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Could someone point me to any function ${ f(x) }$ which is continuous at some interval ${ x \in [x_0; x_1] }$ and can be represented by formula, so that it rises until some point and then quickly "falls" like on image below?

enter image description here

What may cause such behavior?

Edward Ruchevits
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3 Answers3

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How about the function $f(x)=e^{-|x|}$, which has the following graph:

enter image description here

If you don't want it to be symmetric, you could use $f(x)=e^{-|1-e^x|}$, which has the following graph:

enter image description here

If you need the base line to be the same (the left asymptote of the above graph is $\frac{1}{e}$), you could use $$f(x)=e^{-|e^{-ax}-e^x|}$$ where $0<a<1$ is a constant you can vary. For example, with $a=1/10$, we get

enter image description here

Zev Chonoles
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Try something like a log-normal, $$f(x) = \frac{1}{\sqrt{2\pi}\sigma x}\exp\bigg({-\frac{(\log(x) - \mu)^2}{2\sigma^2}\bigg)}$$ For example,

enter image description here

Neal
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How quickly? Piecewise linear functions satisfy this. In general, for real valued functions you can essentially get as nice bumps as necessary by considering $e^{-x^n}$, where the speed of decay can be acquired by increasing $n$. You can also construct bumps which are supported on compact sets and satisfy the above by playing with functions of the form $e^{-\frac{1}{x^2}}$.

anonymous
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    This was my first thought too, but the OP asks for the function to be "represented by formula". Of course, this is a vague notion, but I don't think piecewise-defined functions are what the OP is after. – Zev Chonoles Jan 21 '13 at 17:59
  • Thank you for the answer! I meant single function, not piecewise. – Edward Ruchevits Jan 21 '13 at 21:51