Find an f(x)

Alexmahone

Active member
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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Evgeny.Makarov

Well-known member
MHB Math Scholar
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

Active member
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

Active member
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 ?

Evgeny.Makarov

Well-known member
MHB Math Scholar
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

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

Jester

Well-known member
MHB Math Helper
How about $f = 1 + e^{-x}$?

Evgeny.Makarov

Well-known member
MHB Math Scholar
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).