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I want to find out a continuous function on $[0,L]$, $L$ is a positive number, which looks like this red curve:

wanted shape

The function is always positive.

This function firstly is strictly concave, then the function is strictly convex. Moreover, like the graph shows, this function increases rapidly first and then increases very slowly and then increases very quickly.

But I can not come up one such function. Can you help to find out a very simple continuous function to satisfy this graph ? Thank you!

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$f(x) = x^3 \qquad g(x) = x^5 \qquad h(x)=e^x-1/2x^2-1$ If you want to move the function to right just put $x=t-c$ where $0<c<L$

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  • $\begingroup$ Thank you. I forget to mention that the function should always be positive. $\endgroup$ – congmingniao Jul 2 '16 at 14:40
  • $\begingroup$ Thank you, I build up my own example based on your suggestion. $\endgroup$ – congmingniao Jul 4 '16 at 1:43
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This can be accomplished with a quartic function. For simplicity, I will illustrate with $L = 1$.

This first example has non-negative first derivative everywhere in [0,1], achieving a minimum first derivative value of $0$ at $x = 0.5$. it also switches from concave to convex at $x = 0.5$. $$f1(x) = -0.1187x^4 + 8.5391x^3-12.6306x^2+6.2856x$$ enter image description here

The second example lets the first derivative get slightly negative in the middle, as your example curve actually does (not sure if you intended to allow it to). It is similar to $f1(x)$, but with somewhat perturbed coefficients. $$f2(x) = -1/3x^4 + 8x^3-12x^2+6x$$ enter image description here

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  • $\begingroup$ wow, thank you, Mark. Really appreciate your help. $\endgroup$ – congmingniao Jul 4 '16 at 1:43

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