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

[alexmahone]

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.

 

[Evgeny.Makarov]

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.

 

[alexmahone]

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. :(

 

[Amer]

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 ?

 

[alexmahone]

what is the domain of the function ? all real numbers ?

Yes.

 

[Evgeny.Makarov]

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?

 

[alexmahone]

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 $\displaystyle {x\to\infty}$.

 

[Mathwebster]

How about $f = 1 + e^{-x}$?

 

[Evgeny.Makarov]

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).