Find a Bounded, Decreasing $\displaystyle f(x)$

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Discussion Overview

The discussion revolves around finding a function $\displaystyle f(x)$ such that $\displaystyle \frac{1}{f(x)}$ is defined for all $\displaystyle x$, is bounded, and $\displaystyle f(x)$ is decreasing. The scope includes mathematical reasoning and exploration of function properties.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • Some participants propose that $\displaystyle f(x)$ must be non-zero and separated from zero, implying that there exists an $\displaystyle \varepsilon$ such that $\displaystyle |f(x)| > \varepsilon$ for all $\displaystyle x$.
  • There is a question about the domain of the function, with participants confirming it is all real numbers.
  • One participant expresses difficulty in finding such a function and asks for a description of how such a function behaves.
  • Another participant notes that since $\displaystyle f(x)$ is decreasing, $\displaystyle \frac{1}{f(x)}$ is increasing and must approach a certain limit as $\displaystyle x \to \infty$.
  • A suggestion is made for the function $\displaystyle f = 1 + e^{-x}$ as a potential candidate.
  • There is a reiteration of the behavior of $\displaystyle f(x)$ and the implications of its decreasing nature on $\displaystyle \frac{1}{f(x)}$.

Areas of Agreement / Disagreement

Participants express differing levels of understanding and approaches to finding the function, with no consensus on a specific function or method to achieve the desired properties.

Contextual Notes

Participants discuss the need for $\displaystyle f(x)$ to remain outside a certain band defined by $\{(x,y):|y|\le\varepsilon\}$, but the implications of this condition remain unresolved.

alexmahone
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Find an $\displaystyle f(x)$ such that $\displaystyle \frac{1}{f(x)}$ is defined for all $\displaystyle x$ and is bounded, but $\displaystyle f(x)$ is decreasing.
 
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This is not hard. Obviously, we must have f(x) ≠ 0 and moreover f(x) must be separated from 0, i.e., for some ε we must have |f(x)| > ε for all x.
 
Evgeny.Makarov said:
This is not hard. Obviously, we must have f(x) ≠ 0 and moreover f(x) must be separated from 0, i.e., for some ε we must have |f(x)| > ε for all x.

I'm still not able to find such a function. :(
 
Alexmahone said:
Find an $\displaystyle f(x)$ such that $\displaystyle \frac{1}{f(x)}$ is defined for all $\displaystyle x$ and is bounded, but $\displaystyle f(x)$ is decreasing.

what is the domain of the function ? all real numbers ?
 
Amer said:
what is the domain of the function ? all real numbers ?

Yes.
 
Alexmahone said:
I'm still not able to find such a function.
You can't find a decreasing function whose graph lies outside the band $\{(x,y):|y|\le\varepsilon\}$? If you don't know a precise formula, can you at least describe how such function behaves?
 
Evgeny.Makarov said:
You can't find a decreasing function whose graph lies outside the band $\{(x,y):|y|\le\varepsilon\}$? If you don't know a precise formula, can you at least describe how such function behaves?

Since $\displaystyle f(x)$ is decreasing, $\displaystyle \frac{1}{f(x)}$ is increasing. But $\displaystyle \frac{1}{f(x)}$ is also bounded. So, it must approach a certain limit as $\displaystyle {x\to\infty}$.
 
How about $f = 1 + e^{-x}$?
 
Alexmahone said:
Since $\displaystyle f(x)$ is decreasing, $\displaystyle \frac{1}{f(x)}$ is increasing. But $\displaystyle \frac{1}{f(x)}$ is also bounded. So, it must approach a certain limit as $\displaystyle {x\to\infty}$.
Yes, but I was asking really about f(x). Here are the possible behaviors of f(x).

function.png
 

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