \frac{64e^x}{(8+e^e)^3}\]
lochana
  • lochana
\[f''(x) = \frac{64e^x}{(8+e^x)^3}\]
lochana
  • lochana
is it?
anonymous
  • anonymous
ya
lochana
  • lochana
okay. To find inflection second derivative should be equal to 0 \[\frac{64e^x}{(8+e^x)^3} = 0\]\[64e^x = 0\]
lochana
  • lochana
get it?
anonymous
  • anonymous
yes
lochana
  • lochana
if you want a fraction to be equal to zero, you can't forget denominator(down part of fraction)
lochana
  • 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
  • lochana
give me a second
anonymous
  • anonymous
so there is no inflection point?
lochana
  • lochana
mm..yes. that is what our solution says. but I want to be sure about that. please wait
lochana
  • lochana
Our simplification is wrong
DanJS
  • DanJS
f ' (x) = (8e^x) / (e^x + 8)^2 f '' (x) = [ -8e^x*(e^x - 8)] / (e^x + 8)^3
DanJS
  • DanJS
the second can be zero to test for possible inflection points
DanJS
  • DanJS
the first derivative is never zero, the second derivative is zero at 3*ln(2)
DanJS
  • DanJS
test both sides of that x value and see if the concavity chanvges
anonymous
  • anonymous
I'm lost now
DanJS
  • DanJS
which part
DanJS
  • 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
  • lochana
you should start from your second derivative. and get \[\frac{8x^x(8^2-e^{2x})}{(8+e^x)^4}\]
lochana
  • lochana
tell me when you found it.
lochana
  • lochana
@DanJS is correct.
DanJS
  • DanJS
yeah you can just factor and simplify if you want a bit more
lochana
  • 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
  • DanJS
(8+e^x) is common in the numerator
DanJS
  • DanJS
right
lochana
  • lochana
yes. now you can find solutions to f''(x) = 0 @kdancer03
DanJS
  • DanJS
the non-simplified expression for the derivative doesn't stop you from solving for y '' = 0
lochana
  • lochana
@kdancer03 are you following this?
anonymous
  • anonymous
yes
lochana
  • lochana
now can you tell what value of e^x makes f''(x) = 0?
lochana
  • lochana
well it is e^x = 8
lochana
  • lochana
now find the value for x \[log_{e}e^x = log_{e}8\] \[x = ln8\]
lochana
  • lochana
inflection point is x = ln8 and y = 0.5
DanJS
  • DanJS
pick values either side to confirm and determine the shape of the thing
lochana
  • lochana
then you can use second derivative to find concavity, |dw:1447297744106:dw|
anonymous
  • anonymous
I got it thanks
lochana
  • lochana
good:)
lochana
  • lochana
hope this helps. bye!

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