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

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

Όλα o καλός,

Integrator

Last edited by Integrator; September 5th, 2018 at 10:32 PM.

September 6th, 2018, 10:05 PM   #22
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Quote:
 Originally Posted by babaliaris 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 $\displaystyle i^2=-1$.

All the best,

Integrator

September 7th, 2018, 03:56 AM   #23
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Quote:
 Originally Posted by Integrator Is it correct what WolframAlpha says?
Strictly speaking, no.

 September 7th, 2018, 06:51 AM #24 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,403 Thanks: 2477 Math Focus: Mainly analysis and algebra 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} Graph of WA Solutions

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