September 5th, 2018, 11:01 PM  #21  
Newbie Joined: Aug 2018 From: România Posts: 22 Thanks: 2  Quote:
What we get from condition $\displaystyle f(x)f'(x)>0$? Όλα o καλός, Integrator Last edited by Integrator; September 5th, 2018 at 11:32 PM.  
September 6th, 2018, 11:05 PM  #22 
Newbie Joined: Aug 2018 From: România Posts: 22 Thanks: 2  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, 04:56 AM  #23 
Global Moderator Joined: Dec 2006 Posts: 20,107 Thanks: 1907  
September 7th, 2018, 07:51 AM  #24 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,561 Thanks: 2562 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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