Decreasing function that behaves like $1-x^2$ for small $x$ and like $e^{-x}$ for large $x$ I am trying to find a function with a domain $D = \mathbb{R}_+$, that is behaving like $1-x^2$ for small $x$ and like $e^{-x}$ for large $x$.
Edit: And is monotonically decreasing.
I thought about using $\chi^2$-distribution or a combination of logistic functions with adding a symmetric "normal density function like".
I would be glad if someone could help me. Or is there a general method how one approach such problems. 
 A: Let us consider a function $f\colon\mathbb{R}_+\to\mathbb{R}$ defined by
$$
f(x) = \frac{1+ax+x^2}{1+bx+x^2}e^{-x}
$$
for $a,b\in\mathbb{R}$ to be chosen. We do have the right asymptotics when $x\to\infty$, namely $f(x) \displaystyle\operatorname*{\sim}_{x\to\infty} e^{-x}$ (that was the point of choosing this particular form for $f$). Now, let us pick $a,b$ such that
$$
f(x) \operatorname*{=}_{x\to0^+} 1-x^2 + o(x^2)
$$
Doing a Taylor expansion of $f$ at $0$:
$$
f(x) = 1 + (a - b - 1) x + \left( \frac{1}{2} - a + b - a b + b^2\right) x^2 + o(x^2)
$$
we find that we must have $a-b=1$ and $b+b^2-a (b+1)=-\frac{3}{2}$ for the coefficients of $x$ to cancel and that of $x^2$ to be $1$. Solving, we obtain
$$
a = \frac32,\quad b = \frac12
$$
that is
$$
f(x) = \frac{1+\frac{3}{2}x+x^2}{1+\frac{1}{2}x+x^2}e^{-x}
= \frac{2+3x+2x^2}{2+x+2x^2}e^{-x}
$$
PS: as a bonus, since the denominator never cancels we have that $f$ is actually defined on $\mathbb{R}$, not only $\mathbb{R}_+$; and the function is monotonically decreasing on $(0,\infty)$. (Although, to be honest I was not aiming for that at first, and saw the OP's edit afterwards.)

A: $$e^{-x^2}(1-x^2)+\frac{x^3}{x^3+1}e^{-x}$$
A: Another way: let 
$$f(x)= e^{-g(x)}$$
for some $g(x)$ such that $g(x) \sim x $ as $x\to \infty$ and $g(x) \sim x^2$ as $x \to 0$. For example we can take
$$g(x)=\frac{x^2}{x+1}$$
or, perhaps more elegant (and even), $g(x)= \sqrt{x^2+1/4}-1/2$, which gives:
$$ f(x) = \exp\left( \frac{1-\sqrt{4{{x}^{2}}+1}}{2}\right) $$
with $f(x) = 1 - x^2 + O(x^4)$ around $x=0$

A: Following a comment made above, below the OP's question:
One may want to throw in the question that $f$ is required smooth or at least $C^1$ (note that the function from my other answer is indeed smooth), otherwise the following continuous non-increasing function $f\colon\mathbb{R}\to\mathbb{R}$ does — trivially — the job:
$$
f(x) = \begin{cases}
1-x^2 & 0 \leq x < x_0\\
e^{-x} & x > x_0\\
1 & x < 0
\end{cases}
$$
where $x_0\simeq 0.714$ is the only positive solution to the equation $e^{-x} = 1-x^2$.

