# Calculus maxWatch

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Thread starter 4 years ago
#1
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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4 years ago
#2
(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

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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4 years ago
#3
(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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Thread starter 4 years ago
#4
(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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4 years ago
#5
(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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