I want to find a function that satisfy the following criteria.
$f(x)$ is continuous and strictly positive ($ f(x) > 0 \ \forall \ x \in \mathbb{R}$)
$f'(x) < 0 \ \forall \ x \in [-\infty, -2] \cup [2, \infty ] $
$f'(\pm 2) = 0$
$f''(2\pm1) = 0$
I was playing around with functions on the form $\exp(ax^2 + bx + c)$, but was not able to find something fitting.
A sketch of the function would be something like
Any help would be appreciated. A splitted function expression is fine as well =)
$f \in \mathcal C^2\left(\Bbb R, \Bbb R\right)$
$\forall x \in \Bbb R, f(x) > 0$
$\forall x \in I = ]-\infty, -2[ \cup ]2,+\infty[, f'(x)<0$
$f'(-2)=0=f'(2)$
$f''(1)=0=f''(3)$
Let $l,m,r \in \mathcal C^2\left(]-\infty, -2], \Bbb R\right) \times \mathcal C^2\left([-2,3], \Bbb R\right) \times \mathcal C^2\left([3,+\infty[, \Bbb R\right)$ so that
$f = \left\{\begin{array}{ll}l(x) & \text{if } x \le -2\\m(x) & \text{if } -2\le x \le 2\\r(x) & \text{if } 2 \le x\end{array}\right.$
We also want those functions, their derivative and their second derivative to have the same values at $-2$ and $3$ so that $f$ is $\mathcal C^2$.
Once I had noticed that, I just took polynomials of increasing degree untill I found one that worked as $m$ and had a few additional properties so that I could make sure I would find an $l$ and a $r$. It turned out I could find a polynomial fitting for both $l$ and $m$ and then, since I knew I could use something similar to $x\mapsto e^{-x}$ for $r$, I searched solutions of the for $x\mapsto e^{p(x)}$ with $p$ a polynomial (and I fixed the coefficient of $x^2$ to $1$ because I got a constant answer if I didn't add degree $3$ and if I did, I had too many variables so they all got expressed in terms of that coefficient). Here's the result of my experiments:
