An equation for an extremely convex (curved) function?

Question

Suppose we have four points, $x_1<x_2<x_3<x_4$

Is anyone able to provide me with an equation for a function that is

• nearly flat $x_1,x_2$ (i.e. $f(x_1)/f(x_2) \approx 1)$
• Steep between $x_3,x_4$ (specifically, I want $\frac{f(x_3)}{f(x_4)}<\frac{f(x_1)}{f(x_2)}$)
• strictly increasing, continuous, positive, and convex over the positive reals

For example, suppose $x_1 = 2.2,x_2=5,x_3=7.5,x_4=9$, Is there a convex, strictly increasing, function where $\frac{f(x_3)}{f(x_4)}<\frac{f(x_1)}{f(x_2)}$

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Some comments:

• Im okay with either a general answer or an answer for the specific points i have given
• convexity is not a requirement as long as the function is strictly increasing (I prefer if it is convex though
  • I'd like $f(0)=0$ if possible (but not required)

I've been generating a bunch of points in mathematica and trying to generate a curve, but i haven't had luck.

I have also tried an exponential function (this is the steepest thing I could think of),but then $x_1,x_2,x_3,x_4$ need to satisfy $x_1+x_4 > x_3+x_2$, (which my example points don't satisfy)

This has been bugging me because its so easy to draw such a function, but I haven't been able to come up with an equation.

Answer 1

If $g(x)$ is strictly increasing and convex then $f(x)=e^{g(x)}$ is strictly increasing, positive, and convex. Then the requirement $f(x_4)/f(x_3)>f(x_2)/f(x_1)$ reduces to $g(x_4)-g(x_3)>g(x_2)-g(x_1)$. It is much easier to construct such a function, for example by setting $g(x)=1/(c-x)$ when $x_1\le x\le x_4$ and extending linearly outside this interval.

Answer 2

A simple answer is to define $x_c=\frac 12{x_2+x_3}$, then $$f(x)=e^{k(x-x_c)}$$ The side below $x_c$ will be pretty flat because it is like $e^x$ for $x \lt 0$ while the side above is like $x \gt 0$. Increase $k$ to make it more pronounced.