Is there a function that changes large values ​​a lot and slightly small ones?

Question

Mathematics

Good afternoon. We need a function (or any other mathematical object) that changes large values ​​a lot and slightly small ones.
Those. the larger the value, the more it is subject to change.
Example:
a = 75
b = 0.335

When run through the function it should be:
f(a) = 30
f(b) = 0.334

Answer 1

Main requirement: f'(0)=1. But it is not clear from your description how the function should behave on ∞-u, and here are some options.
The simplest is a linear function kx, k<1 (for example, 0.9x). It is linear everywhere (and on ∞-s).
It can be more complicated: x / (kx² + 1). This beast will shove to zero.
Something in between is possible: k·ln(x/k + 1). tends to ∞, but not in the same way as x.
Another option: 2k sqrt(x/k + 1). Also tends to ∞-and, but faster.
If you need to constant a - then [2a / pi] arctg (x / a). The only thing is that to control the rate of convergence, you will have to substitute approximately the same function (f'(0)=1) instead of x in the arc tangent.

Answer 2

Take, for example, the combination of a linear function and a power function f(x)=x*(1+x^n). From -1 to 1, values ​​remain weak for small n, while |x|>1 values ​​change strongly.

Answer 3

Head-on formula: f(x)=(kx+b)^n
(x is the input value).
Simple rules (heuristics) for finding (implicative) dependencies (complicated ones are the search for all permutations in mathematical operations with all their possible types, giving the required number of solutions):
1 value - equation.
2 (4,6,...) values ​​- first proportion, then system.
3 (5,7,...) values ​​- a system of equations.
Goto 2.
ax1=y1
bx2=y2
y1/y2=(K1(x1/x2)+K2)^K3+K4(x1/x2)+K5
and to get a more accurate result, repeat the process recursively.

Answer 4

Play around with the divider
function def(a){return a - a /10 }

Answer 5

Take the pivot points you know, plot them on an X-Y chart in some Excel/calc, and plot a trend line. The exponential function will do, maybe you'll like the logarithmic one better. But more points are needed. For your two points, you can use a linear function.