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September 5th, 2018, 10:01 PM   #21
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Quote:
Originally Posted by babaliaris View Post
So back to the problem :
Finally:
$\displaystyle
f(x) = \frac{1}{-1 - ke^{-x}}
$
Καλημέρα,

What we get from condition $\displaystyle f(x)-f'(x)>0$?


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Last edited by Integrator; September 5th, 2018 at 10:32 PM.
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September 6th, 2018, 10:05 PM   #22
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Quote:
Originally Posted by babaliaris View Post
Finally:
$\displaystyle
f(x) = \frac{1}{-1 - ke^{-x}}
$
Hello,

From the calculations it follows that $\displaystyle f:\mathbb C$ $\displaystyle \rightarrow \mathbb C$ where $\displaystyle
f(x) = \frac{1}{-1 - ke^{-x}}$ , $\displaystyle f(x)-f'(x)=a^2$ with $\displaystyle a\cdot k\neq 0$ , $\displaystyle a\in \mathbb R$ , $\displaystyle a\cdot k\neq 0$ and thus $\displaystyle x=2i\pi n+\log \bigg (-\frac{ak}{a\mp i}\bigg )$ where $\displaystyle i^2=-1$.

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September 7th, 2018, 03:56 AM   #23
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Is it correct what WolframAlpha says?
Strictly speaking, no.
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September 7th, 2018, 06:51 AM   #24
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I don't know about "strictly". The only way it could be partially true is if $x$ is complex. If $x \in \mathbb R$, the given solutions are false. But in either case, the solution $f(x)=0$ has been omitted. It corresponds to the limiting cases
\begin{align}f(x) &=\lim_{c \to \pm\infty} \frac{e^x}{c+e^x} = 0 \\ f(x) &=\lim_{c \to -\infty} -\frac{e^{c+x}}{e^{c+x}-1} = 0 \end{align}

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