# Function which gradually rises until some point and then quickly “falls”

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?

What may cause such behavior?

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Do you need it to be differentiable at that "point"? – Ian Mateus Jan 21 '13 at 17:57
@IanMateus That is not necessary. – Edward Ruchevits Jan 21 '13 at 21:46

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

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

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

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Thank you! The second modification is what I wanted. – Edward Ruchevits Jan 21 '13 at 21:54

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,

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Thank you, that's interesting! – Edward Ruchevits Jan 21 '13 at 21:53

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 $e^{-\frac{1}{x^2}}$.

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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