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I am trying to define a simple function that is first concave and then convex as shown in the picture below. Since the resulting equation have to be explained/used by a non-technical audience, the function should be ideally as simple as possible, but I have been unable to find any simple form that matches the requirements below.

enter image description here

Any help of ideas would be greatly appreciated.

EDIT: to further clarify

  1. The dashed red line is the constant line $Y=X$
  2. $X_1$ and $X_2$ should be ideally points that I can control within the function.
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  • $\begingroup$ What about $x/[(\sin{x})^2+1.5]$ $\endgroup$
    – Matthew
    Commented Apr 4, 2017 at 9:07
  • $\begingroup$ Why +1.5 and not +1? $\endgroup$ Commented Apr 4, 2017 at 9:13
  • $\begingroup$ 1.5 is my first guess for everything, so I just went with it, but as long as the number is bigger than 0, there should be no problems :) $\endgroup$
    – Matthew
    Commented Apr 4, 2017 at 9:15
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    $\begingroup$ Just to explain my solution. You take the straight line x and want the slope to be not perfect anymore, but to decrease first, then increase again. As the slope is the factor in front of the x, you just need a factor to decrease first and then increase again and the sine has just the inverse property. So you just take the inverse of it, but make sure, you don't divide by 0. $\endgroup$
    – Matthew
    Commented Apr 4, 2017 at 9:19
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    $\begingroup$ But WHAT are the points $x_1$ and $x_2$? Are they the midpoints of the two ranges, where $f(x)>x$ and $f(x)<x$, respectively? Or the points of local extrema of $f(x)-x$ difference (doesn't quite look so from the graph...)? Something else? $\endgroup$
    – CiaPan
    Commented Apr 4, 2017 at 9:44

4 Answers 4

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How about $f(x) = x^3$?

$f''(x) = 6x$ for all $x \in \mathbb{R}$ which means that:

  • $f'' > 0$ for $x > 0$ (concave up)
  • $f'' < 0$ for $x < 0$ (concave down)

You can also shift it up/down and left/right to put the inflection point (currently at $(0,0)$) wherever you want.

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  • $\begingroup$ Just shift it as needed $\endgroup$ Commented Apr 4, 2017 at 9:14
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How about $x^3 - x$ or $(x-2)^3 - x + 8$ or $\sqrt x + e^x/30$?

Plots in Wolfram Alpha:

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$f(x)=(x-1)^3+1$ should do.

Lowest power polynomial with desired properties

EDIT

$f(x)=(x-a)^3+a^3+b*x^2$ should give you some control by adequate choice of a and b.

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Assume $f(0) = 0$ and $f(x)$ at $x=0$ is increasing faster than $x$, that is $f'(0) > 1$.

Suppose $A=x_1>0$ is a midpoint of the range, where $f(x)>x$.
Hence the next point of $f(x)=x$ is $x=2A$.

Let $B=x_2$ be the midpoint of the range of $f(x)<x$.
Of course $B>2A$ and then the next $f(x)=x$ shall happen at $x=2B-2A$.

So we can make a function as $$f(x) = x + kx(x-2A)(x-(2B-2A))$$ with $k>0$.

The slope of the graph at the origin: $$f'(0) = 1 + 4kA(B-A)$$

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