Is there a name for a function that produces this graph?

Question

I am trying to find out if there is a name for the function that produces a graph like shown in the picture.

Answer 1

It’s a pretty generic shape but depending on how wide the flat part is you can get something like that as follows:

Let’s say we want to keep slope of middle part less than $0< \delta \ll 1$ over the interval $(a,b)$.

We see it has two generally equal rising sections. This suggests a function that has a positive, decreasing derivative in the first part and a positive increasing derivative in the second part. This means that any positive, symmetric, convex function will technically work. Assuming it’s centered at 0 we have $(a,b) \to (-c,c), c>0$ (since we assume your graph is symmetric so we’ve shifted the interval to be centered at zero):

$$f’(x)=(kx)^{2n}, n \in \mathbb{N}, k> 0$$

Since $f’ < \delta \; \forall x \in (-c,c)$ we can select $k=\frac{\delta^{\frac{1}{2n}}}{c}$ so this requirement is satisfied.

The parameter $n$ just determines how flat the flat part is

Answer 2

I doubt that the function that produces this graph has a name. However the function $$f(x) = \begin{cases} ax, \; 0 \leq x < x_0 \\ ax_0, \; x_0 \leq x < x_1 \\ e^{x-x_1} + ax_0 -1, \; x \geq x_1 \end{cases}$$ (where $a, x_0, x_1$ are positive constants) produces a graph resembling the one you posted.

Answer 3

I doubt it has a name. But the function $$ f(x) = \begin{cases} \dfrac{3.6}bx & x\le b\\ 3.6+\dfrac{e^{c\left(x-d\right)}-e^{c\left(b-d\right)}}{a\left(e^{c\left(x-d\right)}+e^{c\left(E-d\right)}\right)} & x> b\end{cases}$$ for certain parameter values $a,b,c,d,E$ looks like this: Here's a Desmos graph where you can tweak the parameters yourself. If you didn't mean for the graph to taper off at the end, you can instead use $$ g(x) = \begin{cases} \frac{3.6}bx & x\le b\\3.6+e^{c\left(x-d\right)}-e^{c\left(b-d\right)} & x> b\end{cases}$$ This function's graph (Desmos link) is as follows: