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I would like to find a function for the following graph: enter image description here

I have drawn the graph myself, so not all subtle bends are to be replicated. I have noted the important points the graph should have in the picture (and even those could be tweaked a little bit if it makes things easier).

How would you go about this? I tried setting up a polynomial of degree four and solving a linear equation system, but the system did not have any solutions. I'd be happy for a solution, preferably with an explanation and maybe even the possibility for me to move the points a little bit.

Thank you! :)

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    $\begingroup$ Does the limit as $x\to\infty$ need to be $0$? If so, a polynomial is not the way to go... $\endgroup$
    – 5xum
    Jun 20, 2016 at 8:07
  • $\begingroup$ If you want a polynomial it should have degree $3$. $\endgroup$
    – David
    Jun 20, 2016 at 8:08
  • $\begingroup$ @5xum: Preferably. I could also set (1,0) for the function and set all input data points with x>1 to x=1. Why can't this be achieved by a polynomial? $\endgroup$
    – user1809923
    Jun 20, 2016 at 8:22
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    $\begingroup$ @user1809923 If $p$ is a polynomial of degree at least $1$, then $\lim_{x\to\infty} p(x)$ can only be $\infty$ or $-\infty$. $\endgroup$
    – 5xum
    Jun 20, 2016 at 8:23
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    $\begingroup$ @user1809923 For any $n$ points $(x_k,y_k)$ with all the $x$ coordinates different, there is a unique polynomial of degree at most $n-1$ which passes through those points. If you take the procedure you followed for a quartic and start with a cubic instead, it should work. (Though, as pointed out by 5xum, it may not be an appropriate answer to your problem.) $\endgroup$
    – David
    Jun 20, 2016 at 8:29

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this might work $$\frac{p(x)}{e^{x^2(x-.2)(x-.6)(x-1)}}$$ where $p(x)$ is your polynomial

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  • $\begingroup$ if i see it correctly this function has a clear minimal point between (1.0,0.0) and x to infinity. $\endgroup$
    – user1809923
    Jun 20, 2016 at 12:47

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