smileatyourself
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I have the following function:

1/((e^(-2x))+1)

I need to find its max value.

I differentiated it to get:

[2e^(-2x)]/[(e^-(2x)))+1]^2

if I set this to 0 and solve for x, it doesn't make sense.

Can someone please help?

Thanks
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SamKeene
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(Original post by smileatyourself)
I have the following function:

1/((e^(-2x))+1)

I need to find its max value.

I differentiated it to get:

[2e^(-2x)]/[(e^-(2x)))+1]^2

if I set this to 0 and solve for x, it doesn't make sense.

Can someone please help?

Thanks
Aye, setting it = 0 you end up with

\displaystyle 2e^{-2x}=0

There is no value of x which satisfies this... therefore what can we say about the location of any minima or maxima?
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lazy_fish
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(Original post by smileatyourself)
...
There is no value of x at which the function is a maximum but, instead, you may notice that the derivative is 0 as x tends to something...
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smileatyourself
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(Original post by lazy_fish)
There is no value of x at which the function is a maximum but, instead, you may notice that the derivative is 0 as x tends to something...
yeah, as x -> infinity, the graph tends to a max value of 1/2.
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davros
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(Original post by smileatyourself)
yeah, as x -> infinity, the graph tends to a max value of 1/2.
Note that at x = 0 the function is 1/2. As x approaches infinity, the function approaches the value x = 1 from below but never reaches it. The function does not have a maximum!
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